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\begin{document}

\title{A few classical results on tensor, symmetric and exterior powers}
\author{Darij Grinberg}
\date{Version 0.3 (\today) (not proofread!)}
\maketitle
\tableofcontents

\begin{noncompile}
\textbf{TO-DO\ LIST:}
\end{noncompile}

\subsection{Version}

\begin{vershort}
ver:S
\end{vershort}

\begin{verlong}
ver:L
\end{verlong}

\subsection{Introduction}

In this note, I am going to give proofs to a few results about tensor products
as well as tensor, pseudoexterior, symmetric and exterior powers of
$k$-modules (where $k$ is a commutative ring with $1$). None of the results is
new, as I have seen them used all around literature as if they were well-known
and/or completely trivial. I have not yet found a place where they are
actually proved (though I have not looked far), so I am doing it here.

This note is not completely new: The first four Subsections (\ref{subsect.(X)}%
, \ref{subsect.(X)^n}, \ref{subsect.(X)alg} and \ref{subsect.9l1}) as well as
the proof of Proposition \ref{prop.Q_n} are lifted from my diploma thesis
\cite{dg0}, while Subsections \ref{subsect.9l2} and \ref{subsect.f(X)g} are
translated from an additional section of \cite{dg-hopf} which was written by me.

\subsection{Basic conventions}

Before we come to the actual body of this note, let us fix some conventions to
prevent misunderstandings from happening:

\begin{condition}
In this note, $\mathbb{N}$ will mean the set $\left\{  0,1,2,3,\ldots\right\}
$ (rather than the set $\left\{  1,2,3,\ldots\right\}  $, which is denoted by
$\mathbb{N}$ by various other authors).

For each $n\in\mathbb{N}$, we let $S_{n}$ denote the $n$-th symmetric group
(defined as the group of all permutations of the set $\left\{  1,2,\ldots
,n\right\}  $).
\end{condition}

\begin{condition}
\label{defs.ring}In this note, a \textit{ring} will always mean an associative
ring with $1$. If $k$ is a commutative ring, then a $k$-\textit{algebra} will
mean a (not necessarily commutative, but necessarily associative) $k$-algebra
with $1$. Sometimes we will use the word \textquotedblleft
algebra\textquotedblright\ as an abbreviation for \textquotedblleft%
$k$-algebra\textquotedblright. If $L$ is a $k$-algebra, then a \textit{left
}$L$\textit{-module} is always supposed to be a left $L$-module on which the
unity of $L$ acts as the identity. Whenever we use the tensor product sign
$\otimes$ without an index, we mean $\otimes_{k}$.
\end{condition}

\subsection{\label{subsect.(X)}Tensor products}

The goal of this note is \textit{not} to define tensor products; we assume
that the reader already knows what they are. But let us recall one possible
way to define the tensor product of several $k$-modules (assuming that the
tensor product of \textbf{two} $k$-modules is already defined):

\begin{definition}
\label{defs.(X)mult}Let $k$ be a commutative ring. Let $n\in\mathbb{N}%
$.\newline Now, by induction over $n$, we are going to define a $k$-module
$V_{1}\otimes V_{2}\otimes\cdots\otimes V_{n}$ for any $n$ arbitrary
$k$-modules $V_{1}$, $V_{2}$, $\ldots$, $V_{n}$:\newline\textit{Induction
base:} For $n=0$, we define $V_{1}\otimes V_{2}\otimes\cdots\otimes V_{n}$ as
the $k$-module $k$.\newline\textit{Induction step:} Let $p\in\mathbb{N}$.
Assuming that we have defined a $k$-module $V_{1}\otimes V_{2}\otimes
\cdots\otimes V_{p}$ for any $p$ arbitrary $k$-modules $V_{1}$, $V_{2}$,
$\ldots$, $V_{p}$, we now define a $k$-module $V_{1}\otimes V_{2}\otimes
\cdots\otimes V_{p+1}$ for any $p+1$ arbitrary $k$-modules $V_{1}$, $V_{2}$,
$\ldots$, $V_{p+1}$ by the equation%
\begin{equation}
V_{1}\otimes V_{2}\otimes\cdots\otimes V_{p+1}=V_{1}\otimes\left(
V_{2}\otimes V_{3}\otimes\cdots\otimes V_{p+1}\right)  .
\label{defs.(X)mult.left}%
\end{equation}
Here, $V_{1}\otimes\left(  V_{2}\otimes V_{3}\otimes\cdots\otimes
V_{p+1}\right)  $ is to be understood as the tensor product of the $k$-module
$V_{1}$ with the $k$-module $V_{2}\otimes V_{3}\otimes\cdots\otimes V_{p+1}$
(note that the $k$-module $V_{2}\otimes V_{3}\otimes\cdots\otimes V_{p+1}$ is
already defined because we assumed that we have defined a $k$-module
$V_{1}\otimes V_{2}\otimes\cdots\otimes V_{p}$ for any $p$ arbitrary
$k$-modules $V_{1}$, $V_{2}$, $\ldots$, $V_{p}$). This completes the inductive
definition.\newline Thus we have defined a $k$-module $V_{1}\otimes
V_{2}\otimes\cdots\otimes V_{n}$ for any $n$ arbitrary $k$-modules $V_{1}$,
$V_{2}$, $\ldots$, $V_{n}$ for any $n\in\mathbb{N}$. This $k$-module
$V_{1}\otimes V_{2}\otimes\cdots\otimes V_{n}$ is called the \textit{tensor
product} of the $k$-modules $V_{1}$, $V_{2}$, $\ldots$, $V_{n}$.
\end{definition}

\begin{remark}
\textbf{(a)} Definition \ref{defs.(X)mult} is not the only possible definition
of the tensor product of several $k$-modules. One could obtain a different
definition by replacing the equation (\ref{defs.(X)mult.left}) by%
\[
V_{1}\otimes V_{2}\otimes\cdots\otimes V_{p+1}=\left(  V_{1}\otimes
V_{2}\otimes\cdots\otimes V_{p}\right)  \otimes V_{p+1}.
\]
This definition would have given us a \textit{different} $k$-module
$V_{1}\otimes V_{2}\otimes\cdots\otimes V_{n}$ for any $n$ arbitrary
$k$-modules $V_{1}$, $V_{2}$, $\ldots$, $V_{n}$ for any $n\in\mathbb{N}$ than
the one defined in Definition \ref{defs.(X)mult}. However, this $k$-module
would still be \textit{canonically isomorphic} to the one defined in
Definition \ref{defs.(X)mult}, and thus it is commonly considered to be ``more
or less the same $k$-module''.\newline There is yet another definition of
$V_{1}\otimes V_{2}\otimes\cdots\otimes V_{n}$, which proceeds by taking the
free $k$-module on the set $V_{1}\times V_{2}\times\cdots\times V_{n}$ and
factoring it modulo a certain submodule. This definition gives yet
\textit{another} $k$-module $V_{1}\otimes V_{2}\otimes\cdots\otimes V_{n}$,
but this module is also canonically isomorphic to the $k$-module $V_{1}\otimes
V_{2}\otimes\cdots\otimes V_{n}$ defined in Definition \ref{defs.(X)mult}, and
thus can be considered to be ``more or less the same $k$-module''.\newline%
\textbf{(b)} Definition \ref{defs.(X)mult}, applied to $n=1$, defines the
tensor product of \textit{one} $k$-module $V_{1}$ as $V_{1}\otimes k$. This
takes some getting used to, since it seems more natural to define the tensor
product of one $k$-module $V_{1}$ simply as $V_{1}$. But this isn't really
different because there is a canonical isomorphism of $k$-modules $V_{1}\cong
V_{1}\otimes k$, so most people consider $V_{1}$ to be ``more or less the same
$k$-module'' as $V_{1}\otimes k$.
\end{remark}

\begin{condition}
\label{conv.left-ind}A remark about notation is appropriate at this
point:\newline There are two different conflicting notions of a ``pure
tensor'' in a tensor product $V_{1}\otimes V_{2}\otimes\cdots\otimes V_{n}$ of
$n$ arbitrary $k$-modules $V_{1}$, $V_{2}$, $\ldots$, $V_{n}$, where $n\geq1$.
The one notion defines a ``pure tensor'' as an element of the form $v\otimes
T$ for some $v\in V_{1}$ and some $T\in V_{2}\otimes V_{3}\otimes\cdots\otimes
V_{n}$\ \ \ \ {\footnotemark}. The other notion defines a ``pure tensor'' as
an element of the form $v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}$ for some
$\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V_{1}\times V_{2}\times
\cdots\times V_{n}$. These two notions are not equivalent. In this note, we
are going to yield right of way to the second of these notions, i. e. we are
going to define a pure tensor in $V_{1}\otimes V_{2}\otimes\cdots\otimes
V_{n}$ as an element of the form $v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}$ for some $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V_{1}\times
V_{2}\times\cdots\times V_{n}$. The first notion, however, will also be used -
but we will not call it a ``pure tensor'' but rather a ``left-induced
tensor''. Thus we define a \textit{left-induced tensor} in $V_{1}\otimes
V_{2}\otimes\cdots\otimes V_{n}$ as an element of the form $v\otimes T$ for
some $v\in V_{1}$ and some $T\in V_{2}\otimes V_{3}\otimes\cdots\otimes V_{n}%
$.\newline We note that the $k$-module $V_{1}\otimes V_{2}\otimes\cdots\otimes
V_{n}$ is generated by its left-induced tensors, but also generated by its
pure tensors.
\end{condition}

\footnotetext{In fact, if we look at Definition \ref{defs.(X)mult}, we see
that the $k$-module $V_{1}\otimes V_{2}\otimes\cdots\otimes V_{n}$ was defined
as $V_{1}\otimes\left(  V_{2}\otimes V_{3}\otimes\cdots\otimes V_{n}\right)
$, so it is the $k$-module $A\otimes B$ where $A=V_{1}$ and $B=V_{2}\otimes
V_{3}\otimes\cdots\otimes V_{n}$. Since the usual definition of a pure tensor
in $A\otimes B$ defines it as an element of the form $v\otimes T$ for some
$v\in A$ and $T\in B$, it thus is logical to say that a pure tensor in
$V_{1}\otimes V_{2}\otimes\cdots\otimes V_{n}$ means an element of the form
$v\otimes T$ for $v\in V_{1}$ and $T\in V_{2}\otimes V_{3}\otimes\cdots\otimes
V_{n}$.}

We also recall the definition of the tensor product of several $k$-module
homomorphisms (assuming that the notion of the tensor product of \textbf{two}
$k$-module homomorphisms is already defined):

\begin{definition}
\label{defs.f_1(X)f_2}Let $k$ be a commutative ring. Let $n\in\mathbb{N}%
$.\newline Now, by induction over $n$, we are going to define a $k$-module
homomorphism $f_{1}\otimes f_{2}\otimes\cdots\otimes f_{n}:V_{1}\otimes
V_{2}\otimes\cdots\otimes V_{n}\rightarrow W_{1}\otimes W_{2}\otimes
\cdots\otimes W_{n}$ whenever $V_{1}$, $V_{2}$, $\ldots$, $V_{n}$ are $n$
arbitrary $k$-modules, $W_{1}$, $W_{2}$, $\ldots$, $W_{n}$ are $n$ arbitrary
$k$-modules, and $f_{1}:V_{1}\rightarrow W_{1}$, $f_{2}:V_{2}\rightarrow
W_{2}$, $\ldots$, $f_{n}:V_{n}\rightarrow W_{n}$ are $n$ arbitrary $k$-module
homomorphisms:\newline\textit{Induction base:} For $n=0$, we define
$f_{1}\otimes f_{2}\otimes\cdots\otimes f_{n}$ as the identity map
$\operatorname*{id}:k\rightarrow k$.\newline\textit{Induction step:} Let
$p\in\mathbb{N}$. Assume that we have defined a $k$-module homomorphism
$f_{1}\otimes f_{2}\otimes\cdots\otimes f_{p}:V_{1}\otimes V_{2}\otimes
\cdots\otimes V_{p}\rightarrow W_{1}\otimes W_{2}\otimes\cdots\otimes W_{p}$
whenever $V_{1}$, $V_{2}$, $\ldots$, $V_{p}$ are $p$ arbitrary $k$-modules,
$W_{1}$, $W_{2}$, $\ldots$, $W_{p}$ are $p$ arbitrary $k$-modules, and
$f_{1}:V_{1}\rightarrow W_{1}$, $f_{2}:V_{2}\rightarrow W_{2}$, $\ldots$,
$f_{p}:V_{p}\rightarrow W_{p}$ are $p$ arbitrary $k$-module homomorphisms. Now
let us define a $k$-module homomorphism $f_{1}\otimes f_{2}\otimes
\cdots\otimes f_{p+1}:V_{1}\otimes V_{2}\otimes\cdots\otimes V_{p+1}%
\rightarrow W_{1}\otimes W_{2}\otimes\cdots\otimes W_{p+1}$ whenever $V_{1}$,
$V_{2}$, $\ldots$, $V_{p+1}$ are $p+1$ arbitrary $k$-modules, $W_{1}$, $W_{2}%
$, $\ldots$, $W_{p+1}$ are $p+1$ arbitrary $k$-modules, and $f_{1}%
:V_{1}\rightarrow W_{1}$, $f_{2}:V_{2}\rightarrow W_{2}$, $\ldots$,
$f_{p+1}:V_{p+1}\rightarrow W_{p+1}$ are $p+1$ arbitrary $k$-module
homomorphisms. Namely, we define this homomorphism $f_{1}\otimes f_{2}%
\otimes\cdots\otimes f_{p+1}$ to be $f_{1}\otimes\left(  f_{2}\otimes
f_{3}\otimes\cdots\otimes f_{p+1}\right)  $.\newline Here, $f_{1}%
\otimes\left(  f_{2}\otimes f_{3}\otimes\cdots\otimes f_{p+1}\right)  $ is to
be understood as the tensor product of the $k$-module homomorphism
$f_{1}:V_{1}\rightarrow W_{1}$ with the $k$-module homomorphism $f_{2}\otimes
f_{3}\otimes\cdots\otimes f_{p+1}:V_{2}\otimes V_{3}\otimes\cdots\otimes
V_{p+1}\rightarrow W_{2}\otimes W_{3}\otimes\cdots\otimes W_{p+1}$ (note that
the $k$-module homomorphism $f_{2}\otimes f_{3}\otimes\cdots\otimes
f_{p+1}:V_{2}\otimes V_{3}\otimes\cdots\otimes V_{p+1}\rightarrow W_{2}\otimes
W_{3}\otimes\cdots\otimes W_{p+1}$ is already defined (because we assumed that
we have defined a $k$-module homomorphism $f_{1}\otimes f_{2}\otimes
\cdots\otimes f_{p}:V_{1}\otimes V_{2}\otimes\cdots\otimes V_{p}\rightarrow
W_{1}\otimes W_{2}\otimes\cdots\otimes W_{p}$ whenever $V_{1}$, $V_{2}$,
$\ldots$, $V_{p}$ are $p$ arbitrary $k$-modules, $W_{1}$, $W_{2}$, $\ldots$,
$W_{p}$ are $p$ arbitrary $k$-modules, and $f_{1}:V_{1}\rightarrow W_{1}$,
$f_{2}:V_{2}\rightarrow W_{2}$, $\ldots$, $f_{p}:V_{p}\rightarrow W_{p}$ are
$p$ arbitrary $k$-module homomorphisms)). This completes the inductive
definition.\newline Thus we have defined a $k$-module homomorphism
$f_{1}\otimes f_{2}\otimes\cdots\otimes f_{n}:V_{1}\otimes V_{2}\otimes
\cdots\otimes V_{n}\rightarrow W_{1}\otimes W_{2}\otimes\cdots\otimes W_{n}$
whenever $V_{1}$, $V_{2}$, $\ldots$, $V_{n}$ are $n$ arbitrary $k$-modules,
$W_{1}$, $W_{2}$, $\ldots$, $W_{n}$ are $n$ arbitrary $k$-modules, and
$f_{1}:V_{1}\rightarrow W_{1}$, $f_{2}:V_{2}\rightarrow W_{2}$, $\ldots$,
$f_{n}:V_{n}\rightarrow W_{n}$ are $n$ arbitrary $k$-module homomorphisms.
This $k$-module homomorphism $f_{1}\otimes f_{2}\otimes\cdots\otimes f_{n}$ is
called the \textit{tensor product} of the $k$-module homomorphisms $f_{1}$,
$f_{2}$, $\ldots$, $f_{n}$.
\end{definition}

Finally let us agree on a rather harmless abuse of notation:

\begin{condition}
\label{conv.V(X)k}Let $k$ be a commutative ring. Let $V$ be a $k$%
-module.\newline We are going to identify the three $k$-modules $V\otimes k$,
$k\otimes V$ and $V$ with each other (due to the canonical isomorphisms
$V\rightarrow V\otimes k$ and $V\rightarrow k\otimes V$).
\end{condition}

\subsection{\label{subsect.(X)^n}Tensor powers of $k$-modules}

Next we define a particular case of tensor products of $k$-modules, namely the
tensor powers. Here is the classical definition of this notion:

\begin{definition}
\label{defs.(X)^n}Let $k$ be a commutative ring. Let $n\in\mathbb{N}$. For any
$k$-module $V$, we define a $k$-module $V^{\otimes n}$ by $V^{\otimes
n}=\underbrace{V\otimes V\otimes\cdots\otimes V}_{n\text{ times}}$. This
$k$-module $V^{\otimes n}$ is called the $n$\textit{-th tensor power} of the
$k$-module $V$.
\end{definition}

\begin{remark}
\label{rmk.(X)^0}Let $k$ be a commutative ring, and let $V$ be a $k$-module.
Then, $V^{\otimes0}=k$ (because $V^{\otimes n}=\underbrace{V\otimes
V\otimes\cdots\otimes V}_{0\text{ times}}=\left(  \text{tensor product of zero
}k\text{-modules}\right)  =k$ according to the induction base of Definition
\ref{defs.(X)mult}) and $V^{\otimes1}=V\otimes k$ (because $V^{\otimes
1}=\underbrace{V\otimes V\otimes\cdots\otimes V}_{1\text{ times}}=V\otimes k$
according to the induction step of Definition \ref{defs.(X)mult}). Since we
identify $V\otimes k$ with $V$, we thus have $V^{\otimes1}=V$.
\end{remark}

\begin{condition}
\label{conv.f^(X)n}Let $k$ be a commutative ring. Let $n\in\mathbb{N}$. Let
$V$ and $V^{\prime}$ be $k$-modules, and let $f:V\rightarrow V^{\prime}$ be a
$k$-module homomorphism. Then, $f^{\otimes n}$ denotes the $k$-module
homomorphism $\underbrace{f\otimes f\otimes\cdots\otimes f}_{n\text{ times}%
}:\underbrace{V\otimes V\otimes\cdots\otimes V}_{n\text{ times}}%
\rightarrow\underbrace{V^{\prime}\otimes V^{\prime}\otimes\cdots\otimes
V^{\prime}}_{n\text{ times}}$. Since $\underbrace{V\otimes V\otimes
\cdots\otimes V}_{n\text{ times}}=V^{\otimes n}$ and $\underbrace{V^{\prime
}\otimes V^{\prime}\otimes\cdots\otimes V^{\prime}}_{n\text{ times}}%
=V^{\prime\otimes n}$, this $f^{\otimes n}$ is thus a $k$-module homomorphism
from $V^{\otimes n}$ to $V^{\prime\otimes n}$.
\end{condition}

\subsection{\label{subsect.(X)alg}The tensor algebra}

First let us agree on a convention which simplifies working with direct sums:

\begin{condition}
\label{conv.(++).inj}Let $k$ be a commutative ring. Let $S$ be a set. For
every $s\in S$, let $V_{s}$ be a $k$-module. For every $t\in S$, we are going
to identify the $k$-module $V_{t}$ with the image of $V_{t}$ under the
canonical injection $V_{t}\rightarrow\bigoplus\limits_{s\in S}V_{s}$. This is
an abuse of notation, but a relatively harmless one. It allows us to consider
$V_{t}$ as a $k$-submodule of the direct sum $\bigoplus\limits_{s\in S}V_{s}$.
\end{condition}

Secondly, we make a convention that simplifies working with the tensor powers
of a $k$-module:

\begin{condition}
\label{conv.(X)^n.ident}Let $k$ be a commutative ring. For every $k$-module
$V$, every $n\in\mathbb{N}$ and every $i\in\left\{  0,1,\ldots,n\right\}  $,
we are going to identify the $k$-module $V^{\otimes i}\otimes V^{\otimes
\left(  n-i\right)  }$ with the $k$-module $V^{\otimes n}$ (using the
canonical isomorphism $V^{\otimes i}\otimes V^{\otimes\left(  n-i\right)
}\cong V^{\otimes n}$). In other words, for every $k$-module $V$, every
$a\in\mathbb{N}$ and every $b\in\mathbb{N}$, we are going to identify the
$k$-module $V^{\otimes a}\otimes V^{\otimes b}$ with the $k$-module
$V^{\otimes\left(  a+b\right)  }$.
\end{condition}

The tensor powers $V^{\otimes n}$ of a $k$-module $V$ can be combined to a
$k$-module $\otimes V$ which turns out to have an algebra structure: that of
the so-called tensor algebra. Let us recall its definition (which can easily
shown to be well-defined):

\begin{definition}
\label{defs.tensor}Let $k$ be a commutative ring.\newline\textbf{(a)} Let $V$
be a $k$-module. The \textit{tensor algebra} $\otimes V$ of $V$ over $k$ is
defined to be the $k$-algebra formed by the $k$-module $\bigoplus
\limits_{i\in\mathbb{N}}V^{\otimes i}=V^{\otimes0}\oplus V^{\otimes1}\oplus
V^{\otimes2}\oplus\cdots$ equipped with a multiplication which is defined by%
\begin{equation}
\left(
\begin{array}
[c]{l}%
\left(  a_{i}\right)  _{i\in\mathbb{N}}\cdot\left(  b_{i}\right)
_{i\in\mathbb{N}}=\left(  \sum\limits_{i=0}^{n}a_{i}\otimes b_{n-i}\right)
_{n\in\mathbb{N}}\\
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  a_{i}\right)  _{i\in\mathbb{N}%
}\in\bigoplus\limits_{i\in\mathbb{N}}V^{\otimes i}\text{ and }\left(
b_{i}\right)  _{i\in\mathbb{N}}\in\bigoplus\limits_{i\in\mathbb{N}}V^{\otimes
i}%
\end{array}
\right)  \label{defs.tensor.mult}%
\end{equation}
(where for every $n\in\mathbb{N}$ and every $i\in\left\{  0,1,\ldots
,n\right\}  $, the tensor $a_{i}\otimes b_{n-i}\in V^{\otimes i}\otimes
V^{\otimes\left(  n-i\right)  }$ is considered as an element of $V^{\otimes
n}$ due to the canonical identification $V^{\otimes i}\otimes V^{\otimes
\left(  n-i\right)  }\cong V^{\otimes n}$ which was defined in Convention
\ref{conv.(X)^n.ident}).\newline The $k$-module $\otimes V$ itself (without
the $k$-algebra structure) is called the \textit{tensor module} of
$V$.\newline\textbf{(b)} Let $V$ and $W$ be two $k$-modules, and let
$f:V\rightarrow W$ be a $k$-module homomorphism. The $k$-module homomorphisms
$f^{\otimes i}:V^{\otimes i}\rightarrow W^{\otimes i}$ for all $i\in
\mathbb{N}$ can be combined together to a $k$-module homomorphism from
$V^{\otimes0}\oplus V^{\otimes1}\oplus V^{\otimes2}\oplus\cdots$ to
$W^{\otimes0}\oplus W^{\otimes1}\oplus W^{\otimes2}\oplus\cdots$. This
homomorphism is called $\otimes f$. Since $V^{\otimes0}\oplus V^{\otimes
1}\oplus V^{\otimes2}\oplus\cdots=\otimes V$ and $W^{\otimes0}\oplus
W^{\otimes1}\oplus W^{\otimes2}\oplus\cdots=\otimes W$, we see that this
homomorphism $\otimes f$ is a $k$-module homomorphism from $\otimes V$ to
$\otimes W$. Moreover, it follows easily from (\ref{defs.tensor.mult}) that
this $\otimes f$ is actually a $k$-algebra homomorphism from $\otimes V$ to
$\otimes W$.\newline\textbf{(c)} Let $V$ be a $k$-module. Then, according to
Convention \ref{conv.(++).inj}, we consider $V^{\otimes n}$ as a $k$-submodule
of the direct sum $\bigoplus\limits_{i\in\mathbb{N}}V^{\otimes i}=\otimes V$
for every $n\in\mathbb{N}$. In particular, every element of $k$ is considered
to be an element of $\otimes V$ by means of the canonical embedding
$k=V^{\otimes0}\subseteq\otimes V$, and every element of $V$ is considered to
be an element of $\otimes V$ by means of the canonical embedding
$V=V^{\otimes1}\subseteq\otimes V$. The element $1\in k\subseteq\otimes V$ is
easily seen to be the unity of the tensor algebra $\otimes V$.
\end{definition}

\begin{remark}
The formula (\ref{defs.tensor.mult}) (which defines the multiplication on the
tensor algebra $\otimes V$) is often put in words by saying that
\textquotedblleft the multiplication in the tensor algebra $\otimes V$ is
given by the tensor product\textquotedblright. This informal statement tempts
many authors (including myself in \cite{dg1}) to use the sign $\otimes$ for
multiplication in the algebra $\otimes V$, that is, to write $u\otimes v$ for
the product of any two elements $u$ and $v$ of the tensor algebra $\otimes V$.
This notation, however, can collide with the notation $u\otimes v$ for the
tensor product of two vectors $u$ and $v$ in a $k$-module.{\footnotemark} Due
to this possibility of collision, we are \textit{not} going to use the sign
$\otimes$ for multiplication in the algebra $\otimes V$ in this paper. Instead
we will use the sign $\cdot$ for this multiplication. However, due to
(\ref{defs.tensor.mult}), we still have%
\begin{equation}
\left(  a\cdot b=a\otimes b\ \ \ \ \ \ \ \ \ \ \text{for any }n\in
\mathbb{N}\text{, any }m\in\mathbb{N}\text{, any }a\in V^{\otimes n}\text{ and
any }b\in V^{\otimes m}\right)  , \label{*=(X)}%
\end{equation}
where $a\otimes b$ is considered to be an element of $V^{\otimes\left(
n+m\right)  }$ by means of the identification of $V^{\otimes n}\otimes
V^{\otimes m}$ with $V^{\otimes\left(  n+m\right)  }$.

The $k$-algebra $\otimes V$ is also denoted by $T\left(  V\right)  $ by many authors.
\end{remark}

\footnotetext{For example, if $z$ is a vector in the $k$-module $V$, then we
can define two elements $u$ and $v$ of $\otimes V$ by $u=1+z$ and $v=1-z$
(where $1$ and $z$ are considered to be elements of $\otimes V$ according to
Definition \ref{defs.tensor} \textbf{(c)}), and while the product of these
elements $u$ and $v$ in $\otimes V$ is the element $\left(  1+z\right)
\cdot\left(  1-z\right)  =1\cdot1-1\cdot z+1\cdot z-z\otimes z=1-z\otimes
z\in\otimes V$, the tensor product of these elements $u$ and $v$ is the
element $\left(  1+z\right)  \otimes\left(  1-z\right)  $ of $\left(  k\oplus
V\right)  \otimes\left(  k\oplus V\right)  \cong k\oplus V\oplus
V\oplus\left(  V\otimes V\right)  $, which is a different element of a totally
different $k$-module. So if we would use one and the same notation $u\otimes
v$ for both the product of $u$ and $v$ in $\otimes V$ and the tensor product
of $u$ and $v$ in $\left(  k\oplus V\right)  \otimes\left(  k\oplus V\right)
$, we would have ambiguous notations.}

\subsection{\label{subsect.9l1}A variation on the nine lemma}

The following fact is one of several algebraic statements related to the nine
lemma, but having both weaker assertions and weaker conditions. We record it
here to use it later:

\begin{proposition}
\label{prop.9l}Let $k$ be a commutative ring. Let $A$, $B$, $C$ and $D$ be
$k$-modules, and let $x:A\rightarrow B$, $y:A\rightarrow C$, $z:B\rightarrow
D$ and $w:C\rightarrow D$ be $k$-linear maps such that the diagram%
\begin{equation}%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%A \ar[r]^x \ar[d]_y & B \ar[d]^z \\
%C \ar[r]_w & D
%}} }%
%BeginExpansion
\xymatrix{
A \ar[r]^x \ar[d]_y & B \ar[d]^z \\
C \ar[r]_w & D
}
%EndExpansion
\label{prop.9l.diag}%
\end{equation}
commutes. Assume that $\operatorname*{Ker}z\subseteq x\left(
\operatorname*{Ker}y\right)  $. Further assume that $y$ is surjective. Then,
$\operatorname*{Ker}w=y\left(  \operatorname*{Ker}x\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.9l}.]We know that the diagram%
\[%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%A \ar[r]^x \ar[d]_y & B \ar[d]^z \\
%C \ar[r]_w & D
%}}}%
%BeginExpansion
\xymatrix{
A \ar[r]^x \ar[d]_y & B \ar[d]^z \\
C \ar[r]_w & D
}%
%EndExpansion
\]
commutes. In other words, $w\circ y=z\circ x$.

We have%
\begin{align*}
w\left(  y\left(  \operatorname*{Ker}x\right)  \right)   &
=\underbrace{\left(  w\circ y\right)  }_{=z\circ x}\left(  \operatorname*{Ker}%
x\right)  =\left(  z\circ x\right)  \left(  \operatorname*{Ker}x\right)
=z\left(  \underbrace{x\left(  \operatorname*{Ker}x\right)  }_{=0}\right)
=z\left(  0\right)  =0\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }z\text{ is }k\text{-linear}%
\right)  ,
\end{align*}
and thus $y\left(  \operatorname*{Ker}x\right)  \subseteq\operatorname*{Ker}%
w$. We will now prove that $\operatorname*{Ker}w\subseteq y\left(
\operatorname*{Ker}x\right)  $:

Let $c\in\operatorname*{Ker}w$ be arbitrary. Then, $w\left(  c\right)  =0$.
Now, since $y$ is surjective, there exists some $a\in A$ such that $c=y\left(
a\right)  $. Consider this $a$. Then,
\[
0=w\left(  \underbrace{c}_{=y\left(  a\right)  }\right)  =w\left(  y\left(
a\right)  \right)  =\underbrace{\left(  w\circ y\right)  }_{=z\circ x}\left(
a\right)  =\left(  z\circ x\right)  \left(  a\right)  =z\left(  x\left(
a\right)  \right)  ,
\]
so that $x\left(  a\right)  \in\operatorname*{Ker}z\subseteq x\left(
\operatorname*{Ker}y\right)  $. Thus, there exists some $a^{\prime}%
\in\operatorname*{Ker}y$ such that $x\left(  a\right)  =x\left(  a^{\prime
}\right)  $. Consider this $a^{\prime}$. Since $x$ is $k$-linear, we have
$x\left(  a-a^{\prime}\right)  =\underbrace{x\left(  a\right)  }_{=x\left(
a^{\prime}\right)  }-x\left(  a^{\prime}\right)  =x\left(  a^{\prime}\right)
-x\left(  a^{\prime}\right)  =0$, so that $a-a^{\prime}\in\operatorname*{Ker}%
x$. Thus, $y\left(  a-a^{\prime}\right)  \in y\left(  \operatorname*{Ker}%
x\right)  $. But since%
\begin{align*}
y\left(  a-a^{\prime}\right)   &  =\underbrace{y\left(  a\right)  }%
_{=c}-\underbrace{y\left(  a^{\prime}\right)  }_{=0\text{ (since }a^{\prime
}\in\operatorname*{Ker}y\text{)}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}y\text{ is }k\text{-linear}\right) \\
&  =c-0=c,
\end{align*}
this rewrites as $c\in y\left(  \operatorname*{Ker}x\right)  $.

We have thus shown that every $c\in\operatorname*{Ker}w$ satisfies $c\in
y\left(  \operatorname*{Ker}x\right)  $. Thus, $\operatorname*{Ker}w\subseteq
y\left(  \operatorname*{Ker}x\right)  $. Combined with $y\left(
\operatorname*{Ker}x\right)  \subseteq\operatorname*{Ker}w$, this yields
$\operatorname*{Ker}w=y\left(  \operatorname*{Ker}x\right)  $. This proves
Proposition \ref{prop.9l}.
\end{proof}

Note that we would not lose any generality if we would replace $k$ by
$\mathbb{Z}$ in the statement of Proposition \ref{prop.9l}, because every
$k$-module is an abelian group, i. e., a $\mathbb{Z}$-module (with additional
structure). We could actually generalize Proposition \ref{prop.9l} by
replacing ``$k$-modules'' by ``groups'' (not necessarily abelian), but we will
not have any use for Proposition \ref{prop.9l} in this generality here.

\subsection{\label{subsect.9l2}Another diagram theorem about the nine lemma
configuration}

The next fact we will use is, again, about the nine lemma configuration:

\begin{proposition}
\label{prop.9l2}Let $k$ be a ring. Let
\begin{equation}%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{5pc}}}%
%BeginExpansion
\xymatrixcolsep{5pc}%
%EndExpansion%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%A_1 \ar[r]^{a_1} \ar[d]_{u_1}   & A_2 \ar[r]^{a_2} \ar[d]_{u_2}
%& A_3 \ar[r] \ar[d]_{u_3}                   & 0 \\
%B_1 \ar[r]^{b_1} \ar[d]_{v_1}   & B_2 \ar[r]^{b_2} \ar[d]_{v_2}
%& B_3 \ar[r] \ar[d]_{v_3}                   & 0 \\
%C_1 \ar[r]^{c_1}              & C_2 \ar[r]^{c_2}
%& C_3 \ar[r]                              & 0
%}} }%
%BeginExpansion
\xymatrix{
A_1 \ar[r]^{a_1} \ar[d]_{u_1}   & A_2 \ar[r]^{a_2} \ar[d]_{u_2}
& A_3 \ar[r] \ar[d]_{u_3}                   & 0 \\
B_1 \ar[r]^{b_1} \ar[d]_{v_1}   & B_2 \ar[r]^{b_2} \ar[d]_{v_2}
& B_3 \ar[r] \ar[d]_{v_3}                   & 0 \\
C_1 \ar[r]^{c_1}              & C_2 \ar[r]^{c_2}
& C_3 \ar[r]                              & 0
}
%EndExpansion
\label{prop.9l2.diag.real}%
\end{equation}
be a commutative diagram of $k$-left modules. Assume that every row of the
diagram (\ref{prop.9l2.diag}) is an exact sequence, and that every column of
the diagram (\ref{prop.9l2.diag}) is an exact sequence. Then,%
\[
\operatorname*{Ker}\left(  c_{2}\circ v_{2}\right)  =\operatorname*{Ker}%
\left(  v_{3}\circ b_{2}\right)  =b_{1}\left(  B_{1}\right)  +u_{2}\left(
A_{2}\right)  .
\]

\end{proposition}

Actually we will show something a bit stronger:

\begin{proposition}
\label{prop.9l3}Let $k$ be a ring. Let
\begin{equation}%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{5pc}}}%
%BeginExpansion
\xymatrixcolsep{5pc}%
%EndExpansion%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%A_1 \ar[r]^{a_1} \ar[d]_{u_1}   & A_2 \ar[r]^{a_2} \ar[d]_{u_2}
%& A_3 \ar[d]_{u_3}                  \\
%B_1 \ar[r]^{b_1} \ar[d]_{v_1}   & B_2 \ar[r]^{b_2} \ar[d]_{v_2}
%& B_3 \ar[d]_{v_3}                   \\
%C_1 \ar[r]^{c_1}              & C_2 \ar[r]^{c_2}
%& C_3
%}} }%
%BeginExpansion
\xymatrix{
A_1 \ar[r]^{a_1} \ar[d]_{u_1}   & A_2 \ar[r]^{a_2} \ar[d]_{u_2}
& A_3 \ar[d]_{u_3}                  \\
B_1 \ar[r]^{b_1} \ar[d]_{v_1}   & B_2 \ar[r]^{b_2} \ar[d]_{v_2}
& B_3 \ar[d]_{v_3}                   \\
C_1 \ar[r]^{c_1}              & C_2 \ar[r]^{c_2}
& C_3
}
%EndExpansion
\label{prop.9l2.diag}%
\end{equation}
be a commutative diagram of $k$-left modules. Assume that every row of the
diagram (\ref{prop.9l2.diag}) is an exact sequence, and that every column of
the diagram (\ref{prop.9l2.diag}) is an exact sequence. Also assume that
$a_{2}$ is surjective. Then,%
\[
\operatorname*{Ker}\left(  c_{2}\circ v_{2}\right)  =\operatorname*{Ker}%
\left(  v_{3}\circ b_{2}\right)  =b_{1}\left(  B_{1}\right)  +u_{2}\left(
A_{2}\right)  .
\]

\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.9l3}.]Since (\ref{prop.9l2.diag}) is a
commutative diagram, we have $c_{2}\circ v_{2}=v_{3}\circ b_{2}$ and
$b_{2}\circ u_{2}=u_{3}\circ a_{2}$.

Since every row of the diagram (\ref{prop.9l2.diag}) is an exact sequence, we
have $b_{2}\circ b_{1}=0$.

Since every column of the diagram (\ref{prop.9l2.diag}) is an exact sequence,
we have $v_{3}\circ u_{3}=0$.

From $c_{2}\circ v_{2}=v_{3}\circ b_{2}$, we conclude $\operatorname*{Ker}%
\left(  c_{2}\circ v_{2}\right)  =\operatorname*{Ker}\left(  v_{3}\circ
b_{2}\right)  $. Thus, it only remains to prove that $\operatorname*{Ker}%
\left(  v_{3}\circ b_{2}\right)  =b_{1}\left(  B_{1}\right)  +u_{2}\left(
A_{2}\right)  $. Since $b_{1}\left(  B_{1}\right)  +u_{2}\left(  A_{2}\right)
\subseteq\operatorname*{Ker}\left(  v_{3}\circ b_{2}\right)  $ is obvious
(because%
\begin{align*}
&  \left(  v_{3}\circ b_{2}\right)  \left(  b_{1}\left(  B_{1}\right)
+u_{2}\left(  A_{2}\right)  \right) \\
&  =v_{3}\left(  b_{2}\left(  b_{1}\left(  B_{1}\right)  +u_{2}\left(
A_{2}\right)  \right)  \right)  =\underbrace{v_{3}\left(  b_{2}\left(
b_{1}\left(  B_{1}\right)  \right)  \right)  }_{=v_{3}\left(  \left(
b_{2}\circ b_{1}\right)  \left(  B_{1}\right)  \right)  }+\underbrace{v_{3}%
\left(  b_{2}\left(  u_{2}\left(  A_{2}\right)  \right)  \right)  }_{=\left(
v_{3}\circ b_{2}\circ u_{2}\right)  \left(  A_{2}\right)  }\\
&  =v_{3}\left(  \underbrace{\left(  b_{2}\circ b_{1}\right)  }_{=0}\left(
B_{1}\right)  \right)  +\left(  v_{3}\circ\underbrace{b_{2}\circ u_{2}%
}_{=u_{3}\circ a_{2}}\right)  \left(  A_{2}\right)  =\underbrace{v_{3}\left(
0\left(  B_{1}\right)  \right)  }_{=0}+\left(  \underbrace{v_{3}\circ u_{3}%
}_{=0}\circ a_{2}\right)  \left(  A_{2}\right) \\
&  =0+\underbrace{\left(  0\circ a_{2}\right)  \left(  A_{2}\right)  }_{=0}=0
\end{align*}
), we must now only show that $\operatorname*{Ker}\left(  v_{3}\circ
b_{2}\right)  \subseteq b_{1}\left(  B_{1}\right)  +u_{2}\left(  A_{2}\right)
$.

Let $t\in\operatorname*{Ker}\left(  v_{3}\circ b_{2}\right)  $ be arbitrary.
Then, $\left(  v_{3}\circ b_{2}\right)  \left(  t\right)  =0$, so that
$v_{3}\left(  b_{2}\left(  t\right)  \right)  =\left(  v_{3}\circ
b_{2}\right)  \left(  t\right)  =0$ and thus $b_{2}\left(  t\right)
\in\operatorname*{Ker}v_{3}=u_{3}\left(  A_{3}\right)  $ (because every column
of the diagram (\ref{prop.9l2.diag}) is an exact sequence). Thus, there exists
some $x\in A_{3}$ such that $b_{2}\left(  t\right)  =u_{3}\left(  x\right)  $.
Consider this $x$. Since $a_{2}:A_{2}\rightarrow A_{3}$ is surjective, we have
$x=a_{2}\left(  x^{\prime}\right)  $ for some $x^{\prime}\in A_{2}$. Consider
this $x^{\prime}$. Now,%
\[
b_{2}\left(  t-u_{2}\left(  x^{\prime}\right)  \right)  =\underbrace{b_{2}%
\left(  t\right)  }_{=u_{3}\left(  x\right)  }-\underbrace{b_{2}\left(
u_{2}\left(  x^{\prime}\right)  \right)  }_{=\left(  b_{2}\circ u_{2}\right)
\left(  x^{\prime}\right)  }=u_{3}\left(  x\right)  -\underbrace{\left(
b_{2}\circ u_{2}\right)  }_{=u_{3}\circ a_{2}}\left(  x^{\prime}\right)
=u_{3}\left(  x\right)  -u_{3}\left(  \underbrace{a_{2}\left(  x^{\prime
}\right)  }_{=x}\right)  =0,
\]
so that $t-u_{2}\left(  x^{\prime}\right)  \in\operatorname*{Ker}b_{2}%
=b_{1}\left(  B_{1}\right)  $ (because every row of the diagram
(\ref{prop.9l2.diag}) is an exact sequence). Thus, $t=\underbrace{t-u_{2}%
\left(  x^{\prime}\right)  }_{\in b_{1}\left(  B_{1}\right)  }%
+\underbrace{u_{2}\left(  x^{\prime}\right)  }_{\in u_{2}\left(  A_{2}\right)
}\in b_{1}\left(  B_{1}\right)  +u_{2}\left(  A_{2}\right)  $.

We thus have shown that every $t\in\operatorname*{Ker}\left(  v_{3}\circ
b_{2}\right)  $ satisfies $t\in b_{1}\left(  B_{1}\right)  +u_{2}\left(
A_{2}\right)  $. Consequently, $\operatorname*{Ker}\left(  v_{3}\circ
b_{2}\right)  \subseteq b_{1}\left(  B_{1}\right)  +u_{2}\left(  A_{2}\right)
$. Combined with $b_{1}\left(  B_{1}\right)  +u_{2}\left(  A_{2}\right)
\subseteq\operatorname*{Ker}\left(  v_{3}\circ b_{2}\right)  $, this yields
$\operatorname*{Ker}\left(  v_{3}\circ b_{2}\right)  =b_{1}\left(
B_{1}\right)  +u_{2}\left(  A_{2}\right)  $. Combined with
$\operatorname*{Ker}\left(  c_{2}\circ v_{2}\right)  =\operatorname*{Ker}%
\left(  v_{3}\circ b_{2}\right)  $, this now completes the proof of
Proposition \ref{prop.9l3}.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.9l2}.]Since the diagram
(\ref{prop.9l2.diag.real}) is commutative, the diagram (\ref{prop.9l2.diag})
must also be commutative (because the diagram (\ref{prop.9l2.diag}) is a
subdiagram of the diagram (\ref{prop.9l2.diag.real})). Also, the map $a_{2}$
is surjective (since every row of the diagram (\ref{prop.9l2.diag.real}) is an
exact sequence). Therefore, we can apply Proposition \ref{prop.9l3}, and
conclude that $\operatorname*{Ker}\left(  c_{2}\circ v_{2}\right)
=\operatorname*{Ker}\left(  v_{3}\circ b_{2}\right)  =b_{1}\left(
B_{1}\right)  +u_{2}\left(  A_{2}\right)  $. This proves Proposition
\ref{prop.9l2}.
\end{proof}

\subsection{\label{subsect.f(X)g}$\operatorname*{Ker}\left(  f\otimes
g\right)  $ when $f$ and $g$ are surjective}

\begin{theorem}
\label{thm.f(X)g}Let $k$ be a commutative ring. Let $V$, $W$, $V^{\prime}$ and
$W^{\prime}$ be four $k$-modules. Let $f:V\rightarrow V^{\prime}$ and
$g:W\rightarrow W^{\prime}$ be two surjective $k$-linear maps. Let $i_{V}$ be
the canonical inclusion $\operatorname*{Ker}f\rightarrow V$. Let $i_{W}$ be
the canonical inclusion $\operatorname*{Ker}g\rightarrow W$. Then,%
\[
\operatorname*{Ker}\left(  f\otimes g\right)  =\left(  i_{V}\otimes
\operatorname*{id}\right)  \left(  \left(  \operatorname*{Ker}f\right)
\otimes W\right)  +\left(  \operatorname*{id}\otimes i_{W}\right)  \left(
V\otimes\left(  \operatorname*{Ker}g\right)  \right)  .
\]

\end{theorem}

\begin{remark}
\textbf{(a)} In Theorem \ref{thm.f(X)g}, the condition that $f$ and $g$ be
surjective cannot be removed (otherwise, $V=\mathbb{Z}$, $W=\mathbb{Z}$,
$V^{\prime}=\mathbb{Z}\diagup4\mathbb{Z}$, $W^{\prime}=\mathbb{Z}%
\diagup4\mathbb{Z}$, $f=\left(  x\mapsto\overline{2x}\right)  $, $g=\left(
x\mapsto\overline{2x}\right)  $ would be a counterexample), but it can be
replaced by some other conditions (see Lemma \ref{lem.f(X)g} and Corollary
\ref{cor.f(X)g}). (Here is a more complicated counterexample to show that
having only $g$ surjective is not yet enough: $V=\mathbb{Z}$, $W=\mathbb{Z}%
\oplus\left(  \mathbb{Z}\diagup4\mathbb{Z}\right)  $, $V^{\prime}=\mathbb{Z}$,
$W^{\prime}=\mathbb{Z}\diagup4\mathbb{Z}$, $f=\left(  x\mapsto2x\right)  $,
$g=\left(  \left(  x,\alpha\right)  \mapsto\overline{2x}+\alpha\right)
$.)\newline\textbf{(b)} If the $k$-module $V$ is flat in Theorem
\ref{thm.f(X)g}, then the map $\operatorname*{id}\otimes i_{W}$ is injective
(as can be easily seen), and therefore many people prefer to identify the
image $\left(  \operatorname*{id}\otimes i_{W}\right)  \left(  V\otimes\left(
\operatorname*{Ker}g\right)  \right)  $ with $V\otimes\left(
\operatorname*{Ker}g\right)  $. Similarly, the image $\left(  i_{V}%
\otimes\operatorname*{id}\right)  \left(  \left(  \operatorname*{Ker}f\right)
\otimes W\right)  $ can be identified with $\left(  \operatorname*{Ker}%
f\right)  \otimes W$ when the $k$-module $W$ is flat. It is common among
algebraists to perform these identifications when $k$ is a field (because when
$k$ is a field, both $k$-modules $V$ and $W$ are flat), and sometimes even
when $k$ is not, but we will not perform these identifications here.
\end{remark}

\begin{proof}
[Proof of Theorem \ref{thm.f(X)g}.]The sequence%
\[%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%0 \ar[r] & \Ker f \ar[r]^-{i_V} & V \ar[r]^{f} & V^{\prime} \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrix{
0 \ar[r] & \Ker f \ar[r]^-{i_V} & V \ar[r]^{f} & V^{\prime} \ar[r] & 0
}%
%EndExpansion
\]
is exact (since $i_{V}$ is the inclusion map $\operatorname*{Ker}f\rightarrow
V$, while $f$ is surjective). Since the tensor product is right exact, this
yields that%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{the sequences}\\%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{4pc}
%\xymatrix{
%\left(\Ker f\right)\otimes\left(\Ker g\right) \ar[r]^-{i_V\otimes\id}
%& V \otimes\left(\Ker g\right) \ar[r]^{f\otimes\id}
%& V^{\prime} \otimes\left(\Ker g\right) \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrixcolsep{4pc}
\xymatrix{
\left(\Ker f\right)\otimes\left(\Ker g\right) \ar[r]^-{i_V\otimes\id}
& V \otimes\left(\Ker g\right) \ar[r]^{f\otimes\id}
& V^{\prime} \otimes\left(\Ker g\right) \ar[r] & 0
}%
%EndExpansion
\text{,}\\%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{4pc}
%\xymatrix{
%\left(\Ker f\right)\otimes W \ar[r]^-{i_V\otimes\id} & V \otimes
%W \ar[r]^{f\otimes\id}
%& V^{\prime} \otimes W \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrixcolsep{4pc}
\xymatrix{
\left(\Ker f\right)\otimes W \ar[r]^-{i_V\otimes\id} & V \otimes
W \ar[r]^{f\otimes\id}
& V^{\prime} \otimes W \ar[r] & 0
}%
%EndExpansion
\text{ and}\\%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{4pc}
%\xymatrix{
%\left(\Ker f\right)\otimes W^{\prime} \ar[r]^-{i_V\otimes\id} & V \otimes
%W^{\prime}
%\ar[r]^{f\otimes\id}
%& V^{\prime} \otimes W^{\prime} \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrixcolsep{4pc}
\xymatrix{
\left(\Ker f\right)\otimes W^{\prime} \ar[r]^-{i_V\otimes\id} & V \otimes
W^{\prime}
\ar[r]^{f\otimes\id}
& V^{\prime} \otimes W^{\prime} \ar[r] & 0
}%
%EndExpansion
\text{ are exact}%
\end{array}
\right)  . \label{pf.f(X)g.1}%
\end{equation}


On the other hand, the sequence%
\[%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%0 \ar[r] & \Ker g \ar[r]^-{i_W} & W \ar[r]^{g} & W^{\prime} \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrix{
0 \ar[r] & \Ker g \ar[r]^-{i_W} & W \ar[r]^{g} & W^{\prime} \ar[r] & 0
}%
%EndExpansion
\]
is exact (since $i_{W}$ is the inclusion map $\operatorname*{Ker}g\rightarrow
W$, while $g$ is surjective). Since the tensor product is right exact, this
yields that%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{the sequences}\\%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{4pc}
%\xymatrix{
%\left(\Ker f\right)\otimes\left(\Ker g\right) \ar[r]^-{\id\otimes i_W}
%& \left(\Ker f\right) \otimes W \ar[r]^{\id\otimes g}
%& \left(\Ker f\right) \otimes W^{\prime} \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrixcolsep{4pc}
\xymatrix{
\left(\Ker f\right)\otimes\left(\Ker g\right) \ar[r]^-{\id\otimes i_W}
& \left(\Ker f\right) \otimes W \ar[r]^{\id\otimes g}
& \left(\Ker f\right) \otimes W^{\prime} \ar[r] & 0
}%
%EndExpansion
\text{,}\\%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{4pc}
%\xymatrix{
%V\otimes\left(\Ker g\right) \ar[r]^-{\id\otimes i_W} & V \otimes W \ar
%[r]^{\id\otimes g}
%& V \otimes W^{\prime} \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrixcolsep{4pc}
\xymatrix{
V\otimes\left(\Ker g\right) \ar[r]^-{\id\otimes i_W} & V \otimes W \ar
[r]^{\id\otimes g}
& V \otimes W^{\prime} \ar[r] & 0
}%
%EndExpansion
\text{ and}\\%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{4pc}
%\xymatrix{
%V^{\prime} \otimes\left(\Ker g\right) \ar[r]^-{\id\otimes i_W} & V^{\prime}
%\otimes W \ar[r]^{\id\otimes g}
%& V^{\prime} \otimes W^{\prime} \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrixcolsep{4pc}
\xymatrix{
V^{\prime} \otimes\left(\Ker g\right) \ar[r]^-{\id\otimes i_W} & V^{\prime}
\otimes W \ar[r]^{\id\otimes g}
& V^{\prime} \otimes W^{\prime} \ar[r] & 0
}%
%EndExpansion
\text{ are exact}%
\end{array}
\right)  . \label{pf.f(X)g.2}%
\end{equation}


Now, the diagram%
\begin{equation}%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{5pc}}}%
%BeginExpansion
\xymatrixcolsep{5pc}%
%EndExpansion%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%\left(\Ker f\right) \otimes\left(\Ker g\right) \ar[r]^{\id\otimes i_W}
%\ar[d]_{i_V\otimes\id}
%& \left(\Ker f\right)\otimes W \ar[r]^{\id\otimes g} \ar[d]_{i_V\otimes\id}
%& \left(\Ker f\right)\otimes W^{\prime} \ar[r] \ar[d]_{i_V\otimes\id}
%& 0 \\
%V\otimes\left(\Ker g\right) \ar[r]^{\id\otimes i_W} \ar[d]_{ f\otimes\id}
%& V\otimes W \ar[r]^{\operatorname*{id}\otimes g}
%\ar[d]_{ f\otimes\id}
%& V\otimes W^{\prime} \ar[r] \ar[d]_{ f\otimes\id}
%& 0 \\
%V^{\prime}\otimes\left(\Ker g\right) \ar[r]^{\id\otimes i_W}
%& V^{\prime}\otimes W \ar[r]^{\id\otimes g}
%& V^{\prime}\otimes W^{\prime} \ar[r]                              & 0
%}} }%
%BeginExpansion
\xymatrix{
\left(\Ker f\right) \otimes\left(\Ker g\right) \ar[r]^{\id\otimes i_W}
\ar[d]_{i_V\otimes\id}
& \left(\Ker f\right)\otimes W \ar[r]^{\id\otimes g} \ar[d]_{i_V\otimes\id}
& \left(\Ker f\right)\otimes W^{\prime} \ar[r] \ar[d]_{i_V\otimes\id}
& 0 \\
V\otimes\left(\Ker g\right) \ar[r]^{\id\otimes i_W} \ar[d]_{ f\otimes\id}
& V\otimes W \ar[r]^{\operatorname*{id}\otimes g}
\ar[d]_{ f\otimes\id}
& V\otimes W^{\prime} \ar[r] \ar[d]_{ f\otimes\id}
& 0 \\
V^{\prime}\otimes\left(\Ker g\right) \ar[r]^{\id\otimes i_W}
& V^{\prime}\otimes W \ar[r]^{\id\otimes g}
& V^{\prime}\otimes W^{\prime} \ar[r]                              & 0
}
%EndExpansion
\label{pf.f(X)g.3}%
\end{equation}
is commutative (because%
\begin{align*}
\left(  i_{V}\otimes\operatorname*{id}\right)  \circ\left(  \operatorname*{id}%
\otimes i_{W}\right)   &  =i_{V}\otimes i_{W}=\left(  \operatorname*{id}%
\otimes i_{W}\right)  \circ\left(  i_{V}\otimes\operatorname*{id}\right)  ;\\
\left(  i_{V}\otimes\operatorname*{id}\right)  \circ\left(  \operatorname*{id}%
\otimes g\right)   &  =i_{V}\otimes g=\left(  \operatorname*{id}\otimes
g\right)  \circ\left(  i_{V}\otimes\operatorname*{id}\right)  ;\\
\left(  f\otimes\operatorname*{id}\right)  \circ\left(  \operatorname*{id}%
\otimes i_{W}\right)   &  =f\otimes i_{W}=\left(  \operatorname*{id}\otimes
i_{W}\right)  \circ\left(  f\otimes\operatorname*{id}\right)  ;\\
\left(  f\otimes\operatorname*{id}\right)  \circ\left(  \operatorname*{id}%
\otimes g\right)   &  =f\otimes g=\left(  \operatorname*{id}\otimes g\right)
\circ\left(  f\otimes\operatorname*{id}\right)
\end{align*}
). Every row of this diagram is an exact sequence (due to (\ref{pf.f(X)g.2})),
and every column of this diagram is an exact sequence (due to
(\ref{pf.f(X)g.1})). Thus, Proposition \ref{prop.9l2} (applied to the diagram
(\ref{pf.f(X)g.3}) instead of the diagram (\ref{prop.9l2.diag.real})) yields
that%
\begin{align*}
\operatorname*{Ker}\left(  \left(  \operatorname*{id}\otimes g\right)
\circ\left(  f\otimes\operatorname*{id}\right)  \right)   &
=\operatorname*{Ker}\left(  \left(  f\otimes\operatorname*{id}\right)
\circ\left(  \operatorname*{id}\otimes g\right)  \right) \\
&  =\left(  \operatorname*{id}\otimes i_{W}\right)  \left(  V\otimes\left(
\operatorname*{Ker}g\right)  \right)  +\left(  i_{V}\otimes\operatorname*{id}%
\right)  \left(  \left(  \operatorname*{Ker}f\right)  \otimes W\right)  .
\end{align*}
Thus,%
\begin{align*}
\operatorname*{Ker}\underbrace{\left(  f\otimes g\right)  }_{=\left(
f\otimes\operatorname*{id}\right)  \circ\left(  \operatorname*{id}\otimes
g\right)  }  &  =\operatorname*{Ker}\left(  \left(  f\otimes\operatorname*{id}%
\right)  \circ\left(  \operatorname*{id}\otimes g\right)  \right) \\
&  =\left(  \operatorname*{id}\otimes i_{W}\right)  \left(  V\otimes\left(
\operatorname*{Ker}g\right)  \right)  +\left(  i_{V}\otimes\operatorname*{id}%
\right)  \left(  \left(  \operatorname*{Ker}f\right)  \otimes W\right) \\
&  =\left(  i_{V}\otimes\operatorname*{id}\right)  \left(  \left(
\operatorname*{Ker}f\right)  \otimes W\right)  +\left(  \operatorname*{id}%
\otimes i_{W}\right)  \left(  V\otimes\left(  \operatorname*{Ker}g\right)
\right)  .
\end{align*}
This proves Theorem \ref{thm.f(X)g}.
\end{proof}

Let us notice a corollary of this theorem:

\begin{corollary}
\label{cor.f(X)g}Let $k$ be a commutative ring. Let $V$, $W$, $V^{\prime}$ and
$W^{\prime}$ be four $k$-modules. Let $f:V\rightarrow V^{\prime}$ and
$g:W\rightarrow W^{\prime}$ be two $k$-linear maps. Assume that $f\left(
V\right)  $ and $W^{\prime}$ are flat $k$-modules. Let $i_{V}$ be the
canonical inclusion $\operatorname*{Ker}f\rightarrow V$. Let $i_{W}$ be the
canonical inclusion $\operatorname*{Ker}g\rightarrow W$. Then,%
\[
\operatorname*{Ker}\left(  f\otimes g\right)  =\left(  i_{V}\otimes
\operatorname*{id}\right)  \left(  \left(  \operatorname*{Ker}f\right)
\otimes W\right)  +\left(  \operatorname*{id}\otimes i_{W}\right)  \left(
V\otimes\left(  \operatorname*{Ker}g\right)  \right)  .
\]

\end{corollary}

Note that this Corollary \ref{cor.f(X)g} does not require $f$ or $g$ to be
surjective, but instead it requires $f\left(  V\right)  $ and $W^{\prime}$ to
be flat (which is always satisfied if $k$ is a field, for example).

To show this, we first prove:

\begin{lemma}
\label{lem.f(X)g}Let $k$ be a commutative ring. Let $V$, $W$, $V^{\prime}$ and
$W^{\prime}$ be four $k$-modules. Let $f:V\rightarrow V^{\prime}$ and
$g:W\rightarrow W^{\prime}$ be two $k$-linear maps. Assume that $V^{\prime}$
is a flat $k$-module, and that $f$ is surjective. Let $i_{V}$ be the canonical
inclusion $\operatorname*{Ker}f\rightarrow V$. Let $i_{W}$ be the canonical
inclusion $\operatorname*{Ker}g\rightarrow W$. Then,%
\[
\operatorname*{Ker}\left(  f\otimes g\right)  =\left(  i_{V}\otimes
\operatorname*{id}\right)  \left(  \left(  \operatorname*{Ker}f\right)
\otimes W\right)  +\left(  \operatorname*{id}\otimes i_{W}\right)  \left(
V\otimes\left(  \operatorname*{Ker}g\right)  \right)  .
\]

\end{lemma}

\begin{lemma}
\label{lem.flat}Let $k$ be a commutative ring. Let $V$, $W$ and $A$ be three
$k$-modules such that the $k$-module $A$ is flat. Let $i:V\rightarrow W$ be an
injective $k$-module homomorphism.\newline\textbf{(a)} The $k$-module
homomorphism $\operatorname*{id}\otimes i:A\otimes V\rightarrow A\otimes W$ is
injective.\newline\textbf{(b)} The $k$-module homomorphism $i\otimes
\operatorname*{id}:V\otimes A\rightarrow W\otimes A$ is injective.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.flat}.]Let $p$ be the canonical projection
$p:W\rightarrow W\diagup\left(  i\left(  V\right)  \right)  $. Then, $p$ is a
surjective $k$-module homomorphism, and $\operatorname*{Ker}p=i\left(
V\right)  $. Thus,%
\begin{equation}%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{4pc}
%\xymatrix{
%0 \ar[r] & V \ar[r]^-{i} & W \ar[r]^-{p} & W\diagup\left(i\left(V\right
%)\right) \ar[r] & 0
%}} }%
%BeginExpansion
\xymatrixcolsep{4pc}
\xymatrix{
0 \ar[r] & V \ar[r]^-{i} & W \ar[r]^-{p} & W\diagup\left(i\left(V\right
)\right) \ar[r] & 0
}
%EndExpansion
\label{pf.flat.1}%
\end{equation}
is a short exact sequence (since $p$ is surjective, since $i$ is injective,
and since $\operatorname*{Ker}p=i\left(  V\right)  $). Since tensoring with
$A$ is an exact functor (because $A$ is a flat $k$-module), this yields that%
\[%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{4pc}
%\xymatrix{
%0 \ar[r] & A \otimes V \ar[r]^-{\id\otimes i} & A \otimes W \ar[r]^-{\id
%\otimes p} & A \otimes\left( W\diagup\left(i\left(V\right)\right) \right
%) \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrixcolsep{4pc}
\xymatrix{
0 \ar[r] & A \otimes V \ar[r]^-{\id\otimes i} & A \otimes W \ar[r]^-{\id
\otimes p} & A \otimes\left( W\diagup\left(i\left(V\right)\right) \right
) \ar[r] & 0
}%
%EndExpansion
\]
is a short exact sequence. Therefore, $\operatorname*{id}\otimes i:A\otimes
V\rightarrow A\otimes W$ is injective. This proves Lemma \ref{lem.flat}
\textbf{(a)}.

Since (\ref{pf.flat.1}) is a short exact sequence, and since tensoring with
$A$ is an exact functor (because $A$ is a flat $k$-module), we find that%
\[%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{4pc}
%\xymatrix{
%0 \ar[r] &  V \otimes A \ar[r]^-{i \otimes\id} & W \otimes A \ar[r]^-{p
%\otimes\id} & \left( W\diagup\left(i\left(V\right)\right) \right) \otimes
%A \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrixcolsep{4pc}
\xymatrix{
0 \ar[r] &  V \otimes A \ar[r]^-{i \otimes\id} & W \otimes A \ar[r]^-{p
\otimes\id} & \left( W\diagup\left(i\left(V\right)\right) \right) \otimes
A \ar[r] & 0
}%
%EndExpansion
\]
is a short exact sequence. Therefore, $i\otimes\operatorname*{id}:V\otimes
A\rightarrow W\otimes A$ is injective. This proves Lemma \ref{lem.flat}
\textbf{(b)}.
\end{proof}

\begin{proof}
[Proof of Lemma \ref{lem.f(X)g}.]Define a $k$-linear map $g_{1}:W\rightarrow
g\left(  W\right)  $ by%
\[
\left(  g_{1}\left(  w\right)  =g\left(  w\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }w\in W\right)
\]
(this is well-defined since $g\left(  w\right)  \in g\left(  W\right)  $ for
every $w\in W$). Let $m_{W}$ be the canonical inclusion $g\left(  W\right)
\rightarrow W^{\prime}$. Clearly, every $w\in W$ satisfies%
\begin{align*}
\left(  m_{W}\circ g_{1}\right)  \left(  w\right)   &  =m_{W}\left(
g_{1}\left(  w\right)  \right)  =g_{1}\left(  w\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }m_{W}\text{ is the canonical
inclusion}\right) \\
&  =g\left(  w\right)  .
\end{align*}
Thus, $m_{W}\circ g_{1}=g$.

Also, $g_{1}$ is surjective, because every $x\in g\left(  W\right)  $
satisfies $x\in g_{1}\left(  W\right)  $\ \ \ \ \footnote{\textit{Proof.} Let
$x\in g\left(  W\right)  $ be arbitrary. Then, there exists some $w\in W$ such
that $x=g\left(  w\right)  $. Consider this $w$. Then, $x=g\left(  w\right)
=g_{1}\left(  w\right)  \in g_{1}\left(  W\right)  $, qed.}. Also,
\[
\operatorname{Ker}g=\left\{  w\in W\ \mid\ \underbrace{g\left(  w\right)
}_{=g_{1}\left(  w\right)  }=0\right\}  =\left\{  w\in W\ \mid\ g_{1}\left(
w\right)  =0\right\}  =\operatorname{Ker}g_{1}.
\]
Thus, $i_{W}$ is the canonical inclusion $\operatorname{Ker}g_{1}\rightarrow
W$ (since $i_{W}$ is the canonical inclusion $\operatorname*{Ker}g\rightarrow
W$). Thus, Theorem \ref{thm.f(X)g} (applied to $g\left(  W\right)  $ and
$g_{1}$ instead of $W^{\prime}$ and $g$) shows that%
\begin{align}
\operatorname*{Ker}\left(  f\otimes g_{1}\right)   &  =\left(  i_{V}%
\otimes\operatorname*{id}\right)  \left(  \left(  \operatorname*{Ker}f\right)
\otimes W\right)  +\left(  \operatorname*{id}\otimes i_{W}\right)  \left(
V\otimes\underbrace{\left(  \operatorname*{Ker}g_{1}\right)  }%
_{=\operatorname{Ker}g}\right) \nonumber\\
&  =\left(  i_{V}\otimes\operatorname*{id}\right)  \left(  \left(
\operatorname*{Ker}f\right)  \otimes W\right)  +\left(  \operatorname*{id}%
\otimes i_{W}\right)  \left(  V\otimes\left(  \operatorname*{Ker}g\right)
\right)  . \label{pf.f(X)g.lem.1}%
\end{align}


Now, $m_{W}$ is injective (since $m_{W}$ is a canonical inclusion). Thus,
applying Lemma \ref{lem.flat} \textbf{(a)} to $m_{W}$, $g\left(  W\right)  $,
$W^{\prime}$ and $V^{\prime}$ instead of $i$, $V$, $W$ and $A$, we obtain that
the map $\operatorname{id}\otimes m_{W}:V^{\prime}\otimes g\left(  W\right)
\rightarrow V^{\prime}\otimes W^{\prime}$ is injective. In other words,
$\operatorname{Ker}\left(  \operatorname{id}\otimes m_{W}\right)  =0$.

Now, it is known that whenever $A$, $B$, $C$, $A^{\prime}$, $B^{\prime}$,
$C^{\prime}$ are six $k$-modules, and $\alpha:A\rightarrow B$, $\beta
:B\rightarrow C$, $\gamma:A^{\prime}\rightarrow B^{\prime}$ and $\delta
:B^{\prime}\rightarrow C^{\prime}$ are four $k$-linear maps, then $\left(
\beta\otimes\delta\right)  \circ\left(  \alpha\otimes\gamma\right)  =\left(
\beta\circ\alpha\right)  \otimes\left(  \delta\circ\gamma\right)  $. Applying
this fact to $A=V$, $B=V^{\prime}$, $C=V^{\prime}$, $A^{\prime}=W$,
$B^{\prime}=g\left(  W\right)  $, $C^{\prime}=W^{\prime}$, $\alpha=f$,
$\beta=\operatorname{id}$, $\gamma=g_{1}$ and $\delta=m_{W}$, we obtain%
\[
\left(  \operatorname{id}\otimes m_{W}\right)  \circ\left(  f\otimes
g_{1}\right)  =\underbrace{\left(  \operatorname{id}\circ f\right)  }%
_{=f}\otimes\left(  \underbrace{m_{W}\circ g_{1}}_{=g}\right)  =f\otimes g.
\]


Now, let $x\in\operatorname{Ker}\left(  f\otimes g_{1}\right)  $ be arbitrary.
Then, $\left(  f\otimes g_{1}\right)  \left(  x\right)  =0$. Now,
\begin{align*}
\underbrace{\left(  f\otimes g\right)  }_{=\left(  \operatorname{id}\otimes
m_{W}\right)  \circ\left(  f\otimes g_{1}\right)  }\left(  x\right)   &
=\left(  \left(  \operatorname{id}\otimes m_{W}\right)  \circ\left(  f\otimes
g_{1}\right)  \right)  \left(  x\right)  =\left(  \operatorname{id}\otimes
m_{W}\right)  \left(  \underbrace{\left(  f\otimes g_{1}\right)  \left(
x\right)  }_{=0}\right) \\
&  =\left(  \operatorname{id}\otimes m_{W}\right)  \left(  0\right)  =0,
\end{align*}
so that $x\in\operatorname{Ker}\left(  f\otimes g\right)  $. Thus we have seen
that every $x\in\operatorname{Ker}\left(  f\otimes g_{1}\right)  $ satisfies
$x\in\operatorname{Ker}\left(  f\otimes g\right)  $. In other words,
$\operatorname{Ker}\left(  f\otimes g_{1}\right)  \subseteq\operatorname{Ker}%
\left(  f\otimes g\right)  $.

On the other hand, let $y\in\operatorname{Ker}\left(  f\otimes g\right)  $ be
arbitrary. Then,%
\[
\left(  \operatorname{id}\otimes m_{W}\right)  \left(  \left(  f\otimes
g_{1}\right)  \left(  y\right)  \right)  =\left(  \underbrace{\left(
\operatorname{id}\otimes m_{W}\right)  \circ\left(  f\otimes g_{1}\right)
}_{=f\otimes g}\right)  \left(  y\right)  =\left(  f\otimes g\right)  \left(
y\right)  =0
\]
(since $y\in\operatorname{Ker}\left(  f\otimes g\right)  $), so that $\left(
f\otimes g_{1}\right)  \left(  y\right)  \in\operatorname{Ker}\left(
\operatorname{id}\otimes m_{W}\right)  =0$. Thus, $\left(  f\otimes
g_{1}\right)  \left(  y\right)  =0$, so that $y\in\operatorname{Ker}\left(
f\otimes g_{1}\right)  $. Thus we have shown that every $y\in
\operatorname{Ker}\left(  f\otimes g\right)  $ satisfies $y\in
\operatorname{Ker}\left(  f\otimes g_{1}\right)  $. In other words,
$\operatorname{Ker}\left(  f\otimes g\right)  \subseteq\operatorname{Ker}%
\left(  f\otimes g_{1}\right)  $.

Combined with $\operatorname{Ker}\left(  f\otimes g_{1}\right)  \subseteq
\operatorname{Ker}\left(  f\otimes g\right)  $, this yields
$\operatorname{Ker}\left(  f\otimes g\right)  =\operatorname{Ker}\left(
f\otimes g_{1}\right)  $. Thus, (\ref{pf.f(X)g.lem.1}) becomes%
\[
\operatorname{Ker}\left(  f\otimes g\right)  =\left(  i_{V}\otimes
\operatorname*{id}\right)  \left(  \left(  \operatorname*{Ker}f\right)
\otimes W\right)  +\left(  \operatorname*{id}\otimes i_{W}\right)  \left(
V\otimes\left(  \operatorname*{Ker}g\right)  \right)  .
\]
This proves Lemma \ref{lem.f(X)g}.
\end{proof}

\begin{proof}
[Proof of Corollary \ref{cor.f(X)g}.]Define a $k$-linear map $f_{1}%
:V\rightarrow f\left(  V\right)  $ by%
\[
\left(  f_{1}\left(  v\right)  =f\left(  v\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }v\in V\right)
\]
(this is well-defined since $f\left(  v\right)  \in f\left(  V\right)  $ for
every $v\in V$). Let $m_{V}$ be the canonical inclusion $f\left(  V\right)
\rightarrow V^{\prime}$. Clearly, every $v\in V$ satisfies%
\begin{align*}
\left(  m_{V}\circ f_{1}\right)  \left(  v\right)   &  =m_{V}\left(
f_{1}\left(  v\right)  \right)  =f_{1}\left(  v\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }m_{V}\text{ is the canonical
inclusion}\right) \\
&  =f\left(  v\right)  .
\end{align*}
Thus, $m_{V}\circ f_{1}=f$.

Also, $f_{1}$ is surjective, because every $x\in f\left(  V\right)  $
satisfies $x\in f_{1}\left(  V\right)  $\ \ \ \ \footnote{\textit{Proof.} Let
$x\in f\left(  V\right)  $ be arbitrary. Then, there exists some $v\in V$ such
that $x=f\left(  v\right)  $. Consider this $v$. Then, $x=f\left(  v\right)
=f_{1}\left(  v\right)  \in f_{1}\left(  V\right)  $, qed.}. Also,
\[
\operatorname{Ker}f=\left\{  v\in V\ \mid\ \underbrace{f\left(  v\right)
}_{=f_{1}\left(  v\right)  }=0\right\}  =\left\{  v\in V\ \mid\ f_{1}\left(
v\right)  =0\right\}  =\operatorname{Ker}f_{1}.
\]
Thus, $i_{V}$ is the canonical inclusion $\operatorname{Ker}f_{1}\rightarrow
V$ (since $i_{V}$ is the canonical inclusion $\operatorname*{Ker}f\rightarrow
V$). Thus, Lemma \ref{lem.f(X)g} (applied to $f\left(  V\right)  $ and $f_{1}$
instead of $V^{\prime}$ and $f$) shows that%
\begin{align}
\operatorname*{Ker}\left(  f_{1}\otimes g\right)   &  =\left(  i_{V}%
\otimes\operatorname*{id}\right)  \left(  \underbrace{\left(
\operatorname*{Ker}f_{1}\right)  }_{=\operatorname*{Ker}f}\otimes W\right)
+\left(  \operatorname*{id}\otimes i_{W}\right)  \left(  V\otimes\left(
\operatorname*{Ker}g\right)  \right) \nonumber\\
&  =\left(  i_{V}\otimes\operatorname*{id}\right)  \left(  \left(
\operatorname*{Ker}f\right)  \otimes W\right)  +\left(  \operatorname*{id}%
\otimes i_{W}\right)  \left(  V\otimes\left(  \operatorname*{Ker}g\right)
\right)  . \label{pf.f(X)g.cor.1}%
\end{align}


Now, $m_{V}$ is injective (since $m_{V}$ is a canonical inclusion). Thus,
applying Lemma \ref{lem.flat} \textbf{(b)} to $m_{V}$, $f\left(  V\right)  $,
$V^{\prime}$ and $W^{\prime}$ instead of $i$, $V$, $W$ and $A$, we obtain that
the map $m_{V}\otimes\operatorname*{id}:f\left(  V\right)  \otimes W^{\prime
}\rightarrow V^{\prime}\otimes W^{\prime}$ is injective. In other words,
$\operatorname{Ker}\left(  m_{V}\otimes\operatorname*{id}\right)  =0$.

Now, it is known that whenever $A$, $B$, $C$, $A^{\prime}$, $B^{\prime}$,
$C^{\prime}$ are six $k$-modules, and $\alpha:A\rightarrow B$, $\beta
:B\rightarrow C$, $\gamma:A^{\prime}\rightarrow B^{\prime}$ and $\delta
:B^{\prime}\rightarrow C^{\prime}$ are four $k$-linear maps, then $\left(
\beta\otimes\delta\right)  \circ\left(  \alpha\otimes\gamma\right)  =\left(
\beta\circ\alpha\right)  \otimes\left(  \delta\circ\gamma\right)  $. Applying
this fact to $A=V$, $B=f\left(  V\right)  $, $C=V^{\prime}$, $A^{\prime}=W$,
$B^{\prime}=W^{\prime}$, $C^{\prime}=W^{\prime}$, $\alpha=f_{1}$, $\beta
=m_{V}$, $\gamma=g$ and $\delta=\operatorname{id}$, we obtain%
\[
\left(  m_{V}\otimes\operatorname{id}\right)  \circ\left(  f_{1}\otimes
g\right)  =\underbrace{\left(  m_{V}\circ f_{1}\right)  }_{=f}\otimes
\underbrace{\left(  \operatorname{id}\circ g\right)  }_{=g}=f\otimes g.
\]


Now, let $x\in\operatorname{Ker}\left(  f_{1}\otimes g\right)  $ be arbitrary.
Then, $\left(  f_{1}\otimes g\right)  \left(  x\right)  =0$. Now,
\begin{align*}
\underbrace{\left(  f\otimes g\right)  }_{=\left(  m_{V}\otimes
\operatorname{id}\right)  \circ\left(  f_{1}\otimes g\right)  }\left(
x\right)   &  =\left(  \left(  m_{V}\otimes\operatorname{id}\right)
\circ\left(  f_{1}\otimes g\right)  \right)  \left(  x\right)  =\left(
m_{V}\otimes\operatorname{id}\right)  \left(  \underbrace{\left(  f_{1}\otimes
g\right)  \left(  x\right)  }_{=0}\right) \\
&  =\left(  m_{V}\otimes\operatorname{id}\right)  \left(  0\right)  =0,
\end{align*}
so that $x\in\operatorname{Ker}\left(  f\otimes g\right)  $. Thus we have seen
that every $x\in\operatorname{Ker}\left(  f_{1}\otimes g\right)  $ satisfies
$x\in\operatorname{Ker}\left(  f\otimes g\right)  $. In other words,
$\operatorname{Ker}\left(  f_{1}\otimes g\right)  \subseteq\operatorname{Ker}%
\left(  f\otimes g\right)  $.

On the other hand, let $y\in\operatorname{Ker}\left(  f\otimes g\right)  $ be
arbitrary. Then,%
\[
\left(  m_{V}\otimes\operatorname{id}\right)  \left(  \left(  f_{1}\otimes
g\right)  \left(  y\right)  \right)  =\left(  \underbrace{\left(  m_{V}%
\otimes\operatorname{id}\right)  \circ\left(  f_{1}\otimes g\right)
}_{=f\otimes g}\right)  \left(  y\right)  =\left(  f\otimes g\right)  \left(
y\right)  =0
\]
(since $y\in\operatorname{Ker}\left(  f\otimes g\right)  $), so that $\left(
f_{1}\otimes g\right)  \left(  y\right)  \in\operatorname{Ker}\left(
m_{V}\otimes\operatorname{id}\right)  =0$. Thus, $\left(  f_{1}\otimes
g\right)  \left(  y\right)  =0$, so that $y\in\operatorname{Ker}\left(
f_{1}\otimes g\right)  $. Thus we have shown that every $y\in
\operatorname{Ker}\left(  f\otimes g\right)  $ satisfies $y\in
\operatorname{Ker}\left(  f_{1}\otimes g\right)  $. In other words,
$\operatorname{Ker}\left(  f\otimes g\right)  \subseteq\operatorname{Ker}%
\left(  f_{1}\otimes g\right)  $.

Combined with $\operatorname{Ker}\left(  f_{1}\otimes g\right)  \subseteq
\operatorname{Ker}\left(  f\otimes g\right)  $, this yields
$\operatorname{Ker}\left(  f\otimes g\right)  =\operatorname{Ker}\left(
f_{1}\otimes g\right)  $. Thus, (\ref{pf.f(X)g.cor.1}) becomes%
\[
\operatorname{Ker}\left(  f\otimes g\right)  =\left(  i_{V}\otimes
\operatorname*{id}\right)  \left(  \left(  \operatorname*{Ker}f\right)
\otimes W\right)  +\left(  \operatorname*{id}\otimes i_{W}\right)  \left(
V\otimes\left(  \operatorname*{Ker}g\right)  \right)  .
\]
This proves Corollary \ref{cor.f(X)g}.
\end{proof}

We notice a triviality on tensor products of surjective maps:

\begin{lemma}
\label{lem.f(X)g.sur}Let $k$ be a commutative ring. Let $V$, $W$, $V^{\prime}$
and $W^{\prime}$ be four $k$-modules. Let $f:V\rightarrow V^{\prime}$ and
$g:W\rightarrow W^{\prime}$ be two surjective $k$-linear maps. Then, the map
$f\otimes g:V\otimes W\rightarrow V^{\prime}\otimes W^{\prime}$ is surjective.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.f(X)g.sur}.]Let $T\in V^{\prime}\otimes W^{\prime}$
be arbitrary. Then, we can write the tensor $T$ in the form $T=\sum
\limits_{i=1}^{n}\alpha_{i}\otimes\beta_{i}$ for some $n\in\mathbb{N}$, some
elements $\alpha_{1}$, $\alpha_{2}$, $\ldots$, $\alpha_{n}$ of $V^{\prime}$
and some elements $\beta_{1}$, $\beta_{2}$, $\ldots$, $\beta_{n}$ of
$W^{\prime}$. Consider this $n$, these $\alpha_{1}$, $\alpha_{2}$, $\ldots$,
$\alpha_{n}$ and these $\beta_{1}$, $\beta_{2}$, $\ldots$, $\beta_{n}$.

For every $i\in\left\{  1,2,\ldots,n\right\}  $, there exists some $v_{i}\in
V$ such that $\alpha_{i}=f\left(  v_{i}\right)  $ (since $f$ is surjective).
Consider this $v_{i}$.

For every $i\in\left\{  1,2,\ldots,n\right\}  $, there exists some $w_{i}\in
W$ such that $\beta_{i}=g\left(  w_{i}\right)  $ (since $g$ is surjective).
Consider this $w_{i}$.

Now,%
\begin{align*}
T  &  =\sum\limits_{i=1}^{n}\underbrace{\alpha_{i}}_{=f\left(  v_{i}\right)
}\otimes\underbrace{\beta_{i}}_{=g\left(  w_{i}\right)  }=\sum\limits_{i=1}%
^{n}\underbrace{f\left(  v_{i}\right)  \otimes g\left(  w_{i}\right)
}_{=\left(  f\otimes g\right)  \left(  v_{i}\otimes w_{i}\right)  }%
=\sum\limits_{i=1}^{n}\left(  f\otimes g\right)  \left(  v_{i}\otimes
w_{i}\right) \\
&  =\left(  f\otimes g\right)  \left(  \sum\limits_{i=1}^{n}v_{i}\otimes
w_{i}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\otimes g\text{ is
}k\text{-linear}\right) \\
&  \in\left(  f\otimes g\right)  \left(  V\otimes W\right)  .
\end{align*}


So we have proven that every $T\in V^{\prime}\otimes W^{\prime}$ satisfies
$T\in\left(  f\otimes g\right)  \left(  V\otimes W\right)  $. Thus, $f\otimes
g$ is surjective, so that Lemma \ref{lem.f(X)g.sur} is proven.
\end{proof}

\subsection{\label{subsect.n}Extension to $n$ modules}

We can trivially generalize Lemma \ref{lem.f(X)g.sur} to several $k$-modules:

\begin{lemma}
\label{lem.f_i.sur}Let $k$ be a commutative ring. Let $n\in\mathbb{N}$. For
any $i\in\left\{  1,2,\ldots,n\right\}  $, let $V_{i}$ and $V_{i}^{\prime}$ be
two $k$-modules, and let $f_{i}:V_{i}\rightarrow V_{i}^{\prime}$ be a
surjective $k$-module homomorphism. Then, the map $f_{1}\otimes f_{2}%
\otimes\cdots\otimes f_{n}:V_{1}\otimes V_{2}\otimes\cdots\otimes
V_{n}\rightarrow V_{1}^{\prime}\otimes V_{2}^{\prime}\otimes\cdots\otimes
V_{n}^{\prime}$ is surjective.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.f_i.sur}.]We are going to prove Lemma
\ref{lem.f_i.sur} by induction over $n$:

\textit{Induction base:} For $n=0$, Lemma \ref{lem.f_i.sur} holds (because for
$n=0$, the map $f_{1}\otimes f_{2}\otimes\cdots\otimes f_{n}:V_{1}\otimes
V_{2}\otimes\cdots\otimes V_{n}\rightarrow V_{1}^{\prime}\otimes V_{2}%
^{\prime}\otimes\cdots\otimes V_{n}^{\prime}$ is the identity map
$\operatorname{id}:k\rightarrow k$ and therefore surjective). Thus, the
induction base is complete.

\textit{Induction step:} Let $p\in\mathbb{N}$ be arbitrary. Assume that Lemma
\ref{lem.f_i.sur} holds for $n=p$.

Now let us prove that Lemma \ref{lem.f_i.sur} holds for $n=p+1$. So let
$V_{i}$ and $V_{i}^{\prime}$ be two $k$-modules for every $i\in\left\{
1,2,\ldots,p+1\right\}  $, and let $f_{i}:V_{i}\rightarrow V_{i}^{\prime}$ be
a surjective $k$-module homomorphism for every $i\in\left\{  1,2,\ldots
,p+1\right\}  $.

According to Definition \ref{defs.f_1(X)f_2}, we have $f_{1}\otimes
f_{2}\otimes\cdots\otimes f_{p+1}=f_{1}\otimes\left(  f_{2}\otimes
f_{3}\otimes\cdots\otimes f_{p+1}\right)  $.

We know that $V_{i}$ and $V_{i}^{\prime}$ are two $k$-modules for every
$i\in\left\{  1,2,\ldots,p+1\right\}  $. Thus, $V_{i}$ and $V_{i}^{\prime}$
are two $k$-modules for every $i\in\left\{  2,3,\ldots,p+1\right\}  $.
Substituting $i+1$ for $i$ in this fact, we obtain that $V_{i+1}$ and
$V_{i+1}^{\prime}$ are two $k$-modules for every $i\in\left\{  1,2,\ldots
,p\right\}  $.

We know that $f_{i}:V_{i}\rightarrow V_{i}^{\prime}$ is a surjective
$k$-module homomorphism for every $i\in\left\{  1,2,\ldots,p+1\right\}  $.
Thus, $f_{i}:V_{i}\rightarrow V_{i}^{\prime}$ is a surjective $k$-module
homomorphism for every $i\in\left\{  2,3,\ldots,p+1\right\}  $. Substituting
$i+1$ for $i$ in this fact, we obtain that $f_{i+1}$ is a surjective
$k$-module homomorphism for every $i\in\left\{  1,2,\ldots,p\right\}  $.

Applying Lemma \ref{lem.f_i.sur} to $p$, $V_{i+1}$, $V_{i+1}^{\prime}$ and
$f_{i+1}$ instead of $n$, $V_{i}$, $V_{i}^{\prime}$ and $f_{i}$ (this is
allowed, because we have assumed that Lemma \ref{lem.f_i.sur} holds for
$n=p$), we see that the map $f_{2}\otimes f_{3}\otimes\cdots\otimes f_{p+1}$
is surjective.

We know that $f_{i}:V_{i}\rightarrow V_{i}^{\prime}$ is a surjective
$k$-module homomorphism for every $i\in\left\{  1,2,\ldots,p+1\right\}  $.
Applying this to $i=1$, we conclude that $f_{1}:V_{1}\rightarrow V_{1}%
^{\prime}$ is a surjective $k$-module homomorphism.

Applying Lemma \ref{lem.f(X)g.sur} to $V=V_{1}$, $V^{\prime}=V_{1}^{\prime}$,
$W=V_{2}\otimes V_{3}\otimes\cdots\otimes V_{p+1}$, $W^{\prime}=V_{2}^{\prime
}\otimes V_{3}^{\prime}\otimes\cdots\otimes V_{p+1}^{\prime}$, $f=f_{1}$ and
$g=f_{2}\otimes f_{3}\otimes\cdots\otimes f_{p+1}$, we now conclude that the
map $f_{1}\otimes\left(  f_{2}\otimes f_{3}\otimes\cdots\otimes f_{p+1}%
\right)  $ is surjective. Since $f_{1}\otimes f_{2}\otimes\cdots\otimes
f_{p+1}=f_{1}\otimes\left(  f_{2}\otimes f_{3}\otimes\cdots\otimes
f_{p+1}\right)  $, this yields that the map $f_{1}\otimes f_{2}\otimes
\cdots\otimes f_{p+1}$ is surjective.

We have thus proven that if $V_{i}$ and $V_{i}^{\prime}$ are two $k$-modules
for every $i\in\left\{  1,2,\ldots,p+1\right\}  $, and $f_{i}:V_{i}\rightarrow
V_{i}^{\prime}$ is a surjective $k$-module homomorphism for every
$i\in\left\{  1,2,\ldots,p+1\right\}  $, then the map $f_{1}\otimes
f_{2}\otimes\cdots\otimes f_{p+1}:V_{1}\otimes V_{2}\otimes\cdots\otimes
V_{p+1}\rightarrow V_{1}^{\prime}\otimes V_{2}^{\prime}\otimes\cdots\otimes
V_{p+1}^{\prime}$ is surjective. In other words, we have proven that Lemma
\ref{lem.f_i.sur} holds for $n=p+1$. This completes the induction step, and
thus Lemma \ref{lem.f_i.sur} is proven.
\end{proof}

Now let us extend Theorem \ref{thm.f(X)g} to $n$ modules:

\begin{theorem}
\label{thm.f_i}Let $k$ be a commutative ring. Let $n\in\mathbb{N}$. For any
$i\in\left\{  1,2,\ldots,n\right\}  $, let $V_{i}$ and $V_{i}^{\prime}$ be two
$k$-modules, and let $f_{i}:V_{i}\rightarrow V_{i}^{\prime}$ be a surjective
$k$-module homomorphism. For any $i\in\left\{  1,2,\ldots,n\right\}  $, let
$\mathfrak{i}_{i}$ be the canonical inclusion $\operatorname*{Ker}%
f_{i}\rightarrow V_{i}$. Then,%
\begin{align}
&  \operatorname*{Ker}\left(  f_{1}\otimes f_{2}\otimes\cdots\otimes
f_{n}\right) \nonumber\\
&  =\sum_{i=1}^{n}\left(  \underbrace{\operatorname*{id}\otimes
\operatorname*{id}\otimes\cdots\otimes\operatorname*{id}}_{i-1\text{ times}%
}\otimes\mathfrak{i}_{i}\otimes\underbrace{\operatorname*{id}\otimes
\operatorname*{id}\otimes\cdots\otimes\operatorname*{id}}_{n-i\text{ times}%
}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  V_{1}\otimes V_{2}%
\otimes\cdots\otimes V_{i-1}\otimes\left(  \operatorname*{Ker}f_{i}\right)
\otimes V_{i+1}\otimes V_{i+2}\otimes\cdots\otimes V_{n}\right)  .
\label{thm.f_i.eq}%
\end{align}

\end{theorem}

Before we show this, we need an (almost trivial) lemma:

\begin{lemma}
\label{lem.sum(X)}Let $k$ be a commutative ring. Let $n\in\mathbb{N}$. Let $A$
and $B$ be two $k$-modules. For any $i\in\left\{  1,2,\ldots,n\right\}  $, let
$B_{i}$ be a $k$-submodule of $B$. Let $B^{\prime}$ be the $k$-submodule
$\sum\limits_{i=1}^{n}B_{i}$ of $B$.\newline For any $k$-module $C$ and any
$k$-submodule $D$ of $C$, we let $\operatorname*{inc}\nolimits_{D,C}$ denote
the canonical inclusion map $D\rightarrow C$.\newline Then,%
\[
\left(  \operatorname*{id}\otimes\operatorname*{inc}\nolimits_{B^{\prime}%
,B}\right)  \left(  A\otimes B^{\prime}\right)  =\sum_{i=1}^{n}\left(
\operatorname*{id}\otimes\operatorname*{inc}\nolimits_{B_{i},B}\right)
\left(  A\otimes B_{i}\right)
\]
(as $k$-submodules of $A\otimes B$).
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sum(X)}.]Since $\sum\limits_{i=1}^{n}B_{i}=B^{\prime
}$, we have $B_{i}\subseteq B^{\prime}$ for every $i\in\left\{  1,2,\ldots
,n\right\}  $.

The maps $\operatorname*{inc}\nolimits_{B_{i},B}:B_{i}\rightarrow B$ for all
$i\in\left\{  1,2,\ldots,n\right\}  $ give rise to a map $\sum\limits_{i=1}%
^{n}\operatorname*{inc}\nolimits_{B_{i},B}:\bigoplus\limits_{i=1}^{n}%
B_{i}\rightarrow B$. Similarly, the maps $\operatorname*{id}\otimes
\operatorname*{inc}\nolimits_{B_{i},B}:A\otimes B_{i}\rightarrow A\otimes B$
for all $i\in\left\{  1,2,\ldots,n\right\}  $ give rise to a map
$\sum\limits_{i=1}^{n}\operatorname*{id}\otimes\operatorname*{inc}%
\nolimits_{B_{i},B}:\bigoplus\limits_{i=1}^{n}\left(  A\otimes B_{i}\right)
\rightarrow A\otimes B$.

Since the tensor product is known to commute with direct sums, there is a
canonical $k$-module isomorphism $A\otimes\left(  \bigoplus\limits_{i=1}%
^{n}B_{i}\right)  \rightarrow\bigoplus\limits_{i=1}^{n}\left(  A\otimes
B_{i}\right)  $. Denote this isomorphism by $I$. By the universal property of
this $I$, the diagram%
\[%
%TCIMACRO{\TeXButton{proof of lem.sum(X) upper right triangle}{\xymatrixrowsep
%{4pc}
%\xymatrixcolsep{4pc}
%\xymatrix{
%A \otimes\left( \bigoplus\limits_{i=1}^n B_i\right) \ar[r]^I \ar
%[dr]_{\id\otimes\left(\sum\limits_{i=1}^n \inc_{B_i,B}\right)} &  \bigoplus
%\limits_{i=1}^n \left(A\otimes B_i\right) \ar[d]^{\sum\limits_{i=1}%
%^n \left(\id\otimes\inc_{B_i,B}\right)} \\
%& A\otimes B
%}}}%
%BeginExpansion
\xymatrixrowsep{4pc}
\xymatrixcolsep{4pc}
\xymatrix{
A \otimes\left( \bigoplus\limits_{i=1}^n B_i\right) \ar[r]^I \ar
[dr]_{\id\otimes\left(\sum\limits_{i=1}^n \inc_{B_i,B}\right)} &  \bigoplus
\limits_{i=1}^n \left(A\otimes B_i\right) \ar[d]^{\sum\limits_{i=1}%
^n \left(\id\otimes\inc_{B_i,B}\right)} \\
& A\otimes B
}%
%EndExpansion
\]
commutes. In other words,%
\[
\operatorname*{id}\otimes\left(  \sum\limits_{i=1}^{n}\operatorname*{inc}%
\nolimits_{B_{i},B}\right)  =\left(  \sum_{i=1}^{n}\left(  \operatorname*{id}%
\otimes\operatorname*{inc}\nolimits_{B_{i},B}\right)  \right)  \circ I,
\]
so that%
\begin{align}
\left(  \operatorname*{id}\otimes\left(  \sum\limits_{i=1}^{n}%
\operatorname*{inc}\nolimits_{B_{i},B}\right)  \right)  \left(  A\otimes
\left(  \bigoplus\limits_{i=1}^{n}B_{i}\right)  \right)   &  =\left(  \left(
\sum_{i=1}^{n}\left(  \operatorname*{id}\otimes\operatorname*{inc}%
\nolimits_{B_{i},B}\right)  \right)  \circ I\right)  \left(  A\otimes\left(
\bigoplus\limits_{i=1}^{n}B_{i}\right)  \right) \nonumber\\
&  =\left(  \sum_{i=1}^{n}\left(  \operatorname*{id}\otimes\operatorname*{inc}%
\nolimits_{B_{i},B}\right)  \right)  \underbrace{\left(  I\left(
A\otimes\left(  \bigoplus\limits_{i=1}^{n}B_{i}\right)  \right)  \right)
}_{\substack{=\bigoplus\limits_{i=1}^{n}\left(  A\otimes B_{i}\right)
\\\text{(since }I\text{ is an isomorphism)}}}\nonumber\\
&  =\left(  \sum_{i=1}^{n}\left(  \operatorname*{id}\otimes\operatorname*{inc}%
\nolimits_{B_{i},B}\right)  \right)  \left(  \bigoplus\limits_{i=1}^{n}\left(
A\otimes B_{i}\right)  \right) \nonumber\\
&  =\sum_{i=1}^{n}\left(  \operatorname*{id}\otimes\operatorname*{inc}%
\nolimits_{B_{i},B}\right)  \left(  A\otimes B_{i}\right)  .
\label{pf.sum(X).up}%
\end{align}


On the other hand, the maps $\operatorname*{inc}\nolimits_{B_{i},B^{\prime}%
}:B_{i}\rightarrow B^{\prime}$ for all $i\in\left\{  1,2,\ldots,n\right\}  $
(these maps are well-defined since $B_{i}\subseteq B^{\prime}$ for every
$i\in\left\{  1,2,\ldots,n\right\}  $) give rise to a map $\sum\limits_{i=1}%
^{n}\operatorname*{inc}\nolimits_{B_{i},B^{\prime}}:\bigoplus\limits_{i=1}%
^{n}B_{i}\rightarrow B^{\prime}$. Every $i\in\left\{  1,2,\ldots,n\right\}  $
satisfies $\operatorname*{inc}\nolimits_{B_{i},B}=\operatorname*{inc}%
\nolimits_{B^{\prime},B}\circ\operatorname*{inc}\nolimits_{B_{i},B^{\prime}}$,
so that%
\[
\sum\limits_{i=1}^{n}\operatorname*{inc}\nolimits_{B_{i},B^{\prime}}%
=\sum\limits_{i=1}^{n}\left(  \operatorname*{inc}\nolimits_{B^{\prime},B}%
\circ\operatorname*{inc}\nolimits_{B_{i},B^{\prime}}\right)
=\operatorname*{inc}\nolimits_{B^{\prime},B}\circ\left(  \sum\limits_{i=1}%
^{n}\operatorname*{inc}\nolimits_{B_{i},B^{\prime}}\right)  .
\]
Hence,%
\begin{align*}
\operatorname*{id}\otimes\sum\limits_{i=1}^{n}\operatorname*{inc}%
\nolimits_{B_{i},B^{\prime}}  &  =\operatorname*{id}\otimes\left(
\operatorname*{inc}\nolimits_{B^{\prime},B}\circ\left(  \sum\limits_{i=1}%
^{n}\operatorname*{inc}\nolimits_{B_{i},B^{\prime}}\right)  \right) \\
&  =\left(  \operatorname*{id}\otimes\operatorname*{inc}\nolimits_{B^{\prime
},B}\right)  \circ\left(  \operatorname*{id}\otimes\left(  \sum\limits_{i=1}%
^{n}\operatorname*{inc}\nolimits_{B_{i},B^{\prime}}\right)  \right)
\end{align*}
as maps from $A\otimes\left(  \bigoplus\limits_{i=1}^{n}B_{i}\right)  $ to
$A\otimes B$ (where $\operatorname*{id}$ always denotes $\operatorname*{id}%
\nolimits_{A}$). Hence,%
\begin{align}
&  \left(  \operatorname*{id}\otimes\left(  \sum\limits_{i=1}^{n}%
\operatorname*{inc}\nolimits_{B_{i},B}\right)  \right)  \left(  A\otimes
\left(  \bigoplus\limits_{i=1}^{n}B_{i}\right)  \right) \nonumber\\
&  =\left(  \left(  \operatorname*{id}\otimes\operatorname*{inc}%
\nolimits_{B^{\prime},B}\right)  \circ\left(  \operatorname*{id}\otimes\left(
\sum\limits_{i=1}^{n}\operatorname*{inc}\nolimits_{B_{i},B^{\prime}}\right)
\right)  \right)  \left(  A\otimes\left(  \bigoplus\limits_{i=1}^{n}%
B_{i}\right)  \right) \nonumber\\
&  =\left(  \operatorname*{id}\otimes\operatorname*{inc}\nolimits_{B^{\prime
},B}\right)  \left(  \left(  \operatorname*{id}\otimes\left(  \sum
\limits_{i=1}^{n}\operatorname*{inc}\nolimits_{B_{i},B^{\prime}}\right)
\right)  \left(  A\otimes\left(  \bigoplus\limits_{i=1}^{n}B_{i}\right)
\right)  \right)  . \label{pf.sum(X).down}%
\end{align}
But since $\operatorname*{inc}\nolimits_{B_{i},B^{\prime}}$ is the inclusion
map $B_{i}\rightarrow B^{\prime}$ for every $i\in\left\{  1,2,\ldots
,n\right\}  $, it is clear that the image of the map $\sum\limits_{i=1}%
^{n}\operatorname*{inc}\nolimits_{B_{i},B^{\prime}}:\bigoplus\limits_{i=1}%
^{n}B_{i}\rightarrow B^{\prime}$ is $\sum\limits_{i=1}^{n}B_{i}=B^{\prime}$.
In other words, the map $\sum\limits_{i=1}^{n}\operatorname*{inc}%
\nolimits_{B_{i},B^{\prime}}:\bigoplus\limits_{i=1}^{n}B_{i}\rightarrow
B^{\prime}$ is surjective. Hence, the map $\operatorname*{id}\otimes\left(
\sum\limits_{i=1}^{n}\operatorname*{inc}\nolimits_{B_{i},B^{\prime}}\right)
:A\otimes\left(  \bigoplus\limits_{i=1}^{n}B_{i}\right)  \rightarrow A\otimes
B^{\prime}$ is surjective as well (by Lemma \ref{lem.f(X)g.sur}, applied to
$V=A$, $V^{\prime}=A$, $W=\bigoplus\limits_{i=1}^{n}B_{i}$, $W^{\prime
}=B^{\prime}$, $f=\operatorname*{id}$ and $g=\sum\limits_{i=1}^{n}%
\operatorname*{inc}\nolimits_{B_{i},B^{\prime}}$), so that $\left(
\operatorname*{id}\otimes\left(  \sum\limits_{i=1}^{n}\operatorname*{inc}%
\nolimits_{B_{i},B^{\prime}}\right)  \right)  \left(  A\otimes\left(
\bigoplus\limits_{i=1}^{n}B_{i}\right)  \right)  =A\otimes B^{\prime}$. Thus,
(\ref{pf.sum(X).down}) simplifies to%
\[
\left(  \operatorname*{id}\otimes\left(  \sum\limits_{i=1}^{n}%
\operatorname*{inc}\nolimits_{B_{i},B}\right)  \right)  \left(  A\otimes
\left(  \bigoplus\limits_{i=1}^{n}B_{i}\right)  \right)  =\left(
\operatorname*{id}\otimes\operatorname*{inc}\nolimits_{B^{\prime},B}\right)
\left(  A\otimes B^{\prime}\right)  .
\]
Compared with (\ref{pf.sum(X).up}), we obtain
\[
\sum_{i=1}^{n}\left(  \operatorname*{id}\otimes\operatorname*{inc}%
\nolimits_{B_{i},B}\right)  \left(  A\otimes B_{i}\right)  =\left(
\operatorname*{id}\otimes\operatorname*{inc}\nolimits_{B^{\prime},B}\right)
\left(  A\otimes B^{\prime}\right)  .
\]
This proves Lemma \ref{lem.sum(X)}.
\end{proof}

Another lemma:

\begin{lemma}
\label{lem.im(X)}Let $A$, $B$ and $C$ be three $k$-modules. Let
$f:B\rightarrow C$ be a $k$-module map. Then,%
\[
\left(  \operatorname*{id}\otimes f\right)  \left(  A\otimes B\right)
=\left(  \operatorname*{id}\otimes\operatorname*{inc}\nolimits_{f\left(
B\right)  ,C}\right)  \left(  A\otimes\left(  f\left(  B\right)  \right)
\right)
\]
as $k$-submodules of $A\otimes C$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.im(X)}.]We define a map $f^{\prime}:B\rightarrow
f\left(  B\right)  $ by%
\[
\left(  f^{\prime}\left(  x\right)  =f\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in B\right)  .
\]
(This map is well-defined, since $f\left(  x\right)  \in f\left(  B\right)  $
for every $x\in B$.) Then, every $x\in B$ satisfies%
\[
f\left(  x\right)  =f^{\prime}\left(  x\right)  =\operatorname*{inc}%
\nolimits_{f\left(  B\right)  ,C}\left(  f^{\prime}\left(  x\right)  \right)
=\left(  \operatorname*{inc}\nolimits_{f\left(  B\right)  ,C}\circ f^{\prime
}\right)  \left(  x\right)  .
\]
Thus, $f=\operatorname*{inc}\nolimits_{f\left(  B\right)  ,C}\circ f^{\prime}%
$. Hence,%
\[
\operatorname*{id}\otimes f=\operatorname*{id}\otimes\left(
\operatorname*{inc}\nolimits_{f\left(  B\right)  ,C}\circ f^{\prime}\right)
=\left(  \operatorname*{id}\otimes\operatorname*{inc}\nolimits_{f\left(
B\right)  ,C}\right)  \circ\left(  \operatorname*{id}\otimes f^{\prime
}\right)  ,
\]
where $\operatorname*{id}$ means $\operatorname*{id}\nolimits_{A}$. Thus,%
\begin{align}
\left(  \operatorname*{id}\otimes f\right)  \left(  A\otimes B\right)   &
=\left(  \left(  \operatorname*{id}\otimes\operatorname*{inc}%
\nolimits_{f\left(  B\right)  ,C}\right)  \circ\left(  \operatorname*{id}%
\otimes f^{\prime}\right)  \right)  \left(  A\otimes B\right) \nonumber\\
&  =\left(  \operatorname*{id}\otimes\operatorname*{inc}\nolimits_{f\left(
B\right)  ,C}\right)  \left(  \left(  \operatorname*{id}\otimes f^{\prime
}\right)  \left(  A\otimes B\right)  \right)  . \label{pf.im(X).1}%
\end{align}


Now, the map $f^{\prime}:B\rightarrow f\left(  B\right)  $ is
surjective\footnote{\textit{Proof.} Let $y\in f\left(  B\right)  $. Then,
there exists some $x\in B$ such that $y=f\left(  x\right)  $ (by the
definition of $f\left(  B\right)  $). Consider this $x$. Then, $f^{\prime
}\left(  x\right)  =f\left(  x\right)  =y$.
\par
Hence, we have shown that for every $y\in f\left(  B\right)  $, there exists
some $x\in B$ such that $y=f^{\prime}\left(  x\right)  $. In other words, the
map $f^{\prime}:B\rightarrow f\left(  B\right)  $ is surjective, qed.}, so
that the map $\operatorname*{id}\otimes f^{\prime}:A\otimes B\rightarrow
A\otimes\left(  f\left(  B\right)  \right)  $ is surjective as well (by Lemma
\ref{lem.f(X)g.sur}, applied to $A$, $A$, $B$, $f\left(  B\right)  $,
$\operatorname*{id}$, $f^{\prime}$ instead of $V$, $V^{\prime}$, $W$,
$W^{\prime}$, $f$, $g$, respectively). Thus, $\left(  \operatorname*{id}%
\otimes f^{\prime}\right)  \left(  A\otimes B\right)  =A\otimes\left(
f\left(  B\right)  \right)  $. Hence, (\ref{pf.im(X).1}) becomes%
\[
\left(  \operatorname*{id}\otimes f\right)  \left(  A\otimes B\right)
=\left(  \operatorname*{id}\otimes\operatorname*{inc}\nolimits_{f\left(
B\right)  ,C}\right)  \left(  A\otimes\left(  f\left(  B\right)  \right)
\right)  .
\]
This proves Lemma \ref{lem.im(X)}.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.f_i}.]We are going to prove Theorem \ref{thm.f_i}
by induction over $n$:

\textit{Induction base:} For $n=0$, Theorem \ref{thm.f_i} holds (because for
$n=0$, the map $f_{1}\otimes f_{2}\otimes\cdots\otimes f_{n}:V_{1}\otimes
V_{2}\otimes\cdots\otimes V_{n}\rightarrow V_{1}^{\prime}\otimes V_{2}%
^{\prime}\otimes\cdots\otimes V_{n}^{\prime}$ is the identity map
$\operatorname{id}:k\rightarrow k$ and therefore its kernel
$\operatorname*{Ker}\left(  f_{1}\otimes f_{2}\otimes\cdots\otimes
f_{n}\right)  $ is $0$, while the right hand side of (\ref{thm.f_i.eq}) is
also $0$ when $n=0$). Thus, the induction base is complete.

\textit{Induction step:} Let $p\in\mathbb{N}$ be arbitrary. Assume that
Theorem \ref{thm.f_i} holds for $n=p$.

Now let us prove that Theorem \ref{thm.f_i} holds for $n=p+1$. So let $V_{i}$
and $V_{i}^{\prime}$ be two $k$-modules for every $i\in\left\{  1,2,\ldots
,p+1\right\}  $, and let $f_{i}:V_{i}\rightarrow V_{i}^{\prime}$ be a
surjective $k$-module homomorphism for every $i\in\left\{  1,2,\ldots
,p+1\right\}  $.

According to Definition \ref{defs.f_1(X)f_2}, we have $f_{1}\otimes
f_{2}\otimes\cdots\otimes f_{p+1}=f_{1}\otimes\left(  f_{2}\otimes
f_{3}\otimes\cdots\otimes f_{p+1}\right)  $.

We know that $f_{i}:V_{i}\rightarrow V_{i}^{\prime}$ is a surjective
$k$-module homomorphism for every $i\in\left\{  1,2,\ldots,p+1\right\}  $.
Thus, $f_{i}:V_{i}\rightarrow V_{i}^{\prime}$ is a surjective $k$-module
homomorphism for every $i\in\left\{  2,3,\ldots,p+1\right\}  $. Substituting
$i+1$ for $i$ in this fact, we obtain that $f_{i+1}$ is a surjective
$k$-module homomorphism for every $i\in\left\{  1,2,\ldots,p\right\}  $. Thus,
Lemma \ref{lem.f_i.sur} (applied to $p$, $V_{i+1}$, $V_{i+1}^{\prime}$ and
$f_{i+1}$ instead of $n$, $V_{i}$, $V_{i}^{\prime}$ and $f_{i}$) yields that
the map $f_{2}\otimes f_{3}\otimes\cdots\otimes f_{p+1}$ is surjective. On the
other hand, the map $f_{1}$ is surjective (since $f_{i}$ is surjective for
every $i\in\left\{  1,2,\ldots,p+1\right\}  $).

Now let $V=V_{1}$, $V^{\prime}=W_{1}$, $f=f_{1}$ and $i_{V}=\mathfrak{i}_{1}$.
Then, $f$ is a surjective $k$-linear map (since $f_{1}$ is a surjective
$k$-linear map), and $i_{V}$ is the canonical inclusion $\operatorname*{Ker}%
f\rightarrow V$ (since $i_{V}=\mathfrak{i}_{1}$ is the canonical inclusion
$\operatorname*{Ker}f_{1}\rightarrow V_{1}$).

Further let $W=V_{2}\otimes V_{3}\otimes\cdots\otimes V_{p+1}$, $W^{\prime
}=W_{2}\otimes W_{3}\otimes\cdots\otimes W_{p+1}$ and $g=f_{2}\otimes
f_{3}\otimes\cdots\otimes f_{p+1}$. Then, $g$ is a surjective $k$-linear map
(since $f_{2}\otimes f_{3}\otimes\cdots\otimes f_{p+1}$ is a surjective
$k$-linear map). Let $i_{W}$ be the canonical inclusion $\operatorname*{Ker}%
g\rightarrow W$. Then, Theorem \ref{thm.f(X)g} yields%
\begin{equation}
\operatorname*{Ker}\left(  f\otimes g\right)  =\left(  i_{V}\otimes
\operatorname*{id}\right)  \left(  \left(  \operatorname*{Ker}f\right)
\otimes W\right)  +\left(  \operatorname*{id}\otimes i_{W}\right)  \left(
V\otimes\left(  \operatorname*{Ker}g\right)  \right)  . \label{pf.f_i.4}%
\end{equation}


Note that $V=V_{1}$ and $W=V_{2}\otimes V_{3}\otimes\cdots\otimes V_{p+1}$
yield $V\otimes W=V_{1}\otimes\left(  V_{2}\otimes V_{3}\otimes\cdots\otimes
V_{p+1}\right)  =V_{1}\otimes V_{2}\otimes\cdots\otimes V_{p+1}$.

We now define some more abbreviations. For every $i\in\left\{  1,2,\ldots
,p+1\right\}  $, let $K_{i}$ denote the $k$-module%
\[
V_{1}\otimes V_{2}\otimes\cdots\otimes V_{i-1}\otimes\left(
\operatorname*{Ker}f_{i}\right)  \otimes V_{i+1}\otimes V_{i+2}\otimes
\cdots\otimes V_{p+1}.
\]
For every $i\in\left\{  1,2,\ldots,p+1\right\}  $, let $\kappa_{i}$ denote the
$k$-linear map%
\[
\underbrace{\operatorname*{id}\otimes\operatorname*{id}\otimes\cdots
\otimes\operatorname*{id}}_{i-1\text{ times}}\otimes\mathfrak{i}_{i}%
\otimes\underbrace{\operatorname*{id}\otimes\operatorname*{id}\otimes
\cdots\otimes\operatorname*{id}}_{p+1-i\text{ times}}:K_{i}\rightarrow
V\otimes W
\]
(this is well-defined since $K_{i}=V_{1}\otimes V_{2}\otimes\cdots\otimes
V_{i-1}\otimes\left(  \operatorname*{Ker}f_{i}\right)  \otimes V_{i+1}\otimes
V_{i+2}\otimes\cdots\otimes V_{p+1}$ and $V\otimes W=V_{1}\otimes V_{2}%
\otimes\cdots\otimes V_{p+1}$).

For every $i\in\left\{  2,3,\ldots,p+1\right\}  $, let $M_{i}$ denote the
$k$-module%
\[
V_{2}\otimes V_{3}\otimes\cdots\otimes V_{i-1}\otimes\left(
\operatorname*{Ker}f_{i}\right)  \otimes V_{i+1}\otimes V_{i+2}\otimes
\cdots\otimes V_{p+1}.
\]
For every $i\in\left\{  2,3,\ldots,p+1\right\}  $, let $\mu_{i}$ denote the
$k$-linear map%
\[
\underbrace{\operatorname*{id}\otimes\operatorname*{id}\otimes\cdots
\otimes\operatorname*{id}}_{i-2\text{ times}}\otimes\mathfrak{i}_{i}%
\otimes\underbrace{\operatorname*{id}\otimes\operatorname*{id}\otimes
\cdots\otimes\operatorname*{id}}_{p+1-i\text{ times}}:M_{i}\rightarrow W
\]
(this is well-defined since $M_{i}=V_{2}\otimes V_{3}\otimes\cdots\otimes
V_{i-1}\otimes\left(  \operatorname*{Ker}f_{i}\right)  \otimes V_{i+1}\otimes
V_{i+2}\otimes\cdots\otimes V_{p+1}$ and $W=V_{2}\otimes V_{3}\otimes
\cdots\otimes V_{p+1}$).

For any $k$-module $C$ and any $k$-submodule $D$ of $C$, we let
$\operatorname*{inc}\nolimits_{D,C}$ denote the canonical inclusion map
$D\rightarrow C$. Then,%
\[
\operatorname*{inc}\nolimits_{\operatorname*{Ker}f_{i},V_{i}}=\left(
\text{the canonical inclusion map }\operatorname*{Ker}f_{i}\rightarrow
V_{i}\right)  =\mathfrak{i}_{i}%
\]
for every $i\in\left\{  1,2,\ldots,p+1\right\}  $. On the other hand,%
\[
\operatorname*{inc}\nolimits_{\operatorname*{Ker}g,W}=\left(  \text{the
canonical inclusion map }\operatorname*{Ker}g\rightarrow W\right)  =i_{W}%
\]
(by definition of $i_{W}$).

Applying Theorem \ref{thm.f_i} to $p$, $V_{i+1}$, $V_{i+1}^{\prime}$ and
$f_{i+1}$ instead of $n$, $V_{i}$, $V_{i}^{\prime}$ and $f_{i}$ (this is
allowed, because we have assumed that Theorem \ref{thm.f_i} holds for $n=p$),
we see that%
\begin{align*}
&  \operatorname*{Ker}\left(  f_{2}\otimes f_{3}\otimes\cdots\otimes
f_{p+1}\right) \\
&  =\sum_{i=1}^{p}\left(  \underbrace{\operatorname*{id}\otimes
\operatorname*{id}\otimes\cdots\otimes\operatorname*{id}}_{i-1\text{ times}%
}\otimes\mathfrak{i}_{i+1}\otimes\underbrace{\operatorname*{id}\otimes
\operatorname*{id}\otimes\cdots\otimes\operatorname*{id}}_{p-i\text{ times}%
}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  V_{2}\otimes V_{3}%
\otimes\cdots\otimes V_{i}\otimes\left(  \operatorname*{Ker}f_{i+1}\right)
\otimes V_{i+2}\otimes V_{i+3}\otimes\cdots\otimes V_{p+1}\right) \\
&  =\sum_{i=2}^{p+1}\underbrace{\left(  \underbrace{\operatorname*{id}%
\otimes\operatorname*{id}\otimes\cdots\otimes\operatorname*{id}}_{i-2\text{
times}}\otimes\mathfrak{i}_{i}\otimes\underbrace{\operatorname*{id}%
\otimes\operatorname*{id}\otimes\cdots\otimes\operatorname*{id}}_{p-i+1\text{
times}}\right)  }_{=\mu_{i}}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underbrace{\left(  V_{2}\otimes
V_{3}\otimes\cdots\otimes V_{i-1}\otimes\left(  \operatorname*{Ker}%
f_{i}\right)  \otimes V_{i+1}\otimes V_{i+2}\otimes\cdots\otimes
V_{p+1}\right)  }_{=M_{i}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }i\text{ for
}i+1\text{ in the sum}\right) \\
&  =\sum_{i=2}^{p+1}\mu_{i}\left(  M_{i}\right)  .
\end{align*}
Since $f_{2}\otimes f_{3}\otimes\cdots\otimes f_{p+1}=g$, this rewrites as%
\begin{equation}
\operatorname*{Ker}g=\sum_{i=2}^{p+1}\mu_{i}\left(  M_{i}\right)  =\sum
_{i=1}^{p}\mu_{i+1}\left(  M_{i+1}\right)  \label{pf.f_i.5}%
\end{equation}
(here, we substituted $i$ for $i-1$). But now it is easy to see that%
\begin{equation}
\left(  \operatorname*{id}\otimes i_{W}\right)  \left(  V\otimes\left(
\operatorname*{Ker}g\right)  \right)  =\sum_{i=2}^{p+1}\kappa_{i}\left(
K_{i}\right)  . \label{pf.f_i.6}%
\end{equation}


\textit{Proof of (\ref{pf.f_i.6}).} Let $i\in\left\{  2,3,\ldots,p+1\right\}
$. Then,%
\begin{align*}
\kappa_{i}  &  =\underbrace{\operatorname*{id}\otimes\operatorname*{id}%
\otimes\cdots\otimes\operatorname*{id}}_{i-1\text{ times}}\otimes
\mathfrak{i}_{i}\otimes\underbrace{\operatorname*{id}\otimes\operatorname*{id}%
\otimes\cdots\otimes\operatorname*{id}}_{p+1-i\text{ times}}\\
&  =\operatorname*{id}\otimes\underbrace{\left(
\underbrace{\operatorname*{id}\otimes\operatorname*{id}\otimes\cdots
\otimes\operatorname*{id}}_{i-2\text{ times}}\otimes\mathfrak{i}_{i}%
\otimes\underbrace{\operatorname*{id}\otimes\operatorname*{id}\otimes
\cdots\otimes\operatorname*{id}}_{p+1-i\text{ times}}\right)  }_{=\mu_{i}%
}=\operatorname*{id}\otimes\mu_{i}%
\end{align*}
and%
\begin{align*}
K_{i}  &  =V_{1}\otimes V_{2}\otimes\cdots\otimes V_{i-1}\otimes\left(
\operatorname*{Ker}f_{i}\right)  \otimes V_{i+1}\otimes V_{i+2}\otimes
\cdots\otimes V_{p+1}\\
&  =\underbrace{V_{1}}_{=V}\otimes\underbrace{\left(  V_{2}\otimes
V_{3}\otimes\cdots\otimes V_{i-1}\otimes\left(  \operatorname*{Ker}%
f_{i}\right)  \otimes V_{i+1}\otimes V_{i+2}\otimes\cdots\otimes
V_{p+1}\right)  }_{=M_{i}}\\
&  =V\otimes M_{i}.
\end{align*}
Thus,%
\[
\kappa_{i}\left(  K_{i}\right)  =\left(  \operatorname*{id}\otimes\mu
_{i}\right)  \left(  V\otimes M_{i}\right)  =\left(  \operatorname*{id}%
\otimes\operatorname*{inc}\nolimits_{\mu_{i}\left(  M_{i}\right)  ,W}\right)
\left(  V\otimes\left(  \mu_{i}\left(  M_{i}\right)  \right)  \right)
\]
(by Lemma \ref{lem.im(X)}, applied to $A=V$, $B=M_{i}$, $C=W$ and $f=\mu_{i}$).

But since $\mu_{i}\left(  M_{i}\right)  \subseteq\operatorname*{Ker}g$ (by
(\ref{pf.f_i.5})), we have
\begin{align*}
\operatorname*{inc}\nolimits_{\mu_{i}\left(  M_{i}\right)  ,W}  &
=\underbrace{\operatorname*{inc}\nolimits_{\operatorname*{Ker}g,W}}_{=i_{W}%
}\circ\operatorname*{inc}\nolimits_{\mu_{i}\left(  M_{i}\right)
,\operatorname*{Ker}g}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since any three }k\text{-modules }A\text{, }B\text{, }C\text{ such that
}A\subseteq B\subseteq C\text{ satisfy}\\
\operatorname*{inc}\nolimits_{A,C}=\operatorname*{inc}\nolimits_{B,C}%
\circ\operatorname*{inc}\nolimits_{A,B}%
\end{array}
\right) \\
&  =i_{W}\circ\operatorname*{inc}\nolimits_{\mu_{i}\left(  M_{i}\right)
,\operatorname*{Ker}g}.
\end{align*}


Now forget that we fixed $i\in\left\{  2,3,\ldots,p+1\right\}  $. Due to
(\ref{pf.f_i.5}), we can apply Lemma \ref{lem.sum(X)} to $n=p$, $A=V$, $B=W$,
$B^{\prime}=\operatorname*{Ker}g$ and $B_{i}=\mu_{i+1}\left(  M_{i+1}\right)
$.\ Applying it yields that%
\begin{align*}
\left(  \operatorname*{id}\otimes\operatorname*{inc}%
\nolimits_{\operatorname*{Ker}g,W}\right)  \left(  V\otimes\left(
\operatorname*{Ker}g\right)  \right)   &  =\sum_{i=1}^{p}\left(
\operatorname*{id}\otimes\operatorname*{inc}\nolimits_{\mu_{i+1}\left(
M_{i+1}\right)  ,W}\right)  \left(  V\otimes\left(  \mu_{i+1}\left(
M_{i+1}\right)  \right)  \right) \\
&  =\sum\limits_{i=2}^{p+1}\underbrace{\left(  \operatorname*{id}%
\otimes\operatorname*{inc}\nolimits_{\mu_{i}\left(  M_{i}\right)  ,W}\right)
\left(  V\otimes\left(  \mu_{i}\left(  M_{i}\right)  \right)  \right)
}_{=\kappa_{i}\left(  K_{i}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }i\text{ for
}i+1\text{ in the sum}\right) \\
&  =\sum_{i=2}^{p+1}\kappa_{i}\left(  K_{i}\right)  ,
\end{align*}
and thus (\ref{pf.f_i.6}) is proven.

On the other hand, we defined the $k$-module $K_{i}$ as
\[
V_{1}\otimes V_{2}\otimes\cdots\otimes V_{i-1}\otimes\left(
\operatorname*{Ker}f_{i}\right)  \otimes V_{i+1}\otimes V_{i+2}\otimes
\cdots\otimes V_{p+1}%
\]
for every $i\in\left\{  1,2,\ldots,p+1\right\}  $. Applied to $i=1$, this
yields%
\begin{equation}
K_{1}=\left(  \operatorname*{Ker}f_{1}\right)  \otimes V_{2}\otimes
V_{3}\otimes\cdots\otimes V_{p+1}=\left(  \operatorname*{Ker}\underbrace{f_{1}%
}_{=f}\right)  \otimes\underbrace{\left(  V_{2}\otimes V_{3}\otimes
\cdots\otimes V_{p+1}\right)  }_{=W}=\left(  \operatorname*{Ker}f\right)
\otimes W. \label{pf.f_i.10}%
\end{equation}


We further defined the map $\kappa_{i}$ as%
\[
\underbrace{\operatorname*{id}\otimes\operatorname*{id}\otimes\cdots
\otimes\operatorname*{id}}_{i-1\text{ times}}\otimes\mathfrak{i}_{i}%
\otimes\underbrace{\operatorname*{id}\otimes\operatorname*{id}\otimes
\cdots\otimes\operatorname*{id}}_{p+1-i\text{ times}}:K_{i}\rightarrow
V\otimes W
\]
for every $i\in\left\{  1,2,\ldots,p+1\right\}  $. Applied to $i=1$, this
yields%
\begin{equation}
\kappa_{1}=\mathfrak{i}_{1}\otimes\underbrace{\operatorname*{id}%
\otimes\operatorname*{id}\otimes\cdots\otimes\operatorname*{id}}_{p+1-1\text{
times}}=\underbrace{\mathfrak{i}_{1}}_{=i_{V}}\otimes\underbrace{\left(
\underbrace{\operatorname*{id}\otimes\operatorname*{id}\otimes\cdots
\otimes\operatorname*{id}}_{p+1-1\text{ times}}\right)  }_{=\operatorname*{id}%
}=i_{V}\otimes\operatorname*{id}. \label{pf.f_i.11}%
\end{equation}


Now, (\ref{pf.f_i.4}) becomes%
\begin{align*}
\operatorname*{Ker}\left(  f\otimes g\right)   &  =\underbrace{\left(
i_{V}\otimes\operatorname*{id}\right)  }_{=\kappa_{1}\text{ (by
(\ref{pf.f_i.11}))}}\left(  \underbrace{\left(  \operatorname*{Ker}f\right)
\otimes W}_{=K_{1}\text{ (by (\ref{pf.f_i.10}))}}\right)  +\underbrace{\left(
\operatorname*{id}\otimes i_{W}\right)  \left(  V\otimes\left(
\operatorname*{Ker}g\right)  \right)  }_{=\sum\limits_{i=2}^{p+1}\kappa
_{i}\left(  K_{i}\right)  \text{ (by (\ref{pf.f_i.6}))}}\\
&  =\kappa_{1}\left(  K_{1}\right)  +\sum\limits_{i=2}^{p+1}\kappa_{i}\left(
K_{i}\right)  =\sum\limits_{i=1}^{p+1}\kappa_{i}\left(  K_{i}\right)  .
\end{align*}
Since%
\[
\underbrace{f}_{=f_{1}}\otimes\underbrace{g}_{=f_{2}\otimes f_{3}\otimes
\cdots\otimes f_{p+1}}=f_{1}\otimes\left(  f_{2}\otimes f_{3}\otimes
\cdots\otimes f_{p+1}\right)  =f_{1}\otimes f_{2}\otimes\cdots\otimes
f_{p+1},
\]
this rewrites as%
\begin{align*}
&  \operatorname*{Ker}\left(  f_{1}\otimes f_{2}\otimes\cdots\otimes
f_{p+1}\right) \\
&  =\sum\limits_{i=1}^{p+1}\underbrace{\kappa_{i}}%
_{=\underbrace{\operatorname*{id}\otimes\operatorname*{id}\otimes\cdots
\otimes\operatorname*{id}}_{i-1\text{ times}}\otimes\mathfrak{i}_{i}%
\otimes\underbrace{\operatorname*{id}\otimes\operatorname*{id}\otimes
\cdots\otimes\operatorname*{id}}_{p+1-i\text{ times}}}\left(
\underbrace{K_{i}}_{=V_{1}\otimes V_{2}\otimes\cdots\otimes V_{i-1}%
\otimes\left(  \operatorname*{Ker}f_{i}\right)  \otimes V_{i+1}\otimes
V_{i+2}\otimes\cdots\otimes V_{p+1}}\right) \\
&  =\sum\limits_{i=1}^{p+1}\left(  \underbrace{\operatorname*{id}%
\otimes\operatorname*{id}\otimes\cdots\otimes\operatorname*{id}}_{i-1\text{
times}}\otimes\mathfrak{i}_{i}\otimes\underbrace{\operatorname*{id}%
\otimes\operatorname*{id}\otimes\cdots\otimes\operatorname*{id}}_{p+1-i\text{
times}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  V_{1}\otimes V_{2}%
\otimes\cdots\otimes V_{i-1}\otimes\left(  \operatorname*{Ker}f_{i}\right)
\otimes V_{i+1}\otimes V_{i+2}\otimes\cdots\otimes V_{p+1}\right)  .
\end{align*}


We thus have proven that (\ref{thm.f_i.eq}) holds for $n=p+1$. This completes
the induction step. Thus, the induction proof of Theorem \ref{thm.f_i} is complete.
\end{proof}

\subsection{\label{subsect.tensalg}The tensor algebra case}

Before we actually come to the tensor algebra, let us bring Theorem
\ref{thm.f_i} to a nicer form when all the $f_{i}$ are equal:

\begin{theorem}
\label{thm.f_i.V}Let $k$ be a commutative ring. Let $n\in\mathbb{N}$. Let $V$
and $V^{\prime}$ be two $k$-modules, and let $f:V\rightarrow V^{\prime}$ be a
surjective $k$-module homomorphism. Let $\mathfrak{i}$ be the canonical
inclusion $\operatorname*{Ker}f\rightarrow V$. Then,%
\begin{equation}
\operatorname*{Ker}\left(  f^{\otimes n}\right)  =\sum_{i=1}^{n}\left(
\operatorname*{id}\nolimits_{V^{\otimes\left(  i-1\right)  }}\otimes
\mathfrak{i}\otimes\operatorname*{id}\nolimits_{V^{\otimes\left(  n-i\right)
}}\right)  \left(  V^{\otimes\left(  i-1\right)  }\otimes\left(
\operatorname*{Ker}f\right)  \otimes V^{\otimes\left(  n-i\right)  }\right)  .
\label{thm.f_i.V.eq}%
\end{equation}

\end{theorem}

Notice that the left hand side of the equation (\ref{thm.f_i.V.eq}) is a
subset of $V^{\otimes n}$, while the $i$-th addend on the right hand side is a
subset of $V^{\otimes\left(  i-1\right)  }\otimes V\otimes V^{\otimes\left(
n-i\right)  }$. To make sense of the equation (\ref{thm.f_i.V.eq}), the set
$V^{\otimes n}$ thus must be equal to $V^{\otimes\left(  i-1\right)  }\otimes
V\otimes V^{\otimes\left(  n-i\right)  }$ for every $i\in\left\{
1,2,\ldots,n\right\}  $. Fortunately, this is guaranteed by Convention
\ref{conv.(X)^n.ident}\ \ \ \ \footnote{In fact, using Convention
\ref{conv.(X)^n.ident}, we have%
\begin{align*}
&  V^{\otimes\left(  i-1\right)  }\otimes\underbrace{V}_{=V^{\otimes1}\text{
(by Remark \ref{rmk.(X)^0})}}\otimes V^{\otimes\left(  n-i\right)  }\\
&  =\underbrace{V^{\otimes\left(  i-1\right)  }\otimes V^{\otimes1}%
}_{=V^{\otimes\left(  i-1+1\right)  }\text{ (by Convention
\ref{conv.(X)^n.ident})}}\otimes V^{\otimes\left(  n-i\right)  }\\
&  =\underbrace{V^{\otimes\left(  i-1+1\right)  }}_{=V^{\otimes i}}\otimes
V^{\otimes\left(  n-i\right)  }=V^{\otimes i}\otimes V^{\otimes\left(
n-i\right)  }=V^{\otimes\left(  i+n-i\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{by Convention \ref{conv.(X)^n.ident}}\right) \\
&  =V^{\otimes n}.
\end{align*}
}.

For the proof of Theorem \ref{thm.f_i.V}, we will use one more convention:

\begin{condition}
\label{conv.AxBxC}Let $k$ be a commutative ring, and let $A$, $B$ and $C$ be
three $k$-modules. Then, we identify the $k$-module $\left(  A\otimes
B\right)  \otimes C$ with the $k$-module $A\otimes\left(  B\otimes C\right)  $
by means of the $k$-module isomorphism%
\begin{align*}
\left(  A\otimes B\right)  \otimes C  &  \rightarrow A\otimes\left(  B\otimes
C\right)  ,\\
\left(  a\otimes b\right)  \otimes c  &  \mapsto a\otimes\left(  b\otimes
c\right)  .
\end{align*}

\end{condition}

Note that we will only use Convention \ref{conv.AxBxC} in the proof of Theorem
\ref{thm.f_i.V}, but nowhere else in this text.

\begin{remark}
\label{rmk.AxBxC}As a consequence of Convention \ref{conv.AxBxC}, it can be
easily seen that the tensor product $V_{1}\otimes V_{2}\otimes\cdots\otimes
V_{n}$ of any $k$-modules $V_{1}$, $V_{2}$, $\ldots$, $V_{n}$ can be computed
by means of any bracketing. For instance, when $n=4$, this means that%
\begin{align*}
V_{1}\otimes\left(  V_{2}\otimes\left(  V_{3}\otimes V_{4}\right)  \right)
&  =V_{1}\otimes\left(  \left(  V_{2}\otimes V_{3}\right)  \otimes
V_{4}\right)  =\left(  V_{1}\otimes V_{2}\right)  \otimes\left(  V_{3}\otimes
V_{4}\right) \\
&  =\left(  V_{1}\otimes\left(  V_{2}\otimes V_{3}\right)  \right)  \otimes
V_{4}=\left(  \left(  V_{1}\otimes V_{2}\right)  \otimes V_{3}\right)  \otimes
V_{4}%
\end{align*}
for any four $k$-modules $V_{1}$, $V_{2}$, $V_{3}$, $V_{4}$.
\end{remark}

\begin{remark}
Convention \ref{conv.AxBxC} is compatible with Convention
\ref{conv.(X)^n.ident}. In fact, Conventions \ref{conv.AxBxC} and
\ref{conv.V(X)k} combined make Convention \ref{conv.(X)^n.ident} redundant, in
the following sense: If we identify $\left(  A\otimes B\right)  \otimes C$
with $A\otimes\left(  B\otimes C\right)  $ for all $k$-modules $A$, $B$ and
$C$ (as in Convention \ref{conv.AxBxC}), and identify $V\otimes k$, $k\otimes
V$ and $V$ for all $k$-modules $V$ (as in Convention \ref{conv.V(X)k}), then
automatically $V^{\otimes a}\otimes V^{\otimes b}$ becomes identical with
$V^{\otimes\left(  a+b\right)  }$ for all $k$-modules $V$ and $a\in\mathbb{N}$
and $b\in\mathbb{N}$ (and this identification is the same as the one given in
Convention \ref{conv.(X)^n.ident}).
\end{remark}

\begin{proof}
[Proof of Theorem \ref{thm.f_i.V}.]During this proof, we are going to use
Convention \ref{conv.AxBxC}.

Now, we apply Theorem \ref{thm.f_i} to $V_{i}=V$, $V_{i}^{\prime}=V^{\prime}$,
$f_{i}=f$ and $\mathfrak{i}_{i}=\mathfrak{i}$. As a result, we obtain%
\begin{align}
&  \operatorname*{Ker}\left(  \underbrace{f\otimes f\otimes\cdots\otimes
f}_{n\text{ times}}\right) \nonumber\\
&  =\sum_{i=1}^{n}\left(  \underbrace{\operatorname*{id}\otimes
\operatorname*{id}\otimes\cdots\otimes\operatorname*{id}}_{i-1\text{ times}%
}\otimes\mathfrak{i}\otimes\underbrace{\operatorname*{id}\otimes
\operatorname*{id}\otimes\cdots\otimes\operatorname*{id}}_{n-i\text{ times}%
}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \underbrace{V\otimes
V\otimes\cdots\otimes V}_{i-1\text{ times}}\otimes\left(  \operatorname*{Ker}%
f\right)  \otimes\underbrace{V\otimes V\otimes\cdots\otimes V}_{n-i\text{
times}}\right)  . \label{pf.f_i.V.1}%
\end{align}
Now, since we are using Convention \ref{conv.AxBxC}, we can write%
\begin{align*}
&  \underbrace{V\otimes V\otimes\cdots\otimes V}_{i-1\text{ times}}%
\otimes\left(  \operatorname*{Ker}f\right)  \otimes\underbrace{V\otimes
V\otimes\cdots\otimes V}_{n-i\text{ times}}\\
&  =\underbrace{\left(  \underbrace{V\otimes V\otimes\cdots\otimes
V}_{i-1\text{ times}}\right)  }_{=V^{\otimes\left(  i-1\right)  }}%
\otimes\left(  \operatorname*{Ker}f\right)  \otimes\underbrace{\left(
\underbrace{V\otimes V\otimes\cdots\otimes V}_{n-i\text{ times}}\right)
}_{=V^{\otimes\left(  n-i\right)  }}\\
&  =V^{\otimes\left(  i-1\right)  }\otimes\left(  \operatorname*{Ker}f\right)
\otimes V^{\otimes\left(  n-i\right)  }%
\end{align*}
and correspondingly%
\begin{align*}
&  \underbrace{\operatorname*{id}\otimes\operatorname*{id}\otimes\cdots
\otimes\operatorname*{id}}_{i-1\text{ times}}\otimes\mathfrak{i}%
\otimes\underbrace{\operatorname*{id}\otimes\operatorname*{id}\otimes
\cdots\otimes\operatorname*{id}}_{n-i\text{ times}}\\
&  =\underbrace{\left(  \underbrace{\operatorname*{id}\otimes
\operatorname*{id}\otimes\cdots\otimes\operatorname*{id}}_{i-1\text{ times}%
}\right)  }_{=\operatorname*{id}\nolimits_{V^{\otimes\left(  i-1\right)  }}%
}\otimes\mathfrak{i}\otimes\underbrace{\left(  \underbrace{\operatorname*{id}%
\otimes\operatorname*{id}\otimes\cdots\otimes\operatorname*{id}}_{n-i\text{
times}}\right)  }_{=\operatorname*{id}\nolimits_{V^{\otimes\left(  n-i\right)
}}}\\
&  =\operatorname*{id}\nolimits_{V^{\otimes\left(  i-1\right)  }}%
\otimes\mathfrak{i}\otimes\operatorname*{id}\nolimits_{V^{\otimes\left(
n-i\right)  }}%
\end{align*}
for every $i\in\left\{  1,2,\ldots,n\right\}  $. Now, since $f^{\otimes
n}=\underbrace{f\otimes f\otimes\cdots\otimes f}_{n\text{ times}}$, we have%
\begin{align*}
\operatorname*{Ker}\left(  f^{\otimes n}\right)   &  =\operatorname*{Ker}%
\left(  \underbrace{f\otimes f\otimes\cdots\otimes f}_{n\text{ times}}\right)
\\
&  =\sum_{i=1}^{n}\underbrace{\left(  \underbrace{\operatorname*{id}%
\otimes\operatorname*{id}\otimes\cdots\otimes\operatorname*{id}}_{i-1\text{
times}}\otimes\mathfrak{i}\otimes\underbrace{\operatorname*{id}\otimes
\operatorname*{id}\otimes\cdots\otimes\operatorname*{id}}_{n-i\text{ times}%
}\right)  }_{=\operatorname*{id}\nolimits_{V^{\otimes\left(  i-1\right)  }%
}\otimes\mathfrak{i}\otimes\operatorname*{id}\nolimits_{V^{\otimes\left(
n-i\right)  }}}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underbrace{\left(
\underbrace{V\otimes V\otimes\cdots\otimes V}_{i-1\text{ times}}\otimes\left(
\operatorname*{Ker}f\right)  \otimes\underbrace{V\otimes V\otimes\cdots\otimes
V}_{n-i\text{ times}}\right)  }_{=V^{\otimes\left(  i-1\right)  }%
\otimes\left(  \operatorname*{Ker}f\right)  \otimes V^{\otimes\left(
n-i\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.f_i.V.1})}\right) \\
&  =\sum_{i=1}^{n}\left(  \operatorname*{id}\nolimits_{V^{\otimes\left(
i-1\right)  }}\otimes\mathfrak{i}\otimes\operatorname*{id}%
\nolimits_{V^{\otimes\left(  n-i\right)  }}\right)  \left(  V^{\otimes\left(
i-1\right)  }\otimes\left(  \operatorname*{Ker}f\right)  \otimes
V^{\otimes\left(  n-i\right)  }\right)  .
\end{align*}
This proves Theorem \ref{thm.f_i.V}.
\end{proof}

Now, our claim about the tensor algebra:

\begin{theorem}
\label{thm.(X)f}Let $k$ be a commutative ring. Let $V$ and $V^{\prime}$ be two
$k$-modules, and let $f:V\rightarrow V^{\prime}$ be a surjective $k$-module
homomorphism. Then, the kernel of the map $\otimes f:\otimes V\rightarrow
\otimes V^{\prime}$ is%
\[
\operatorname*{Ker}\left(  \otimes f\right)  =\left(  \otimes V\right)
\cdot\left(  \operatorname*{Ker}f\right)  \cdot\left(  \otimes V\right)  .
\]
Here, $\operatorname*{Ker}f$ is considered a $k$-submodule of $\otimes V$ by
means of the inclusion $\operatorname*{Ker}f\subseteq V=V^{\otimes1}%
\subseteq\otimes V$.
\end{theorem}

We are going to derive this theorem from Theorem \ref{thm.f_i.V}. For this we
need the following lemma:

\begin{lemma}
\label{lem.(X)f}Let $k$ be a commutative ring. Let $n\in\mathbb{N}$. Let
$i\in\left\{  0,1,\ldots,n\right\}  $. Let $V$ be a $k$-module, and let $W$ be
a $k$-submodule of $V$. Let $\mathfrak{i}$ be the canonical inclusion
$W\rightarrow V$. Then,%
\[
\left(  \operatorname*{id}\nolimits_{V^{\otimes\left(  i-1\right)  }}%
\otimes\mathfrak{i}\otimes\operatorname*{id}\nolimits_{V^{\otimes\left(
n-i\right)  }}\right)  \left(  V^{\otimes\left(  i-1\right)  }\otimes W\otimes
V^{\otimes\left(  n-i\right)  }\right)  =V^{\otimes\left(  i-1\right)  }\cdot
W\cdot V^{\otimes\left(  n-i\right)  },
\]
where we identify $V^{\otimes n}$ with a $k$-submodule of $\otimes V$ as in
Definition \ref{defs.tensor} \textbf{(c)}.
\end{lemma}

To prove this lemma, we make a convention:

\begin{condition}
\label{conv.<>}\textbf{(a)} Whenever $k$ is a commutative ring, $M$ is a
$k$-module, and $S$ is a subset of $M$, we denote by $\left\langle
S\right\rangle $ the $k$-submodule of $M$ generated by the elements of $S$.
This $k$-submodule $\left\langle S\right\rangle $ is called the $k$%
\textit{-linear span} (or simply the $k$\textit{-span}) of $S$.\newline%
\textbf{(b)} Whenever $k$ is a commutative ring, $M$ is a $k$-module, $\Phi$
is a set, and $P:\Phi\rightarrow M$ is a map (not necessarily a linear map),
we denote by $\left\langle P\left(  v\right)  \ \mid\ v\in\Phi\right\rangle $
the $k$-submodule $\left\langle \left\{  P\left(  v\right)  \ \mid\ v\in
\Phi\right\}  \right\rangle $ of $M$. (In other words, $\left\langle P\left(
v\right)  \ \mid\ v\in\Phi\right\rangle $ is the $k$-submodule of $M$
generated by the elements $P\left(  v\right)  $ for all $v\in\Phi$.)
\end{condition}

Note that some authors use the notation $\left\langle S\right\rangle $ for
various other things (e. g., the two-sided ideal generated by $S$, or the Lie
subalgebra generated by $S$), but \textit{we will only use it for the }%
$k$\textit{-submodule generated by }$S$ \textit{(as defined in Convention
\ref{conv.<>} \textbf{(a)})}.

The following fact was proven in \cite{dg0}, \S 1.7 (but is basically trivial):

\begin{proposition}
\label{prop.<>}Let $k$ be a commutative ring. Let $M$ be a $k$-module. Let $S$
be a subset of $M$.\newline\textbf{(a)} Let $Q$ be a $k$-submodule of $M$ such
that $S\subseteq Q$. Then, $\left\langle S\right\rangle \subseteq Q$%
.\newline\textbf{(b)} Let $R$ be a $k$-module, and $f:M\rightarrow R$ be a
$k$-module homomorphism. Then, $f\left(  \left\langle S\right\rangle \right)
=\left\langle f\left(  S\right)  \right\rangle $.
\end{proposition}

Now let us come to the proof of Lemma \ref{lem.(X)f}:

\begin{proof}
[Proof of Lemma \ref{lem.(X)f}.]The tensor product $V^{\otimes\left(
i-1\right)  }\otimes W\otimes V^{\otimes\left(  n-i\right)  }$ is generated
(as a $k$-module) by its pure tensors. In other words,%
\begin{align*}
V^{\otimes\left(  i-1\right)  }\otimes W\otimes V^{\otimes\left(  n-i\right)
}  &  =\left\langle u\otimes v\otimes w\ \mid\ \left(  u,v,w\right)  \in
V^{\otimes\left(  i-1\right)  }\times W\times V^{\otimes\left(  n-i\right)
}\right\rangle \\
&  =\left\langle \left\{  u\otimes v\otimes w\ \mid\ \left(  u,v,w\right)  \in
V^{\otimes\left(  i-1\right)  }\times W\times V^{\otimes\left(  n-i\right)
}\right\}  \right\rangle
\end{align*}
Thus,%
\begin{align*}
&  \left(  \operatorname*{id}\nolimits_{V^{\otimes\left(  i-1\right)  }%
}\otimes\mathfrak{i}\otimes\operatorname*{id}\nolimits_{V^{\otimes\left(
n-i\right)  }}\right)  \left(  V^{\otimes\left(  i-1\right)  }\otimes W\otimes
V^{\otimes\left(  n-i\right)  }\right) \\
&  =\left(  \operatorname*{id}\nolimits_{V^{\otimes\left(  i-1\right)  }%
}\otimes\mathfrak{i}\otimes\operatorname*{id}\nolimits_{V^{\otimes\left(
n-i\right)  }}\right)  \left(  \left\langle \left\{  u\otimes v\otimes
w\ \mid\ \left(  u,v,w\right)  \in V^{\otimes\left(  i-1\right)  }\times
W\times V^{\otimes\left(  n-i\right)  }\right\}  \right\rangle \right) \\
&  =\left\langle \left(  \operatorname*{id}\nolimits_{V^{\otimes\left(
i-1\right)  }}\otimes\mathfrak{i}\otimes\operatorname*{id}%
\nolimits_{V^{\otimes\left(  n-i\right)  }}\right)  \left(  \left\{  u\otimes
v\otimes w\ \mid\ \left(  u,v,w\right)  \in V^{\otimes\left(  i-1\right)
}\times W\times V^{\otimes\left(  n-i\right)  }\right\}  \right)
\right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Proposition \ref{prop.<>} \textbf{(b)}, applied to }M=V^{\otimes
\left(  i-1\right)  }\otimes W\otimes V^{\otimes\left(  n-i\right)  }%
\text{,}\\
R=V^{\otimes n}\text{, }f=\operatorname*{id}\nolimits_{V^{\otimes\left(
i-1\right)  }}\otimes\mathfrak{i}\otimes\operatorname*{id}%
\nolimits_{V^{\otimes\left(  n-i\right)  }}\text{ and}\\
S=\left\{  u\otimes v\otimes w\ \mid\ \left(  u,v,w\right)  \in V^{\otimes
\left(  i-1\right)  }\times W\times V^{\otimes\left(  n-i\right)  }\right\}
\end{array}
\right)  .
\end{align*}
Since%
\begin{align*}
&  \left(  \operatorname*{id}\nolimits_{V^{\otimes\left(  i-1\right)  }%
}\otimes\mathfrak{i}\otimes\operatorname*{id}\nolimits_{V^{\otimes\left(
n-i\right)  }}\right)  \left(  \left\{  u\otimes v\otimes w\ \mid\ \left(
u,v,w\right)  \in V^{\otimes\left(  i-1\right)  }\times W\times V^{\otimes
\left(  n-i\right)  }\right\}  \right) \\
&  =\left\{  \underbrace{\left(  \operatorname*{id}\nolimits_{V^{\otimes
\left(  i-1\right)  }}\otimes\mathfrak{i}\otimes\operatorname*{id}%
\nolimits_{V^{\otimes\left(  n-i\right)  }}\right)  \left(  u\otimes v\otimes
w\right)  }_{=\operatorname*{id}\nolimits_{V^{\otimes\left(  i-1\right)  }%
}\left(  u\right)  \otimes\mathfrak{i}\left(  v\right)  \otimes
\operatorname*{id}\nolimits_{V^{\otimes\left(  n-i\right)  }}\left(  w\right)
}\ \mid\ \left(  u,v,w\right)  \in V^{\otimes\left(  i-1\right)  }\times
W\times V^{\otimes\left(  n-i\right)  }\right\} \\
&  =\left\{  \underbrace{\operatorname*{id}\nolimits_{V^{\otimes\left(
i-1\right)  }}\left(  u\right)  }_{=u}\otimes\underbrace{\mathfrak{i}\left(
v\right)  }_{\substack{=v\text{ (since }\mathfrak{i}\text{ is the}%
\\\text{inclusion map)}}}\otimes\underbrace{\operatorname*{id}%
\nolimits_{V^{\otimes\left(  n-i\right)  }}\left(  w\right)  }_{=w}%
\ \mid\ \left(  u,v,w\right)  \in V^{\otimes\left(  i-1\right)  }\times
W\times V^{\otimes\left(  n-i\right)  }\right\} \\
&  =\left\{  u\otimes v\otimes w\ \mid\ \left(  u,v,w\right)  \in
V^{\otimes\left(  i-1\right)  }\times W\times V^{\otimes\left(  n-i\right)
}\right\}  ,
\end{align*}
this rewrites as%
\begin{align}
&  \left(  \operatorname*{id}\nolimits_{V^{\otimes\left(  i-1\right)  }%
}\otimes\mathfrak{i}\otimes\operatorname*{id}\nolimits_{V^{\otimes\left(
n-i\right)  }}\right)  \left(  V^{\otimes\left(  i-1\right)  }\otimes W\otimes
V^{\otimes\left(  n-i\right)  }\right) \nonumber\\
&  =\left\langle \left\{  u\otimes v\otimes w\ \mid\ \left(  u,v,w\right)  \in
V^{\otimes\left(  i-1\right)  }\times W\times V^{\otimes\left(  n-i\right)
}\right\}  \right\rangle \nonumber\\
&  =\left\langle u\otimes v\otimes w\ \mid\ \left(  u,v,w\right)  \in
V^{\otimes\left(  i-1\right)  }\times W\times V^{\otimes\left(  n-i\right)
}\right\rangle . \label{pf.(X)f.1}%
\end{align}
On the other hand,%
\begin{equation}
V^{\otimes\left(  i-1\right)  }\cdot W\cdot V^{\otimes\left(  n-i\right)
}=\left\langle u\cdot v\cdot w\ \mid\ \left(  u,v,w\right)  \in V^{\otimes
\left(  i-1\right)  }\times W\times V^{\otimes\left(  n-i\right)
}\right\rangle , \label{pf.(X)f.2}%
\end{equation}
where the $\cdot$ sign stands for multiplication inside the $k$-algebra
$\otimes V$. But for every $\left(  u,v,w\right)  \in V^{\otimes\left(
i-1\right)  }\times W\times V^{\otimes\left(  n-i\right)  }$, we have $u\cdot
v\cdot w=u\otimes v\otimes w$\ \ \ \ \footnote{\textit{Proof.} We have $u\in
V^{\otimes\left(  i-1\right)  }$ and $v\in W\subseteq V=V^{\otimes1}$. Hence,
$u\cdot v=u\otimes v$ (by (\ref{*=(X)}), applied to $u$, $v$, $i-1$ and $1$
instead of $a$, $b$, $n$ and $m$) and thus $u\cdot v=\underbrace{u}_{\in
V^{\otimes\left(  i-1\right)  }}\otimes\underbrace{v}_{\in V^{\otimes1}}\in
V^{\otimes\left(  i-1\right)  }\otimes V^{\otimes1}=V^{\otimes i}$. Combined
with $w\in V^{\otimes\left(  n-i\right)  }$, this leads to $\left(  u\otimes
v\right)  \cdot w=\left(  u\otimes v\right)  \otimes w$ (by (\ref{*=(X)}),
applied to $u\otimes v$, $w$, $i$ and $n-i$ instead of $a$, $b$, $n$ and $m$).
Thus, $\underbrace{u\cdot v}_{=u\otimes v}\cdot w=\left(  u\otimes v\right)
\cdot w=\left(  u\otimes v\right)  \otimes w=u\otimes v\otimes w$, qed.}, so
that (\ref{pf.(X)f.2}) becomes%
\begin{align*}
V^{\otimes\left(  i-1\right)  }\cdot W\cdot V^{\otimes\left(  n-i\right)  }
&  =\left\langle u\otimes v\otimes w\ \mid\ \left(  u,v,w\right)  \in
V^{\otimes\left(  i-1\right)  }\times W\times V^{\otimes\left(  n-i\right)
}\right\rangle \\
&  =\left(  \operatorname*{id}\nolimits_{V^{\otimes\left(  i-1\right)  }%
}\otimes\mathfrak{i}\otimes\operatorname*{id}\nolimits_{V^{\otimes\left(
n-i\right)  }}\right)  \left(  V^{\otimes\left(  i-1\right)  }\otimes W\otimes
V^{\otimes\left(  n-i\right)  }\right)
\end{align*}
(by (\ref{pf.(X)f.1})). This proves Lemma \ref{lem.(X)f}.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.(X)f}.]Let $\mathfrak{i}$ be the canonical
inclusion $\operatorname*{Ker}f\rightarrow V$.

We have%
\begin{align}
\left.  \otimes V\right.   &  =\bigoplus\limits_{i\in\mathbb{N}}V^{\otimes
i}=\sum\limits_{i\in\mathbb{N}}V^{\otimes i}\ \ \ \ \ \ \ \ \ \ \left(
\text{since direct sums are sums}\right) \label{(X)V=sum_i}\\
&  =\sum\limits_{j\in\mathbb{N}}V^{\otimes j}\ \ \ \ \ \ \ \ \ \ \left(
\text{here, we renamed the index }i\text{ as }j\right)  . \label{(X)V=sum_j}%
\end{align}


But the map $\otimes f$ is defined as the direct sum of the $k$-module
homomorphisms $f^{\otimes i}:V^{\otimes i}\rightarrow W^{\otimes i}$ for all
$i\in\mathbb{N}$. Hence,%
\begin{align*}
\operatorname*{Ker}\left(  \otimes f\right)   &  =\bigoplus\limits_{i\in
\mathbb{N}}\operatorname*{Ker}\left(  f^{\otimes i}\right)  =\sum
\limits_{i\in\mathbb{N}}\operatorname*{Ker}\left(  f^{\otimes i}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since direct sums are sums}\right) \\
&  =\sum\limits_{n\in\mathbb{N}}\operatorname*{Ker}\left(  f^{\otimes
n}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index
}i\text{ as }n\text{ in the sum}\right) \\
&  =\sum\limits_{n\in\mathbb{N}}\sum_{i=1}^{n}\underbrace{\left(
\operatorname*{id}\nolimits_{V^{\otimes\left(  i-1\right)  }}\otimes
\mathfrak{i}\otimes\operatorname*{id}\nolimits_{V^{\otimes\left(  n-i\right)
}}\right)  \left(  V^{\otimes\left(  i-1\right)  }\otimes\left(
\operatorname*{Ker}f\right)  \otimes V^{\otimes\left(  n-i\right)  }\right)
}_{\substack{=V^{\otimes\left(  i-1\right)  }\cdot\left(  \operatorname*{Ker}%
f\right)  \cdot V^{\otimes\left(  n-i\right)  }\\\text{(by Lemma
\ref{lem.(X)f}, applied to }W=\operatorname*{Ker}f\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{thm.f_i.V.eq})}\right) \\
&  =\sum\limits_{n\in\mathbb{N}}\sum_{i=1}^{n}V^{\otimes\left(  i-1\right)
}\cdot\left(  \operatorname*{Ker}f\right)  \cdot V^{\otimes\left(  n-i\right)
}=\sum\limits_{n\in\mathbb{N}}\sum_{i=0}^{n-1}V^{\otimes i}\cdot\left(
\operatorname*{Ker}f\right)  \cdot\underbrace{V^{\otimes\left(  n-\left(
i+1\right)  \right)  }}_{=V^{\otimes\left(  n-1-i\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }i+1\text{ for
}i\text{ in the second sum}\right) \\
&  =\sum\limits_{n\in\mathbb{N}}\sum_{i=0}^{n-1}V^{\otimes i}\cdot\left(
\operatorname*{Ker}f\right)  \cdot V^{\otimes\left(  n-1-i\right)  }\\
&  =\sum\limits_{\substack{n\in\mathbb{N};\\n\geq1}}\sum_{i=0}^{n-1}V^{\otimes
i}\cdot\left(  \operatorname*{Ker}f\right)  \cdot V^{\otimes\left(
n-1-i\right)  }+\underbrace{\sum_{i=0}^{0-1}V^{\otimes i}\cdot\left(
\operatorname*{Ker}f\right)  \cdot V^{\otimes\left(  0-1-i\right)  }%
}_{=\left(  \text{empty sum}\right)  =0}\\
&  =\sum\limits_{\substack{n\in\mathbb{N};\\n\geq1}}\sum_{i=0}^{n-1}V^{\otimes
i}\cdot\left(  \operatorname*{Ker}f\right)  \cdot V^{\otimes\left(
n-1-i\right)  }=\underbrace{\sum\limits_{n\in\mathbb{N}}\sum_{i=0}^{n}}%
_{=\sum\limits_{i\in\mathbb{N}}\sum\limits_{\substack{n\in\mathbb{N};\\n\geq
i}}}V^{\otimes i}\cdot\left(  \operatorname*{Ker}f\right)  \cdot
V^{\otimes\left(  n-i\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }n\text{ for
}n-1\text{ in the first sum}\right) \\
&  =\sum\limits_{i\in\mathbb{N}}\sum\limits_{\substack{n\in\mathbb{N};\\n\geq
i}}V^{\otimes i}\cdot\left(  \operatorname*{Ker}f\right)  \cdot V^{\otimes
\left(  n-i\right)  }=\sum\limits_{i\in\mathbb{N}}\sum\limits_{j\in\mathbb{N}%
}V^{\otimes i}\cdot\left(  \operatorname*{Ker}f\right)  \cdot V^{\otimes j}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }j\text{ for
}n-i\text{ in the second sum}\right) \\
&  =\underbrace{\left(  \sum\limits_{i\in\mathbb{N}}V^{\otimes i}\right)
}_{=\otimes V}\cdot\left(  \operatorname*{Ker}f\right)  \cdot
\underbrace{\left(  \sum\limits_{j\in\mathbb{N}}V^{\otimes j}\right)
}_{=\otimes V}=\left(  \otimes V\right)  \cdot\left(  \operatorname*{Ker}%
f\right)  \cdot\left(  \otimes V\right)  .
\end{align*}
This proves Theorem \ref{thm.(X)f}.
\end{proof}

\subsection{\label{subsect.exter}The pseudoexterior algebra}

We are now going to introduce the pseudoexterior algebra
$\operatorname*{Exter} V$ of a $k$-module $V$. There are two ways to do this:
one is by constructing $\operatorname*{Exter} V$ as a direct sum of
pseudoexterior powers $\operatorname*{Exter} ^{n}V$ (so these pseudoexterior
powers must be defined first); the other is by directly constructing
$\operatorname*{Exter} V$ as a quotient of the tensor algebra $\otimes V$
modulo a certain two-sided ideal (and then we can construct the pseudoexterior
powers $\operatorname*{Exter} ^{n}V$ as homogeneous components of this
$\operatorname*{Exter} V$). It is not immediately clear (although not
difficult) to prove that these two ways yield one and the same (up to
canonical isomorphism) $k$-algebra $\operatorname*{Exter} V$. We are going to
reconcile these two ways by first proving some properties of the two-sided
ideal that we want to factor the tensor algebra $\otimes V$ by; once these are
shown, it will be easy to see that both definitions of $\operatorname*{Exter}
V$ are the same. We delay the definition of $\operatorname*{Exter} V$ until
that moment. So let us first define the pseudoexterior powers
$\operatorname*{Exter} ^{n}V$:

\begin{definition}
\label{defs.ext-pow}Let $k$ be a commutative ring. Let $V$ be a $k$-module.
Let $n\in\mathbb{N}$.\newline Let $Q_{n}\left(  V\right)  $ be the
$k$-submodule%
\[
\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-\left(  -1\right)
^{\sigma}v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(  2\right)
}\otimes\cdots\otimes v_{\sigma\left(  n\right)  }\ \mid\ \left(  \left(
v_{1},v_{2},\ldots,v_{n}\right)  ,\sigma\right)  \in V^{n}\times
S_{n}\right\rangle
\]
of the $k$-module $V^{\otimes n}$ (where we are using Convention
\ref{conv.<>}, and are denoting the $n$-th symmetric group by $S_{n}%
$).\newline The factor $k$-module $V^{\otimes n}\diagup Q_{n}\left(  V\right)
$ is called the $n$\textit{-th pseudoexterior power} of the $k$-module $V$ and
will be denoted by $\operatorname*{Exter} ^{n}V$. We denote by
$\operatorname*{exter}\nolimits_{V,n}$ the canonical projection $V^{\otimes
n}\rightarrow V^{\otimes n}\diagup Q_{n}\left(  V\right)
=\operatorname*{Exter} ^{n}V$. Clearly, this map $\operatorname*{exter}%
\nolimits_{V,n}$ is a surjective $k$-module homomorphism.
\end{definition}

\begin{conclusion}
\label{warn.pseudo}This $n$-th pseudoexterior power $\operatorname*{Exter}%
\nolimits^{n}V$ is called \textbf{pseudo}exterior for a reason: it is not
exactly the same as the $n$-th exterior power $\wedge^{n}V$ (which we will
introduce in Definition \ref{defs.wedge-pow}). While the difference between
$\operatorname*{Exter}\nolimits^{n}V$ and $\wedge^{n}V$ is not that large (in
particular, they are identic when $2$ is invertible in $k$, as Theorem
\ref{thm.Q=R} will show), this difference exists and should not be
forgotten.\newline Most literature only works with the $n$-th exterior power
$\wedge^{n}V$, because the $n$-th pseudoexterior power $\operatorname*{Exter}%
\nolimits^{n}V$ is much less interesting in the general case. \textbf{However}%
, a number of texts which are only concerned with the case when $2$ is
invertible in $k$ define the $n$-th exterior power $\wedge^{n}V$ by our
Definition \ref{defs.ext-pow}; i. e., what they call the $n$-th exterior power
$\wedge^{n}V$ is what we call the $n$-th pseudoexterior power
$\operatorname*{Exter}\nolimits^{n}V$. Fortunately this does not conflict with
our notation as long as $2$ is invertible in $k$ (because when $2$ is
invertible in $k$, Theorem \ref{thm.Q=R} \textbf{(c)} yields $\wedge
^{n}V=\operatorname*{Exter}\nolimits^{n}V$).
\end{conclusion}

This is not the pseudoexterior algebra $\operatorname*{Exter} V$ yet, but only
the $n$-th pseudoexterior power $\operatorname*{Exter} ^{n}V$; we will compose
the pseudoexterior algebra from these later. First, here is an alternative
description of the module $Q_{n}\left(  V\right)  $ from this definition:

\begin{proposition}
\label{prop.Q_n}Let $k$ be a commutative ring. Let $V$ be a $k$-module. Let
$n\in\mathbb{N}$.\newline Then,%
\[
Q_{n}\left(  V\right)  =\sum_{i=1}^{n-1}\left\langle v_{1}\otimes v_{2}%
\otimes\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau
_{i}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)
}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\rangle ,
\]
where $\tau_{i}$ denotes the transposition $\left(  i,i+1\right)  \in S_{n}$.
\end{proposition}

This proposition is classical and can be concluded from the definition of
$Q_{n}\left(  V\right)  $ and the fact that the transpositions $\tau_{1}$,
$\tau_{2}$, $\ldots$, $\tau_{n-1}$ generate the symmetric group $S_{n}$. Here
are the details of this proof:

\begin{proof}
[Proof of Proposition \ref{prop.Q_n}.]Let $T$ denote the subset%
\[
\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-\left(  -1\right)
^{\sigma}v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(  2\right)
}\otimes\cdots\otimes v_{\sigma\left(  n\right)  }\ \mid\ \left(  \left(
v_{1},v_{2},\ldots,v_{n}\right)  ,\sigma\right)  \in V^{n}\times
S_{n}\right\}
\]
of $V^{\otimes n}$. Then,%
\begin{align*}
\left\langle T\right\rangle  &  =\left\langle \left\{  v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}-\left(  -1\right)  ^{\sigma}v_{\sigma\left(
1\right)  }\otimes v_{\sigma\left(  2\right)  }\otimes\cdots\otimes
v_{\sigma\left(  n\right)  }\ \mid\ \left(  \left(  v_{1},v_{2},\ldots
,v_{n}\right)  ,\sigma\right)  \in V^{n}\times S_{n}\right\}  \right\rangle \\
&  =\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-\left(
-1\right)  ^{\sigma}v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(
2\right)  }\otimes\cdots\otimes v_{\sigma\left(  n\right)  }\ \mid\ \left(
\left(  v_{1},v_{2},\ldots,v_{n}\right)  ,\sigma\right)  \in V^{n}\times
S_{n}\right\rangle \\
&  =Q_{n}\left(  V\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Definition
\ref{defs.ext-pow}}\right)  .
\end{align*}
On the other hand,%
\begin{align}
&  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-\left(  -1\right)  ^{\sigma
}v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(  2\right)  }\otimes
\cdots\otimes v_{\sigma\left(  n\right)  }\in T\label{pf.symm-alg.1}\\
&  \ \ \ \ \ \ \ \ \ \ \text{for every }\left(  \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  ,\sigma\right)  \in V^{n}\times S_{n}\nonumber
\end{align}
(since%
\[
T=\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-\left(  -1\right)
^{\sigma}v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(  2\right)
}\otimes\cdots\otimes v_{\sigma\left(  n\right)  }\ \mid\ \left(  \left(
v_{1},v_{2},\ldots,v_{n}\right)  ,\sigma\right)  \in V^{n}\times
S_{n}\right\}
\]
).

On the other hand, let $Z$ denote the $k$-submodule%
\[
\sum_{i=1}^{n-1}\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)
}\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\rangle
\]
of $V^{\otimes n}$. Then,%
\begin{equation}
\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau_{\mathbf{I}%
}\left(  1\right)  }\otimes v_{\tau_{\mathbf{I}}\left(  2\right)  }%
\otimes\cdots\otimes v_{\tau_{\mathbf{I}}\left(  n\right)  }\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\rangle \subseteq Z
\label{pf.symm-alg.2}%
\end{equation}
for every $\mathbf{I}\in\left\{  1,2,\ldots,n-1\right\}  $. Now,%
\begin{align*}
&  \left\{  w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}+w_{\tau_{\mathbf{I}%
}\left(  1\right)  }\otimes w_{\tau_{\mathbf{I}}\left(  2\right)  }%
\otimes\cdots\otimes w_{\tau_{\mathbf{I}}\left(  n\right)  }\ \mid\ \left(
w_{1},w_{2},\ldots,w_{n}\right)  \in V^{n}\right\} \\
&  =\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau_{\mathbf{I}%
}\left(  1\right)  }\otimes v_{\tau_{\mathbf{I}}\left(  2\right)  }%
\otimes\cdots\otimes v_{\tau_{\mathbf{I}}\left(  n\right)  }\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed }\left(  w_{1}%
,w_{2},\ldots,w_{n}\right)  \text{ as }\left(  v_{1},v_{2},\ldots
,v_{n}\right)  \right) \\
&  \subseteq\left\langle \left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}+v_{\tau_{\mathbf{I}}\left(  1\right)  }\otimes v_{\tau_{\mathbf{I}%
}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{\mathbf{I}}\left(  n\right)
}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\}
\right\rangle \\
&  =\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau
_{\mathbf{I}}\left(  1\right)  }\otimes v_{\tau_{\mathbf{I}}\left(  2\right)
}\otimes\cdots\otimes v_{\tau_{\mathbf{I}}\left(  n\right)  }\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\rangle \subseteq Z
\end{align*}
for every $\mathbf{I}\in\left\{  1,2,\ldots,n-1\right\}  $. Thus,%
\begin{align}
&  w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}+w_{\tau_{\mathbf{I}}\left(
1\right)  }\otimes w_{\tau_{\mathbf{I}}\left(  2\right)  }\otimes\cdots\otimes
w_{\tau_{\mathbf{I}}\left(  n\right)  }\in Z\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \text{for every }\left(  w_{1},w_{2},\ldots
,w_{n}\right)  \in V^{n}\text{ and every }\mathbf{I}\in\left\{  1,2,\ldots
,n-1\right\}  . \label{pf.symm-alg.3}%
\end{align}


We are now going to show that $Z=\left\langle T\right\rangle $.

First, let us prove that $Z\subseteq\left\langle T\right\rangle $. In fact,
every $i\in\left\{  1,2,\ldots,n-1\right\}  $ satisfies%
\[
\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau_{i}\left(
1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes
v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)
\in V^{n}\right\}  \subseteq\left\langle T\right\rangle
\]
(since every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$ satisfies%
\begin{align*}
&  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+\underbrace{v_{\tau_{i}\left(
1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes
v_{\tau_{i}\left(  n\right)  }}_{=-\left(  -1\right)  v_{\tau_{i}\left(
1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes
v_{\tau_{i}\left(  n\right)  }}\\
&  =v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-\underbrace{\left(
-1\right)  }_{\substack{=\left(  -1\right)  ^{\tau_{i}}\\\text{(since }%
\tau_{i}\text{ is a transposition,}\\\text{so that }\left(  -1\right)
^{\tau_{i}}=-1\text{)}}}v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau
_{i}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }\\
&  =v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-\left(  -1\right)  ^{\tau
_{i}}v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)
}\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }\\
&  \in T\ \ \ \ \ \ \ \ \ \ \left(  \text{due to (\ref{pf.symm-alg.1}),
applied to }\left(  \left(  v_{1},v_{2},\ldots,v_{n}\right)  ,\tau_{i}\right)
\text{ instead of }\ \left(  \left(  v_{1},v_{2},\ldots,v_{n}\right)
,\sigma\right)  \right) \\
&  \subseteq\left\langle T\right\rangle
\end{align*}
). Now,%
\begin{align*}
Z  &  =\sum_{i=1}^{n-1}\underbrace{\left\langle v_{1}\otimes v_{2}%
\otimes\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau
_{i}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)
}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\rangle
}_{\subseteq\left\langle T\right\rangle }\\
&  \subseteq\sum_{i=1}^{n-1}\left\langle T\right\rangle \subseteq\left\langle
T\right\rangle \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left\langle
T\right\rangle \text{ is a }k\text{-module}\right)  .
\end{align*}


Now, let us show that $\left\langle T\right\rangle \subseteq Z$. To that aim,
we will show that $T\subseteq Z$.

In fact, let $\left(  \left(  v_{1},v_{2},\ldots,v_{n}\right)  ,\sigma\right)
\in V^{n}\times S_{n}$ be arbitrary. Then, $\left(  v_{1},v_{2},\ldots
,v_{n}\right)  \in V^{n}$ and $\sigma\in S_{n}$. Now, it is known that every
element of the symmetric group $S_{n}$ can be written as a product of some
transpositions from the set $\left\{  \tau_{1},\tau_{2},\ldots,\tau
_{n-1}\right\}  $. Applying this to the element $\sigma\in S_{n}$, we conclude
that $\sigma$ can be written as a product of some transpositions from the set
$\left\{  \tau_{1},\tau_{2},\ldots,\tau_{n-1}\right\}  $. In other words,
there exists a natural number $m\in\mathbb{N}$ and a sequence $\left(
i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{  1,2,\ldots,n-1\right\}  ^{m}$
such that $\sigma=\tau_{i_{1}}\tau_{i_{2}}\cdots\tau_{i_{m}}$. Consider this
$m$ and this $\left(  i_{1},i_{2},\ldots,i_{m}\right)  $. For every
$j\in\left\{  0,1,\ldots,m\right\}  $, let $\sigma_{j}$ denote the permutation
$\tau_{i_{1}}\tau_{i_{2}}\cdots\tau_{i_{j}}\in S_{n}$. Thus, $\sigma_{0}%
=\tau_{i_{1}}\tau_{i_{2}}\cdots\tau_{i_{0}}=\left(  \text{empty product}%
\right)  =\operatorname{id}$ and $\sigma_{m}=\tau_{i_{1}}\tau_{i_{2}}%
\cdots\tau_{i_{m}}=\sigma$. Moreover, every $j\in\left\{  1,2,\ldots
,m\right\}  $ satisfies $\left(  -1\right)  ^{\sigma_{j-1}}v_{\sigma
_{j-1}\left(  1\right)  }\otimes v_{\sigma_{j-1}\left(  2\right)  }%
\otimes\cdots\otimes v_{\sigma_{j-1}\left(  n\right)  }-\left(  -1\right)
^{\sigma_{j}}v_{\sigma_{j}\left(  1\right)  }\otimes v_{\sigma_{j}\left(
2\right)  }\otimes\cdots\otimes v_{\sigma_{j}\left(  n\right)  }\in
Z$.\ \ \ \ \footnote{\textit{Proof.} Let $j\in\left\{  1,2,\ldots,m\right\}  $
be arbitrary. Then,
\[
\underbrace{\sigma_{j-1}}_{\substack{=\tau_{i_{1}}\tau_{i_{2}}\cdots
\tau_{i_{j-1}}\\\text{(by the formula }\sigma_{j}=\tau_{i_{1}}\tau_{i_{2}%
}\cdots\tau_{i_{j}}\text{,}\\\text{applied to }j-1\text{ instead of }%
j\text{)}}}\tau_{i_{j}}=\tau_{i_{1}}\tau_{i_{2}}\cdots\tau_{i_{j-1}}%
\tau_{i_{j}}=\tau_{i_{1}}\tau_{i_{2}}\cdots\tau_{i_{j}}=\sigma_{j},
\]
so that $\sigma_{j}=\sigma_{j-1}\tau_{i_{j}}$. Denote $i_{j}$ by $\mathbf{I}$.
Then, $\sigma_{j}=\sigma_{j-1}\tau_{i_{j}}$ rewrites as $\sigma_{j}%
=\sigma_{j-1}\tau_{\mathbf{I}}$. Thus,%
\[
\left(  -1\right)  ^{\sigma_{j}}=\left(  -1\right)  ^{\sigma_{j-1}%
\tau_{\mathbf{I}}}=\left(  -1\right)  ^{\sigma_{j-1}}\underbrace{\left(
-1\right)  ^{\tau_{\mathbf{I}}}}_{=-1\text{ (since }\tau_{\mathbf{I}}\text{ is
a transposition)}}=-\left(  -1\right)  ^{\sigma_{j-1}}.
\]
\par
Now, define an $n$-tuple $\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in V^{n}$
by $\left(  w_{\rho}=v_{\sigma_{j-1}\left(  \rho\right)  }\text{ for every
}\rho\in\left\{  1,2,\ldots,n\right\}  \right)  $. Then, $\left(  w_{1}%
,w_{2},\ldots,w_{n}\right)  =\left(  v_{\sigma_{j-1}\left(  1\right)
},v_{\sigma_{j-1}\left(  2\right)  },\ldots,v_{\sigma_{j-1}\left(  n\right)
}\right)  $, so that $w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}%
=v_{\sigma_{j-1}\left(  1\right)  }\otimes v_{\sigma_{j-1}\left(  2\right)
}\otimes\cdots\otimes v_{\sigma_{j-1}\left(  n\right)  }$. On the other hand,
every $\xi\in\left\{  1,2,\ldots,n\right\}  $ satisfies
\begin{align*}
w_{\tau_{\mathbf{I}}\left(  \xi\right)  }  &  =v_{\sigma_{j-1}\left(
\tau_{\mathbf{I}}\left(  \xi\right)  \right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{by the formula }w_{\rho}=v_{\sigma_{j-1}\left(  \rho\right)  }\text{,
applied to }\rho=\tau_{\mathbf{I}}\left(  \xi\right)  \right) \\
&  =v_{\sigma_{j}\left(  \xi\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\sigma_{j-1}\left(  \tau_{\mathbf{I}}\left(  \xi\right)  \right)
=\underbrace{\left(  \sigma_{j-1}\tau_{\mathbf{I}}\right)  }_{=\sigma_{j}%
}\left(  \xi\right)  =\sigma_{j}\left(  \xi\right)  \right)  .
\end{align*}
Thus, $\left(  w_{\tau_{\mathbf{I}}\left(  1\right)  },w_{\tau_{\mathbf{I}%
}\left(  2\right)  },\ldots,w_{\tau_{\mathbf{I}}\left(  n\right)  }\right)
=\left(  v_{\sigma_{j}\left(  1\right)  },v_{\sigma_{j}\left(  2\right)
},\ldots,v_{\sigma_{j}\left(  n\right)  }\right)  $, so that $w_{\tau
_{\mathbf{I}}\left(  1\right)  }\otimes w_{\tau_{\mathbf{I}}\left(  2\right)
}\otimes\cdots\otimes w_{\tau_{\mathbf{I}}\left(  n\right)  }=v_{\sigma
_{j}\left(  1\right)  }\otimes v_{\sigma_{j}\left(  2\right)  }\otimes
\cdots\otimes v_{\sigma_{j}\left(  n\right)  }$.
\par
Now,%
\begin{align*}
&  \left(  -1\right)  ^{\sigma_{j-1}}\underbrace{v_{\sigma_{j-1}\left(
1\right)  }\otimes v_{\sigma_{j-1}\left(  2\right)  }\otimes\cdots\otimes
v_{\sigma_{j-1}\left(  n\right)  }}_{=w_{1}\otimes w_{2}\otimes\cdots\otimes
w_{n}}-\underbrace{\left(  -1\right)  ^{\sigma_{j}}}_{=-\left(  -1\right)
^{\sigma_{j-1}}}\underbrace{v_{\sigma_{j}\left(  1\right)  }\otimes
v_{\sigma_{j}\left(  2\right)  }\otimes\cdots\otimes v_{\sigma_{j}\left(
n\right)  }}_{=w_{\tau_{\mathbf{I}}\left(  1\right)  }\otimes w_{\tau
_{\mathbf{I}}\left(  2\right)  }\otimes\cdots\otimes w_{\tau_{\mathbf{I}%
}\left(  n\right)  }}\\
&  =\left(  -1\right)  ^{\sigma_{j-1}}w_{1}\otimes w_{2}\otimes\cdots\otimes
w_{n}-\left(  -\left(  -1\right)  ^{\sigma_{j-1}}\right)  w_{\tau_{\mathbf{I}%
}\left(  1\right)  }\otimes w_{\tau_{\mathbf{I}}\left(  2\right)  }%
\otimes\cdots\otimes w_{\tau_{\mathbf{I}}\left(  n\right)  }\\
&  =\left(  -1\right)  ^{\sigma_{j-1}}\underbrace{\left(  w_{1}\otimes
w_{2}\otimes\cdots\otimes w_{n}+w_{\tau_{\mathbf{I}}\left(  1\right)  }\otimes
w_{\tau_{\mathbf{I}}\left(  2\right)  }\otimes\cdots\otimes w_{\tau
_{\mathbf{I}}\left(  n\right)  }\right)  }_{\in Z\text{ (by
(\ref{pf.symm-alg.3}))}}\in Z\ \ \ \ \ \ \ \ \ \ \left(  \text{since }Z\text{
is a }k\text{-module}\right)  ,
\end{align*}
qed.} Thus,%
\[
\sum_{j=1}^{m}\underbrace{\left(  \left(  -1\right)  ^{\sigma_{j-1}}%
v_{\sigma_{j-1}\left(  1\right)  }\otimes v_{\sigma_{j-1}\left(  2\right)
}\otimes\cdots\otimes v_{\sigma_{j-1}\left(  n\right)  }-\left(  -1\right)
^{\sigma_{j}}v_{\sigma_{j}\left(  1\right)  }\otimes v_{\sigma_{j}\left(
2\right)  }\otimes\cdots\otimes v_{\sigma_{j}\left(  n\right)  }\right)
}_{\in Z}\in\sum_{j=1}^{m}Z\subseteq Z
\]
(since $Z$ is a $k$-module). Since%
\begin{align}
&  \sum_{j=1}^{m}\left(  \left(  -1\right)  ^{\sigma_{j-1}}v_{\sigma
_{j-1}\left(  1\right)  }\otimes v_{\sigma_{j-1}\left(  2\right)  }%
\otimes\cdots\otimes v_{\sigma_{j-1}\left(  n\right)  }-\left(  -1\right)
^{\sigma_{j}}v_{\sigma_{j}\left(  1\right)  }\otimes v_{\sigma_{j}\left(
2\right)  }\otimes\cdots\otimes v_{\sigma_{j}\left(  n\right)  }\right)
\nonumber\\
&  =v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-\left(  -1\right)  ^{\sigma
}v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(  2\right)  }\otimes
\cdots\otimes v_{\sigma\left(  n\right)  } \label{pf.symm-alg.4}%
\end{align}
\footnote{\textit{Proof of (\ref{pf.symm-alg.4}).} We distinguish between two
cases: the case when $m>0$, and the case when $m=0$.
\par
In the case when $m>0$, we have%
\begin{align*}
&  \sum_{j=1}^{m}\left(  \left(  -1\right)  ^{\sigma_{j-1}}v_{\sigma
_{j-1}\left(  1\right)  }\otimes v_{\sigma_{j-1}\left(  2\right)  }%
\otimes\cdots\otimes v_{\sigma_{j-1}\left(  n\right)  }-\left(  -1\right)
^{\sigma_{j}}v_{\sigma_{j}\left(  1\right)  }\otimes v_{\sigma_{j}\left(
2\right)  }\otimes\cdots\otimes v_{\sigma_{j}\left(  n\right)  }\right) \\
&  =\sum_{j=1}^{m}\left(  -1\right)  ^{\sigma_{j-1}}v_{\sigma_{j-1}\left(
1\right)  }\otimes v_{\sigma_{j-1}\left(  2\right)  }\otimes\cdots\otimes
v_{\sigma_{j-1}\left(  n\right)  }-\sum_{j=1}^{m}\left(  -1\right)
^{\sigma_{j}}v_{\sigma_{j}\left(  1\right)  }\otimes v_{\sigma_{j}\left(
2\right)  }\otimes\cdots\otimes v_{\sigma_{j}\left(  n\right)  }\\
&  =\underbrace{\sum_{j=0}^{m-1}\left(  -1\right)  ^{\sigma_{j}}v_{\sigma
_{j}\left(  1\right)  }\otimes v_{\sigma_{j}\left(  2\right)  }\otimes
\cdots\otimes v_{\sigma_{j}\left(  n\right)  }}_{=\left(  -1\right)
^{\sigma_{0}}v_{\sigma_{0}\left(  1\right)  }\otimes v_{\sigma_{0}\left(
2\right)  }\otimes\cdots\otimes v_{\sigma_{0}\left(  n\right)  }%
+\sum\limits_{j=1}^{m-1}\left(  -1\right)  ^{\sigma_{j}}v_{\sigma_{j}\left(
1\right)  }\otimes v_{\sigma_{j}\left(  2\right)  }\otimes\cdots\otimes
v_{\sigma_{j}\left(  n\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ -\underbrace{\sum_{j=1}^{m}\left(  -1\right)
^{\sigma_{j}}v_{\sigma_{j}\left(  1\right)  }\otimes v_{\sigma_{j}\left(
2\right)  }\otimes\cdots\otimes v_{\sigma_{j}\left(  n\right)  }}%
_{=\sum\limits_{j=1}^{m-1}\left(  -1\right)  ^{\sigma_{j}}v_{\sigma_{j}\left(
1\right)  }\otimes v_{\sigma_{j}\left(  2\right)  }\otimes\cdots\otimes
v_{\sigma_{j}\left(  n\right)  }+\left(  -1\right)  ^{\sigma_{m}}v_{\sigma
_{m}\left(  1\right)  }\otimes v_{\sigma_{m}\left(  2\right)  }\otimes
\cdots\otimes v_{\sigma_{m}\left(  n\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }j\text{ for
}j-1\text{ in the first sum}\right) \\
&  =\left(  \left(  -1\right)  ^{\sigma_{0}}v_{\sigma_{0}\left(  1\right)
}\otimes v_{\sigma_{0}\left(  2\right)  }\otimes\cdots\otimes v_{\sigma
_{0}\left(  n\right)  }+\sum\limits_{j=1}^{m-1}\left(  -1\right)  ^{\sigma
_{j}}v_{\sigma_{j}\left(  1\right)  }\otimes v_{\sigma_{j}\left(  2\right)
}\otimes\cdots\otimes v_{\sigma_{j}\left(  n\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  \sum\limits_{j=1}^{m-1}\left(  -1\right)
^{\sigma_{j}}v_{\sigma_{j}\left(  1\right)  }\otimes v_{\sigma_{j}\left(
2\right)  }\otimes\cdots\otimes v_{\sigma_{j}\left(  n\right)  }+\left(
-1\right)  ^{\sigma_{m}}v_{\sigma_{m}\left(  1\right)  }\otimes v_{\sigma
_{m}\left(  2\right)  }\otimes\cdots\otimes v_{\sigma_{m}\left(  n\right)
}\right) \\
&  =\underbrace{\left(  -1\right)  ^{\sigma_{0}}}_{=1\text{ (since }\sigma
_{0}=\operatorname*{id}\text{)}}\underbrace{v_{\sigma_{0}\left(  1\right)
}\otimes v_{\sigma_{0}\left(  2\right)  }\otimes\cdots\otimes v_{\sigma
_{0}\left(  n\right)  }}_{\substack{=v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\\\text{(since }\sigma_{0}=\operatorname{id}\text{ yields}\\v_{\sigma
_{0}\left(  1\right)  }\otimes v_{\sigma_{0}\left(  2\right)  }\otimes
\cdots\otimes v_{\sigma_{0}\left(  n\right)  }=v_{\operatorname{id}\left(
1\right)  }\otimes v_{\operatorname{id}\left(  2\right)  }\otimes\cdots\otimes
v_{\operatorname{id}\left(  n\right)  }\\=v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}\text{)}}}-\underbrace{\left(  -1\right)  ^{\sigma_{m}}%
}_{\substack{=\left(  -1\right)  ^{\sigma}\\\text{(since }\sigma_{m}%
=\sigma\text{)}}}\underbrace{v_{\sigma_{m}\left(  1\right)  }\otimes
v_{\sigma_{m}\left(  2\right)  }\otimes\cdots\otimes v_{\sigma_{m}\left(
n\right)  }}_{\substack{=v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(
2\right)  }\otimes\cdots\otimes v_{\sigma\left(  n\right)  }\\\text{(since
}\sigma_{m}=\sigma\text{)}}}\\
&  =v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-\left(  -1\right)  ^{\sigma
}v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(  2\right)  }\otimes
\cdots\otimes v_{\sigma\left(  n\right)  },
\end{align*}
so that (\ref{pf.symm-alg.4}) is proven in the case when $m>0$.
\par
In the case when $m=0$, we have%
\begin{align*}
&  \sum_{j=1}^{m}\left(  \left(  -1\right)  ^{\sigma_{j-1}}v_{\sigma
_{j-1}\left(  1\right)  }\otimes v_{\sigma_{j-1}\left(  2\right)  }%
\otimes\cdots\otimes v_{\sigma_{j-1}\left(  n\right)  }-\left(  -1\right)
^{\sigma_{j}}v_{\sigma_{j}\left(  1\right)  }\otimes v_{\sigma_{j}\left(
2\right)  }\otimes\cdots\otimes v_{\sigma_{j}\left(  n\right)  }\right) \\
&  =\left(  \text{empty sum}\right)  =0\\
&  =\underbrace{\left(  -1\right)  ^{\operatorname*{id}}}_{=1}%
\underbrace{v_{\operatorname{id}\left(  1\right)  }\otimes
v_{\operatorname{id}\left(  2\right)  }\otimes\cdots\otimes
v_{\operatorname{id}\left(  n\right)  }}_{=v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}}-\underbrace{\left(  -1\right)  ^{\operatorname*{id}%
}v_{\operatorname{id}\left(  1\right)  }\otimes v_{\operatorname{id}\left(
2\right)  }\otimes\cdots\otimes v_{\operatorname*{id}\left(  n\right)  }%
}_{\substack{=\left(  -1\right)  ^{\sigma}v_{\sigma\left(  1\right)  }\otimes
v_{\sigma\left(  2\right)  }\otimes\cdots\otimes v_{\sigma\left(  n\right)
}\\\text{(since }m=0\text{ leads to }\sigma_{m}=\sigma_{0}\text{, so
that}\\\operatorname{id}=\sigma_{0}=\sigma_{m}=\sigma\text{)}}}\\
&  =v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-\left(  -1\right)  ^{\sigma
}v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(  2\right)  }\otimes
\cdots\otimes v_{\sigma\left(  n\right)  },
\end{align*}
so that (\ref{pf.symm-alg.4}) is proven in the case when $m=0$.
\par
Thus, (\ref{pf.symm-alg.4}) is proven in both possible cases. This completes
the proof of (\ref{pf.symm-alg.4}).}, this rewrites as%
\[
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-\left(  -1\right)  ^{\sigma
}v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(  2\right)  }\otimes
\cdots\otimes v_{\sigma\left(  n\right)  }\in Z.
\]


We have thus shown that every $\left(  \left(  v_{1},v_{2},\ldots
,v_{n}\right)  ,\sigma\right)  \in V^{n}\times S_{n}$ satisfies $v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}-\left(  -1\right)  ^{\sigma}v_{\sigma\left(
1\right)  }\otimes v_{\sigma\left(  2\right)  }\otimes\cdots\otimes
v_{\sigma\left(  n\right)  }\in Z$. Thus,%
\[
\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-\left(  -1\right)
^{\sigma}v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(  2\right)
}\otimes\cdots\otimes v_{\sigma\left(  n\right)  }\ \mid\ \left(  \left(
v_{1},v_{2},\ldots,v_{n}\right)  ,\sigma\right)  \in V^{n}\times
S_{n}\right\}  \subseteq Z.
\]
Since $\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-\left(
-1\right)  ^{\sigma}v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(
2\right)  }\otimes\cdots\otimes v_{\sigma\left(  n\right)  }\ \mid\ \left(
\left(  v_{1},v_{2},\ldots,v_{n}\right)  ,\sigma\right)  \in V^{n}\times
S_{n}\right\}  =T$, this rewrites as $T\subseteq Z$. Proposition \ref{prop.<>}
\textbf{(a)} (applied to $V^{\otimes n}$, $T$ and $Z$ instead of $M$, $S$ and
$Q$) thus yields that $\left\langle T\right\rangle \subseteq Z$. Combined with
$Z\subseteq\left\langle T\right\rangle $, this yields $Z=\left\langle
T\right\rangle $.

We now have%
\begin{align*}
&  \sum_{i=1}^{n-1}\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)
}\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\rangle \\
&  =Z=\left\langle T\right\rangle =Q_{n}\left(  V\right)  .
\end{align*}
This proves Proposition \ref{prop.Q_n}.
\end{proof}

A trivial corollary from Proposition \ref{prop.Q_n}:

\begin{corollary}
\label{coro.Q_2}Let $k$ be a commutative ring. Let $V$ be a $k$-module. Then,%
\[
Q_{2}\left(  V\right)  =\left\langle v_{1}\otimes v_{2}+v_{2}\otimes
v_{1}\ \mid\ \left(  v_{1},v_{2}\right)  \in V^{2}\right\rangle .
\]

\end{corollary}

\begin{proof}
[Proof of Corollary \ref{coro.Q_2}.]Applying Proposition \ref{prop.Q_n} to
$n=2$, we obtain%
\begin{align*}
Q_{2}\left(  V\right)   &  =\sum_{i=1}^{2-1}\left\langle v_{1}\otimes
v_{2}+v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)
}\ \mid\ \left(  v_{1},v_{2}\right)  \in V^{2}\right\rangle \\
&  =\left\langle v_{1}\otimes v_{2}+v_{\tau_{1}\left(  1\right)  }\otimes
v_{\tau_{1}\left(  2\right)  }\ \mid\ \left(  v_{1},v_{2}\right)  \in
V^{2}\right\rangle =\left\langle v_{1}\otimes v_{2}+v_{2}\otimes v_{1}%
\ \mid\ \left(  v_{1},v_{2}\right)  \in V^{2}\right\rangle
\end{align*}
(since $\tau_{1}\left(  1\right)  =2$ and $\tau_{1}\left(  2\right)  =1$).
This proves Corollary \ref{coro.Q_2}.
\end{proof}

Next something very basic:

\begin{lemma}
\label{lem.ABC}Let $k$ be a commutative. Let $P$ be a $k$-algebra.\newline%
\textbf{(a)} Let $X$ and $Y$ be two sets, and $a:X\rightarrow P$ and
$b:Y\rightarrow P$ be two maps. Then,%
\[
\left\langle a\left(  x\right)  \ \mid\ x\in X\right\rangle \cdot\left\langle
b\left(  y\right)  \ \mid\ y\in Y\right\rangle =\left\langle a\left(
x\right)  b\left(  y\right)  \ \mid\ \left(  x,y\right)  \in X\times
Y\right\rangle .
\]
\newline\textbf{(b)} Let $X$, $Y$ and $Z$ be three sets, and $a:X\rightarrow
P$, $b:Y\rightarrow P$ and $c:Z\rightarrow P$ be three maps. Then,%
\begin{align*}
&  \left\langle a\left(  x\right)  \ \mid\ x\in X\right\rangle \cdot
\left\langle b\left(  y\right)  \ \mid\ y\in Y\right\rangle \cdot\left\langle
c\left(  z\right)  \ \mid\ z\in Z\right\rangle \\
&  =\left\langle a\left(  x\right)  b\left(  y\right)  c\left(  z\right)
\ \mid\ \left(  x,y,z\right)  \in X\times Y\times Z\right\rangle .
\end{align*}

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.ABC}.]\textbf{(a)} Let $X^{\prime}=\left\langle
a\left(  x\right)  \ \mid\ x\in X\right\rangle $ and $Y^{\prime}=\left\langle
b\left(  y\right)  \ \mid\ y\in Y\right\rangle $.

We will now prove that $X^{\prime}Y^{\prime}\subseteq\left\langle a\left(
x\right)  b\left(  y\right)  \ \mid\ \left(  x,y\right)  \in X\times
Y\right\rangle $ and \newline$\left\langle a\left(  x\right)  b\left(
y\right)  \ \mid\ \left(  x,y\right)  \in X\times Y\right\rangle \subseteq
X^{\prime}Y^{\prime}$.

\textit{Proof of }$X^{\prime}Y^{\prime}\subseteq\left\langle a\left(
x\right)  b\left(  y\right)  \ \mid\ \left(  x,y\right)  \in X\times
Y\right\rangle $\textit{:}

By the definition of the product of two $k$-submodules, we have%
\begin{equation}
X^{\prime}Y^{\prime}=\left\langle pq\ \mid\ \left(  p,q\right)  \in X^{\prime
}\times Y^{\prime}\right\rangle =\left\langle \left\{  pq\ \mid\ \left(
p,q\right)  \in X^{\prime}\times Y^{\prime}\right\}  \right\rangle .
\label{pf.ABC.1}%
\end{equation}


Now, let $\left(  p,q\right)  \in X^{\prime}\times Y^{\prime}$ be arbitrary.
Then, $p\in X^{\prime}=\left\langle a\left(  x\right)  \ \mid\ x\in
X\right\rangle $, so that we can find some $n\in\mathbb{N}$, some elements
$x_{1}$, $x_{2}$, $\ldots$, $x_{n}$ of $X$ and some elements $\lambda_{1}$,
$\lambda_{2}$, $\ldots$, $\lambda_{n}$ of $k$ such that $p=\sum\limits_{i=1}%
^{n}\lambda_{i}a\left(  x_{i}\right)  $. Consider this $n$, these $x_{1}$,
$x_{2}$, $\ldots$, $x_{n}$ and these $\lambda_{1}$, $\lambda_{2}$, $\ldots$,
$\lambda_{n}$.

Since $\left(  p,q\right)  \in X^{\prime}\times Y^{\prime}$, we have $q\in
Y^{\prime}=\left\langle b\left(  y\right)  \ \mid\ y\in Y\right\rangle $, so
that we can find some $m\in\mathbb{N}$, some elements $y_{1}$, $y_{2}$,
$\ldots$, $y_{m}$ of $Y$ and some elements $\mu_{1}$, $\mu_{2}$, $\ldots$,
$\mu_{m}$ of $k$ such that $q=\sum\limits_{j=1}^{m}\mu_{j}b\left(
y_{j}\right)  $. Consider this $m$, these $y_{1}$, $y_{2}$, $\ldots$, $y_{m}$
and these $\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu_{m}$.

Since $p=\sum\limits_{i=1}^{n}\lambda_{i}a\left(  x_{i}\right)  $ and
$q=\sum\limits_{j=1}^{m}\mu_{j}b\left(  y_{j}\right)  $, we have%
\begin{align*}
pq  &  =\sum\limits_{i=1}^{n}\lambda_{i}a\left(  x_{i}\right)  \cdot
\sum\limits_{j=1}^{m}\mu_{j}b\left(  y_{j}\right)  =\sum\limits_{i=1}^{n}%
\sum\limits_{j=1}^{m}\lambda_{i}\mu_{j}\underbrace{a\left(  x_{i}\right)
b\left(  y_{j}\right)  }_{\substack{\in\left\{  a\left(  x\right)  b\left(
y\right)  \ \mid\ \left(  x,y\right)  \in X\times Y\right\}  \\\subseteq
\left\langle \left\{  a\left(  x\right)  b\left(  y\right)  \ \mid\ \left(
x,y\right)  \in X\times Y\right\}  \right\rangle \\=\left\langle a\left(
x\right)  b\left(  y\right)  \ \mid\ \left(  x,y\right)  \in X\times
Y\right\rangle }}\\
&  \subseteq\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}\left\langle a\left(
x\right)  b\left(  y\right)  \ \mid\ \left(  x,y\right)  \in X\times
Y\right\rangle \\
&  \subseteq\left\langle a\left(  x\right)  b\left(  y\right)  \ \mid\ \left(
x,y\right)  \in X\times Y\right\rangle \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\left\langle a\left(  x\right)  b\left(  y\right)  \ \mid\ \left(
x,y\right)  \in X\times Y\right\rangle \text{ is a }k\text{-module}\right)  .
\end{align*}
Since this holds for all $\left(  p,q\right)  \in X^{\prime}\times Y^{\prime}%
$, we have thus proven that%
\[
\left\{  pq\ \mid\ \left(  p,q\right)  \in X^{\prime}\times Y^{\prime
}\right\}  \subseteq\left\langle a\left(  x\right)  b\left(  y\right)
\ \mid\ \left(  x,y\right)  \in X\times Y\right\rangle .
\]
Therefore, Proposition \ref{prop.<>} \textbf{(a)} (applied to $P$, $\left\{
pq\ \mid\ \left(  p,q\right)  \in X^{\prime}\times Y^{\prime}\right\}  $ and
\newline$\left\langle a\left(  x\right)  b\left(  y\right)  \ \mid\ \left(
x,y\right)  \in X\times Y\right\rangle $ instead of $M$, $S$ and $Q$) yields
that%
\[
\left\langle \left\{  pq\ \mid\ \left(  p,q\right)  \in X^{\prime}\times
Y^{\prime}\right\}  \right\rangle \subseteq\left\langle a\left(  x\right)
b\left(  y\right)  \ \mid\ \left(  x,y\right)  \in X\times Y\right\rangle .
\]
Combined with (\ref{pf.ABC.1}), this yields%
\[
X^{\prime}Y^{\prime}\subseteq\left\langle a\left(  x\right)  b\left(
y\right)  \ \mid\ \left(  x,y\right)  \in X\times Y\right\rangle .
\]
We have thus proven that $X^{\prime}Y^{\prime}\subseteq\left\langle a\left(
x\right)  b\left(  y\right)  \ \mid\ \left(  x,y\right)  \in X\times
Y\right\rangle $.

\textit{Proof of }$\left\langle a\left(  x\right)  b\left(  y\right)
\ \mid\ \left(  x,y\right)  \in X\times Y\right\rangle \subseteq X^{\prime
}Y^{\prime}$\textit{:} We have%
\[
X^{\prime}=\left\langle a\left(  x\right)  \ \mid\ x\in X\right\rangle
=\left\langle \left\{  a\left(  x\right)  \ \mid\ x\in X\right\}
\right\rangle \supseteq\left\{  a\left(  x\right)  \ \mid\ x\in X\right\}  .
\]
Thus, $a\left(  x\right)  \in X^{\prime}$ for every $x\in X$. Similarly,
$b\left(  y\right)  \in Y^{\prime}$ for every $y\in Y$.

Now, let $\left(  x,y\right)  \in X\times Y$ be arbitrary. Then, $x\in X$ and
$y\in Y^{\prime}$, so that $a\left(  x\right)  \in X^{\prime}$ and $b\left(
y\right)  \in Y^{\prime}$ (as we just have seen). Hence, $a\left(  x\right)
b\left(  y\right)  \in X^{\prime}Y^{\prime}$.

We have thus shown that $a\left(  x\right)  b\left(  y\right)  \in X^{\prime
}Y^{\prime}$ for every $\left(  x,y\right)  \in X\times Y$. In other words,
$\left\{  a\left(  x\right)  b\left(  y\right)  \ \mid\ \left(  x,y\right)
\in X\times Y\right\}  \subseteq X^{\prime}Y^{\prime}$. Therefore, Proposition
\ref{prop.<>} \textbf{(a)} (applied to $P$, $\left\{  a\left(  x\right)
b\left(  y\right)  \ \mid\ \left(  x,y\right)  \in X\times Y\right\}  $ and
$X^{\prime}Y^{\prime}$ instead of $M$, $S$ and $Q$) yields that%
\[
\left\langle \left\{  a\left(  x\right)  b\left(  y\right)  \ \mid\ \left(
x,y\right)  \in X\times Y\right\}  \right\rangle \subseteq X^{\prime}%
Y^{\prime}.
\]
Since $\left\langle \left\{  a\left(  x\right)  b\left(  y\right)
\ \mid\ \left(  x,y\right)  \in X\times Y\right\}  \right\rangle =\left\langle
a\left(  x\right)  b\left(  y\right)  \ \mid\ \left(  x,y\right)  \in X\times
Y\right\rangle $, we have thus proven $\left\langle a\left(  x\right)
b\left(  y\right)  \ \mid\ \left(  x,y\right)  \in X\times Y\right\rangle
\subseteq X^{\prime}Y^{\prime}$.

Combined with $X^{\prime}Y^{\prime}\subseteq\left\langle a\left(  x\right)
b\left(  y\right)  \ \mid\ \left(  x,y\right)  \in X\times Y\right\rangle $,
this yields that $X^{\prime}Y^{\prime}=\left\langle a\left(  x\right)
b\left(  y\right)  \ \mid\ \left(  x,y\right)  \in X\times Y\right\rangle $.
Since $X^{\prime}=\left\langle a\left(  x\right)  \ \mid\ x\in X\right\rangle
$ and $Y^{\prime}=\left\langle b\left(  y\right)  \ \mid\ y\in Y\right\rangle
$, this rewrites as follows:%
\[
\left\langle a\left(  x\right)  \ \mid\ x\in X\right\rangle \cdot\left\langle
b\left(  y\right)  \ \mid\ y\in Y\right\rangle =\left\langle a\left(
x\right)  b\left(  y\right)  \ \mid\ \left(  x,y\right)  \in X\times
Y\right\rangle .
\]
Thus, Lemma \ref{lem.ABC} \textbf{(a)} is proven.

\textbf{(b)} Define a map $d:X\times Y\rightarrow P$ by%
\[
\left(  d\left(  x,y\right)  =a\left(  x\right)  b\left(  y\right)  \text{ for
all }\left(  x,y\right)  \in X\times Y\right)  .
\]
Now, Lemma \ref{lem.ABC} \textbf{(a)} yields%
\begin{align*}
\left\langle a\left(  x\right)  \ \mid\ x\in X\right\rangle \cdot\left\langle
b\left(  y\right)  \ \mid\ y\in Y\right\rangle  &  =\left\langle
\underbrace{a\left(  x\right)  b\left(  y\right)  }_{=d\left(  x,y\right)
}\ \mid\ \left(  x,y\right)  \in X\times Y\right\rangle \\
&  =\left\langle d\left(  x,y\right)  \ \mid\ \left(  x,y\right)  \in X\times
Y\right\rangle =\left\langle d\left(  x\right)  \ \mid\ x\in X\times
Y\right\rangle
\end{align*}
(here, we renamed the index $\left(  x,y\right)  $ as $x$). Also,
$\left\langle c\left(  z\right)  \ \mid\ z\in Z\right\rangle =\left\langle
c\left(  y\right)  \ \mid\ y\in Z\right\rangle $ (here, we renamed the index
$z$ as $y$). But Lemma \ref{lem.ABC} \textbf{(a)} (applied to $X\times Y$,
$Z$, $d$ and $c$ instead of $X$, $Y$, $a$ and $b$) yields%
\[
\left\langle d\left(  x\right)  \ \mid\ x\in X\times Y\right\rangle
\cdot\left\langle c\left(  y\right)  \ \mid\ y\in Z\right\rangle =\left\langle
d\left(  x\right)  \cdot c\left(  y\right)  \ \mid\ \left(  x,y\right)
\in\left(  X\times Y\right)  \times Z\right\rangle .
\]
Thus,%
\begin{align*}
&  \underbrace{\left\langle a\left(  x\right)  \ \mid\ x\in X\right\rangle
\cdot\left\langle b\left(  y\right)  \ \mid\ y\in Y\right\rangle
}_{=\left\langle d\left(  x\right)  \ \mid\ x\in X\times Y\right\rangle }%
\cdot\underbrace{\left\langle c\left(  z\right)  \ \mid\ z\in Z\right\rangle
}_{=\left\langle c\left(  y\right)  \ \mid\ y\in Z\right\rangle }\\
&  =\left\langle d\left(  x\right)  \ \mid\ x\in X\times Y\right\rangle
\cdot\left\langle c\left(  y\right)  \ \mid\ y\in Z\right\rangle =\left\langle
d\left(  x\right)  \cdot c\left(  y\right)  \ \mid\ \left(  x,y\right)
\in\left(  X\times Y\right)  \times Z\right\rangle \\
&  =\left\langle d\left(  t\right)  \cdot c\left(  z\right)  \ \mid\ \left(
t,z\right)  \in\left(  X\times Y\right)  \times Z\right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index }\left(
x,y\right)  \text{ as }\left(  t,z\right)  \right) \\
&  =\left\langle d\left(  x,y\right)  \cdot c\left(  z\right)  \ \mid\ \left(
\left(  x,y\right)  ,z\right)  \in\left(  X\times Y\right)  \times
Z\right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index }\left(
t,z\right)  \text{ as }\left(  \left(  x,y\right)  ,z\right)  \right) \\
&  =\left\langle \underbrace{d\left(  x,y\right)  }_{=a\left(  x\right)
b\left(  y\right)  }\cdot c\left(  z\right)  \ \mid\ \left(  x,y,z\right)  \in
X\times Y\times Z\right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted the triple }\left(
x,y,z\right)  \text{ for the pair }\left(  \left(  x,y\right)  ,z\right)
\right) \\
&  =\left\langle a\left(  x\right)  b\left(  y\right)  c\left(  z\right)
\ \mid\ \left(  x,y,z\right)  \in X\times Y\times Z\right\rangle .
\end{align*}
This proves Lemma \ref{lem.ABC} \textbf{(b)}.
\end{proof}

The following lemma will help us in making use of Proposition \ref{prop.Q_n}:

\begin{lemma}
\label{lem.Q_n}Let $k$ be a commutative ring. Let $V$ be a $k$-module. Let
$n\in\mathbb{N}$. Let $i\in\left\{  1,2,\ldots,n-1\right\}  $.\newline Then,%
\begin{align*}
&  \left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau
_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }\otimes
\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{n}\right\rangle \\
&  =V^{\otimes\left(  i-1\right)  }\cdot\left(  Q_{2}\left(  V\right)
\right)  \cdot V^{\otimes\left(  n-1-i\right)  },
\end{align*}
where $\tau_{i}$ denotes the transposition $\left(  i,i+1\right)  \in S_{n}$.
Here, we consider $V^{\otimes n}$ as a $k$-submodule of $\otimes V$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.Q_n}.]Define a map $a:V^{i-1}\rightarrow
V^{\otimes\left(  i-1\right)  }$ by%
\[
\left(  a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  =v_{1}\otimes v_{2}%
\otimes\cdots\otimes v_{i-1}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(
v_{1},v_{2},\ldots,v_{i-1}\right)  \in V^{i-1}\right)  .
\]
Define a map $b:V^{2}\rightarrow V^{\otimes2}$ by%
\[
\left(  b\left(  v_{i},v_{i+1}\right)  =v_{i}\otimes v_{i+1}+v_{i+1}\otimes
v_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  v_{i},v_{i+1}\right)  \in
V^{2}\right)  .
\]
Define a map $c:V^{n-1-i}\rightarrow V^{\otimes\left(  n-1-i\right)  }$ by%
\[
\left(  c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)  =v_{i+2}\otimes
v_{i+3}\otimes\cdots\otimes v_{n}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(
v_{i+2},v_{i+3},\ldots,v_{n}\right)  \in V^{n-1-i}\right)  .
\]
Since $V^{\otimes\left(  i-1\right)  }$, $V^{\otimes2}$ and $V^{\otimes\left(
n-1-i\right)  }$ are $k$-submodules of $\otimes V$, we can consider all three
maps $a$, $b$ and $c$ as maps to the set $\otimes V$.

It is now easy to see that every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{n}$ satisfies%
\[
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)
}\otimes v_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(
n\right)  }=a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \cdot b\left(
v_{i},v_{i+1}\right)  \cdot c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)  ,
\]
where the multiplication on the right hand side is the multiplication in the
tensor algebra $\otimes V$.\ \ \ \ \footnote{\textit{Proof.} Let $\left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$. Then, recalling Convention
\ref{conv.(X)^n.ident}, we have%
\begin{align}
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}  &  =\underbrace{\left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{i-1}\right)  }_{=a\left(
v_{1},v_{2},\ldots,v_{i-1}\right)  }\otimes\left(  v_{i}\otimes v_{i+1}%
\right)  \otimes\underbrace{\left(  v_{i+2}\otimes v_{i+3}\otimes\cdots\otimes
v_{n}\right)  }_{=c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)  }\nonumber\\
&  =a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \otimes\left(  v_{i}\otimes
v_{i+1}\right)  \otimes c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)  .
\label{pf.Q_n.2}%
\end{align}
\par
On the other hand, every $j\in\left\{  1,2,\ldots,i-1\right\}  $ satisfies
$\tau_{i}\left(  j\right)  =j$ (since $\tau_{i}$ is the transposition $\left(
i,i+1\right)  $) and thus $v_{\tau_{i}\left(  j\right)  }=v_{j}$. In other
words, we have the equalities $v_{\tau_{i}\left(  1\right)  }=v_{1}$,
$v_{\tau_{i}\left(  2\right)  }=v_{2}$, $\ldots$, $v_{\tau_{i}\left(
i-1\right)  }=v_{i-1}$. Taking the tensor product of these equalities yields%
\[
v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }%
\otimes\cdots\otimes v_{\tau_{i}\left(  i-1\right)  }=v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{i-1}=a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)
.
\]
\par
Every $j\in\left\{  i+2,i+3,\ldots,n\right\}  $ satisfies $\tau_{i}\left(
j\right)  =j$ (since $\tau_{i}$ is the transposition $\left(  i,i+1\right)  $)
and thus $v_{\tau_{i}\left(  j\right)  }=v_{j}$. In other words, we have the
equalities $v_{\tau_{i}\left(  i+2\right)  }=v_{i+2}$, $v_{\tau_{i}\left(
i+3\right)  }=v_{i+3}$, $\ldots$, $v_{\tau_{i}\left(  n\right)  }=v_{n}$.
Taking the tensor product of these equalities yields%
\[
v_{\tau_{i}\left(  i+2\right)  }\otimes v_{\tau_{i}\left(  i+3\right)
}\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }=v_{i+2}\otimes
v_{i+3}\otimes\cdots\otimes v_{n}=c\left(  v_{i+2},v_{i+3},\ldots
,v_{n}\right)  .
\]
\par
Since $\tau_{i}$ is the transposition $\left(  i,i+1\right)  $, we have
$\tau_{i}\left(  i\right)  =i+1$ and $\tau_{i}\left(  i+1\right)  =i$. These
equalities yield $v_{\tau_{i}\left(  i\right)  }=v_{i+1}$ and $v_{\tau
_{i}\left(  i+1\right)  }=v_{i}$, respectively.
\par
Now,%
\begin{align*}
&  v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }%
\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }\\
&  =\underbrace{\left(  v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau
_{i}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(  i-1\right)
}\right)  }_{=a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  }\otimes\left(
\underbrace{v_{\tau_{i}\left(  i\right)  }}_{=v_{i+1}}\otimes
\underbrace{v_{\tau_{i}\left(  i+1\right)  }}_{=v_{i}}\right)  \otimes
\underbrace{\left(  v_{\tau_{i}\left(  i+2\right)  }\otimes v_{\tau_{i}\left(
i+3\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }\right)
}_{=c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)  }\\
&  =a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \otimes\left(  v_{i+1}\otimes
v_{i}\right)  \otimes c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)  .
\end{align*}
Adding this to (\ref{pf.Q_n.2}), we get%
\begin{align*}
&  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)
}\otimes v_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(
n\right)  }\\
&  =a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \otimes\left(  v_{i}\otimes
v_{i+1}\right)  \otimes c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)
\otimes\left(  v_{i+1}\otimes v_{i}\right)  \otimes c\left(  v_{i+2}%
,v_{i+3},\ldots,v_{n}\right) \\
&  =a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \otimes\underbrace{\left(
v_{i}\otimes v_{i+1}+v_{i+1}\otimes v_{i}\right)  }_{=b\left(  v_{i}%
,v_{i+1}\right)  }\otimes c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right) \\
&  =a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \otimes b\left(  v_{i}%
,v_{i+1}\right)  \otimes c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)  .
\end{align*}
\par
On the other hand, (\ref{*=(X)}) (applied to $a\left(  v_{1},v_{2}%
,\ldots,v_{i-1}\right)  $, $b\left(  v_{i},v_{i+1}\right)  $, $i-1$ and $2$
instead of $a$, $b$, $n$ and $m$) yields%
\[
a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \cdot b\left(  v_{i}%
,v_{i+1}\right)  =a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \otimes b\left(
v_{i},v_{i+1}\right)  .
\]
Also, (\ref{*=(X)}) (applied to $a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)
\cdot b\left(  v_{i},v_{i+1}\right)  $, $c\left(  v_{i+2},v_{i+3},\ldots
,v_{n}\right)  $, $i+1$ and $n-1-i$ instead of $a$, $b$, $n$ and $m$) yields%
\begin{align*}
&  a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \cdot b\left(  v_{i}%
,v_{i+1}\right)  \cdot c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right) \\
&  =\underbrace{\left(  a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \cdot
b\left(  v_{i},v_{i+1}\right)  \right)  }_{=a\left(  v_{1},v_{2}%
,\ldots,v_{i-1}\right)  \otimes b\left(  v_{i},v_{i+1}\right)  }\otimes
c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right) \\
&  =a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \otimes b\left(  v_{i}%
,v_{i+1}\right)  \otimes c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right) \\
&  =v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)
}\otimes v_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(
n\right)  },
\end{align*}
qed.} Thus,%
\begin{align}
&  \left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau
_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }\otimes
\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{n}\right\rangle \nonumber\\
&  =\left\langle a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \cdot b\left(
v_{i},v_{i+1}\right)  \cdot c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\rangle
\nonumber\\
&  =\left\langle a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \cdot b\left(
v_{i},v_{i+1}\right)  \cdot c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)
\right. \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left.  \ \mid\ \left(  \left(  v_{1},v_{2}%
,\ldots,v_{i-1}\right)  ,\left(  v_{i},v_{i+1}\right)  ,\left(  v_{i+2}%
,v_{i+3},\ldots,v_{n}\right)  \right)  \in V^{i-1}\times V^{2}\times
V^{n-1-i}\right\rangle \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted the triple }\left(  \left(  v_{1},v_{2}%
,\ldots,v_{i-1}\right)  ,\left(  v_{i},v_{i+1}\right)  ,\left(  v_{i+2}%
,v_{i+3},\ldots,v_{n}\right)  \right) \\
\text{ for the }n\text{-tuple }\left(  v_{1},v_{2},\ldots,v_{n}\right)
\end{array}
\right) \nonumber\\
&  =\left\langle a\left(  x\right)  b\left(  y\right)  c\left(  z\right)
\ \mid\ \left(  x,y,z\right)  \in V^{i-1}\times V^{2}\times V^{n-1-i}%
\right\rangle \label{pf.Q_n.7}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed }\left(  \left(
v_{1},v_{2},\ldots,v_{i-1}\right)  ,\left(  v_{i},v_{i+1}\right)  ,\left(
v_{i+2},v_{i+3},\ldots,v_{n}\right)  \right)  \text{ as }\left(  x,y,z\right)
\right)  .\nonumber
\end{align}


But Lemma \ref{lem.ABC} \textbf{(b)} (applied to $X=V^{i-1}$, $Y=V^{2}$,
$Z=V^{n-1-i}$ and $P=\otimes V$) yields%
\begin{align*}
&  \left\langle a\left(  x\right)  \ \mid\ x\in V^{i-1}\right\rangle
\cdot\left\langle b\left(  y\right)  \ \mid\ y\in V^{2}\right\rangle
\cdot\left\langle c\left(  z\right)  \ \mid\ z\in V^{n-1-i}\right\rangle \\
&  =\left\langle a\left(  x\right)  b\left(  y\right)  c\left(  z\right)
\ \mid\ \left(  x,y,z\right)  \in V^{i-1}\times V^{2}\times V^{n-1-i}%
\right\rangle .
\end{align*}
Compared to (\ref{pf.Q_n.7}), this yields%
\begin{align}
&  \left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau
_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }\otimes
\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{n}\right\rangle \nonumber\\
&  =\left\langle a\left(  x\right)  \ \mid\ x\in V^{i-1}\right\rangle
\cdot\left\langle b\left(  y\right)  \ \mid\ y\in V^{2}\right\rangle
\cdot\left\langle c\left(  z\right)  \ \mid\ z\in V^{n-1-i}\right\rangle .
\label{pf.Q_n.9}%
\end{align}
But%
\begin{align*}
\left\langle a\left(  x\right)  \ \mid\ x\in V^{i-1}\right\rangle  &
=\left\langle \underbrace{a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)
}_{=v_{1}\otimes v_{2}\otimes\cdots\otimes v_{i-1}}\ \mid\ \left(  v_{1}%
,v_{2},\ldots,v_{i-1}\right)  \in V^{i-1}\right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed }x\text{ as }\left(
v_{1},v_{2},\ldots,v_{i-1}\right)  \right) \\
&  =\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{i-1}\ \mid\ \left(
v_{1},v_{2},\ldots,v_{i-1}\right)  \in V^{i-1}\right\rangle =V^{\otimes\left(
i-1\right)  }%
\end{align*}
(since the $k$-module $V^{\otimes\left(  i-1\right)  }$ is generated by its
pure tensors, i. e., by tensors of the form $v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{i-1}$ with $\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \in
V^{i-1}$). Also,%
\begin{align*}
\left\langle c\left(  z\right)  \ \mid\ z\in V^{n-1-i}\right\rangle  &
=\left\langle \underbrace{c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)
}_{=v_{i+2}\otimes v_{i+3}\otimes\cdots\otimes v_{n}}\ \mid\ \left(
v_{i+2},v_{i+3},\ldots,v_{n}\right)  \in V^{n-1-i}\right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed }z\text{ as }\left(
v_{i+2},v_{i+3},\ldots,v_{n}\right)  \right) \\
&  =\left\langle v_{i+2}\otimes v_{i+3}\otimes\cdots\otimes v_{n}%
\ \mid\ \left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)  \in V^{n-1-i}%
\right\rangle =V^{\otimes\left(  n-1-i\right)  }%
\end{align*}
(since the $k$-module $V^{\otimes\left(  n-1-i\right)  }$ is generated by its
pure tensors, i. e., by tensors of the form $v_{i+2}\otimes v_{i+3}%
\otimes\cdots\otimes v_{n}$ with $\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)
\in V^{n-1-i}$). Also,%
\begin{align*}
\left\langle b\left(  y\right)  \ \mid\ y\in V^{2}\right\rangle  &
=\left\langle \underbrace{b\left(  v_{i},v_{i+1}\right)  }_{=v_{i}\otimes
v_{i+1}+v_{i+1}\otimes v_{i}}\ \mid\ \left(  v_{i},v_{i+1}\right)  \in
V^{2}\right\rangle \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed }y\text{
as }\left(  v_{i},v_{i+1}\right)  \right) \\
&  =\left\langle v_{i}\otimes v_{i+1}+v_{i+1}\otimes v_{i}\ \mid\ \left(
v_{i},v_{i+1}\right)  \in V^{2}\right\rangle \\
&  =\left\langle v_{1}\otimes v_{2}+v_{2}\otimes v_{1}\ \mid\ \left(
v_{1},v_{2}\right)  \in V^{2}\right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed }\left(  v_{i}%
,v_{i+1}\right)  \text{ as }\left(  v_{1},v_{2}\right)  \right) \\
&  =Q_{2}\left(  V\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Corollary
\ref{coro.Q_2}}\right)  .
\end{align*}
Thus, (\ref{pf.Q_n.9}) becomes%
\begin{align*}
&  \left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau
_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }\otimes
\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{n}\right\rangle \\
&  =\underbrace{\left\langle a\left(  x\right)  \ \mid\ x\in V^{i-1}%
\right\rangle }_{=V^{\otimes\left(  i-1\right)  }}\cdot
\underbrace{\left\langle b\left(  y\right)  \ \mid\ y\in V^{2}\right\rangle
}_{=Q_{2}\left(  V\right)  }\cdot\underbrace{\left\langle c\left(  z\right)
\ \mid\ z\in V^{n-1-i}\right\rangle }_{=V^{\otimes\left(  n-1-i\right)  }}\\
&  =V^{\otimes\left(  i-1\right)  }\cdot\left(  Q_{2}\left(  V\right)
\right)  \cdot V^{\otimes\left(  n-1-i\right)  },
\end{align*}
so that Lemma \ref{lem.Q_n} is proven.
\end{proof}

This lemma yields:

\begin{corollary}
\label{cor.Q_n}Let $k$ be a commutative ring. Let $V$ be a $k$-module. Let
$n\in\mathbb{N}$.\newline Then,%
\[
Q_{n}\left(  V\right)  =\sum_{i=1}^{n-1}V^{\otimes\left(  i-1\right)  }%
\cdot\left(  Q_{2}\left(  V\right)  \right)  \cdot V^{\otimes\left(
n-1-i\right)  }%
\]
(this is an equality between $k$-submodules of $\otimes V$, where
$Q_{n}\left(  V\right)  $ becomes such a $k$-submodule by means of the
inclusion $Q_{n}\left(  V\right)  \subseteq V^{\otimes n}\subseteq\otimes V$).
Here, the multiplication on the right hand side is multiplication inside the
$k$-algebra $\otimes V$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.Q_n}.]For every $i\in\left\{  1,2,\ldots
,i-1\right\}  $, let $\tau_{i}$ denote the transposition $\left(
i,i+1\right)  \in S_{n}$. Then, by Proposition \ref{prop.Q_n}, we have%
\begin{align*}
Q_{n}\left(  V\right)   &  =\sum_{i=1}^{n-1}\underbrace{\left\langle
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)
}\otimes v_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(
n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}%
\right\rangle }_{=V^{\otimes\left(  i-1\right)  }\cdot\left(  Q_{2}\left(
V\right)  \right)  \cdot V^{\otimes\left(  n-1-i\right)  }\text{ (by Lemma
\ref{lem.Q_n})}}\\
&  =\sum_{i=1}^{n-1}V^{\otimes\left(  i-1\right)  }\cdot\left(  Q_{2}\left(
V\right)  \right)  \cdot V^{\otimes\left(  n-1-i\right)  }.
\end{align*}
Thus, Corollary \ref{cor.Q_n} is proven.
\end{proof}

We now claim that:

\begin{theorem}
\label{thm.Q}Let $k$ be a commutative ring. Let $V$ be a $k$-module. We know
that $Q_{n}\left(  V\right)  $ is a $k$-submodule of $V^{\otimes n}$ for every
$n\in\mathbb{N}$. Thus, $\bigoplus\limits_{n\in\mathbb{N}}Q_{n}\left(
V\right)  $ is a $k$-submodule of $\bigoplus\limits_{n\in\mathbb{N}}V^{\otimes
n}=\otimes V$. This $k$-submodule satisfies%
\[
\bigoplus\limits_{n\in\mathbb{N}}Q_{n}\left(  V\right)  =\left(  \otimes
V\right)  \cdot\left(  Q_{2}\left(  V\right)  \right)  \cdot\left(  \otimes
V\right)  .
\]

\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.Q}.]Working inside $\otimes V$, we have%
\begin{align*}
\bigoplus\limits_{n\in\mathbb{N}}Q_{n}\left(  V\right)   &  =\sum
\limits_{n\in\mathbb{N}}\sum_{i=1}^{n-1}V^{\otimes\left(  i-1\right)  }%
\cdot\left(  Q_{2}\left(  V\right)  \right)  \cdot V^{\otimes\left(
n-1-i\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by Corollary \ref{cor.Q_n}%
}\right) \\
&  =\sum\limits_{n\in\mathbb{N}}\underbrace{\sum\limits_{i=0}^{n-2}}%
_{=\sum\limits_{\substack{i\in\mathbb{N};\\n-2\geq i}}=\sum
\limits_{\substack{i\in\mathbb{N};\\n\geq i+2}}}V^{\otimes i}\cdot\left(
Q_{2}\left(  V\right)  \right)  \cdot\underbrace{V^{\otimes\left(  n-1-\left(
i+1\right)  \right)  }}_{=V^{\otimes\left(  n-1-i-1\right)  }=V^{\otimes
\left(  n-\left(  i+2\right)  \right)  }}\\
&  =\underbrace{\sum\limits_{n\in\mathbb{N}}\sum\limits_{\substack{i\in
\mathbb{N};\\n\geq i+2}}}_{=\sum\limits_{i\in\mathbb{N}}\sum
\limits_{\substack{n\in\mathbb{N};\\n\geq i+2}}}V^{\otimes i}\cdot\left(
Q_{2}\left(  V\right)  \right)  \cdot V^{\otimes\left(  n-\left(  i+2\right)
\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }i\text{ for
}i-1\text{ in the second sum}\right) \\
&  =\sum\limits_{i\in\mathbb{N}}\sum\limits_{\substack{n\in\mathbb{N};\\n\geq
i+2}}V^{\otimes i}\cdot\left(  Q_{2}\left(  V\right)  \right)  \cdot
V^{\otimes\left(  n-\left(  i+2\right)  \right)  }=\sum\limits_{i\in
\mathbb{N}}V^{\otimes i}\cdot\left(  Q_{2}\left(  V\right)  \right)  \cdot
\sum\limits_{\substack{n\in\mathbb{N};\\n\geq i+2}}V^{\otimes\left(  n-\left(
i+2\right)  \right)  }\\
&  =\sum\limits_{i\in\mathbb{N}}V^{\otimes i}\cdot\left(  Q_{2}\left(
V\right)  \right)  \cdot\sum\limits_{j\in\mathbb{N}}V^{\otimes j}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }j\text{ for
}n-\left(  i+2\right)  \text{ in the second sum}\right) \\
&  =\underbrace{\left(  \sum\limits_{i\in\mathbb{N}}V^{\otimes i}\right)
}_{=\otimes V\text{ (by (\ref{(X)V=sum_i}))}}\cdot\left(  Q_{2}\left(
V\right)  \right)  \cdot\underbrace{\left(  \sum\limits_{j\in\mathbb{N}%
}V^{\otimes j}\right)  }_{=\otimes V\text{ (by (\ref{(X)V=sum_j}))}}\\
&  =\left(  \otimes V\right)  \cdot\left(  Q_{2}\left(  V\right)  \right)
\cdot\left(  \otimes V\right)  .
\end{align*}
This proves Theorem \ref{thm.Q}.
\end{proof}

Now we can finally define the pseudoexterior algebra:

\begin{definition}
\label{defs.ext-alg}Let $k$ be a commutative ring. Let $V$ be a $k$%
-module.\newline By Theorem \ref{thm.Q}, the two $k$-submodules $\bigoplus
\limits_{n\in\mathbb{N}}Q_{n}\left(  V\right)  $ and $\left(  \otimes
V\right)  \cdot\left(  Q_{2}\left(  V\right)  \right)  \cdot\left(  \otimes
V\right)  $ of $\otimes V$ are identic (where $\bigoplus\limits_{n\in
\mathbb{N}}Q_{n}\left(  V\right)  $ becomes a $k$-submodule of $\otimes V$ in
the same way as explained in Theorem \ref{thm.Q}). We denote these two identic
$k$-submodules by $Q\left(  V\right)  $. In other words, we define $Q\left(
V\right)  $ by%
\[
Q\left(  V\right)  =\bigoplus\limits_{n\in\mathbb{N}}Q_{n}\left(  V\right)
=\left(  \otimes V\right)  \cdot\left(  Q_{2}\left(  V\right)  \right)
\cdot\left(  \otimes V\right)  .
\]
Since $Q\left(  V\right)  =\left(  \otimes V\right)  \cdot\left(  Q_{2}\left(
V\right)  \right)  \cdot\left(  \otimes V\right)  $, it is clear that
$Q\left(  V\right)  $ is a two-sided ideal of the $k$-algebra $\otimes
V$.\newline Now we define a $k$-module $\operatorname*{Exter} V$ as the direct
sum $\bigoplus\limits_{n\in\mathbb{N}}\operatorname*{Exter} ^{n}V$. Then,%
\begin{align*}
\left.  \operatorname*{Exter} V\right.   &  =\bigoplus\limits_{n\in\mathbb{N}%
}\underbrace{\operatorname*{Exter} ^{n}V}_{=V^{\otimes n}\diagup Q_{n}\left(
V\right)  }=\bigoplus\limits_{n\in\mathbb{N}}\left(  V^{\otimes n}\diagup
Q_{n}\left(  V\right)  \right)  \cong\underbrace{\left(  \bigoplus
\limits_{n\in\mathbb{N}}V^{\otimes n}\right)  }_{=\otimes V}\diagup
\underbrace{\left(  \bigoplus\limits_{n\in\mathbb{N}}Q_{n}\left(  V\right)
\right)  }_{=Q\left(  V\right)  }\\
&  =\left(  \otimes V\right)  \diagup Q\left(  V\right)  .
\end{align*}
This is a canonical isomorphism, so we will use it to identify
$\operatorname*{Exter} V$ with $\left(  \otimes V\right)  \diagup Q\left(
V\right)  $. Since $Q\left(  V\right)  $ is a two-sided ideal of the
$k$-algebra $\otimes V$, the quotient $k$-module $\left(  \otimes V\right)
\diagup Q\left(  V\right)  $ canonically becomes a $k$-algebra. Since
$\operatorname*{Exter} V=\left(  \otimes V\right)  \diagup Q\left(  V\right)
$, this means that $\operatorname*{Exter} V$ becomes a $k$-algebra. We refer
to this $k$-algebra as the \textit{pseudoexterior algebra} of the $k$-module
$V$.\newline We denote by $\operatorname*{exter}\nolimits_{V}$ the canonical
projection $\otimes V\rightarrow\left(  \otimes V\right)  \diagup Q\left(
V\right)  =\operatorname*{Exter} V$. Clearly, this map $\operatorname*{exter}%
\nolimits_{V}$ is a surjective $k$-algebra homomorphism. Besides, due to
$\otimes V=\bigoplus\limits_{n\in\mathbb{N}}V^{\otimes n}$ and $Q\left(
V\right)  =\bigoplus\limits_{n\in\mathbb{N}}Q_{n}\left(  V\right)  $, it is
clear that the canonical projection $\otimes V\rightarrow\left(  \otimes
V\right)  \diagup Q\left(  V\right)  $ is the direct sum of the canonical
projections $V^{\otimes n}\rightarrow V^{\otimes n}\diagup Q_{n}\left(
V\right)  $ over all $n\in\mathbb{N}$. Since the canonical projection $\otimes
V\rightarrow\left(  \otimes V\right)  \diagup Q\left(  V\right)  $ is the map
$\operatorname*{exter}\nolimits_{V}$, whereas the canonical projection
$V^{\otimes n}\rightarrow V^{\otimes n}\diagup Q_{n}\left(  V\right)  $ is the
map $\operatorname*{exter}\nolimits_{V,n}$, this rewrites as follows: The map
$\operatorname*{exter}\nolimits_{V}$ is the direct sum of the maps
$\operatorname*{exter}\nolimits_{V,n}$ over all $n\in\mathbb{N}$.
\end{definition}

We now prove a first, almost trivial result about the $Q\left(  V\right)  $:

\begin{lemma}
\label{lem.Q(f)}Let $k$ be a commutative ring. Let $V$ and $W$ be two
$k$-modules. Let $f:V\rightarrow W$ be a $k$-module homomorphism.\newline%
\textbf{(a)} Then, the $k$-algebra homomorphism $\otimes f:\otimes
V\rightarrow\otimes W$ satisfies $\left(  \otimes f\right)  \left(  Q\left(
V\right)  \right)  \subseteq Q\left(  W\right)  $. Also, for every
$n\in\mathbb{N}$, the $k$-module homomorphism $f^{\otimes n}:V^{\otimes
n}\rightarrow W^{\otimes n}$ satisfies $f^{\otimes n}\left(  Q_{n}\left(
V\right)  \right)  \subseteq Q_{n}\left(  W\right)  $.\newline\textbf{(b)}
Assume that $f$ is surjective. Then, the $k$-algebra homomorphism $\otimes
f:\otimes V\rightarrow\otimes W$ satisfies $\left(  \otimes f\right)  \left(
Q\left(  V\right)  \right)  =Q\left(  W\right)  $. Also, for every
$n\in\mathbb{N}$, the $k$-module homomorphism $f^{\otimes n}:V^{\otimes
n}\rightarrow W^{\otimes n}$ satisfies $f^{\otimes n}\left(  Q_{n}\left(
V\right)  \right)  =Q_{n}\left(  W\right)  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.Q(f)}.]\textbf{(a)} Fix some $n\in\mathbb{N}$. For
every $i\in\left\{  1,2,\ldots,n-1\right\}  $, let $\tau_{i}$ denote the
transposition $\left(  i,i+1\right)  \in S_{n}$. Then, the definition of
$Q_{n}\left(  V\right)  $ yields%
\begin{equation}
Q_{n}\left(  V\right)  =\sum_{i=1}^{n-1}\left\langle v_{1}\otimes v_{2}%
\otimes\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau
_{i}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)
}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\rangle ,
\label{pf.Q(f).1}%
\end{equation}
whereas the definition of $Q_{n}\left(  W\right)  $ yields%
\begin{equation}
Q_{n}\left(  W\right)  =\sum_{i=1}^{n-1}\left\langle w_{1}\otimes w_{2}%
\otimes\cdots\otimes w_{n}+w_{\tau_{i}\left(  1\right)  }\otimes w_{\tau
_{i}\left(  2\right)  }\otimes\cdots\otimes w_{\tau_{i}\left(  n\right)
}\ \mid\ \left(  w_{1},w_{2},\ldots,w_{n}\right)  \in W^{n}\right\rangle .
\label{pf.Q(f).2}%
\end{equation}


Now it is easy to see that every $i\in\left\{  1,2,\ldots,n-1\right\}  $
satisfies%
\begin{align}
&  f^{\otimes n}\left(  \left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)
}\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\rangle \right) \nonumber\\
&  \subseteq\left\langle w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}%
+w_{\tau_{i}\left(  1\right)  }\otimes w_{\tau_{i}\left(  2\right)  }%
\otimes\cdots\otimes w_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  w_{1}%
,w_{2},\ldots,w_{n}\right)  \in W^{n}\right\rangle . \label{pf.Q(f).4}%
\end{align}
\footnote{\textit{Proof.} Fix some $i\in\left\{  1,2,\ldots,n-1\right\}  $.
Let $S$ be the set%
\[
\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau_{i}\left(
1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes
v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)
\in V^{n}\right\}  .
\]
It is clear that every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$
satisfies
\begin{align*}
&  f^{\otimes n}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}%
+v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }%
\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }\right) \\
&  =f\left(  v_{1}\right)  \otimes f\left(  v_{2}\right)  \otimes\cdots\otimes
f\left(  v_{n}\right)  +f\left(  v_{\tau_{i}\left(  1\right)  }\right)
\otimes f\left(  v_{\tau_{i}\left(  2\right)  }\right)  \otimes\cdots\otimes
f\left(  v_{\tau_{i}\left(  n\right)  }\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }f^{\otimes n}\right) \\
&  \in\left\{  w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}+w_{\tau_{i}\left(
1\right)  }\otimes w_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes
w_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  w_{1},w_{2},\ldots,w_{n}\right)
\in W^{n}\right\} \\
&  \subseteq\left\langle \left\{  w_{1}\otimes w_{2}\otimes\cdots\otimes
w_{n}+w_{\tau_{i}\left(  1\right)  }\otimes w_{\tau_{i}\left(  2\right)
}\otimes\cdots\otimes w_{\tau_{i}\left(  n\right)  }\ \mid\ \left(
w_{1},w_{2},\ldots,w_{n}\right)  \in W^{n}\right\}  \right\rangle \\
&  =\left\langle w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}+w_{\tau
_{i}\left(  1\right)  }\otimes w_{\tau_{i}\left(  2\right)  }\otimes
\cdots\otimes w_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  w_{1},w_{2}%
,\ldots,w_{n}\right)  \in W^{n}\right\rangle .
\end{align*}
In other words,%
\begin{align*}
&  \left\{  f^{\otimes n}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)
}\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }\right)  \ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\} \\
&  \subseteq\left\langle w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}%
+w_{\tau_{i}\left(  1\right)  }\otimes w_{\tau_{i}\left(  2\right)  }%
\otimes\cdots\otimes w_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  w_{1}%
,w_{2},\ldots,w_{n}\right)  \in W^{n}\right\rangle .
\end{align*}
Since%
\begin{align*}
&  \left\{  f^{\otimes n}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)
}\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }\right)  \ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\} \\
&  =f^{\otimes n}\left(  \underbrace{\left\{  v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(
2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\}  }_{=S}\right)  =f^{\otimes
n}\left(  S\right)  ,
\end{align*}
this rewrites as%
\[
f^{\otimes n}\left(  S\right)  \subseteq\left\langle w_{1}\otimes w_{2}%
\otimes\cdots\otimes w_{n}+w_{\tau_{i}\left(  1\right)  }\otimes w_{\tau
_{i}\left(  2\right)  }\otimes\cdots\otimes w_{\tau_{i}\left(  n\right)
}\ \mid\ \left(  w_{1},w_{2},\ldots,w_{n}\right)  \in W^{n}\right\rangle .
\]
By Proposition \ref{prop.<>} \textbf{(a)} (applied to $f^{\otimes n}\left(
S\right)  $, $W^{\otimes n}$ and \newline$\left\langle w_{1}\otimes
w_{2}\otimes\cdots\otimes w_{n}+w_{\tau_{i}\left(  1\right)  }\otimes
w_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes w_{\tau_{i}\left(
n\right)  }\ \mid\ \left(  w_{1},w_{2},\ldots,w_{n}\right)  \in W^{n}%
\right\rangle $ instead of $S$, $M$ and $Q$), this yields
\[
\left\langle f^{\otimes n}\left(  S\right)  \right\rangle \subseteq
\left\langle w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}+w_{\tau_{i}\left(
1\right)  }\otimes w_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes
w_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  w_{1},w_{2},\ldots,w_{n}\right)
\in W^{n}\right\rangle .
\]
But by Proposition \ref{prop.<>} \textbf{(b)} (applied to $f^{\otimes n}$,
$V^{\otimes n}$ and $W^{\otimes n}$ instead of $f$, $M$ and $R$), we have
$f^{\otimes n}\left(  \left\langle S\right\rangle \right)  =\left\langle
f^{\otimes n}\left(  S\right)  \right\rangle $. Thus,
\[
f^{\otimes n}\left(  \left\langle S\right\rangle \right)  =\left\langle
f^{\otimes n}\left(  S\right)  \right\rangle \subseteq\left\langle
w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}+w_{\tau_{i}\left(  1\right)
}\otimes w_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes w_{\tau_{i}\left(
n\right)  }\ \mid\ \left(  w_{1},w_{2},\ldots,w_{n}\right)  \in W^{n}%
\right\rangle .
\]
Since
\begin{align*}
\left\langle S\right\rangle  &  =\left\langle \left\{  v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes
v_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(
n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\}
\right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{because }S=\left\{  v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes
v_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(
n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\}
\right) \\
&  =\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau
_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }\otimes
\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{n}\right\rangle ,
\end{align*}
this becomes%
\begin{align*}
&  f^{\otimes n}\left(  \left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)
}\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\rangle \right) \\
&  \subseteq\left\langle w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}%
+w_{\tau_{i}\left(  1\right)  }\otimes w_{\tau_{i}\left(  2\right)  }%
\otimes\cdots\otimes w_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  w_{1}%
,w_{2},\ldots,w_{n}\right)  \in W^{n}\right\rangle ,
\end{align*}
qed.} Now, (\ref{pf.Q(f).1}) yields%
\begin{align*}
&  f^{\otimes n}\left(  Q_{n}\left(  V\right)  \right) \\
&  =f^{\otimes n}\left(  \sum_{i=1}^{n-1}\left\langle v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes
v_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(
n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}%
\right\rangle \right) \\
&  =\sum_{i=1}^{n-1}\underbrace{f^{\otimes n}\left(  \left\langle v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes
v_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(
n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}%
\right\rangle \right)  }_{\subseteq\left\langle w_{1}\otimes w_{2}%
\otimes\cdots\otimes w_{n}+w_{\tau_{i}\left(  1\right)  }\otimes w_{\tau
_{i}\left(  2\right)  }\otimes\cdots\otimes w_{\tau_{i}\left(  n\right)
}\ \mid\ \left(  w_{1},w_{2},\ldots,w_{n}\right)  \in W^{n}\right\rangle
\text{ (by (\ref{pf.Q(f).4}))}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }f^{\otimes n}\text{ is
linear}\right) \\
&  \subseteq\sum_{i=1}^{n-1}\left\langle w_{1}\otimes w_{2}\otimes
\cdots\otimes w_{n}+w_{\tau_{i}\left(  1\right)  }\otimes w_{\tau_{i}\left(
2\right)  }\otimes\cdots\otimes w_{\tau_{i}\left(  n\right)  }\ \mid\ \left(
w_{1},w_{2},\ldots,w_{n}\right)  \in W^{n}\right\rangle \\
&  =Q_{n}\left(  W\right)  .
\end{align*}
Thus, we have shown that%
\begin{equation}
\text{for every }n\in\mathbb{N}\text{, we have }f^{\otimes n}\left(
Q_{n}\left(  V\right)  \right)  \subseteq Q_{n}\left(  W\right)  .
\label{pf.Q(f).5}%
\end{equation}


Now forget that we fixed $n$. Since the map $\otimes f$ is the direct sum of
the maps $f^{\otimes n}:V^{\otimes n}\rightarrow W^{\otimes n}$ for all
$n\in\mathbb{N}$, we have $\left(  \otimes f\right)  \left(  x\right)
=f^{\otimes n}\left(  x\right)  $ for every $n\in\mathbb{N}$ and every $x\in
V^{\otimes n}$. Thus, for every $n\in\mathbb{N}$, we have $\left(  \otimes
f\right)  \left(  Q_{n}\left(  V\right)  \right)  =\left\{
\underbrace{\left(  \otimes f\right)  \left(  x\right)  }%
_{\substack{=f^{\otimes n}\left(  x\right)  \\\text{(since }x\in Q_{n}\left(
V\right)  \subseteq V^{\otimes n}\text{)}}}\ \mid\ x\in Q_{n}\left(  V\right)
\right\}  =\left\{  f^{\otimes n}\left(  x\right)  \ \mid\ x\in Q_{n}\left(
V\right)  \right\}  =f^{\otimes n}\left(  Q_{n}\left(  V\right)  \right)  $.

The definition of $Q\left(  W\right)  $ yields $Q\left(  W\right)
=\bigoplus\limits_{n\in\mathbb{N}}Q_{n}\left(  W\right)  $. Since direct sums
are sums, this rewrites as $Q\left(  W\right)  =\sum\limits_{n\in\mathbb{N}%
}Q_{n}\left(  W\right)  $.

Now, $Q\left(  V\right)  =\bigoplus\limits_{n\in\mathbb{N}}Q_{n}\left(
V\right)  =\sum\limits_{n\in\mathbb{N}}Q_{n}\left(  V\right)  $ (since direct
sums are sums) and thus%
\[
\left(  \otimes f\right)  \left(  Q\left(  V\right)  \right)  =\left(  \otimes
f\right)  \left(  \sum\limits_{n\in\mathbb{N}}Q_{n}\left(  V\right)  \right)
=\sum\limits_{n\in\mathbb{N}}\underbrace{\left(  \otimes f\right)  \left(
Q_{n}\left(  V\right)  \right)  }_{\substack{=f^{\otimes n}\left(
Q_{n}\left(  V\right)  \right)  \subseteq Q_{n}\left(  W\right)  \\\text{(by
(\ref{pf.Q(f).5}))}}}\subseteq\sum\limits_{n\in\mathbb{N}}Q_{n}\left(
W\right)  =Q\left(  W\right)  .
\]
This completes the proof of Lemma \ref{lem.Q(f)} \textbf{(a).}

\textbf{(b)} Fix some $n\in\mathbb{N}$. For every $i\in\left\{  1,2,\ldots
,n-1\right\}  $, it is easy to prove (using the surjectivity of $f$) that%
\begin{align}
&  f^{\otimes n}\left(  \left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)
}\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\rangle \right) \nonumber\\
&  =\left\langle w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}+w_{\tau
_{i}\left(  1\right)  }\otimes w_{\tau_{i}\left(  2\right)  }\otimes
\cdots\otimes w_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  w_{1},w_{2}%
,\ldots,w_{n}\right)  \in W^{n}\right\rangle . \label{pf.Q(f).8}%
\end{align}


\textit{Proof of (\ref{pf.Q(f).8}).} Fix some $i\in\left\{  1,2,\ldots
,n-1\right\}  $. Then, every $\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in
W^{n}$ satisfies
\begin{align*}
&  w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}+w_{\tau_{i}\left(  1\right)
}\otimes w_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes w_{\tau_{i}\left(
n\right)  }\\
&  \in f^{\otimes n}\left(  \left\langle v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(
2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\rangle \right)  .
\end{align*}
\footnote{\textit{Proof.} Let $\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in
W^{n}$. For every $i\in\left\{  1,2,\ldots,n\right\}  $, there exists some
$z_{i}\in V$ such that $w_{i}=f\left(  z_{i}\right)  $ (since $f$ is
surjective). Fix such a $z_{i}$ for each $i\in\left\{  1,2,\ldots,n\right\}
$. Then, $w_{1}=f\left(  z_{1}\right)  $, $w_{2}=f\left(  z_{2}\right)  $,
$\ldots$, $w_{n}=f\left(  z_{n}\right)  $. Taking the tensor product of these
equalities, we get $w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}=f\left(
z_{1}\right)  \otimes f\left(  z_{2}\right)  \otimes\cdots\otimes f\left(
z_{n}\right)  $. Also, since $w_{i}=f\left(  z_{i}\right)  $ for each
$i\in\left\{  1,2,\ldots,n\right\}  $, we have $w_{\tau_{i}\left(  1\right)
}=f\left(  z_{\tau_{i}\left(  1\right)  }\right)  $, $w_{\tau_{i}\left(
2\right)  }=f\left(  z_{\tau_{i}\left(  2\right)  }\right)  $, $\ldots$,
$w_{\tau_{i}\left(  n\right)  }=f\left(  z_{\tau_{i}\left(  n\right)
}\right)  $. Taking the tensor product of these equalities, we get
$w_{\tau_{i}\left(  1\right)  }\otimes w_{\tau_{i}\left(  2\right)  }%
\otimes\cdots\otimes w_{\tau_{i}\left(  n\right)  }=f\left(  z_{\tau
_{i}\left(  1\right)  }\right)  \otimes f\left(  z_{\tau_{i}\left(  2\right)
}\right)  \otimes\cdots\otimes f\left(  z_{\tau_{i}\left(  n\right)  }\right)
$. Now, by the definition of $f^{\otimes n}$, we have%
\begin{align*}
&  f^{\otimes n}\left(  z_{1}\otimes z_{2}\otimes\cdots\otimes z_{n}%
+z_{\tau_{i}\left(  1\right)  }\otimes z_{\tau_{i}\left(  2\right)  }%
\otimes\cdots\otimes z_{\tau_{i}\left(  n\right)  }\right) \\
&  =\underbrace{f\left(  z_{1}\right)  \otimes f\left(  z_{2}\right)
\otimes\cdots\otimes f\left(  z_{n}\right)  }_{=w_{1}\otimes w_{2}%
\otimes\cdots\otimes w_{n}}+\underbrace{f\left(  z_{\tau_{i}\left(  1\right)
}\right)  \otimes f\left(  z_{\tau_{i}\left(  2\right)  }\right)
\otimes\cdots\otimes f\left(  z_{\tau_{i}\left(  n\right)  }\right)
}_{=w_{\tau_{i}\left(  1\right)  }\otimes w_{\tau_{i}\left(  2\right)
}\otimes\cdots\otimes w_{\tau_{i}\left(  n\right)  }}\\
&  =w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}+w_{\tau_{i}\left(  1\right)
}\otimes w_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes w_{\tau_{i}\left(
n\right)  },
\end{align*}
so that%
\begin{align*}
&  w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}+w_{\tau_{i}\left(  1\right)
}\otimes w_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes w_{\tau_{i}\left(
n\right)  }\\
&  =f^{\otimes n}\left(  \underbrace{z_{1}\otimes z_{2}\otimes\cdots\otimes
z_{n}+z_{\tau_{i}\left(  1\right)  }\otimes z_{\tau_{i}\left(  2\right)
}\otimes\cdots\otimes z_{\tau_{i}\left(  n\right)  }}_{\substack{\in\left\{
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)
}\otimes v_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(
n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\}
\\\subseteq\left\langle \left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)
}\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\}  \right\rangle
\\=\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau
_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }\otimes
\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{n}\right\rangle }}\right) \\
&  \in f^{\otimes n}\left(  \left\langle v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(
2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\rangle \right)  ,
\end{align*}
qed.} In other words,%
\begin{align*}
&  \left\{  w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}+w_{\tau_{i}\left(
1\right)  }\otimes w_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes
w_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  w_{1},w_{2},\ldots,w_{n}\right)
\in W^{n}\right\} \\
&  \subseteq f^{\otimes n}\left(  \left\langle v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(
2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\rangle \right)  .
\end{align*}
Applying Proposition \ref{prop.<>} \textbf{(a)} to \newline$\left\{
w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}+w_{\tau_{i}\left(  1\right)
}\otimes w_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes w_{\tau_{i}\left(
n\right)  }\ \mid\ \left(  w_{1},w_{2},\ldots,w_{n}\right)  \in W^{n}\right\}
$, $W^{\otimes n}$ and $f^{\otimes n}\left(  \left\langle v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes
v_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(
n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}%
\right\rangle \right)  $ instead of $S$, $M$ and $Q$), we conclude from this
that
\begin{align*}
&  \left\langle \left\{  w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}%
+w_{\tau_{i}\left(  1\right)  }\otimes w_{\tau_{i}\left(  2\right)  }%
\otimes\cdots\otimes w_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  w_{1}%
,w_{2},\ldots,w_{n}\right)  \in W^{n}\right\}  \right\rangle \\
&  \subseteq f^{\otimes n}\left(  \left\langle v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(
2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\rangle \right)  .
\end{align*}
Thus,%
\begin{align*}
&  \left\langle w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}+w_{\tau
_{i}\left(  1\right)  }\otimes w_{\tau_{i}\left(  2\right)  }\otimes
\cdots\otimes w_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  w_{1},w_{2}%
,\ldots,w_{n}\right)  \in W^{n}\right\rangle \\
&  =\left\langle \left\{  w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}%
+w_{\tau_{i}\left(  1\right)  }\otimes w_{\tau_{i}\left(  2\right)  }%
\otimes\cdots\otimes w_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  w_{1}%
,w_{2},\ldots,w_{n}\right)  \in W^{n}\right\}  \right\rangle \\
&  \subseteq f^{\otimes n}\left(  \left\langle v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(
2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\rangle \right)  .
\end{align*}
Combined with (\ref{pf.Q(f).4}), this yields (\ref{pf.Q(f).8}). Thus, we have
proven (\ref{pf.Q(f).8}).

In the proof of Lemma \ref{lem.Q(f)} \textbf{(a)}, we have used
(\ref{pf.Q(f).4}) to show that $f^{\otimes n}\left(  Q_{n}\left(  V\right)
\right)  \subseteq Q_{n}\left(  W\right)  $. In the same way, we can use
(\ref{pf.Q(f).8}) to prove that $f^{\otimes n}\left(  Q_{n}\left(  V\right)
\right)  =Q_{n}\left(  W\right)  $ (in the situation of Lemma \ref{lem.Q(f)}
\textbf{(b)}).

So we have shown that%
\begin{equation}
\text{for every }n\in\mathbb{N}\text{, we have }f^{\otimes n}\left(
Q_{n}\left(  V\right)  \right)  =Q_{n}\left(  W\right)  . \label{pf.Q(f).9}%
\end{equation}
In the proof of Lemma \ref{lem.Q(f)} \textbf{(a)}, we have used
(\ref{pf.Q(f).5}) to conclude that $\left(  \otimes f\right)  \left(  Q\left(
V\right)  \right)  \subseteq Q\left(  W\right)  $. In the same way, we can use
(\ref{pf.Q(f).9}) to conclude that $\left(  \otimes f\right)  \left(  Q\left(
V\right)  \right)  =Q\left(  W\right)  $ (in the situation of Lemma
\ref{lem.Q(f)} \textbf{(b)}). This completes the proof of Lemma \ref{lem.Q(f)}
\textbf{(b)}.
\end{proof}

[\textit{Remark.} The above proof of Lemma \ref{lem.Q(f)} uses the definition
of $Q\left(  V\right)  $ as $\bigoplus\limits_{n\in\mathbb{N}}Q_{n}\left(
V\right)  $. We could just as well have proven Lemma \ref{lem.Q(f)} using the
definition of $Q\left(  V\right)  $ as $\left(  \otimes V\right)  \cdot\left(
Q_{2}\left(  V\right)  \right)  \cdot\left(  \otimes V\right)  $.]

The pseudoexterior algebra is (just as most other constructions we did above)
functorial in $V$. This means that:

\begin{definition}
\label{def.ext-functor}Let $k$ be a commutative ring. Let $V$ and $W$ be two
$k$-modules. Let $f:V\rightarrow W$ be a $k$-module homomorphism. Then, the
$k$-algebra homomorphism $\otimes f:\otimes V\rightarrow\otimes W$ satisfies
$\left(  \otimes f\right)  \left(  Q\left(  V\right)  \right)  \subseteq
Q\left(  W\right)  $ (by Lemma \ref{lem.Q(f)} \textbf{(a)}), and thus gives
rise to a $k$-algebra homomorphism $\left(  \otimes V\right)  \diagup Q\left(
V\right)  \rightarrow\left(  \otimes W\right)  \diagup Q\left(  W\right)  $.
This latter $k$-algebra homomorphism will be denoted by $\operatorname*{Exter}%
f$. Since $\left(  \otimes V\right)  \diagup Q\left(  V\right)
=\operatorname*{Exter}V$ and $\left(  \otimes W\right)  \diagup Q\left(
W\right)  =\operatorname*{Exter}W$, this homomorphism $\operatorname*{Exter}%
f:\left(  \otimes V\right)  \diagup Q\left(  V\right)  \rightarrow\left(
\otimes W\right)  \diagup Q\left(  W\right)  $ is actually a homomorphism from
$\operatorname*{Exter}V$ to $\operatorname*{Exter}W$.\newline By the
construction of $\operatorname*{Exter}f$, the diagram%
\begin{equation}%
%TCIMACRO{\TeXButton{exter f from otimes f}{\xymatrixcolsep{4pc}
%\xymatrix{
%\otimes V \ar[r]^{\otimes f}\ar[d]_{\operatorname*{exter}_V} & \otimes
%W \ar[d]^{\operatorname*{exter}_W} \\
%\operatorname*{Exter}  V \ar[r]_{\operatorname*{Exter}  f} & \operatorname
%*{Exter}  W
%}} }%
%BeginExpansion
\xymatrixcolsep{4pc}
\xymatrix{
\otimes V \ar[r]^{\otimes f}\ar[d]_{\operatorname*{exter}_V} & \otimes
W \ar[d]^{\operatorname*{exter}_W} \\
\operatorname*{Exter}  V \ar[r]_{\operatorname*{Exter}  f} & \operatorname
*{Exter}  W
}
%EndExpansion
\label{def.ext-functor.diag}%
\end{equation}
commutes (since $\operatorname*{exter}\nolimits_{V}$ is the canonical
projection $\otimes V\rightarrow\operatorname*{Exter}V$ and since
$\operatorname*{exter}\nolimits_{W}$ is the canonical projection $\otimes
W\rightarrow\operatorname*{Exter}W$).
\end{definition}

As a consequence of Lemma \ref{lem.Q(f)} \textbf{(b)}, we have:

\begin{proposition}
\label{prop.ext-surj}Let $k$ be a commutative ring. Let $V$ and $W$ be two
$k$-modules. Let $f:V\rightarrow W$ be a surjective $k$-module homomorphism.
Then:\newline\textbf{(a)} The $k$-module homomorphism $f^{\otimes
n}:V^{\otimes n}\rightarrow W^{\otimes n}$ is surjective for every
$n\in\mathbb{N}$.\newline\textbf{(b)} The $k$-algebra homomorphism $\otimes
f:\otimes V\rightarrow\otimes W$ is surjective.\newline\textbf{(c)} The
$k$-algebra homomorphism $\operatorname*{Exter} f:\operatorname*{Exter}
V\rightarrow\operatorname*{Exter} W$ is surjective.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.ext-surj}.]\textbf{(a)} Let $n\in\mathbb{N}$.
Lemma \ref{lem.f_i.sur} (applied to $f_{i}=f$) yields that the map
$\underbrace{f\otimes f\otimes\cdots\otimes f}_{n\text{ times}}$ is
surjective. Since $f^{\otimes n}=\underbrace{f\otimes f\otimes\cdots\otimes
f}_{n\text{ times}}$, this yields that the map $f^{\otimes n}$ is surjective.
This proves Proposition \ref{prop.ext-surj} \textbf{(a)}.

\textbf{(b)} We defined the map $\otimes f$ as the direct sum of the maps
$f^{\otimes n}$ for all $n\in\mathbb{N}$. Since the maps $f^{\otimes n}$ are
surjective (by Proposition \ref{prop.ext-surj} \textbf{(a)}), this yields that
the map $\otimes f$ is surjective (since the direct sum of surjective maps is
always surjective). This proves Proposition \ref{prop.ext-surj} \textbf{(b)}.

\textbf{(c)} Since the diagram (\ref{def.ext-functor.diag}) commutes, we have
$\operatorname*{exter}\nolimits_{W}\circ\left(  \otimes f\right)  =\left(
\operatorname*{Exter}f\right)  \circ\operatorname*{exter}\nolimits_{V}$. Now,
the map $\operatorname*{exter}\nolimits_{W}$ is surjective (since it is the
canonical projection $\otimes W\rightarrow\operatorname*{Exter}W$), and the
map $\otimes f$ is surjective (by Proposition \ref{prop.ext-surj}
\textbf{(b)}). Hence, the map $\operatorname*{exter}\nolimits_{W}\circ\left(
\otimes f\right)  $ is surjective (since the composition of surjective maps is
always surjective). Since $\operatorname*{exter}\nolimits_{W}\circ\left(
\otimes f\right)  =\left(  \operatorname*{Exter}f\right)  \circ
\operatorname*{exter}\nolimits_{V}$, this yields that the map $\left(
\operatorname*{Exter}f\right)  \circ\operatorname*{exter}\nolimits_{V}$ is
surjective. Hence, the map $\operatorname*{Exter}f$ is surjective (because if
$\alpha$ and $\beta$ are two maps such that the composition $\alpha\circ\beta$
is surjective, then $\alpha$ must itself be surjective). This proves
Proposition \ref{prop.ext-surj} \textbf{(c)}.
\end{proof}

\subsection{\label{subsect.Ker(exter)}The kernel of $\operatorname*{Exter} f$}

We will now formulate a result about the kernel of $\operatorname*{Exter} f$
for a $k$-module map $f$ (similar to Theorem \ref{thm.(X)f}, but with a twist):

\begin{theorem}
\label{thm.Ker(exter)}Let $k$ be a commutative ring. Let $V$ and $V^{\prime}$
be two $k$-modules, and let $f:V\rightarrow V^{\prime}$ be a surjective
$k$-module homomorphism. Then, the kernel of the map $\operatorname*{Exter}%
f:\operatorname*{Exter}V\rightarrow\operatorname*{Exter}V^{\prime}$ is%
\begin{align*}
\operatorname*{Ker}\left(  \operatorname*{Exter}f\right)   &  =\left(
\operatorname*{Exter}V\right)  \cdot\operatorname*{exter}\nolimits_{V}\left(
\operatorname*{Ker}f\right)  \cdot\left(  \operatorname*{Exter}V\right)
=\left(  \operatorname*{Exter}V\right)  \cdot\operatorname*{exter}%
\nolimits_{V}\left(  \operatorname*{Ker}f\right) \\
&  =\operatorname*{exter}\nolimits_{V}\left(  \operatorname*{Ker}f\right)
\cdot\left(  \operatorname*{Exter}V\right)  .
\end{align*}
Here, $\operatorname*{Ker}f$ is considered a $k$-submodule of $\otimes V$ by
means of the inclusion $\operatorname*{Ker}f\subseteq V=V^{\otimes1}%
\subseteq\otimes V$.
\end{theorem}

Note that the $\operatorname*{Ker}\left(  \operatorname*{Exter} V\right)
=\left(  \operatorname*{Exter} V\right)  \cdot\operatorname*{exter}%
\nolimits_{V}\left(  \operatorname*{Ker}f\right)  \cdot\left(
\operatorname*{Exter} V\right)  $ part of this theorem will be a rather quick
application of Proposition \ref{prop.9l} to the results of Theorem
\ref{thm.(X)f} and Proposition \ref{prop.ext-surj} \textbf{(c)}. It is
slightly less clear how to show the $\left(  \operatorname*{Exter} V\right)
\cdot\operatorname*{exter}\nolimits_{V}\left(  \operatorname*{Ker}f\right)
\cdot\left(  \operatorname*{Exter} V\right)  =\left(  \operatorname*{Exter}
V\right)  \cdot\operatorname*{exter}\nolimits_{V}\left(  \operatorname*{Ker}%
f\right)  =\operatorname*{exter}\nolimits_{V}\left(  \operatorname*{Ker}%
f\right)  \cdot\left(  \operatorname*{Exter} V\right)  $ part. We will do this
using the following lemma:

\begin{lemma}
\label{lem.ideal}Let $k$ be a commutative ring. Let $V$ be a $k$-module. Let
$A$ be a $k$-algebra, and let $\pi:\otimes V\rightarrow A$ be a surjective
$k$-algebra homomorphism. Let $M$ be a $k$-submodule of $A$ such that
$M\cdot\pi\left(  V\right)  \subseteq M$. Then, $M$ is a right ideal of $A$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.ideal}.]We claim that for every $n\in\mathbb{N}$, we
have%
\begin{equation}
M\cdot\pi\left(  V^{\otimes n}\right)  \subseteq M\text{.} \label{pf.ideal.1}%
\end{equation}


\textit{Proof of (\ref{pf.ideal.1}).} We are going to prove (\ref{pf.ideal.1})
by induction over $n$:

\textit{Induction base:} For $n=0$, we have%
\[
M\cdot\pi\left(  \underbrace{V^{\otimes n}}_{=V^{\otimes0}=k=k\cdot1}\right)
=M\cdot\underbrace{\pi\left(  k\cdot1\right)  }_{\substack{=k\cdot\pi\left(
1\right)  \\\text{(since }\pi\text{ is }k\text{-linear)}}}=M\cdot
k\cdot\underbrace{\pi\left(  1\right)  }_{\substack{=1\text{ (since }\pi\text{
is a}\\k\text{-algebra homomorphism)}}}=M\cdot k\cdot1=M
\]
(since $M$ is a $k$-module). Thus, (\ref{pf.ideal.1}) is true for $n=0$. This
completes the induction base.

\textit{Induction step:} Let $m\in\mathbb{N}$. Assume that (\ref{pf.ideal.1})
holds for $n=m$. We now must prove that (\ref{pf.ideal.1}) holds for $n=m+1$.

Since (\ref{pf.ideal.1}) holds for $n=m$, we have $M\cdot\pi\left(  V^{\otimes
m}\right)  \subseteq M$. Since $V^{\otimes\left(  m+1\right)  }=V\cdot
V^{\otimes m}\ \ \ \ $\footnote{\textit{Proof.} Every $v\in V$ and $w\in
V^{\otimes m}$ satisfy $v\cdot w=v\otimes w$ (by (\ref{*=(X)}), applied to
$n=1$, $a=v$ and $b=w$). In other words, every $\left(  v,w\right)  \in
V\times V^{\otimes m}$ satisfies $v\cdot w=v\otimes w$. Thus,%
\[
V\cdot V^{\otimes m}=\left\langle \underbrace{v\cdot w}_{=v\otimes w}%
\ \mid\ \left(  v,w\right)  \in V\times V^{\otimes m}\right\rangle
=\left\langle v\otimes w\ \mid\ \left(  v,w\right)  \in V\times V^{\otimes
m}\right\rangle .
\]
Compared with $V^{\otimes\left(  m+1\right)  }=V\otimes V^{\otimes
m}=\left\langle v\otimes w\ \mid\ \left(  v,w\right)  \in V\times V^{\otimes
m}\right\rangle $ (since a tensor product is generated by its pure tensors),
this yields $V^{\otimes\left(  m+1\right)  }=V\cdot V^{\otimes m}$, qed.}, we
have%
\[
\pi\left(  V^{\otimes\left(  m+1\right)  }\right)  =\pi\left(  V\cdot
V^{\otimes m}\right)  =\pi\left(  V\right)  \cdot\pi\left(  V^{\otimes
m}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi\text{ is a
}k\text{-algebra homomorphism}\right)  ,
\]
so that%
\[
M\cdot\pi\left(  V^{\otimes\left(  m+1\right)  }\right)  =\underbrace{M\cdot
\pi\left(  V\right)  }_{\subseteq M}\cdot\pi\left(  V^{\otimes m}\right)
\subseteq M\cdot\pi\left(  V^{\otimes m}\right)  \subseteq M.
\]
In other words, (\ref{pf.ideal.1}) holds for $n=m+1$. This completes the
induction step.

Thus, the induction proof of (\ref{pf.ideal.1}) is complete.

Now that (\ref{pf.ideal.1}) is proven, we notice that%
\[
\left.  \otimes V\right.  =\bigoplus\limits_{n\in\mathbb{N}}V^{\otimes n}%
=\sum\limits_{n\in\mathbb{N}}V^{\otimes n}\ \ \ \ \ \ \ \ \ \ \left(
\text{since direct sums are sums}\right)  ,
\]
so that%
\[
\pi\left(  \otimes V\right)  =\pi\left(  \sum\limits_{n\in\mathbb{N}%
}V^{\otimes n}\right)  =\sum\limits_{n\in\mathbb{N}}\pi\left(  V^{\otimes
n}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi\text{ is
linear}\right)  .
\]
Since $\pi\left(  \otimes V\right)  =A$ (because $\pi$ is surjective), this
becomes $A=\sum\limits_{n\in\mathbb{N}}\pi\left(  V^{\otimes n}\right)  $.
Thus,%
\[
M\cdot A=M\cdot\sum\limits_{n\in\mathbb{N}}\pi\left(  V^{\otimes n}\right)
=\sum\limits_{n\in\mathbb{N}}\underbrace{M\cdot\pi\left(  V^{\otimes
n}\right)  }_{\subseteq M\text{ (by (\ref{pf.ideal.1}))}}\subseteq
\sum\limits_{n\in\mathbb{N}}M\subseteq M
\]
(since $M$ is a $k$-module). In other words, $M$ is a right ideal of $A$. This
proves Lemma \ref{lem.ideal}.
\end{proof}

\begin{corollary}
\label{cor.exter.W}Let $k$ be a commutative ring. Let $V$ be a $k$-module, and
let $W$ be a $k$-submodule of $V$. Then,%
\[
\left(  \operatorname*{Exter} V\right)  \cdot\operatorname*{exter}%
\nolimits_{V}\left(  W\right)  \cdot\left(  \operatorname*{Exter} V\right)
=\left(  \operatorname*{Exter} V\right)  \cdot\operatorname*{exter}%
\nolimits_{V}\left(  W\right)  =\operatorname*{exter}\nolimits_{V}\left(
W\right)  \cdot\left(  \operatorname*{Exter} V\right)  .
\]
Here, $W$ is considered a $k$-submodule of $\otimes V$ by means of the
inclusion $W\subseteq V=V^{\otimes1}\subseteq\otimes V$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.exter.W}.]\textbf{(i)} We have $V\cdot
W+Q_{2}\left(  V\right)  =W\cdot V+Q_{2}\left(  V\right)  $ (as $k$-submodules
of $\otimes V$).

\textit{Proof.} Let $\left(  v,w\right)  \in V\times W$ be arbitrary. Then,%
\begin{align*}
&  \underbrace{v\cdot w}_{\substack{=v\otimes w\\\text{(by (\ref{(X)V=sum_j}),
applied to}\\a=v\text{, }b=w\text{, }n=1\text{, }m=1\text{)}}%
}+\underbrace{w\cdot v}_{\substack{=w\otimes v\\\text{(by (\ref{(X)V=sum_j}),
applied to}\\a=w\text{, }b=v\text{, }n=1\text{, }m=1\text{)}}}\\
&  =v\otimes w+w\otimes v\in\left\{  v_{1}\otimes v_{2}+v_{2}\otimes
v_{1}\ \mid\ \left(  v_{1},v_{2}\right)  \in V^{2}\right\} \\
&  \subseteq\left\langle \left\{  v_{1}\otimes v_{2}+v_{2}\otimes v_{1}%
\ \mid\ \left(  v_{1},v_{2}\right)  \in V^{2}\right\}  \right\rangle
=\left\langle v_{1}\otimes v_{2}+v_{2}\otimes v_{1}\ \mid\ \left(  v_{1}%
,v_{2}\right)  \in V^{2}\right\rangle =Q_{2}\left(  V\right)
\end{align*}
(by Corollary \ref{coro.Q_2}), so that%
\[
v\cdot w\in Q_{2}\left(  V\right)  -\underbrace{w}_{\in W}\cdot\underbrace{v}%
_{\in V}=Q_{2}\left(  V\right)  -W\cdot V=Q_{2}\left(  V\right)  +W\cdot
V\ \ \ \ \ \ \ \ \ \ \left(  \text{since }W\cdot V\text{ is a }k\text{-module}%
\right)  .
\]
Since this holds for all $\left(  v,w\right)  \in V\times W$, we thus have%
\[
\left\{  v\cdot w\ \mid\ \left(  v,w\right)  \in V\times W\right\}  \subseteq
Q_{2}\left(  V\right)  +W\cdot V.
\]
Applying Proposition \ref{prop.<>} \textbf{(a)} to $\otimes V$, $\left\{
v\cdot w\ \mid\ \left(  v,w\right)  \in V\times W\right\}  $ and $Q_{2}\left(
V\right)  +W\cdot V$ instead of $M$, $S$ and $Q$, we see that this yields%
\[
\left\langle \left\{  v\cdot w\ \mid\ \left(  v,w\right)  \in V\times
W\right\}  \right\rangle \subseteq Q_{2}\left(  V\right)  +W\cdot V.
\]
Since%
\[
\left\langle \left\{  v\cdot w\ \mid\ \left(  v,w\right)  \in V\times
W\right\}  \right\rangle =\left\langle v\cdot w\ \mid\ \left(  v,w\right)  \in
V\times W\right\rangle =V\cdot W,
\]
this rewrites as $V\cdot W\subseteq Q_{2}\left(  V\right)  +W\cdot V$. Thus,%
\begin{align*}
V\cdot W+Q_{2}\left(  V\right)   &  \subseteq\left(  Q_{2}\left(  V\right)
+W\cdot V\right)  +Q_{2}\left(  V\right)  =W\cdot V+\underbrace{Q_{2}\left(
V\right)  +Q_{2}\left(  V\right)  }_{\substack{\subseteq Q_{2}\left(
V\right)  \\\text{(since }Q_{2}\left(  V\right)  \text{ is a }k\text{-module)}%
}}\\
&  \subseteq W\cdot V+Q_{2}\left(  V\right)  .
\end{align*}
Combining this with the fact that $W\cdot V+Q_{2}\left(  V\right)  \subseteq
V\cdot W+Q_{2}\left(  V\right)  $ (which can be proven completely
analogously), we obtain that $V\cdot W+Q_{2}\left(  V\right)  =W\cdot
V+Q_{2}\left(  V\right)  $. This proves \textbf{(i)}.

\textbf{(ii)} We have $\operatorname*{exter}\nolimits_{V}\left(  V\right)
\cdot\operatorname*{exter}\nolimits_{V}\left(  W\right)
=\operatorname*{exter}\nolimits_{V}\left(  W\right)  \cdot
\operatorname*{exter}\nolimits_{V}\left(  V\right)  $ (as $k$-submodules of
$\operatorname*{Exter}V$).

\textit{Proof.} Since $\operatorname*{exter}\nolimits_{V}$ is a $k$-algebra
homomorphism, we have%
\[
\operatorname*{exter}\nolimits_{V}\left(  V\cdot W+Q_{2}\left(  V\right)
\right)  =\operatorname*{exter}\nolimits_{V}\left(  V\right)  \cdot
\operatorname*{exter}\nolimits_{V}\left(  W\right)  +\operatorname*{exter}%
\nolimits_{V}\left(  Q_{2}\left(  V\right)  \right)  .
\]
But since $\operatorname*{exter}\nolimits_{V}\left(  Q_{2}\left(  V\right)
\right)  =0$ (because $\operatorname*{exter}\nolimits_{V}$ is the canonical
projection $\otimes V\rightarrow\left(  \otimes V\right)  \diagup Q\left(
V\right)  $, and thus $Q\left(  V\right)  =\operatorname*{Ker}%
\operatorname*{exter}\nolimits_{V}$, so that $Q_{2}\left(  V\right)
\subseteq\bigoplus\limits_{n\in\mathbb{N}}Q_{n}\left(  V\right)  =Q\left(
V\right)  =\operatorname*{Ker}\operatorname*{exter}\nolimits_{V}$), this
rewrites as%
\[
\operatorname*{exter}\nolimits_{V}\left(  V\cdot W+Q_{2}\left(  V\right)
\right)  =\operatorname*{exter}\nolimits_{V}\left(  V\right)  \cdot
\operatorname*{exter}\nolimits_{V}\left(  W\right)  +0=\operatorname*{exter}%
\nolimits_{V}\left(  V\right)  \cdot\operatorname*{exter}\nolimits_{V}\left(
W\right)  .
\]
Similarly,%
\[
\operatorname*{exter}\nolimits_{V}\left(  W\cdot V+Q_{2}\left(  V\right)
\right)  =\operatorname*{exter}\nolimits_{V}\left(  W\right)  \cdot
\operatorname*{exter}\nolimits_{V}\left(  V\right)  .
\]
Now,%
\begin{align*}
\operatorname*{exter}\nolimits_{V}\left(  V\right)  \cdot\operatorname*{exter}%
\nolimits_{V}\left(  W\right)   &  =\operatorname*{exter}\nolimits_{V}\left(
V\cdot W+Q_{2}\left(  V\right)  \right)  =\operatorname*{exter}\nolimits_{V}%
\left(  W\cdot V+Q_{2}\left(  V\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }V\cdot W+Q_{2}\left(  V\right)
=W\cdot V+Q_{2}\left(  V\right)  \text{ by \textbf{(i)}}\right) \\
&  =\operatorname*{exter}\nolimits_{V}\left(  W\right)  \cdot
\operatorname*{exter}\nolimits_{V}\left(  V\right)  ,
\end{align*}
and thus \textbf{(ii)} is proven.

\textbf{(iii)} We have $\left(  \operatorname*{Exter}V\right)  \cdot
\operatorname*{exter}\nolimits_{V}\left(  W\right)  =\left(
\operatorname*{Exter}V\right)  \cdot\operatorname*{exter}\nolimits_{V}\left(
W\right)  \cdot\left(  \operatorname*{Exter}V\right)  $.

\textit{Proof.} We have%
\[
\left(  \operatorname*{Exter}V\right)  \cdot\underbrace{\operatorname*{exter}%
\nolimits_{V}\left(  W\right)  \cdot\operatorname*{exter}\nolimits_{V}\left(
V\right)  }_{=\operatorname*{exter}\nolimits_{V}\left(  V\right)
\cdot\operatorname*{exter}\nolimits_{V}\left(  W\right)  }=\underbrace{\left(
\operatorname*{Exter}V\right)  \cdot\operatorname*{exter}\nolimits_{V}\left(
V\right)  }_{\subseteq\operatorname*{Exter}V}\cdot\operatorname*{exter}%
\nolimits_{V}\left(  W\right)  \subseteq\left(  \operatorname*{Exter}V\right)
\cdot\operatorname*{exter}\nolimits_{V}\left(  W\right)  .
\]
By Lemma \ref{lem.ideal} (applied to $A=\operatorname*{Exter}V$,
$\pi=\operatorname*{exter}\nolimits_{V}$ and $M=\left(  \operatorname*{Exter}%
V\right)  \cdot\operatorname*{exter}\nolimits_{V}\left(  W\right)  $), this
yields that $\left(  \operatorname*{Exter}V\right)  \cdot\operatorname*{exter}%
\nolimits_{V}\left(  W\right)  $ is a right ideal of $\operatorname*{Exter}V$.
In other words, $\left(  \operatorname*{Exter}V\right)  \cdot
\operatorname*{exter}\nolimits_{V}\left(  W\right)  =\left(
\operatorname*{Exter}V\right)  \cdot\operatorname*{exter}\nolimits_{V}\left(
W\right)  \cdot\left(  \operatorname*{Exter}V\right)  $. This proves
\textbf{(iii)}.

\textbf{(iv)} We have $\operatorname*{exter}\nolimits_{V}\left(  W\right)
\cdot\left(  \operatorname*{Exter}V\right)  =\left(  \operatorname*{Exter}%
V\right)  \cdot\operatorname*{exter}\nolimits_{V}\left(  W\right)
\cdot\left(  \operatorname*{Exter}V\right)  $.

\textit{Proof.} The proof of \textbf{(iv)} is analogous to the proof of
\textbf{(iii)} (but this time we need an analogue of Lemma \ref{lem.ideal} for
left instead of right ideals).

\textbf{(v)} Corollary \ref{cor.exter.W} clearly follows by combining
\textbf{(iii)} and \textbf{(iv)}. The proof of Corollary \ref{cor.exter.W} is
thus complete.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.Ker(exter)}.]Applying the commutative diagram
(\ref{def.ext-functor.diag}) to $W=V^{\prime}$, we obtain the commutative
diagram%
\begin{equation}%
%TCIMACRO{\TeXButton{exter f from otimes f}{\xymatrix{
%\otimes V \ar[r]^{\otimes f}\ar[d]_{\operatorname*{exter}_V} & \otimes
%V^{\prime} \ar[d]^{\operatorname*{exter}_{V^{\prime}}} \\
%\operatorname*{Exter}  V \ar[r]_{\operatorname*{Exter}  f} & \operatorname
%*{Exter}  V^{\prime}
%}}}%
%BeginExpansion
\xymatrix{
\otimes V \ar[r]^{\otimes f}\ar[d]_{\operatorname*{exter}_V} & \otimes
V^{\prime} \ar[d]^{\operatorname*{exter}_{V^{\prime}}} \\
\operatorname*{Exter}  V \ar[r]_{\operatorname*{Exter}  f} & \operatorname
*{Exter}  V^{\prime}
}%
%EndExpansion
. \label{pf.Ker(exter).1}%
\end{equation}
But it is easy to see that%
\[
\operatorname*{Ker}\operatorname*{exter}\nolimits_{V^{\prime}}\subseteq\left(
\otimes f\right)  \left(  \operatorname*{Ker}\operatorname*{exter}%
\nolimits_{V}\right)
\]
\footnote{\textit{Proof.} Since $\operatorname*{exter}\nolimits_{V}$ is the
canonical projection $\otimes V\rightarrow\left(  \otimes V\right)  \diagup
Q\left(  V\right)  $, we have $\operatorname*{Ker}\operatorname*{exter}%
\nolimits_{V}=Q\left(  V\right)  $. Similarly, $\operatorname*{Ker}%
\operatorname*{exter}\nolimits_{V^{\prime}}=Q\left(  V^{\prime}\right)  $. But
Lemma \ref{lem.Q(f)} \textbf{(b)} (applied to $W=V^{\prime}$) yields that
$\left(  \otimes f\right)  \left(  Q\left(  V\right)  \right)  =Q\left(
V^{\prime}\right)  $. Thus,%
\[
\operatorname*{Ker}\operatorname*{exter}\nolimits_{V^{\prime}}=Q\left(
V^{\prime}\right)  =\left(  \otimes f\right)  \left(  \underbrace{Q\left(
V\right)  }_{=\operatorname*{Ker}\operatorname*{exter}\nolimits_{V}}\right)
=\left(  \otimes f\right)  \left(  \operatorname*{Ker}\operatorname*{exter}%
\nolimits_{V}\right)  ,
\]
qed.} and that the map $\operatorname*{exter}\nolimits_{V}$ is surjective
(since $\operatorname*{exter}\nolimits_{V}$ is the canonical projection
$\otimes V\rightarrow\operatorname*{Exter}V$). Hence, we can apply Proposition
\ref{prop.9l} to the commutative diagram (\ref{pf.Ker(exter).1}), and conclude
that $\operatorname*{Ker}\left(  \operatorname*{Exter}f\right)
=\operatorname*{exter}\nolimits_{V}\left(  \operatorname*{Ker}\left(  \otimes
f\right)  \right)  $. Since $\operatorname*{Ker}\left(  \otimes f\right)
=\left(  \otimes V\right)  \cdot\left(  \operatorname*{Ker}f\right)
\cdot\left(  \otimes V\right)  $ by Theorem \ref{thm.(X)f}, this becomes%
\begin{align*}
\operatorname*{Ker}\left(  \operatorname*{Exter}f\right)   &
=\operatorname*{exter}\nolimits_{V}\left(  \left(  \otimes V\right)
\cdot\left(  \operatorname*{Ker}f\right)  \cdot\left(  \otimes V\right)
\right)  =\operatorname*{exter}\nolimits_{V}\left(  \otimes V\right)
\cdot\operatorname*{exter}\nolimits_{V}\left(  \operatorname*{Ker}f\right)
\cdot\operatorname*{exter}\nolimits_{V}\left(  \otimes V\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname*{exter}%
\nolimits_{V}\text{ is a }k\text{-algebra homomorphism}\right)  .
\end{align*}
Since $\operatorname*{exter}\nolimits_{V}\left(  \otimes V\right)
=\operatorname*{Exter}V$ (because $\operatorname*{exter}\nolimits_{V}$ is
surjective), this becomes%
\[
\operatorname*{Ker}\left(  \operatorname*{Exter}f\right)  =\left(
\operatorname*{Exter}V\right)  \cdot\operatorname*{exter}\nolimits_{V}\left(
\operatorname*{Ker}f\right)  \cdot\left(  \operatorname*{Exter}V\right)  .
\]
Combined with the equality%
\[
\left(  \operatorname*{Exter}V\right)  \cdot\operatorname*{exter}%
\nolimits_{V}\left(  \operatorname*{Ker}f\right)  \cdot\left(
\operatorname*{Exter}V\right)  =\left(  \operatorname*{Exter}V\right)
\cdot\operatorname*{exter}\nolimits_{V}\left(  \operatorname*{Ker}f\right)
=\operatorname*{exter}\nolimits_{V}\left(  \operatorname*{Ker}f\right)
\cdot\left(  \operatorname*{Exter}V\right)
\]
(which follows from Corollary \ref{cor.exter.W}, applied to
$W=\operatorname*{Ker}f$), this yields%
\begin{align*}
\operatorname*{Ker}\left(  \operatorname*{Exter}f\right)   &  =\left(
\operatorname*{Exter}V\right)  \cdot\operatorname*{exter}\nolimits_{V}\left(
\operatorname*{Ker}f\right)  \cdot\left(  \operatorname*{Exter}V\right)
=\left(  \operatorname*{Exter}V\right)  \cdot\operatorname*{exter}%
\nolimits_{V}\left(  \operatorname*{Ker}f\right) \\
&  =\operatorname*{exter}\nolimits_{V}\left(  \operatorname*{Ker}f\right)
\cdot\left(  \operatorname*{Exter}V\right)  .
\end{align*}
This proves Theorem \ref{thm.Ker(exter)}.
\end{proof}

Here is a way to rewrite Theorem \ref{thm.Ker(exter)}:

\begin{corollary}
\label{coro.Ker(exter)}Let $k$ be a commutative ring. Let $V$ be a $k$-module.
Let $W$ be a $k$-submodule of $V$, and let $f:V\rightarrow V\diagup W$ be the
canonical projection.\newline\textbf{(a)} Then, the kernel of the map
$\operatorname*{Exter}f:\operatorname*{Exter}V\rightarrow\operatorname*{Exter}%
\left(  V\diagup W\right)  $ is%
\begin{align*}
\operatorname*{Ker}\left(  \operatorname*{Exter}f\right)   &  =\left(
\operatorname*{Exter}V\right)  \cdot\operatorname*{exter}\nolimits_{V}\left(
W\right)  \cdot\left(  \operatorname*{Exter}V\right)  =\left(
\operatorname*{Exter}V\right)  \cdot\operatorname*{exter}\nolimits_{V}\left(
W\right) \\
&  =\operatorname*{exter}\nolimits_{V}\left(  W\right)  \cdot\left(
\operatorname*{Exter}V\right)  .
\end{align*}
Here, $W$ is considered a $k$-submodule of $\otimes V$ by means of the
inclusion $W\subseteq V=V^{\otimes1}\subseteq\otimes V$.\newline\textbf{(b)}
We have%
\[
\left(  \operatorname*{Exter}V\right)  \diagup\left(  \left(
\operatorname*{Exter}V\right)  \cdot\operatorname*{exter}\nolimits_{V}\left(
W\right)  \right)  \cong\operatorname*{Exter}\left(  V\diagup W\right)
\ \ \ \ \ \ \ \ \ \ \text{as }k\text{-modules.}%
\]

\end{corollary}

\begin{proof}
[Proof of Corollary \ref{coro.Ker(exter)}.]Since $f$ is the canonical
projection $V\rightarrow V\diagup W$, it is clear that $f$ is surjective and
that $\operatorname*{Ker}f=W$. Now, we can apply Theorem \ref{thm.Ker(exter)}
to $V^{\prime}=V\diagup W$ and conclude that%
\begin{align*}
\operatorname*{Ker}\left(  \operatorname*{Exter}f\right)   &  =\left(
\operatorname*{Exter}V\right)  \cdot\operatorname*{exter}\nolimits_{V}\left(
\operatorname*{Ker}f\right)  \cdot\left(  \operatorname*{Exter}V\right)
=\left(  \operatorname*{Exter}V\right)  \cdot\operatorname*{exter}%
\nolimits_{V}\left(  \operatorname*{Ker}f\right) \\
&  =\operatorname*{exter}\nolimits_{V}\left(  \operatorname*{Ker}f\right)
\cdot\left(  \operatorname*{Exter}V\right)  .
\end{align*}
Since $\operatorname*{Ker}f=W$, this simplifies to%
\begin{align*}
\operatorname*{Ker}\left(  \operatorname*{Exter}f\right)   &  =\left(
\operatorname*{Exter}V\right)  \cdot\operatorname*{exter}\nolimits_{V}\left(
W\right)  \cdot\left(  \operatorname*{Exter}V\right)  =\left(
\operatorname*{Exter}V\right)  \cdot\operatorname*{exter}\nolimits_{V}\left(
W\right) \\
&  =\operatorname*{exter}\nolimits_{V}\left(  W\right)  \cdot\left(
\operatorname*{Exter}V\right)  .
\end{align*}
This proves Corollary \ref{coro.Ker(exter)} \textbf{(a)}.

Since the map $f:V\rightarrow V\diagup W$ is surjective, the map
$\operatorname*{Exter}f:\operatorname*{Exter}V\rightarrow\operatorname*{Exter}%
\left(  V\diagup W\right)  $ is also surjective (by Proposition
\ref{prop.ext-surj} \textbf{(c)}, applied to $V\diagup W$ instead of $W$), and
thus we have $\left(  \operatorname*{Exter}f\right)  \left(
\operatorname*{Exter}V\right)  =\operatorname*{Exter}\left(  V\diagup
W\right)  $. But by the isomorphism theorem, $\left(  \operatorname*{Exter}%
f\right)  \left(  \operatorname*{Exter}V\right)  \cong\left(
\operatorname*{Exter}V\right)  \diagup\operatorname*{Ker}\left(
\operatorname*{Exter}f\right)  $ as $k$-modules. Thus,%
\begin{align*}
\operatorname*{Exter}\left(  V\diagup W\right)   &  =\left(
\operatorname*{Exter}f\right)  \left(  \operatorname*{Exter}V\right)
\cong\left(  \operatorname*{Exter}V\right)  \diagup
\underbrace{\operatorname*{Ker}\left(  \operatorname*{Exter}f\right)
}_{=\left(  \operatorname*{Exter}V\right)  \cdot\operatorname*{exter}%
\nolimits_{V}\left(  W\right)  }\\
&  =\left(  \operatorname*{Exter}V\right)  \diagup\left(  \left(
\operatorname*{Exter}V\right)  \cdot\operatorname*{exter}\nolimits_{V}\left(
W\right)  \right)  \ \ \ \ \ \ \ \ \ \ \text{as }k\text{-modules.}%
\end{align*}
This proves Corollary \ref{coro.Ker(exter)} \textbf{(b)}.
\end{proof}

\subsection{\label{subsect.Sym}The symmetric algebra}

In the previous two subsections (Subsections \ref{subsect.exter} and
\ref{subsect.Ker(exter)}), we have studied the pseudoexterior algebra
$\operatorname*{Exter} V$ of a $k$-module $V$. Many properties of the
pseudoexterior algebra $\operatorname*{Exter} V$ are shared by its more
well-known analogue - the symmetric algebra $\operatorname*{Sym}V$. Pretty
much all of our above-proven properties of $\operatorname*{Exter} V$ have
analogues for $\operatorname*{Sym}V$. We are now going to formulate these
analogues, without proving them (because their proofs are completely analogous
to the proofs of the properties of $\operatorname*{Exter} V$ that we did
above). First, before we define the symmetric algebra $\operatorname*{Sym}V$,
let us define the symmetric powers $\operatorname*{Sym}^{n}V$:

\begin{definition}
\label{defs.sym-pow}Let $k$ be a commutative ring. Let $V$ be a $k$-module.
Let $n\in\mathbb{N}$.\newline Let $K_{n}\left(  V\right)  $ be the
$k$-submodule%
\[
\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-v_{\sigma\left(
1\right)  }\otimes v_{\sigma\left(  2\right)  }\otimes\cdots\otimes
v_{\sigma\left(  n\right)  }\ \mid\ \left(  \left(  v_{1},v_{2},\ldots
,v_{n}\right)  ,\sigma\right)  \in V^{n}\times S_{n}\right\rangle
\]
of the $k$-module $V^{\otimes n}$ (where we are using Convention
\ref{conv.<>}, and are denoting the $n$-th symmetric group by $S_{n}%
$).\newline The factor $k$-module $V^{\otimes n}\diagup K_{n}\left(  V\right)
$ is called the $n$\textit{-th symmetric power} of the $k$-module $V$ and will
be denoted by $\operatorname*{Sym}^{n}V$. We denote by $\operatorname*{sym}%
\nolimits_{V,n}$ the canonical projection $V^{\otimes n}\rightarrow V^{\otimes
n}\diagup K_{n}\left(  V\right)  =\operatorname*{Sym}\nolimits^{n}V$. Clearly,
this map $\operatorname*{sym}\nolimits_{V,n}$ is a surjective $k$-module homomorphism.
\end{definition}

We should understand these notions $K_{n}\left(  V\right)  $,
$\operatorname*{Sym}\nolimits^{n}V$ and $\operatorname*{sym}\nolimits_{V,n}$
as analogues of the notions $Q_{n}\left(  V\right)  $, $\operatorname*{Exter}
^{n}V$ and $\operatorname*{exter}\nolimits_{V,n}$ from Definition
\ref{defs.ext-pow}, respectively. Here is an analogue of Proposition
\ref{prop.Q_n}:

\begin{proposition}
\label{prop.K_n}Let $k$ be a commutative ring. Let $V$ be a $k$-module. Let
$n\in\mathbb{N}$.\newline Then,%
\[
K_{n}\left(  V\right)  =\sum_{i=1}^{n-1}\left\langle v_{1}\otimes v_{2}%
\otimes\cdots\otimes v_{n}-v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau
_{i}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)
}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\rangle ,
\]
where $\tau_{i}$ denotes the transposition $\left(  i,i+1\right)  \in S_{n}$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.K_n}.]The proof of this Proposition
\ref{prop.K_n} is completely analogous to the proof of Proposition
\ref{prop.Q_n} (up to some replacing of $+$ signs by $-$ signs and some
removal of powers of $-1$) and can be found in \S 5.1 of the long (detailed)
version of \cite{dg0}.
\end{proof}

Here is the analogue of Corollary \ref{coro.Q_2}:

\begin{corollary}
\label{coro.K_2}Let $k$ be a commutative ring. Let $V$ be a $k$-module. Then,%
\[
K_{2}\left(  V\right)  =\left\langle v_{1}\otimes v_{2}-v_{2}\otimes
v_{1}\ \mid\ \left(  v_{1},v_{2}\right)  \in V^{2}\right\rangle .
\]

\end{corollary}

\begin{proof}
[Proof of Corollary \ref{coro.K_2}.]Again, the proof of Corollary
\ref{coro.K_2} is completely analogous to the proof of Corollary
\ref{coro.Q_2}.
\end{proof}

Next, the analogue of Lemma \ref{lem.Q_n}:

\begin{lemma}
\label{lem.K_n}Let $k$ be a commutative ring. Let $V$ be a $k$-module. Let
$n\in\mathbb{N}$. Let $i\in\left\{  1,2,\ldots,n-1\right\}  $.\newline Then,%
\begin{align*}
&  \left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-v_{\tau
_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }\otimes
\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{n}\right\rangle \\
&  =V^{\otimes\left(  i-1\right)  }\cdot\left(  K_{2}\left(  V\right)
\right)  \cdot V^{\otimes\left(  n-1-i\right)  },
\end{align*}
where $\tau_{i}$ denotes the transposition $\left(  i,i+1\right)  \in S_{n}$.
Here, we consider $V^{\otimes n}$ as a $k$-submodule of $\otimes V$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.K_n}.]The proof of Lemma \ref{lem.K_n} is completely
analogous to the proof of Lemma \ref{lem.Q_n}.
\end{proof}

Next, the analogue of Corollary \ref{cor.Q_n}:

\begin{corollary}
\label{cor.K_n}Let $k$ be a commutative ring. Let $V$ be a $k$-module. Let
$n\in\mathbb{N}$.\newline Then,%
\[
K_{n}\left(  V\right)  =\sum_{i=1}^{n-1}V^{\otimes\left(  i-1\right)  }%
\cdot\left(  K_{2}\left(  V\right)  \right)  \cdot V^{\otimes\left(
n-1-i\right)  }%
\]
(this is an equality between $k$-submodules of $\otimes V$, where
$K_{n}\left(  V\right)  $ becomes such a $k$-submodule by means of the
inclusion $K_{n}\left(  V\right)  \subseteq V^{\otimes n}\subseteq\otimes V$).
Here, the multiplication on the right hand side is multiplication inside the
$k$-algebra $\otimes V$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.K_n}.]The proof of Corollary \ref{cor.K_n} is
completely analogous to the proof of Corollary \ref{cor.Q_n}.
\end{proof}

Now the analogue of Theorem \ref{thm.Q}:

\begin{theorem}
\label{thm.K}Let $k$ be a commutative ring. Let $V$ be a $k$-module. We know
that $K_{n}\left(  V\right)  $ is a $k$-submodule of $V^{\otimes n}$ for every
$n\in\mathbb{N}$. Thus, $\bigoplus\limits_{n\in\mathbb{N}}K_{n}\left(
V\right)  $ is a $k$-submodule of $\bigoplus\limits_{n\in\mathbb{N}}V^{\otimes
n}=\otimes V$. This $k$-submodule satisfies%
\[
\bigoplus\limits_{n\in\mathbb{N}}K_{n}\left(  V\right)  =\left(  \otimes
V\right)  \cdot\left(  K_{2}\left(  V\right)  \right)  \cdot\left(  \otimes
V\right)  .
\]

\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.K}.]The proof of Theorem \ref{thm.K} is completely
analogous to the proof of Theorem \ref{thm.Q}.
\end{proof}

Now we can finally define the symmetric algebra, similarly to Definition
\ref{defs.ext-alg}:

\begin{definition}
\label{defs.sym-alg}Let $k$ be a commutative ring. Let $V$ be a $k$%
-module.\newline By Theorem \ref{thm.K}, the two $k$-submodules $\bigoplus
\limits_{n\in\mathbb{N}}K_{n}\left(  V\right)  $ and $\left(  \otimes
V\right)  \cdot\left(  K_{2}\left(  V\right)  \right)  \cdot\left(  \otimes
V\right)  $ of $\otimes V$ are identic (where $\bigoplus\limits_{n\in
\mathbb{N}}K_{n}\left(  V\right)  $ becomes a $k$-submodule of $\otimes V$ in
the same way as explained in Theorem \ref{thm.K}). We denote these two identic
$k$-submodules by $K\left(  V\right)  $. In other words, we define $K\left(
V\right)  $ by%
\[
K\left(  V\right)  =\bigoplus\limits_{n\in\mathbb{N}}K_{n}\left(  V\right)
=\left(  \otimes V\right)  \cdot\left(  K_{2}\left(  V\right)  \right)
\cdot\left(  \otimes V\right)  .
\]
Since $K\left(  V\right)  =\left(  \otimes V\right)  \cdot\left(  K_{2}\left(
V\right)  \right)  \cdot\left(  \otimes V\right)  $, it is clear that
$K\left(  V\right)  $ is a two-sided ideal of the $k$-algebra $\otimes
V$.\newline Now we define a $k$-module $\operatorname*{Sym}V$ as the direct
sum $\bigoplus\limits_{n\in\mathbb{N}}\operatorname*{Sym}\nolimits^{n}V$.
Then,%
\begin{align*}
\operatorname*{Sym}V  &  =\bigoplus\limits_{n\in\mathbb{N}}%
\underbrace{\operatorname*{Sym}\nolimits^{n}V}_{=V^{\otimes n}\diagup
K_{n}\left(  V\right)  }=\bigoplus\limits_{n\in\mathbb{N}}\left(  V^{\otimes
n}\diagup K_{n}\left(  V\right)  \right)  \cong\underbrace{\left(
\bigoplus\limits_{n\in\mathbb{N}}V^{\otimes n}\right)  }_{=\otimes V}%
\diagup\underbrace{\left(  \bigoplus\limits_{n\in\mathbb{N}}K_{n}\left(
V\right)  \right)  }_{=K\left(  V\right)  }\\
&  =\left(  \otimes V\right)  \diagup K\left(  V\right)  .
\end{align*}
This is a canonical isomorphism, so we will use it to identify
$\operatorname*{Sym}V$ with $\left(  \otimes V\right)  \diagup K\left(
V\right)  $. Since $K\left(  V\right)  $ is a two-sided ideal of the
$k$-algebra $\otimes V$, the quotient $k$-module $\left(  \otimes V\right)
\diagup K\left(  V\right)  $ canonically becomes a $k$-algebra. Since
$\operatorname*{Sym}V=\left(  \otimes V\right)  \diagup K\left(  V\right)  $,
this means that $\operatorname*{Sym}V$ becomes a $k$-algebra. We refer to this
$k$-algebra as the \textit{symmetric algebra} of the $k$-module $V$.\newline
We denote by $\operatorname*{sym}\nolimits_{V}$ the canonical projection
$\otimes V\rightarrow\left(  \otimes V\right)  \diagup K\left(  V\right)
=\operatorname*{Sym}V$. Clearly, this map $\operatorname*{sym}\nolimits_{V}$
is a surjective $k$-algebra homomorphism. Besides, due to $\otimes
V=\bigoplus\limits_{n\in\mathbb{N}}V^{\otimes n}$ and $K\left(  V\right)
=\bigoplus\limits_{n\in\mathbb{N}}K_{n}\left(  V\right)  $, it is clear that
the canonical projection $\otimes V\rightarrow\left(  \otimes V\right)
\diagup K\left(  V\right)  $ is the direct sum of the canonical projections
$V^{\otimes n}\rightarrow V^{\otimes n}\diagup K_{n}\left(  V\right)  $ over
all $n\in\mathbb{N}$. Since the canonical projection $\otimes V\rightarrow
\left(  \otimes V\right)  \diagup K\left(  V\right)  $ is the map
$\operatorname*{sym}\nolimits_{V}$, whereas the canonical projection
$V^{\otimes n}\rightarrow V^{\otimes n}\diagup K_{n}\left(  V\right)  $ is the
map $\operatorname*{sym}\nolimits_{V,n}$, this rewrites as follows: The map
$\operatorname*{sym}\nolimits_{V}$ is the direct sum of the maps
$\operatorname*{sym}\nolimits_{V,n}$ over all $n\in\mathbb{N}$.\newline When
$v_{1}$, $v_{2}$, $\ldots$, $v_{n}$ are some elements of $V$, one often
abbreviates the element $\operatorname*{sym}\nolimits_{V}\left(  v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}\right)  $ of $\operatorname*{Sym}V$ by
$v_{1}v_{2}\cdots v_{n}$. (We will not use this abbreviation in this following.)
\end{definition}

We should think of the notions $K\left(  V\right)  $, $\operatorname*{Sym}V$
and $\operatorname*{sym}\nolimits_{V}$ as analogues of the notions $Q\left(
V\right)  $, $\operatorname*{Exter} V$ and $\operatorname*{exter}%
\nolimits_{V}$ from Definition \ref{defs.ext-alg}, respectively. The next
result provides an analogue of Lemma \ref{lem.Q(f)}:

\begin{lemma}
\label{lem.K(f)}Let $k$ be a commutative ring. Let $V$ and $W$ be two
$k$-modules. Let $f:V\rightarrow W$ be a $k$-module homomorphism.\newline%
\textbf{(a)} Then, the $k$-algebra homomorphism $\otimes f:\otimes
V\rightarrow\otimes W$ satisfies $\left(  \otimes f\right)  \left(  K\left(
V\right)  \right)  \subseteq K\left(  W\right)  $. Also, for every
$n\in\mathbb{N}$, the $k$-module homomorphism $f^{\otimes n}:V^{\otimes
n}\rightarrow W^{\otimes n}$ satisfies $f^{\otimes n}\left(  K_{n}\left(
V\right)  \right)  \subseteq K_{n}\left(  W\right)  $.\newline\textbf{(b)}
Assume that $f$ is surjective. Then, the $k$-algebra homomorphism $\otimes
f:\otimes V\rightarrow\otimes W$ satisfies $\left(  \otimes f\right)  \left(
K\left(  V\right)  \right)  =K\left(  W\right)  $. Also, for every
$n\in\mathbb{N}$, the $k$-module homomorphism $f^{\otimes n}:V^{\otimes
n}\rightarrow W^{\otimes n}$ satisfies $f^{\otimes n}\left(  K_{n}\left(
V\right)  \right)  =K_{n}\left(  W\right)  $.
\end{lemma}

The following definition mirrors Definition \ref{def.ext-functor}:

\begin{definition}
\label{def.sym-functor}Let $k$ be a commutative ring. Let $V$ and $W$ be two
$k$-modules. Let $f:V\rightarrow W$ be a $k$-module homomorphism. Then, the
$k$-algebra homomorphism $\otimes f:\otimes V\rightarrow\otimes W$ satisfies
$\left(  \otimes f\right)  \left(  K\left(  V\right)  \right)  \subseteq
K\left(  W\right)  $ (by Lemma \ref{lem.K(f)} \textbf{(a)}), and thus gives
rise to a $k$-algebra homomorphism $\left(  \otimes V\right)  \diagup K\left(
V\right)  \rightarrow\left(  \otimes W\right)  \diagup K\left(  W\right)  $.
This latter $k$-algebra homomorphism will be denoted by $\operatorname*{Sym}%
f$. Since $\left(  \otimes V\right)  \diagup K\left(  V\right)
=\operatorname*{Sym}V$ and $\left(  \otimes W\right)  \diagup K\left(
W\right)  =\operatorname*{Sym}W$, this homomorphism $\operatorname*{Sym}
f:\left(  \otimes V\right)  \diagup K\left(  V\right)  \rightarrow\left(
\otimes W\right)  \diagup K\left(  W\right)  $ is actually a homomorphism from
$\operatorname*{Sym}V$ to $\operatorname*{Sym}W$.\newline By the construction
of $\operatorname*{Sym}f$, the diagram%
\begin{equation}%
%TCIMACRO{\TeXButton{sym f from otimes f}{\xymatrix{
%\otimes V \ar[r]^{\otimes f}\ar[d]_{\operatorname*{sym}_V} & \otimes
%W \ar[d]^{\operatorname*{sym}_W} \\
%\operatorname*{Sym} V \ar[r]_{\operatorname*{Sym} f} & \operatorname*{Sym} W
%}} }%
%BeginExpansion
\xymatrix{
\otimes V \ar[r]^{\otimes f}\ar[d]_{\operatorname*{sym}_V} & \otimes
W \ar[d]^{\operatorname*{sym}_W} \\
\operatorname*{Sym} V \ar[r]_{\operatorname*{Sym} f} & \operatorname*{Sym} W
}
%EndExpansion
\label{def.sym-functor.diag}%
\end{equation}
commutes (since $\operatorname*{sym}\nolimits_{V}$ is the canonical projection
$\otimes V\rightarrow\operatorname*{Sym}V$ and since $\operatorname*{sym}%
\nolimits_{W}$ is the canonical projection $\otimes W\rightarrow
\operatorname*{Sym}W$).
\end{definition}

Needless to say, the notion $\operatorname*{Sym}f$ introduced in this
definition is an analogue of the notion $\operatorname*{Exter} f$ introduced
in Definition \ref{def.ext-functor}.

Here is the analogue of Proposition \ref{prop.ext-surj}:

\begin{proposition}
\label{prop.sym-surj}Let $k$ be a commutative ring. Let $V$ and $W$ be two
$k$-modules. Let $f:V\rightarrow W$ be a surjective $k$-module homomorphism.
Then:\newline\textbf{(a)} The $k$-module homomorphism $f^{\otimes
n}:V^{\otimes n}\rightarrow W^{\otimes n}$ is surjective for every
$n\in\mathbb{N}$.\newline\textbf{(b)} The $k$-algebra homomorphism $\otimes
f:\otimes V\rightarrow\otimes W$ is surjective.\newline\textbf{(c)} The
$k$-algebra homomorphism $\operatorname*{Sym}f:\operatorname*{Sym}%
V\rightarrow\operatorname*{Sym}W$ is surjective.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sym-surj}.]The proof of this Proposition
\ref{prop.sym-surj} is completely analogous to the proof of Proposition
\ref{prop.ext-surj} (and parts \textbf{(a)} and \textbf{(b)} are even the same).
\end{proof}

So much for analogues of the results of Subsection \ref{subsect.exter}. Now
let us formulate the analogues of the results of Subsection
\ref{subsect.Ker(exter)}. First, the analogue of Theorem \ref{thm.Ker(exter)}:

\begin{theorem}
\label{thm.Ker(sym)}Let $k$ be a commutative ring. Let $V$ and $V^{\prime}$ be
two $k$-modules, and let $f:V\rightarrow V^{\prime}$ be a surjective
$k$-module homomorphism. Then, the kernel of the map $\operatorname*{Sym}%
f:\operatorname*{Sym}V\rightarrow\operatorname*{Sym}V^{\prime}$ is%
\begin{align*}
\operatorname*{Ker}\left(  \operatorname*{Sym}f\right)   &  =\left(
\operatorname*{Sym}V\right)  \cdot\operatorname*{sym}\nolimits_{V}\left(
\operatorname*{Ker}f\right)  \cdot\left(  \operatorname*{Sym}V\right)
=\left(  \operatorname*{Sym}V\right)  \cdot\operatorname*{sym}\nolimits_{V}%
\left(  \operatorname*{Ker}f\right) \\
&  =\operatorname*{sym}\nolimits_{V}\left(  \operatorname*{Ker}f\right)
\cdot\left(  \operatorname*{Sym}V\right)  .
\end{align*}
Here, $\operatorname*{Ker}f$ is considered a $k$-submodule of $\otimes V$ by
means of the inclusion $\operatorname*{Ker}f\subseteq V=V^{\otimes1}%
\subseteq\otimes V$.
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.Ker(sym)}.]The proof of this Theorem
\ref{thm.Ker(sym)} is completely analogous to that of Theorem
\ref{thm.Ker(exter)}.
\end{proof}

The analogue of Corollary \ref{cor.exter.W} comes next:

\begin{corollary}
\label{cor.sym.W}Let $k$ be a commutative ring. Let $V$ be a $k$-module, and
let $W$ be a $k$-submodule of $V$. Then,%
\[
\left(  \operatorname*{Sym}V\right)  \cdot\operatorname*{sym}\nolimits_{V}%
\left(  W\right)  \cdot\left(  \operatorname*{Sym}V\right)  =\left(
\operatorname*{Sym}V\right)  \cdot\operatorname*{sym}\nolimits_{V}\left(
W\right)  =\operatorname*{sym}\nolimits_{V}\left(  W\right)  \cdot\left(
\operatorname*{Sym}V\right)  .
\]
Here, $W$ is considered a $k$-submodule of $\otimes V$ by means of the
inclusion $W\subseteq V=V^{\otimes1}\subseteq\otimes V$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.sym.W}.]Expectedly, the proof of Corollary
\ref{cor.sym.W} is analogous to the proof of Corollary \ref{cor.exter.W}.
\end{proof}

Finally, the analogue of Corollary \ref{coro.Ker(exter)}:

\begin{corollary}
\label{coro.Ker(sym)}Let $k$ be a commutative ring. Let $V$ be a $k$-module.
Let $W$ be a $k$-submodule of $V$, and let $f:V\rightarrow V\diagup W$ be the
canonical projection.\newline\textbf{(a)} Then, the kernel of the map
$\operatorname*{Sym}f:\operatorname*{Sym}V\rightarrow\operatorname*{Sym}%
\left(  V\diagup W\right)  $ is%
\[
\operatorname*{Ker}\left(  \operatorname*{Sym}f\right)  =\left(
\operatorname*{Sym}V\right)  \cdot\operatorname*{sym}\nolimits_{V}\left(
W\right)  \cdot\left(  \operatorname*{Sym}V\right)  =\left(
\operatorname*{Sym}V\right)  \cdot\operatorname*{sym}\nolimits_{V}\left(
W\right)  =\operatorname*{sym}\nolimits_{V}\left(  W\right)  \cdot\left(
\operatorname*{Sym}V\right)  .
\]
Here, $W$ is considered a $k$-submodule of $\otimes V$ by means of the
inclusion $W\subseteq V=V^{\otimes1}\subseteq\otimes V$.\newline\textbf{(b)}
We have%
\[
\left(  \operatorname*{Sym}V\right)  \diagup\left(  \left(
\operatorname*{Sym}V\right)  \cdot\operatorname*{sym}\nolimits_{V}\left(
W\right)  \right)  \cong\operatorname*{Sym}\left(  V\diagup W\right)
\ \ \ \ \ \ \ \ \ \ \text{as }k\text{-modules.}%
\]

\end{corollary}

\begin{proof}
[Proof of Corollary \ref{coro.Ker(sym)}.]The proof of Corollary
\ref{coro.Ker(sym)} is analogous to the proof of Corollary
\ref{coro.Ker(exter)}.
\end{proof}

\subsection{\label{subsect.Wedge}The exterior algebra}

Now we are going to study the exterior algebra $\wedge V$ of a $k$-module $V$.
This algebra is rather similar, but not completely analogous to
$\operatorname*{Exter}V$ and $\operatorname*{Sym}V$. We are going to again
formulate properties similar to corresponding properties of
$\operatorname*{Exter}V$ and $\operatorname*{Sym}V $; but this time, some of
these properties will require different proofs, so we will not always be able
to skip their proofs by referring to analogy. Still some of the proofs will be
very similar to the corresponding proofs for $\operatorname*{Exter}V$ we gave
in Subsections \ref{subsect.exter} and \ref{subsect.Ker(exter)} (some others
will be not). First, before we define the exterior algebra $\wedge V$, let us
define the exterior powers $\wedge^{n}V$:

\begin{definition}
\label{defs.wedge-pow}Let $k$ be a commutative ring. Let $V$ be a $k$-module.
Let $n\in\mathbb{N}$.\newline Let $R_{n}\left(  V\right)  $ be the
$k$-submodule%
\[
\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(
\left(  v_{1},v_{2},\ldots,v_{n}\right)  ,\left(  i,j\right)  \right)  \in
V^{n}\times\left\{  1,2,\ldots,n\right\}  ^{2};\ i\neq j;\ v_{i}%
=v_{j}\right\rangle
\]
of the $k$-module $V^{\otimes n}$ (where we are using Convention
\ref{conv.<>}).\newline The factor $k$-module $V^{\otimes n}\diagup
R_{n}\left(  V\right)  $ is called the $n$\textit{-th exterior power} of the
$k$-module $V$ and will be denoted by $\wedge^{n}V$. We denote by
$\operatorname*{wedge}\nolimits_{V,n}$ the canonical projection $V^{\otimes
n}\rightarrow V^{\otimes n}\diagup R_{n}\left(  V\right)  =\wedge^{n}V$.
Clearly, this map $\operatorname*{wedge}\nolimits_{V,n}$ is a surjective
$k$-module homomorphism.
\end{definition}

We should understand these notions $R_{n}\left(  V\right)  $, $\wedge^{n}V$
and $\operatorname*{wedge}\nolimits_{V,n}$ as analogues of the notions
$Q_{n}\left(  V\right)  $, $\operatorname*{Exter}^{n}V$ and
$\operatorname*{exter}\nolimits_{V,n}$ from Definition \ref{defs.ext-pow},
respectively. First something very basic - an analogue of Corollary
\ref{coro.Q_2}:

\begin{corollary}
\label{coro.R_2}Let $k$ be a commutative ring. Let $V$ be a $k$-module. Then,%
\[
R_{2}\left(  V\right)  =\left\langle v\otimes v\ \mid\ v\in V\right\rangle .
\]

\end{corollary}

\begin{proof}
[Proof of Corollary \ref{coro.R_2}.]We have the inclusions%
\[
\left\{  v_{1}\otimes v_{2}\mid\ \left(  \left(  v_{1},v_{2}\right)  ,\left(
i,j\right)  \right)  \in V^{2}\times\left\{  1,2\right\}  ^{2};\ i\neq
j;\ v_{i}=v_{j}\right\}  \subseteq\left\{  v\otimes v\ \mid\ v\in V\right\}
\]
\footnote{\textit{Proof.} Let $p\in\left\{  v_{1}\otimes v_{2}\mid\ \left(
\left(  v_{1},v_{2}\right)  ,\left(  i,j\right)  \right)  \in V^{2}%
\times\left\{  1,2\right\}  ^{2};\ i\neq j;\ v_{i}=v_{j}\right\}  $. Then,
there exists some $\left(  \left(  v_{1},v_{2}\right)  ,\left(  i,j\right)
\right)  \in V^{2}\times\left\{  1,2\right\}  ^{2}$ such that $i\neq j$ and
$v_{i}=v_{j}$ and $p=v_{1}\otimes v_{2}$. Consider this $\left(  \left(
v_{1},v_{2}\right)  ,\left(  i,j\right)  \right)  $. Then, $\left(
i,j\right)  \in\left\{  1,2\right\}  ^{2}$. Since $i\neq j$, this yields that
either $\left(  i=1\text{ and }j=2\right)  $ or $\left(  j=1\text{ and
}i=2\right)  $. In each of these two cases, we have $v_{1}=v_{2}$ (in fact, in
the case $\left(  i=1\text{ and }j=2\right)  $, the equation $v_{i}=v_{j}$
rewrites as $v_{1}=v_{2}$; and in the other case $\left(  j=1\text{ and
}i=2\right)  $, the equation $v_{i}=v_{j}$ rewrites as $v_{2}=v_{1}$, so that
$v_{1}=v_{2}$). Hence, we have $v_{1}=v_{2}$. Thus, $p=\underbrace{v_{1}%
}_{=v_{2}}\otimes v_{2}=v_{2}\otimes v_{2}\in\left\{  v\otimes v\ \mid\ v\in
V\right\}  $.
\par
We have thus shown that $p\in\left\{  v\otimes v\ \mid\ v\in V\right\}  $ for
every $p\in\left\{  v_{1}\otimes v_{2}\mid\ \left(  \left(  v_{1}%
,v_{2}\right)  ,\left(  i,j\right)  \right)  \in V^{2}\times\left\{
1,2\right\}  ^{2};\ i\neq j;\ v_{i}=v_{j}\right\}  $. Hence, $\left\{
v_{1}\otimes v_{2}\mid\ \left(  \left(  v_{1},v_{2}\right)  ,\left(
i,j\right)  \right)  \in V^{2}\times\left\{  1,2\right\}  ^{2};\ i\neq
j;\ v_{i}=v_{j}\right\}  \subseteq\left\{  v\otimes v\ \mid\ v\in V\right\}
$, qed.} and%
\[
\left\{  v\otimes v\ \mid\ v\in V\right\}  \subseteq\left\{  v_{1}\otimes
v_{2}\mid\ \left(  \left(  v_{1},v_{2}\right)  ,\left(  i,j\right)  \right)
\in V^{2}\times\left\{  1,2\right\}  ^{2};\ i\neq j;\ v_{i}=v_{j}\right\}
\]
\footnote{\textit{Proof.} Let $p\in\left\{  v\otimes v\ \mid\ v\in V\right\}
$. Then, there exists $v\in V$ such that $p=v\otimes v$. Consider this $v$.
Then, there exists some $\left(  \left(  v_{1},v_{2}\right)  ,\left(
i,j\right)  \right)  \in V^{2}\times\left\{  1,2\right\}  ^{2}$ with$\ i\neq
j\ $and $v_{i}=v_{j}$ such that $p=v_{1}\otimes v_{2}$ (namely, $\left(
\left(  v_{1},v_{2}\right)  ,\left(  i,j\right)  \right)  =\left(  \left(
v,v\right)  ,\left(  1,2\right)  \right)  $). Hence, $p\in\left\{
v_{1}\otimes v_{2}\mid\ \left(  \left(  v_{1},v_{2}\right)  ,\left(
i,j\right)  \right)  \in V^{2}\times\left\{  1,2\right\}  ^{2};\ i\neq
j;\ v_{i}=v_{j}\right\}  $. Since we have proven this for every $p\in\left\{
v\otimes v\ \mid\ v\in V\right\}  $, we have thus shown that $\left\{
v\otimes v\ \mid\ v\in V\right\}  \subseteq\left\{  v_{1}\otimes v_{2}%
\mid\ \left(  \left(  v_{1},v_{2}\right)  ,\left(  i,j\right)  \right)  \in
V^{2}\times\left\{  1,2\right\}  ^{2};\ i\neq j;\ v_{i}=v_{j}\right\}  $.}.
Combining these two inclusions, we get%
\[
\left\{  v_{1}\otimes v_{2}\mid\ \left(  \left(  v_{1},v_{2}\right)  ,\left(
i,j\right)  \right)  \in V^{2}\times\left\{  1,2\right\}  ^{2};\ i\neq
j;\ v_{i}=v_{j}\right\}  =\left\{  v\otimes v\ \mid\ v\in V\right\}  .
\]
But by the definition of $R_{2}\left(  V\right)  $, we have%
\begin{align*}
R_{2}\left(  V\right)   &  =\left\langle v_{1}\otimes v_{2}\mid\ \left(
\left(  v_{1},v_{2}\right)  ,\left(  i,j\right)  \right)  \in V^{2}%
\times\left\{  1,2\right\}  ^{2};\ i\neq j;\ v_{i}=v_{j}\right\rangle \\
&  =\left\langle \underbrace{\left\{  v_{1}\otimes v_{2}\mid\ \left(  \left(
v_{1},v_{2}\right)  ,\left(  i,j\right)  \right)  \in V^{2}\times\left\{
1,2\right\}  ^{2};\ i\neq j;\ v_{i}=v_{j}\right\}  }_{=\left\{  v\otimes
v\ \mid\ v\in V\right\}  }\right\rangle \\
&  =\left\langle \left\{  v\otimes v\ \mid\ v\in V\right\}  \right\rangle
=\left\langle v\otimes v\ \mid\ v\in V\right\rangle .
\end{align*}
This proves Corollary \ref{coro.R_2}.
\end{proof}

Here is an analogue of Proposition \ref{prop.Q_n}:

\begin{proposition}
\label{prop.R_n}Let $k$ be a commutative ring. Let $V$ be a $k$-module. Let
$n\in\mathbb{N}$.\newline Then,%
\[
R_{n}\left(  V\right)  =\sum_{i=1}^{n-1}\left\langle v_{1}\otimes v_{2}%
\otimes\cdots\otimes v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{n};\ v_{i}=v_{i+1}\right\rangle .
\]

\end{proposition}

While the proof of this proposition is not too much harder than that of
Proposition \ref{prop.Q_n}, it is better understood when split into lemmas.
Here is the first one:

\begin{lemma}
\label{lem.QinR-1}Let $k$ be a commutative ring. Let $V$ be a $k$-module. Let
$n\in\mathbb{N}$. Let $\widetilde{R}_{n}\left(  V\right)  $ denote the
$k$-submodule%
\[
\sum\limits_{i=1}^{n-1}\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n};\ v_{i}%
=v_{i+1}\right\rangle
\]
of $V^{\otimes n}$. Then, $Q_{n}\left(  V\right)  \subseteq\widetilde{R}%
_{n}\left(  V\right)  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.QinR-1}.]\textbf{(i)} Every $i\in\left\{
1,2,\ldots,n-1\right\}  $ satisfies%
\begin{align}
&  \left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau
_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }\otimes
\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{n}\right\rangle \nonumber\\
&  \subseteq\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n};\ v_{i}%
=v_{i+1}\right\rangle , \label{pf.QinR-1.1}%
\end{align}
where $\tau_{i}$ denotes the transposition $\left(  i,i+1\right)  \in S_{n}$.

\textit{Proof.} Fix some $i\in\left\{  1,2,\ldots,n-1\right\}  $. Now let
$\mathfrak{W}$ be the set%
\[
\left\{  w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}\ \mid\ \left(
w_{1},w_{2},\ldots,w_{n}\right)  \in V^{n};\ w_{i}=w_{i+1}\right\}  .
\]
Then,%
\begin{equation}
w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}\in\mathfrak{W}%
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in V^{n}\text{ satisfying }w_{i}=w_{i+1}. \label{pf.QinR-1.5}%
\end{equation}


Fix some arbitrary $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$.
Define a tensor $A\in V^{\otimes\left(  i-1\right)  }$ by $A=v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{i-1}$. Define a tensor $C\in V^{\otimes\left(
n-1-i\right)  }$ by $C=v_{i+2}\otimes v_{i+3}\otimes\cdots\otimes v_{n}$.
Then, recalling Convention \ref{conv.(X)^n.ident}, we have%
\begin{align}
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}  &  =\underbrace{\left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{i-1}\right)  }_{=A}\otimes\left(
v_{i}\otimes v_{i+1}\right)  \otimes\underbrace{\left(  v_{i+2}\otimes
v_{i+3}\otimes\cdots\otimes v_{n}\right)  }_{=C}\nonumber\\
&  =A\otimes\left(  v_{i}\otimes v_{i+1}\right)  \otimes C.
\label{pf.QinR-1.2}%
\end{align}


On the other hand, every $j\in\left\{  1,2,\ldots,i-1\right\}  $ satisfies
$\tau_{i}\left(  j\right)  =j$ (since $\tau_{i}$ is the transposition $\left(
i,i+1\right)  $) and thus $v_{\tau_{i}\left(  j\right)  }=v_{j}$. In other
words, we have the equalities $v_{\tau_{i}\left(  1\right)  }=v_{1}$,
$v_{\tau_{i}\left(  2\right)  }=v_{2}$, $\ldots$, $v_{\tau_{i}\left(
i-1\right)  }=v_{i-1}$. Taking the tensor product of these equalities yields%
\[
v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }%
\otimes\cdots\otimes v_{\tau_{i}\left(  i-1\right)  }=v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{i-1}=A.
\]


Every $j\in\left\{  i+2,i+3,\ldots,n\right\}  $ satisfies $\tau_{i}\left(
j\right)  =j$ (since $\tau_{i}$ is the transposition $\left(  i,i+1\right)  $)
and thus $v_{\tau_{i}\left(  j\right)  }=v_{j}$. In other words, we have the
equalities $v_{\tau_{i}\left(  i+2\right)  }=v_{i+2}$, $v_{\tau_{i}\left(
i+3\right)  }=v_{i+3}$, $\ldots$, $v_{\tau_{i}\left(  n\right)  }=v_{n}$.
Taking the tensor product of these equalities yields%
\[
v_{\tau_{i}\left(  i+2\right)  }\otimes v_{\tau_{i}\left(  i+3\right)
}\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }=v_{i+2}\otimes
v_{i+3}\otimes\cdots\otimes v_{n}=C.
\]


Since $\tau_{i}$ is the transposition $\left(  i,i+1\right)  $, we have
$\tau_{i}\left(  i\right)  =i+1$ and $\tau_{i}\left(  i+1\right)  =i$. These
equalities yield $v_{\tau_{i}\left(  i\right)  }=v_{i+1}$ and $v_{\tau
_{i}\left(  i+1\right)  }=v_{i}$, respectively.

Now,%
\begin{align*}
&  v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }%
\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }\\
&  =\underbrace{\left(  v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau
_{i}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(  i-1\right)
}\right)  }_{=A}\otimes\left(  \underbrace{v_{\tau_{i}\left(  i\right)  }%
}_{=v_{i+1}}\otimes\underbrace{v_{\tau_{i}\left(  i+1\right)  }}_{=v_{i}%
}\right)  \otimes\underbrace{\left(  v_{\tau_{i}\left(  i+2\right)  }\otimes
v_{\tau_{i}\left(  i+3\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(
n\right)  }\right)  }_{=C}\\
&  =A\otimes\left(  v_{i+1}\otimes v_{i}\right)  \otimes C.
\end{align*}
Adding this to (\ref{pf.QinR-1.2}), we get%
\begin{align}
&  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)
}\otimes v_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(
n\right)  }\nonumber\\
&  =A\otimes\left(  v_{i}\otimes v_{i+1}\right)  \otimes C+A\otimes\left(
v_{i+1}\otimes v_{i}\right)  \otimes C\nonumber\\
&  =A\otimes\left(  v_{i}\otimes v_{i+1}+v_{i+1}\otimes v_{i}\right)  \otimes
C. \label{pf.QinR-1.4}%
\end{align}


But it is easy to see that%
\begin{equation}
A\otimes p\otimes p\otimes C\in\mathfrak{W}\ \ \ \ \ \ \ \ \ \ \text{for every
}p\in V. \label{pf.QinR-1.6}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.QinR-1.6}).} Let $p\in V$. Define an
$n$-tuple $\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in V^{n}$ by%
\begin{equation}
\left(  w_{\ell}=\left\{
\begin{array}
[c]{c}%
v_{\ell}\text{, if }\ell<i;\\
p\text{, if }\ell=i;\\
p\text{, if }\ell=i+1;\\
v_{\ell}\text{, if }\ell>i+1
\end{array}
\right.  \ \ \ \ \ \ \ \ \ \ \text{for every }\ell\in\left\{  1,2,\ldots
,n\right\}  \right)  . \label{pf.QinR-1.9}%
\end{equation}
\par
Then, every $\ell\in\left\{  1,2,\ldots,i-1\right\}  $ satisfies $w_{\ell
}=\left\{
\begin{array}
[c]{c}%
v_{\ell}\text{, if }\ell<i;\\
p\text{, if }\ell=i;\\
p\text{, if }\ell=i+1;\\
v_{\ell}\text{, if }\ell>i+1
\end{array}
\right.  =v_{\ell}$ (since $\ell<i$). In other words, we have the equalities
$w_{1}=v_{1}$, $w_{2}=v_{2}$, $\ldots$, $w_{i-1}=v_{i-1}$. Taking the tensor
product of these equalities, we get $w_{1}\otimes w_{2}\otimes\cdots\otimes
w_{i-1}=v_{1}\otimes v_{2}\otimes\cdots\otimes v_{i-1}=A$.
\par
Also, every $\ell\in\left\{  i+2,i+3,\ldots,n\right\}  $ satisfies $w_{\ell
}=\left\{
\begin{array}
[c]{c}%
v_{\ell}\text{, if }\ell<i;\\
p\text{, if }\ell=i;\\
p\text{, if }\ell=i+1;\\
v_{\ell}\text{, if }\ell>i+1
\end{array}
\right.  =v_{\ell}$ (since $\ell>i+1$). In other words, we have the equalities
$w_{i+2}=v_{i+2}$, $w_{i+3}=v_{i+3}$, $\ldots$, $w_{n}=v_{n}$. Taking the
tensor product of these equalities, we get $w_{i+2}\otimes w_{i+3}%
\otimes\cdots\otimes w_{n}=v_{i+2}\otimes v_{i+3}\otimes\cdots\otimes v_{n}%
=C$.
\par
Applying (\ref{pf.QinR-1.9}) to $\ell=i$, we get $w_{i}=\left\{
\begin{array}
[c]{c}%
v_{i}\text{, if }i<i;\\
p\text{, if }i=i;\\
p\text{, if }i=i+1;\\
v_{i}\text{, if }i>i+1
\end{array}
\right.  =p$ (since $i=i$). Applying (\ref{pf.QinR-1.9}) to $\ell=i+1$, we get
$w_{i}=\left\{
\begin{array}
[c]{c}%
v_{i+1}\text{, if }i+1<i;\\
p\text{, if }i+1=i;\\
p\text{, if }i+1=i+1;\\
v_{i+1}\text{, if }i+1>i+1
\end{array}
\right.  =p$ (since $i+1=i+1$).
\par
Now,%
\[
w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}=\underbrace{\left(  w_{1}\otimes
w_{2}\otimes\cdots\otimes w_{i-1}\right)  }_{=A}\otimes\underbrace{w_{i}}%
_{=p}\otimes\underbrace{w_{i+1}}_{=p}\otimes\underbrace{\left(  w_{i+2}\otimes
w_{i+3}\otimes\cdots\otimes w_{n}\right)  }_{=C}=A\otimes p\otimes p\otimes
C.
\]
Since $w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}\in\mathfrak{W}$ (by
(\ref{pf.QinR-1.5})), we thus have $A\otimes p\otimes p\otimes C\in
\mathfrak{W}$. This proves (\ref{pf.QinR-1.6}).} Since $\mathfrak{W}%
\subseteq\left\langle \mathfrak{W}\right\rangle $, this yields%
\begin{equation}
A\otimes p\otimes p\otimes C\in\left\langle \mathfrak{W}\right\rangle
\ \ \ \ \ \ \ \ \ \ \text{for every }p\in V. \label{pf.QinR-1.10}%
\end{equation}


Now,
\begin{align*}
v_{i}\otimes v_{i+1}+v_{i+1}\otimes v_{i}  &  =\underbrace{\left(
v_{i}\otimes v_{i}+v_{i}\otimes v_{i+1}+v_{i+1}\otimes v_{i}+v_{i+1}\otimes
v_{i+1}\right)  }_{=\left(  v_{i}+v_{i+1}\right)  \otimes\left(  v_{i}%
+v_{i+1}\right)  }-v_{i}\otimes v_{i}-v_{i+1}\otimes v_{i+1}\\
&  =\left(  v_{i}+v_{i+1}\right)  \otimes\left(  v_{i}+v_{i+1}\right)
-v_{i}\otimes v_{i}-v_{i+1}\otimes v_{i+1}.
\end{align*}
Thus, (\ref{pf.QinR-1.4}) becomes%
\begin{align*}
&  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)
}\otimes v_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(
n\right)  }\\
&  =A\otimes\underbrace{\left(  v_{i}\otimes v_{i+1}+v_{i+1}\otimes
v_{i}\right)  }_{=\left(  v_{i}+v_{i+1}\right)  \otimes\left(  v_{i}%
+v_{i+1}\right)  -v_{i}\otimes v_{i}-v_{i+1}\otimes v_{i+1}}\otimes C\\
&  =A\otimes\left(  \left(  v_{i}+v_{i+1}\right)  \otimes\left(  v_{i}%
+v_{i+1}\right)  -v_{i}\otimes v_{i}-v_{i+1}\otimes v_{i+1}\right)  \otimes
C\\
&  =\underbrace{A\otimes\left(  v_{i}+v_{i+1}\right)  \otimes\left(
v_{i}+v_{i+1}\right)  \otimes C}_{\in\left\langle \mathfrak{W}\right\rangle
\text{ (by (\ref{pf.QinR-1.10}), applied to }p=v_{i}+v_{i+1}\text{)}%
}-\underbrace{A\otimes v_{i}\otimes v_{i}\otimes C}_{\in\left\langle
\mathfrak{W}\right\rangle \text{ (by (\ref{pf.QinR-1.10}), applied to }%
p=v_{i}\text{)}}-\underbrace{A\otimes v_{i+1}\otimes v_{i+1}\otimes C}%
_{\in\left\langle \mathfrak{W}\right\rangle \text{ (by (\ref{pf.QinR-1.10}),
applied to }p=v_{i+1}\text{)}}\\
&  \in\left\langle \mathfrak{W}\right\rangle -\left\langle \mathfrak{W}%
\right\rangle -\left\langle \mathfrak{W}\right\rangle \subseteq\left\langle
\mathfrak{W}\right\rangle \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\left\langle \mathfrak{W}\right\rangle \text{ is a }k\text{-module}\right)  .
\end{align*}


We now forget that we fixed $\left(  v_{1},v_{2},\ldots,v_{n}\right)  $. What
we have proven is that every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{n}$ satisfies%
\[
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)
}\otimes v_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(
n\right)  }\in\left\langle \mathfrak{W}\right\rangle .
\]
In other words,%
\[
\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau_{i}\left(
1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes
v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)
\in V^{n}\right\}  \subseteq\left\langle \mathfrak{W}\right\rangle .
\]
Hence, Proposition \ref{prop.<>} \textbf{(a)} (applied to $V^{\otimes n}$,
\newline$\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau
_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }\otimes
\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{n}\right\}  $ and $\left\langle \mathfrak{W}%
\right\rangle $ instead of $M$, $S$ and $Q$) yields%
\[
\left\langle \left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau
_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }\otimes
\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{n}\right\}  \right\rangle \subseteq\left\langle
\mathfrak{W}\right\rangle .
\]
Thus,%
\begin{align*}
&  \left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}+v_{\tau
_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }\otimes
\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{n}\right\rangle \\
&  =\left\langle \left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}%
+v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }%
\otimes\cdots\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  \in V^{n}\right\}  \right\rangle \\
&  \subseteq\left\langle \mathfrak{W}\right\rangle =\left\langle \left\{
w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}\ \mid\ \left(  w_{1},w_{2}%
,\ldots,w_{n}\right)  \in V^{n};\ w_{i}=w_{i+1}\right\}  \right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathfrak{W}=\left\{  w_{1}\otimes
w_{2}\otimes\cdots\otimes w_{n}\ \mid\ \left(  w_{1},w_{2},\ldots
,w_{n}\right)  \in V^{n};\ w_{i}=w_{i+1}\right\}  \right) \\
&  =\left\langle w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}\ \mid\ \left(
w_{1},w_{2},\ldots,w_{n}\right)  \in V^{n};\ w_{i}=w_{i+1}\right\rangle \\
&  =\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n};\ v_{i}=v_{i+1}\right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed }\left(  w_{1}%
,w_{2},\ldots,w_{n}\right)  \text{ as }\left(  v_{1},v_{2},\ldots
,v_{n}\right)  \right)  .
\end{align*}
This proves \textbf{(i)}.

\textbf{(ii)} For every $i\in\left\{  1,2,\ldots,n-1\right\}  $, let $\tau
_{i}$ denote the transposition $\left(  i,i+1\right)  \in S_{n}$. By
Proposition \ref{prop.Q_n}, we have%
\begin{align*}
Q_{n}\left(  V\right)   &  =\sum_{i=1}^{n-1}\left\langle v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}+v_{\tau_{i}\left(  1\right)  }\otimes
v_{\tau_{i}\left(  2\right)  }\otimes\cdots\otimes v_{\tau_{i}\left(
n\right)  }\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}%
\right\rangle \\
&  \subseteq\sum\limits_{i=1}^{n-1}\left\langle v_{1}\otimes v_{2}%
\otimes\cdots\otimes v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{n};\ v_{i}=v_{i+1}\right\rangle \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.QinR-1.1})}\right) \\
&  =\widetilde{R}_{n}\left(  V\right)  .
\end{align*}
This proves Lemma \ref{lem.QinR-1}.
\end{proof}

Our next step is the following lemma:

\begin{lemma}
\label{lem.QinR-2}In the situation of Lemma \ref{lem.QinR-1}, we have
$\widetilde{R}_{n}\left(  V\right)  \subseteq R_{n}\left(  V\right)  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.QinR-2}.]First fix some $\mathbf{I}\in\left\{
1,2,\ldots,n-1\right\}  $. Fix some $\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in V^{n}$ satisfying $w_{\mathbf{I}}=w_{\mathbf{I}+1}$. Then, the pair
$\left(  \left(  w_{1},w_{2},\ldots,w_{n}\right)  ,\left(  \mathbf{I}%
,\mathbf{I}+1\right)  \right)  \in V^{n}\times\left\{  1,2,\ldots,n\right\}
^{2}$ satisfies $\mathbf{I}\neq\mathbf{I}+1$ and $v_{\mathbf{I}}%
=v_{\mathbf{I}+1}$. Therefore,%
\begin{align*}
&  w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}\\
&  \in\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(
\left(  v_{1},v_{2},\ldots,v_{n}\right)  ,\left(  i,j\right)  \right)  \in
V^{n}\times\left\{  1,2,\ldots,n\right\}  ^{2};\ i\neq j;\ v_{i}=v_{j}\right\}
\\
&  \subseteq\left\langle \left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\ \mid\ \left(  \left(  v_{1},v_{2},\ldots,v_{n}\right)  ,\left(
i,j\right)  \right)  \in V^{n}\times\left\{  1,2,\ldots,n\right\}
^{2};\ i\neq j;\ v_{i}=v_{j}\right\}  \right\rangle \\
&  =\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(
\left(  v_{1},v_{2},\ldots,v_{n}\right)  ,\left(  i,j\right)  \right)  \in
V^{n}\times\left\{  1,2,\ldots,n\right\}  ^{2};\ i\neq j;\ v_{i}%
=v_{j}\right\rangle \\
&  =R_{n}\left(  V\right)  .
\end{align*}


Now forget that we fixed some $\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in
V^{n}$ satisfying $w_{\mathbf{I}}=w_{\mathbf{I}+1}$. We have thus proven that
every $\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in V^{n}$ satisfying
$w_{\mathbf{I}}=w_{\mathbf{I}+1}$ satisfies $w_{1}\otimes w_{2}\otimes
\cdots\otimes w_{n}\in R_{n}\left(  V\right)  $. In other words, we have
proven that%
\[
\left\{  w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}\ \mid\ \left(
w_{1},w_{2},\ldots,w_{n}\right)  \in V^{n};\ w_{\mathbf{I}}=w_{\mathbf{I}%
+1}\right\}  \subseteq R_{n}\left(  V\right)  .
\]
Hence, Proposition \ref{prop.<>} \textbf{(a)} (applied to $V^{\otimes n}$,
\newline$\left\{  w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}\ \mid\ \left(
w_{1},w_{2},\ldots,w_{n}\right)  \in V^{n};\ w_{\mathbf{I}}=w_{\mathbf{I}%
+1}\right\}  $ and $R_{n}\left(  V\right)  $ instead of $M$, $S$ and $Q$)
yields%
\[
\left\langle \left\{  w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}%
\ \mid\ \left(  w_{1},w_{2},\ldots,w_{n}\right)  \in V^{n};\ w_{\mathbf{I}%
}=w_{\mathbf{I}+1}\right\}  \right\rangle \subseteq R_{n}\left(  V\right)  .
\]
So we have%
\begin{align}
&  \left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n};\ v_{\mathbf{I}}=v_{\mathbf{I}%
+1}\right\rangle \nonumber\\
&  =\left\langle w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}\ \mid\ \left(
w_{1},w_{2},\ldots,w_{n}\right)  \in V^{n};\ w_{\mathbf{I}}=w_{\mathbf{I}%
+1}\right\rangle \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed }\left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  \text{ as }\left(  w_{1},w_{2},\ldots
,w_{n}\right)  \right) \nonumber\\
&  =\left\langle \left\{  w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}%
\ \mid\ \left(  w_{1},w_{2},\ldots,w_{n}\right)  \in V^{n};\ w_{\mathbf{I}%
}=w_{\mathbf{I}+1}\right\}  \right\rangle \subseteq R_{n}\left(  V\right)  .
\label{pf.QinR-2.4}%
\end{align}


Now forget that we fixed some $\mathbf{I}\in\left\{  1,2,\ldots,n-1\right\}
$. We have now proven that every $\mathbf{I}\in\left\{  1,2,\ldots
,n-1\right\}  $ satisfies (\ref{pf.QinR-2.4}). Now,%
\begin{align*}
\widetilde{R}_{n}\left(  V\right)   &  =\sum\limits_{i=1}^{n-1}\left\langle
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{n};\ v_{i}=v_{i+1}\right\rangle \\
&  =\sum\limits_{\mathbf{I}=1}^{n-1}\underbrace{\left\langle v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots
,v_{n}\right)  \in V^{n};\ v_{\mathbf{I}}=v_{\mathbf{I}+1}\right\rangle
}_{\subseteq R_{n}\left(  V\right)  \text{ (by (\ref{pf.QinR-2.4}))}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}i\text{ as }\mathbf{I}\right) \\
&  \subseteq\sum\limits_{\mathbf{I}=1}^{n-1}R_{n}\left(  V\right)  \subseteq
R_{n}\left(  V\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }R_{n}\left(
V\right)  \text{ is a }k\text{-module}\right)  .
\end{align*}
This proves Lemma \ref{lem.QinR-2}.
\end{proof}

Our final lemma is:

\begin{lemma}
\label{lem.QinR-3}In the situation of Lemma \ref{lem.QinR-1}, we have
$R_{n}\left(  V\right)  \subseteq\widetilde{R}_{n}\left(  V\right)  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.QinR-3}.]\textbf{(i)} It is clear that every
$\mathbf{I}\in\left\{  1,2,\ldots,n-1\right\}  $ satisfies%
\begin{align*}
&  \left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n};\ v_{\mathbf{I}}=v_{\mathbf{I}%
+1}\right\} \\
&  \subseteq\left\langle \left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}%
;\ v_{\mathbf{I}}=v_{\mathbf{I}+1}\right\}  \right\rangle \\
&  =\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n};\ v_{\mathbf{I}}=v_{\mathbf{I}%
+1}\right\rangle \\
&  \subseteq\sum\limits_{i=1}^{n-1}\left\langle v_{1}\otimes v_{2}%
\otimes\cdots\otimes v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{n};\ v_{i}=v_{i+1}\right\rangle =\widetilde{R}_{n}\left(  V\right)  .
\end{align*}
In other words, for every $\mathbf{I}\in\left\{  1,2,\ldots,n-1\right\}  $,%
\begin{equation}
\text{every }\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\text{ such
that }v_{\mathbf{I}}=v_{\mathbf{I}+1}\text{ satisfies }v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}\in\widetilde{R}_{n}\left(  V\right)  .
\label{pf.QinR-3.1}%
\end{equation}


\textbf{(ii)} Every $\left(  \left(  w_{1},w_{2},\ldots,w_{n}\right)  ,\left(
\mathbf{I},\mathbf{J}\right)  \right)  \in V^{n}\times\left\{  1,2,\ldots
,n\right\}  ^{2}$ satisfying $\mathbf{I}\neq\mathbf{J}$ and $w_{\mathbf{I}%
}=w_{\mathbf{J}}$ must satisfy $w_{1}\otimes w_{2}\otimes\cdots\otimes
w_{n}\in\widetilde{R}_{n}\left(  V\right)  $.

\textit{Proof.} Fix some $\left(  \left(  w_{1},w_{2},\ldots,w_{n}\right)
,\left(  \mathbf{I},\mathbf{J}\right)  \right)  \in V^{n}\times\left\{
1,2,\ldots,n\right\}  ^{2}$ satisfying $\mathbf{I}\neq\mathbf{J}$ and
$w_{\mathbf{I}}=w_{\mathbf{J}}$.

Then, $\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in V^{n}$ and $\left(
\mathbf{I},\mathbf{J}\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$.

We can WLOG assume that $\mathbf{I}\leq\mathbf{J}$ (since otherwise, we could
just transpose $\mathbf{I}$ with $\mathbf{J}$, and nothing would change
(because each of the conditions $\mathbf{I}\neq\mathbf{J}$ and $w_{\mathbf{I}%
}=w_{\mathbf{J}}$ is clearly symmetric with respect to $\mathbf{I}$ and
$\mathbf{J}$)). So let us assume this. Then, $\mathbf{I}<\mathbf{J}$ (since
$\mathbf{I}\leq\mathbf{J}$ and $\mathbf{I}\neq\mathbf{J}$). Thus,
$\mathbf{I}<\mathbf{J}\leq n$, so that $\mathbf{I}\leq n-1$ (since
$\mathbf{I}$ and $n$ are integers), and thus $\mathbf{I}+1\leq n$. This allows
us to speak of the vector $w_{\mathbf{I}+1}$.

Now, there clearly exists a permutation $\tau\in S_{n}$ such that $\tau\left(
\mathbf{I}\right)  =\mathbf{I}$ and $\tau\left(  \mathbf{I}+1\right)
=\mathbf{J}$.\ \ \ \ \footnote{\textit{Proof.} We distinguish between two
cases:
\par
\textit{Case 1:} We have $\mathbf{J}=\mathbf{I}+1$.
\par
\textit{Case 2:} We have $\mathbf{J}\neq\mathbf{I}+1$.
\par
First consider Case 1. In this case, the permutation $\operatorname*{id}\in
S_{n}$ satisfies $\operatorname*{id}\left(  \mathbf{I}\right)  =\mathbf{I}$
and $\operatorname*{id}\left(  \mathbf{I}+1\right)  =\mathbf{I}+1=\mathbf{J}$.
Hence, in Case 1, there exists a permutation $\tau\in S_{n}$ such that
$\tau\left(  \mathbf{I}\right)  =\mathbf{I}$ and $\tau\left(  \mathbf{I}%
+1\right)  =\mathbf{J}$ (namely, $\tau=\operatorname*{id}$).
\par
Now let us consider Case 2. In this case, $\mathbf{J}\neq\mathbf{I}+1$. Hence,
the transposition $\left(  \mathbf{J},\mathbf{I}+1\right)  \in S_{n}$ is
well-defined, and it satisfies $\left(  \mathbf{J},\mathbf{I}+1\right)
\left(  \mathbf{I}\right)  =\mathbf{I}$ (since $\mathbf{J}\neq\mathbf{I}$ and
$\mathbf{I}+1\neq\mathbf{I}$) and $\left(  \mathbf{J},\mathbf{I}+1\right)
\left(  \mathbf{I}+1\right)  =\mathbf{J}$. Hence, in Case 2, there exists a
permutation $\tau\in S_{n}$ such that $\tau\left(  \mathbf{I}\right)
=\mathbf{I}$ and $\tau\left(  \mathbf{I}+1\right)  =\mathbf{J}$ (namely,
$\tau=\left(  \mathbf{J},\mathbf{I}+1\right)  $).
\par
We have thus proven in each of the two possible cases that there exists a
permutation $\tau\in S_{n}$ such that $\tau\left(  \mathbf{I}\right)
=\mathbf{I}$ and $\tau\left(  \mathbf{I}+1\right)  =\mathbf{J}$.
\par
This completes the proof that there always exists a permutation $\tau\in
S_{n}$ such that $\tau\left(  \mathbf{I}\right)  =\mathbf{I}$ and $\tau\left(
\mathbf{I}+1\right)  =\mathbf{J}$.} Consider such a $\tau$. From $\tau\left(
\mathbf{I}\right)  =\mathbf{I}$, we obtain $w_{\tau\left(  \mathbf{I}\right)
}=w_{\mathbf{I}}=w_{\mathbf{J}}=w_{\tau\left(  \mathbf{I}+1\right)  }$ (since
$\mathbf{J}=\tau\left(  \mathbf{I}+1\right)  $).

Now, since $\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in V^{n}$ and $\tau\in
S_{n}$, we have $\left(  \left(  w_{1},w_{2},\ldots,w_{n}\right)
,\tau\right)  \in V^{n}\times S_{n}$, so that%
\begin{align*}
&  w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}-\left(  -1\right)  ^{\tau
}w_{\tau\left(  1\right)  }\otimes w_{\tau\left(  2\right)  }\otimes
\cdots\otimes w_{\tau\left(  n\right)  }\\
&  \in\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-\left(  -1\right)
^{\sigma}v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(  2\right)
}\otimes\cdots\otimes v_{\sigma\left(  n\right)  }\ \mid\ \left(  \left(
v_{1},v_{2},\ldots,v_{n}\right)  ,\sigma\right)  \in V^{n}\times S_{n}\right\}
\\
&  \subseteq\left\langle \left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}-\left(  -1\right)  ^{\sigma}v_{\sigma\left(  1\right)  }\otimes
v_{\sigma\left(  2\right)  }\otimes\cdots\otimes v_{\sigma\left(  n\right)
}\ \mid\ \left(  \left(  v_{1},v_{2},\ldots,v_{n}\right)  ,\sigma\right)  \in
V^{n}\times S_{n}\right\}  \right\rangle \\
&  =\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-\left(
-1\right)  ^{\sigma}v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(
2\right)  }\otimes\cdots\otimes v_{\sigma\left(  n\right)  }\ \mid\ \left(
\left(  v_{1},v_{2},\ldots,v_{n}\right)  ,\sigma\right)  \in V^{n}\times
S_{n}\right\rangle \\
&  =Q_{n}\left(  V\right)  \subseteq\widetilde{R}_{n}\left(  V\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.QinR-1}}\right)  .
\end{align*}
On the other hand, the $n$-tuple $\left(  w_{\tau\left(  1\right)  }%
,w_{\tau\left(  2\right)  },\ldots,w_{\tau\left(  n\right)  }\right)  \in
V^{n}$ satisfies $w_{\tau\left(  \mathbf{I}\right)  }=w_{\tau\left(
\mathbf{I}+1\right)  }$. Hence, (\ref{pf.QinR-3.1}) (applied to $\left(
v_{1},v_{2},\ldots,v_{n}\right)  =\left(  w_{\tau\left(  1\right)  }%
,w_{\tau\left(  2\right)  },\ldots,w_{\tau\left(  n\right)  }\right)  $)
yields $w_{\tau\left(  1\right)  }\otimes w_{\tau\left(  2\right)  }%
\otimes\cdots\otimes w_{\tau\left(  n\right)  }\in\widetilde{R}_{n}\left(
V\right)  $.

Now,%
\begin{align*}
&  w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}\\
&  =\underbrace{\left(  w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}-\left(
-1\right)  ^{\tau}w_{\tau\left(  1\right)  }\otimes w_{\tau\left(  2\right)
}\otimes\cdots\otimes w_{\tau\left(  n\right)  }\right)  }_{\in\widetilde{R}%
_{n}\left(  V\right)  }+\left(  -1\right)  ^{\tau}\underbrace{w_{\tau\left(
1\right)  }\otimes w_{\tau\left(  2\right)  }\otimes\cdots\otimes
w_{\tau\left(  n\right)  }}_{\in\widetilde{R}_{n}\left(  V\right)  }\\
&  \in\widetilde{R}_{n}\left(  V\right)  +\left(  -1\right)  ^{\tau
}\widetilde{R}_{n}\left(  V\right)  \subseteq\widetilde{R}_{n}\left(
V\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\widetilde{R}_{n}\left(
V\right)  \text{ is a }k\text{-module}\right)  .
\end{align*}
This proves \textbf{(ii)}.

\textbf{(iii)} According to \textbf{(ii)}, every $\left(  \left(  w_{1}%
,w_{2},\ldots,w_{n}\right)  ,\left(  \mathbf{I},\mathbf{J}\right)  \right)
\in V^{n}\times\left\{  1,2,\ldots,n\right\}  ^{2}$ satisfying $\mathbf{I}%
\neq\mathbf{J}$ and $w_{\mathbf{I}}=w_{\mathbf{J}}$ must satisfy $w_{1}\otimes
w_{2}\otimes\cdots\otimes w_{n}\in\widetilde{R}_{n}\left(  V\right)  $.

In other words,%
\begin{align*}
&  \left\{  w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}\ \mid\ \left(
\left(  w_{1},w_{2},\ldots,w_{n}\right)  ,\left(  \mathbf{I},\mathbf{J}%
\right)  \right)  \in V^{n}\times\left\{  1,2,\ldots,n\right\}  ^{2}%
;\ \mathbf{I}\neq\mathbf{J};\ w_{\mathbf{I}}=w_{\mathbf{J}}\right\} \\
&  \subseteq\widetilde{R}_{n}\left(  V\right)  .
\end{align*}
Thus, Proposition \ref{prop.<>} \textbf{(a)} (applied to $V^{\otimes n}$,
\newline$\left\{  w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}\ \mid\ \left(
\left(  w_{1},w_{2},\ldots,w_{n}\right)  ,\left(  \mathbf{I},\mathbf{J}%
\right)  \right)  \in V^{n}\times\left\{  1,2,\ldots,n\right\}  ^{2}%
;\ \mathbf{I}\neq\mathbf{J};\ w_{\mathbf{I}}=w_{\mathbf{J}}\right\}  $ and
$\widetilde{R}_{n}\left(  V\right)  $ instead of $M$, $S$ and $Q$) yields%
\begin{align*}
&  \left\langle \left\{  w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}%
\ \mid\ \left(  \left(  w_{1},w_{2},\ldots,w_{n}\right)  ,\left(
\mathbf{I},\mathbf{J}\right)  \right)  \in V^{n}\times\left\{  1,2,\ldots
,n\right\}  ^{2};\ \mathbf{I}\neq\mathbf{J};\ w_{\mathbf{I}}=w_{\mathbf{J}%
}\right\}  \right\rangle \\
&  \subseteq\widetilde{R}_{n}\left(  V\right)  .
\end{align*}
Now,%
\begin{align*}
&  R_{n}\left(  V\right) \\
&  =\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(
\left(  v_{1},v_{2},\ldots,v_{n}\right)  ,\left(  i,j\right)  \right)  \in
V^{n}\times\left\{  1,2,\ldots,n\right\}  ^{2};\ i\neq j;\ v_{i}%
=v_{j}\right\rangle \\
&  =\left\langle \left\{  w_{1}\otimes w_{2}\otimes\cdots\otimes w_{n}%
\ \mid\ \left(  \left(  w_{1},w_{2},\ldots,w_{n}\right)  ,\left(
\mathbf{I},\mathbf{J}\right)  \right)  \in V^{n}\times\left\{  1,2,\ldots
,n\right\}  ^{2};\ \mathbf{I}\neq\mathbf{J};\ w_{\mathbf{I}}=w_{\mathbf{J}%
}\right\}  \right\rangle \\
&  \subseteq\widetilde{R}_{n}\left(  V\right)  .
\end{align*}
This proves Lemma \ref{lem.QinR-3}.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.R_n}.]Lemma \ref{lem.QinR-3} yields
$R_{n}\left(  V\right)  \subseteq\widetilde{R}_{n}\left(  V\right)  $. Lemma
\ref{lem.QinR-2} yields $\widetilde{R}_{n}\left(  V\right)  \subseteq
R_{n}\left(  V\right)  $. Combining these two inclusions, we obtain
$\widetilde{R}_{n}\left(  V\right)  =R_{n}\left(  V\right)  $. Thus,%
\[
R_{n}\left(  V\right)  =\widetilde{R}_{n}\left(  V\right)  =\sum_{i=1}%
^{n-1}\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n};\ v_{i}=v_{i+1}\right\rangle .
\]
This proves Proposition \ref{prop.R_n}.
\end{proof}

The analogue of Lemma \ref{lem.Q_n} looks as follows:

\begin{lemma}
\label{lem.R_n}Let $k$ be a commutative ring. Let $V$ be a $k$-module. Let
$n\in\mathbb{N}$. Let $i\in\left\{  1,2,\ldots,n-1\right\}  $.\newline Then,%
\begin{align*}
&  \left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n};\ v_{i}=v_{i+1}\right\rangle \\
&  =V^{\otimes\left(  i-1\right)  }\cdot\left(  R_{2}\left(  V\right)
\right)  \cdot V^{\otimes\left(  n-1-i\right)  }.
\end{align*}
Here, we consider $V^{\otimes n}$ as a $k$-submodule of $\otimes V$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.R_n}.]Let $V^{\Delta}$ be the $k$-submodule $\left\{
\left(  v,v\right)  \ \mid\ v\in V\right\}  $ of $V^{2}$. Then, $V^{\Delta
}=\left\{  \left(  v,w\right)  \in V^{2}\ \mid\ v=w\right\}  $. Hence, for
every $\left(  v_{i},v_{i+1}\right)  \in V^{2}$, we have%
\begin{equation}
\left(  v_{i},v_{i+1}\right)  \in V^{\Delta}\text{ if and only if }%
v_{i}=v_{i+1}. \label{pf.R_n.1}%
\end{equation}


Define a map $a:V^{i-1}\rightarrow V^{\otimes\left(  i-1\right)  }$ by%
\[
\left(  a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  =v_{1}\otimes v_{2}%
\otimes\cdots\otimes v_{i-1}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(
v_{1},v_{2},\ldots,v_{i-1}\right)  \in V^{i-1}\right)  .
\]
Define a map $b:V^{\Delta}\rightarrow V^{\otimes2}$ by%
\[
\left(  b\left(  v_{i},v_{i+1}\right)  =v_{i}\otimes v_{i+1}%
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  v_{i},v_{i+1}\right)  \in
V^{\Delta}\right)  .
\]
(Of course, every $\left(  v_{i},v_{i+1}\right)  \in V^{\Delta}$ in fact
satisfies $v_{i}=v_{i+1}$ by the definition of $V^{\Delta}$; but we still use
different letters for $v_{i}$ and $v_{i+1}$ here to make this notation match
another one.) Define a map $c:V^{n-1-i}\rightarrow V^{\otimes\left(
n-1-i\right)  }$ by%
\[
\left(  c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)  =v_{i+2}\otimes
v_{i+3}\otimes\cdots\otimes v_{n}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(
v_{i+2},v_{i+3},\ldots,v_{n}\right)  \in V^{n-1-i}\right)  .
\]
Since $V^{\otimes\left(  i-1\right)  }$, $V^{\otimes2}$ and $V^{\otimes\left(
n-1-i\right)  }$ are $k$-submodules of $\otimes V$, we can consider all three
maps $a$, $b$ and $c$ as maps to the set $\otimes V$.

It is now easy to see that every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{n}$ such that $v_{i}=v_{i+1}$ satisfies $\left(  v_{i},v_{i+1}\right)  \in
V^{\Delta}$ and%
\[
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}=a\left(  v_{1},v_{2}%
,\ldots,v_{i-1}\right)  \cdot b\left(  v_{i},v_{i+1}\right)  \cdot c\left(
v_{i+2},v_{i+3},\ldots,v_{n}\right)  ,
\]
where the multiplication on the right hand side is the multiplication in the
tensor algebra $\otimes V$.\ \ \ \ \footnote{\textit{Proof.} Let $\left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$ be such that $v_{i}=v_{i+1}$.
Then, $\left(  \underbrace{v_{i}}_{=v_{i+1}},v_{i+1}\right)  =\left(
v_{i+1},v_{i+1}\right)  \in\left\{  \left(  v,v\right)  \ \mid\ v\in
V\right\}  =V^{\Delta}$. Recalling Convention \ref{conv.(X)^n.ident}, we have%
\begin{align*}
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}  &  =\underbrace{\left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{i-1}\right)  }_{=a\left(
v_{1},v_{2},\ldots,v_{i-1}\right)  }\otimes\underbrace{\left(  v_{i}\otimes
v_{i+1}\right)  }_{=b\left(  v_{i},v_{i+1}\right)  }\otimes\underbrace{\left(
v_{i+2}\otimes v_{i+3}\otimes\cdots\otimes v_{n}\right)  }_{=c\left(
v_{i+2},v_{i+3},\ldots,v_{n}\right)  }\\
&  =a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \otimes b\left(  v_{i}%
,v_{i+1}\right)  \otimes c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)  .
\end{align*}
\par
On the other hand, (\ref{*=(X)}) (applied to $a\left(  v_{1},v_{2}%
,\ldots,v_{i-1}\right)  $, $b\left(  v_{i},v_{i+1}\right)  $, $i-1$ and $2$
instead of $a$, $b$, $n$ and $m$) yields%
\[
a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \cdot b\left(  v_{i}%
,v_{i+1}\right)  =a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \otimes b\left(
v_{i},v_{i+1}\right)  .
\]
Also, (\ref{*=(X)}) (applied to $a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)
\cdot b\left(  v_{i},v_{i+1}\right)  $, $c\left(  v_{i+2},v_{i+3},\ldots
,v_{n}\right)  $, $i+1$ and $n-1-i$ instead of $a$, $b$, $n$ and $m$) yields%
\begin{align*}
&  a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \cdot b\left(  v_{i}%
,v_{i+1}\right)  \cdot c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right) \\
&  =\underbrace{\left(  a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \cdot
b\left(  v_{i},v_{i+1}\right)  \right)  }_{=a\left(  v_{1},v_{2}%
,\ldots,v_{i-1}\right)  \otimes b\left(  v_{i},v_{i+1}\right)  }\otimes
c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right) \\
&  =a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \otimes b\left(  v_{i}%
,v_{i+1}\right)  \otimes c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)
=v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n},
\end{align*}
qed.} Thus,%
\begin{align}
&  \left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n};\ v_{i}=v_{i+1}\right\rangle
\nonumber\\
&  =\left\langle a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \cdot b\left(
v_{i},v_{i+1}\right)  \cdot c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n};\ v_{i}%
=v_{i+1}\right\rangle \nonumber\\
&  =\left\langle a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \cdot b\left(
v_{i},v_{i+1}\right)  \cdot c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)
\right. \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left.  \ \mid\ \left(  \left(  v_{1},v_{2}%
,\ldots,v_{i-1}\right)  ,\left(  v_{i},v_{i+1}\right)  ,\left(  v_{i+2}%
,v_{i+3},\ldots,v_{n}\right)  \right)  \in V^{i-1}\times V^{2}\times
V^{n-1-i};\ \underbrace{v_{i}=v_{i+1}}_{\substack{\text{by (\ref{pf.R_n.1}),
this is}\\\text{equivalent to}\\\left(  v_{i},v_{i+1}\right)  \in V^{\Delta}%
}}\right\rangle \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted the triple }\left(  \left(  v_{1},v_{2}%
,\ldots,v_{i-1}\right)  ,\left(  v_{i},v_{i+1}\right)  ,\left(  v_{i+2}%
,v_{i+3},\ldots,v_{n}\right)  \right) \\
\text{ for the }n\text{-tuple }\left(  v_{1},v_{2},\ldots,v_{n}\right)
\end{array}
\right) \nonumber\\
&  =\left\langle a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \cdot b\left(
v_{i},v_{i+1}\right)  \cdot c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)
\right. \nonumber\\
&  \left.  \ \mid\ \left(  \left(  v_{1},v_{2},\ldots,v_{i-1}\right)  ,\left(
v_{i},v_{i+1}\right)  ,\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)  \right)
\in V^{i-1}\times V^{2}\times V^{n-1-i};\ \left(  v_{i},v_{i+1}\right)  \in
V^{\Delta}\right\rangle \nonumber\\
&  =\left\langle a\left(  x\right)  b\left(  y\right)  c\left(  z\right)
\ \mid\ \underbrace{\left(  x,y,z\right)  \in V^{i-1}\times V^{2}\times
V^{n-1-i};\ y\in V^{\Delta}}_{\text{this is equivalent to }\left(
x,y,z\right)  \in V^{i-1}\times V^{\Delta}\times V^{n-1-i}}\right\rangle
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed }\left(  \left(
v_{1},v_{2},\ldots,v_{i-1}\right)  ,\left(  v_{i},v_{i+1}\right)  ,\left(
v_{i+2},v_{i+3},\ldots,v_{n}\right)  \right)  \text{ as }\left(  x,y,z\right)
\right) \nonumber\\
&  =\left\langle a\left(  x\right)  b\left(  y\right)  c\left(  z\right)
\ \mid\ \left(  x,y,z\right)  \in V^{i-1}\times V^{\Delta}\times
V^{n-1-i}\right\rangle . \label{pf.R_n.7}%
\end{align}


But Lemma \ref{lem.ABC} \textbf{(b)} (applied to $X=V^{i-1}$, $Y=V^{\Delta}$,
$Z=V^{n-1-i}$ and $P=\otimes V$) yields%
\begin{align*}
&  \left\langle a\left(  x\right)  \ \mid\ x\in V^{i-1}\right\rangle
\cdot\left\langle b\left(  y\right)  \ \mid\ y\in V^{\Delta}\right\rangle
\cdot\left\langle c\left(  z\right)  \ \mid\ z\in V^{n-1-i}\right\rangle \\
&  =\left\langle a\left(  x\right)  b\left(  y\right)  c\left(  z\right)
\ \mid\ \left(  x,y,z\right)  \in V^{i-1}\times V^{\Delta}\times
V^{n-1-i}\right\rangle .
\end{align*}
Compared to (\ref{pf.R_n.7}), this yields%
\begin{align}
&  \left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n};\ v_{i}=v_{i+1}\right\rangle
\nonumber\\
&  =\left\langle a\left(  x\right)  \ \mid\ x\in V^{i-1}\right\rangle
\cdot\left\langle b\left(  y\right)  \ \mid\ y\in V^{\Delta}\right\rangle
\cdot\left\langle c\left(  z\right)  \ \mid\ z\in V^{n-1-i}\right\rangle .
\label{pf.R_n.9}%
\end{align}
But%
\begin{align*}
\left\langle a\left(  x\right)  \ \mid\ x\in V^{i-1}\right\rangle  &
=\left\langle \underbrace{a\left(  v_{1},v_{2},\ldots,v_{i-1}\right)
}_{=v_{1}\otimes v_{2}\otimes\cdots\otimes v_{i-1}}\ \mid\ \left(  v_{1}%
,v_{2},\ldots,v_{i-1}\right)  \in V^{i-1}\right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed }x\text{ as }\left(
v_{1},v_{2},\ldots,v_{i-1}\right)  \right) \\
&  =\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{i-1}\ \mid\ \left(
v_{1},v_{2},\ldots,v_{i-1}\right)  \in V^{i-1}\right\rangle =V^{\otimes\left(
i-1\right)  }%
\end{align*}
(since the $k$-module $V^{\otimes\left(  i-1\right)  }$ is generated by its
pure tensors, i. e., by tensors of the form $v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{i-1}$ with $\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  \in
V^{i-1}$). Also,%
\begin{align*}
\left\langle c\left(  z\right)  \ \mid\ z\in V^{n-1-i}\right\rangle  &
=\left\langle \underbrace{c\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)
}_{=v_{i+2}\otimes v_{i+3}\otimes\cdots\otimes v_{n}}\ \mid\ \left(
v_{i+2},v_{i+3},\ldots,v_{n}\right)  \in V^{n-1-i}\right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed }z\text{ as }\left(
v_{i+2},v_{i+3},\ldots,v_{n}\right)  \right) \\
&  =\left\langle v_{i+2}\otimes v_{i+3}\otimes\cdots\otimes v_{n}%
\ \mid\ \left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)  \in V^{n-1-i}%
\right\rangle =V^{\otimes\left(  n-1-i\right)  }%
\end{align*}
(since the $k$-module $V^{\otimes\left(  n-1-i\right)  }$ is generated by its
pure tensors, i. e., by tensors of the form $v_{i+2}\otimes v_{i+3}%
\otimes\cdots\otimes v_{n}$ with $\left(  v_{i+2},v_{i+3},\ldots,v_{n}\right)
\in V^{n-1-i}$). Also, the map%
\begin{equation}
V\rightarrow V^{\Delta},\ \ \ \ \ \ \ \ \ \ v\mapsto\left(  v,v\right)
\label{pf.R_n.11}%
\end{equation}
is a bijection (this follows easily by the definition of $V^{\Delta}$), and
thus we have%
\begin{align*}
\left\langle b\left(  y\right)  \ \mid\ y\in V^{\Delta}\right\rangle  &
=\left\langle \underbrace{b\left(  v,v\right)  }_{\substack{=v\otimes
v\\\text{(by the definition of }b\text{)}}}\ \mid\ v\in V\right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }\left(  v,v\right)  \text{ for }y\text{,
because}\\
\text{the map (\ref{pf.R_n.11}) is a bijection}%
\end{array}
\right) \\
&  =\left\langle v\otimes v\ \mid\ v\in V\right\rangle =R_{2}\left(  V\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Corollary \ref{coro.R_2}}\right)  .
\end{align*}
Thus, (\ref{pf.R_n.9}) becomes%
\begin{align*}
&  \left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n};\ v_{i}=v_{i+1}\right\rangle \\
&  =\underbrace{\left\langle a\left(  x\right)  \ \mid\ x\in V^{i-1}%
\right\rangle }_{=V^{\otimes\left(  i-1\right)  }}\cdot
\underbrace{\left\langle b\left(  y\right)  \ \mid\ y\in V^{\Delta
}\right\rangle }_{=R_{2}\left(  V\right)  }\cdot\underbrace{\left\langle
c\left(  z\right)  \ \mid\ z\in V^{n-1-i}\right\rangle }_{=V^{\otimes\left(
n-1-i\right)  }}\\
&  =V^{\otimes\left(  i-1\right)  }\cdot\left(  R_{2}\left(  V\right)
\right)  \cdot V^{\otimes\left(  n-1-i\right)  },
\end{align*}
so that Lemma \ref{lem.R_n} is proven.
\end{proof}

Next, the analogue of Corollary \ref{cor.Q_n}:

\begin{corollary}
\label{cor.R_n}Let $k$ be a commutative ring. Let $V$ be a $k$-module. Let
$n\in\mathbb{N}$.\newline Then,%
\[
R_{n}\left(  V\right)  =\sum_{i=1}^{n-1}V^{\otimes\left(  i-1\right)  }%
\cdot\left(  R_{2}\left(  V\right)  \right)  \cdot V^{\otimes\left(
n-1-i\right)  }%
\]
(this is an equality between $k$-submodules of $\otimes V$, where
$R_{n}\left(  V\right)  $ becomes such a $k$-submodule by means of the
inclusion $R_{n}\left(  V\right)  \subseteq V^{\otimes n}\subseteq\otimes V$).
Here, the multiplication on the right hand side is multiplication inside the
$k$-algebra $\otimes V$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.R_n}.]By Proposition \ref{prop.R_n}, we have%
\begin{align*}
R_{n}\left(  V\right)   &  =\sum_{i=1}^{n-1}\underbrace{\left\langle
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{n};\ v_{i}=v_{i+1}\right\rangle }_{=V^{\otimes
\left(  i-1\right)  }\cdot\left(  R_{2}\left(  V\right)  \right)  \cdot
V^{\otimes\left(  n-1-i\right)  }\text{ (by Lemma \ref{lem.R_n})}}\\
&  =\sum_{i=1}^{n-1}V^{\otimes\left(  i-1\right)  }\cdot\left(  R_{2}\left(
V\right)  \right)  \cdot V^{\otimes\left(  n-1-i\right)  }.
\end{align*}
Thus, Corollary \ref{cor.R_n} is proven.
\end{proof}

Now the analogue of Theorem \ref{thm.Q}:

\begin{theorem}
\label{thm.R}Let $k$ be a commutative ring. Let $V$ be a $k$-module. We know
that $R_{n}\left(  V\right)  $ is a $k$-submodule of $V^{\otimes n}$ for every
$n\in\mathbb{N}$. Thus, $\bigoplus\limits_{n\in\mathbb{N}}R_{n}\left(
V\right)  $ is a $k$-submodule of $\bigoplus\limits_{n\in\mathbb{N}}V^{\otimes
n}=\otimes V$. This $k$-submodule satisfies%
\[
\bigoplus\limits_{n\in\mathbb{N}}R_{n}\left(  V\right)  =\left(  \otimes
V\right)  \cdot\left(  R_{2}\left(  V\right)  \right)  \cdot\left(  \otimes
V\right)  .
\]

\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.R}.]The proof of Theorem \ref{thm.R} using
Corollary \ref{cor.R_n} is completely analogous to the proof of Theorem
\ref{thm.Q} using Corollary \ref{cor.Q_n}.
\end{proof}

Now we can finally define the exterior algebra, similarly to Definition
\ref{defs.ext-alg}:

\begin{definition}
\label{defs.wedge-alg}Let $k$ be a commutative ring. Let $V$ be a
$k$-module.\newline By Theorem \ref{thm.R}, the two $k$-submodules
$\bigoplus\limits_{n\in\mathbb{N}}R_{n}\left(  V\right)  $ and $\left(
\otimes V\right)  \cdot\left(  R_{2}\left(  V\right)  \right)  \cdot\left(
\otimes V\right)  $ of $\otimes V$ are identic (where $\bigoplus
\limits_{n\in\mathbb{N}}R_{n}\left(  V\right)  $ becomes a $k$-submodule of
$\otimes V$ in the same way as explained in Theorem \ref{thm.R}). We denote
these two identic $k$-submodules by $R\left(  V\right)  $. In other words, we
define $R\left(  V\right)  $ by%
\[
R\left(  V\right)  =\bigoplus\limits_{n\in\mathbb{N}}R_{n}\left(  V\right)
=\left(  \otimes V\right)  \cdot\left(  R_{2}\left(  V\right)  \right)
\cdot\left(  \otimes V\right)  .
\]
Since $R\left(  V\right)  =\left(  \otimes V\right)  \cdot\left(  R_{2}\left(
V\right)  \right)  \cdot\left(  \otimes V\right)  $, it is clear that
$R\left(  V\right)  $ is a two-sided ideal of the $k$-algebra $\otimes
V$.\newline Now we define a $k$-module $\wedge V$ as the direct sum
$\bigoplus\limits_{n\in\mathbb{N}}\wedge^{n}V$. Then,%
\begin{align*}
\left.  \wedge V\right.   &  =\bigoplus\limits_{n\in\mathbb{N}}%
\underbrace{\wedge^{n}V}_{=V^{\otimes n}\diagup R_{n}\left(  V\right)
}=\bigoplus\limits_{n\in\mathbb{N}}\left(  V^{\otimes n}\diagup R_{n}\left(
V\right)  \right)  \cong\underbrace{\left(  \bigoplus\limits_{n\in\mathbb{N}%
}V^{\otimes n}\right)  }_{=\otimes V}\diagup\underbrace{\left(  \bigoplus
\limits_{n\in\mathbb{N}}R_{n}\left(  V\right)  \right)  }_{=R\left(  V\right)
}\\
&  =\left(  \otimes V\right)  \diagup R\left(  V\right)  .
\end{align*}
This is a canonical isomorphism, so we will use it to identify $\wedge V$ with
$\left(  \otimes V\right)  \diagup R\left(  V\right)  $. Since $R\left(
V\right)  $ is a two-sided ideal of the $k$-algebra $\otimes V$, the quotient
$k$-module $\left(  \otimes V\right)  \diagup R\left(  V\right)  $ canonically
becomes a $k$-algebra. Since $\wedge V=\left(  \otimes V\right)  \diagup
R\left(  V\right)  $, this means that $\wedge V$ becomes a $k$-algebra. We
refer to this $k$-algebra as the \textit{exterior algebra} of the $k$-module
$V$.\newline We denote by $\operatorname*{wedge}\nolimits_{V}$ the canonical
projection $\otimes V\rightarrow\left(  \otimes V\right)  \diagup R\left(
V\right)  =\wedge V$. Clearly, this map $\operatorname*{wedge}\nolimits_{V}$
is a surjective $k$-algebra homomorphism. Besides, due to $\otimes
V=\bigoplus\limits_{n\in\mathbb{N}}V^{\otimes n}$ and $R\left(  V\right)
=\bigoplus\limits_{n\in\mathbb{N}}R_{n}\left(  V\right)  $, it is clear that
the canonical projection $\otimes V\rightarrow\left(  \otimes V\right)
\diagup R\left(  V\right)  $ is the direct sum of the canonical projections
$V^{\otimes n}\rightarrow V^{\otimes n}\diagup R_{n}\left(  V\right)  $ over
all $n\in\mathbb{N}$. Since the canonical projection $\otimes V\rightarrow
\left(  \otimes V\right)  \diagup R\left(  V\right)  $ is the map
$\operatorname*{wedge}\nolimits_{V}$, whereas the canonical projection
$V^{\otimes n}\rightarrow V^{\otimes n}\diagup R_{n}\left(  V\right)  $ is the
map $\operatorname*{wedge}\nolimits_{V,n}$, this rewrites as follows: The map
$\operatorname*{wedge}\nolimits_{V}$ is the direct sum of the maps
$\operatorname*{wedge}\nolimits_{V,n}$ over all $n\in\mathbb{N}$.\newline When
$v_{1}$, $v_{2}$, $\ldots$, $v_{n}$ are some elements of $V$, one often
abbreviates the element $\operatorname*{wedge}\nolimits_{V}\left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  $ of $\wedge V$ by
$v_{1}\wedge v_{2}\wedge\cdots\wedge v_{n}$. (We will not use this
abbreviation in this following.)
\end{definition}

We should think of the notions $R\left(  V\right)  $, $\wedge V$ and
$\operatorname*{wedge}\nolimits_{V}$ as analogues of the notions $Q\left(
V\right)  $, $\operatorname*{Exter}V$ and $\operatorname*{exter}\nolimits_{V}$
from Definition \ref{defs.ext-alg}, respectively. The next result provides an
analogue of Lemma \ref{lem.Q(f)}:

\begin{lemma}
\label{lem.R(f)}Let $k$ be a commutative ring. Let $V$ and $W$ be two
$k$-modules. Let $f:V\rightarrow W$ be a $k$-module homomorphism.\newline%
\textbf{(a)} Then, the $k$-algebra homomorphism $\otimes f:\otimes
V\rightarrow\otimes W$ satisfies $\left(  \otimes f\right)  \left(  R\left(
V\right)  \right)  \subseteq R\left(  W\right)  $. Also, for every
$n\in\mathbb{N}$, the $k$-module homomorphism $f^{\otimes n}:V^{\otimes
n}\rightarrow W^{\otimes n}$ satisfies $f^{\otimes n}\left(  R_{n}\left(
V\right)  \right)  \subseteq R_{n}\left(  W\right)  $.\newline\textbf{(b)}
Assume that $f$ is surjective. Then, the $k$-algebra homomorphism $\otimes
f:\otimes V\rightarrow\otimes W$ satisfies $\left(  \otimes f\right)  \left(
R\left(  V\right)  \right)  =R\left(  W\right)  $. Also, for every
$n\in\mathbb{N}$, the $k$-module homomorphism $f^{\otimes n}:V^{\otimes
n}\rightarrow W^{\otimes n}$ satisfies $f^{\otimes n}\left(  R_{n}\left(
V\right)  \right)  =R_{n}\left(  W\right)  $.
\end{lemma}

We can prove Lemma \ref{lem.R(f)} by imitating the proof of Lemma
\ref{lem.Q(f)} with some minor changes, but let us instead give a different
proof for a change:

\begin{proof}
[Proof of Lemma \ref{lem.R(f)}.]First, let us prepare.

Corollary \ref{coro.R_2} yields $R_{2}\left(  V\right)  =\left\langle v\otimes
v\ \mid\ v\in V\right\rangle $. Corollary \ref{coro.R_2} (applied to $W$
instead of $V$) yields $R_{2}\left(  W\right)  =\left\langle v\otimes
v\ \mid\ v\in W\right\rangle =\left\langle w\otimes w\ \mid\ w\in
W\right\rangle $.

Now, every $v\in V$ satisfies%
\begin{align*}
\left(  \otimes f\right)  \left(  v\otimes v\right)   &  =f\left(  v\right)
\otimes f\left(  v\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition
of }\otimes f\right) \\
&  \in\left\{  w\otimes w\ \mid\ w\in W\right\}  .
\end{align*}
In other words,
\begin{equation}
\left\{  \left(  \otimes f\right)  \left(  v\otimes v\right)  \ \mid\ v\in
V\right\}  \subseteq\left\{  w\otimes w\ \mid\ w\in W\right\}  .
\label{pf.R(f).0}%
\end{equation}
Thus,%
\[
\left(  \otimes f\right)  \left(  \left\{  v\otimes v\ \mid\ v\in V\right\}
\right)  =\left\{  \left(  \otimes f\right)  \left(  v\otimes v\right)
\ \mid\ v\in V\right\}  \subseteq\left\{  w\otimes w\ \mid\ w\in W\right\}  .
\]
But Proposition \ref{prop.<>} \textbf{(b)} (applied to $\otimes V$, $\left\{
v\otimes v\ \mid\ v\in V\right\}  $, $\otimes W$ and $\otimes f$ instead of
$M$, $S$, $R$ and $f$) yields $\left(  \otimes f\right)  \left(  \left\langle
\left\{  v\otimes v\ \mid\ v\in V\right\}  \right\rangle \right)
=\left\langle \left(  \otimes f\right)  \left(  \left\{  v\otimes
v\ \mid\ v\in V\right\}  \right)  \right\rangle $. Now,%
\[
R_{2}\left(  V\right)  =\left\langle v\otimes v\ \mid\ v\in V\right\rangle
=\left\langle \left\{  v\otimes v\ \mid\ v\in V\right\}  \right\rangle ,
\]
so that%
\begin{align}
\left(  \otimes f\right)  \left(  R_{2}\left(  V\right)  \right)   &  =\left(
\otimes f\right)  \left(  \left\langle \left\{  v\otimes v\ \mid\ v\in
V\right\}  \right\rangle \right)  =\left\langle \underbrace{\left(  \otimes
f\right)  \left(  \left\{  v\otimes v\ \mid\ v\in V\right\}  \right)
}_{\subseteq\left\{  w\otimes w\ \mid\ w\in W\right\}  }\right\rangle
\nonumber\\
&  \subseteq\left\langle \left\{  w\otimes w\ \mid\ w\in W\right\}
\right\rangle =\left\langle w\otimes w\ \mid\ w\in W\right\rangle
=R_{2}\left(  W\right)  . \label{pf.R(f).1}%
\end{align}


By Corollary \ref{cor.R_n}, we have%
\begin{equation}
R_{n}\left(  V\right)  =\sum_{i=1}^{n-1}V^{\otimes\left(  i-1\right)  }%
\cdot\left(  R_{2}\left(  V\right)  \right)  \cdot V^{\otimes\left(
n-1-i\right)  } \label{pf.R(f).2}%
\end{equation}
for every $n\in\mathbb{N}$. Corollary \ref{cor.R_n} (applied to $W$ instead of
$V$) yields
\begin{equation}
R_{n}\left(  W\right)  =\sum_{i=1}^{n-1}W^{\otimes\left(  i-1\right)  }%
\cdot\left(  R_{2}\left(  W\right)  \right)  \cdot W^{\otimes\left(
n-1-i\right)  } \label{pf.R(f).3}%
\end{equation}
for every $n\in\mathbb{N}$.

The map $\otimes f$ is the direct sum of the maps $f^{\otimes n}:V^{\otimes
n}\rightarrow W^{\otimes n}$ for $n\in\mathbb{N}$. Hence, for every
$n\in\mathbb{N}$, the restriction $\left(  \otimes f\right)  \mid_{V^{\otimes
n}}$ of the map $\otimes f$ to $V^{\otimes n}$ is the map $f^{\otimes n}$ (at
least if we ignore the technicality that the targets of the maps $\otimes f$
and $f^{\otimes n}$ are different).

It is also clear that%
\begin{equation}
\left(  \otimes f\right)  \left(  V^{\otimes j}\right)  \subseteq W^{\otimes
j}\ \ \ \ \ \ \ \ \ \ \text{for every }j\in\mathbb{N} \label{pf.R(f).4}%
\end{equation}
(since $\otimes f$ is the direct sum of the maps $f^{\otimes n}:V^{\otimes
n}\rightarrow W^{\otimes n}$ for $n\in\mathbb{N}$).

\textbf{(a)} For every $n\in\mathbb{N}$, we have%
\begin{align*}
\left(  \otimes f\right)  \left(  R_{n}\left(  V\right)  \right)   &  =\left(
\otimes f\right)  \left(  \sum_{i=1}^{n-1}V^{\otimes\left(  i-1\right)  }%
\cdot\left(  R_{2}\left(  V\right)  \right)  \cdot V^{\otimes\left(
n-1-i\right)  }\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.R(f).2}%
)}\right) \\
&  =\sum_{i=1}^{n-1}\underbrace{\left(  \otimes f\right)  \left(
V^{\otimes\left(  i-1\right)  }\right)  }_{\subseteq W^{\otimes\left(
i-1\right)  }\text{ (by (\ref{pf.R(f).4}))}}\cdot\underbrace{\left(  \otimes
f\right)  \left(  R_{2}\left(  V\right)  \right)  }_{\subseteq R_{2}\left(
W\right)  \text{ (by (\ref{pf.R(f).1}))}}\cdot\underbrace{\left(  \otimes
f\right)  \left(  V^{\otimes\left(  n-1-i\right)  }\right)  }_{\subseteq
W^{\otimes\left(  n-1-i\right)  }\text{ (by (\ref{pf.R(f).4}))}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\otimes f\text{ is a
}k\text{-algebra homomorphism}\right) \\
&  \subseteq\sum_{i=1}^{n-1}W^{\otimes\left(  i-1\right)  }\cdot\left(
R_{2}\left(  W\right)  \right)  \cdot W^{\otimes\left(  n-1-i\right)  }%
=R_{n}\left(  W\right)  .
\end{align*}
Since $\left(  \otimes f\right)  \left(  R_{n}\left(  V\right)  \right)
=\underbrace{\left(  \left(  \otimes f\right)  \mid_{V^{\otimes n}}\right)
}_{=f^{\otimes n}}\left(  R_{n}\left(  V\right)  \right)  =f^{\otimes
n}\left(  R_{n}\left(  V\right)  \right)  $, this rewrites as $f^{\otimes
n}\left(  R_{n}\left(  V\right)  \right)  \subseteq R_{n}\left(  W\right)  $.

We have $R\left(  V\right)  =\bigoplus\limits_{n\in\mathbb{N}}R_{n}\left(
V\right)  =\sum\limits_{n\in\mathbb{N}}R_{n}\left(  V\right)  $ (because
direct sums are sums) and $R\left(  W\right)  =\sum\limits_{n\in\mathbb{N}%
}R_{n}\left(  W\right)  $ (similarly). Since $R\left(  V\right)
=\sum\limits_{n\in\mathbb{N}}R_{n}\left(  V\right)  $, we have%
\begin{align*}
\left(  \otimes f\right)  \left(  R\left(  V\right)  \right)   &  =\left(
\otimes f\right)  \left(  \sum\limits_{n\in\mathbb{N}}R_{n}\left(  V\right)
\right)  =\sum\limits_{n\in\mathbb{N}}\underbrace{\left(  \otimes f\right)
\left(  R_{n}\left(  V\right)  \right)  }_{\subseteq R_{n}\left(  W\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\otimes f\text{ is }k\text{-linear}%
\right) \\
&  \subseteq\sum\limits_{n\in\mathbb{N}}R_{n}\left(  W\right)  =R\left(
W\right)  .
\end{align*}


This completes the proof of Lemma \ref{lem.R(f)} \textbf{(a)}.

\textbf{(b)} Assume that the map $f$ is surjective.

Every $w\in W$ satisfies $w\otimes w\in\left\{  \left(  \otimes f\right)
\left(  v\otimes v\right)  \ \mid\ v\in V\right\}  $%
.\ \ \ \ \footnote{\textit{Proof.} Let $w\in W$ be arbitrary. Then, there
exists some $z\in V$ such that $w=f\left(  z\right)  $ (since $f$ is
surjective). Consider this $z$. Then, $w\otimes w=f\left(  z\right)  \otimes
f\left(  z\right)  =\left(  \otimes f\right)  \left(  z\otimes z\right)  $ (by
the definition of $\otimes f$), so that $w\otimes w\in\left\{  \left(  \otimes
f\right)  \left(  v\otimes v\right)  \ \mid\ v\in V\right\}  $, qed.} In other
words, $\left\{  w\otimes w\ \mid\ w\in W\right\}  \subseteq\left\{  \left(
\otimes f\right)  \left(  v\otimes v\right)  \ \mid\ v\in V\right\}  $.
Combined with (\ref{pf.R(f).0}), this yields%
\begin{equation}
\left\{  \left(  \otimes f\right)  \left(  v\otimes v\right)  \ \mid\ v\in
V\right\}  =\left\{  w\otimes w\ \mid\ w\in W\right\}  . \label{pf.R(f).7}%
\end{equation}


Now, in the same way as we used (\ref{pf.R(f).0}) to prove (\ref{pf.R(f).1}),
we can use (\ref{pf.R(f).7}) to prove that%
\begin{equation}
\left(  \otimes f\right)  \left(  R_{n}\left(  V\right)  \right)
=R_{n}\left(  W\right)  . \label{pf.R(f).8}%
\end{equation}


For every $n\in\mathbb{N}$, we have $\left(  \otimes f\right)  \left(
V^{\otimes n}\right)  =\underbrace{\left(  \left(  \otimes f\right)
\mid_{V^{\otimes n}}\right)  }_{=f^{\otimes n}}\left(  V^{\otimes n}\right)
=f^{\otimes n}\left(  V^{\otimes n}\right)  =W^{\otimes n}$ (since $f^{\otimes
n}$ is surjective by Proposition \ref{prop.ext-surj} \textbf{(a)}). Renaming
$n$ as $j$ in this statement, we see that%
\begin{equation}
\left(  \otimes f\right)  \left(  V^{\otimes j}\right)  \subseteq W^{\otimes
j}\ \ \ \ \ \ \ \ \ \ \text{for every }j\in\mathbb{N}. \label{pf.R(f).9}%
\end{equation}


Now, for every $n\in\mathbb{N}$, we have%
\begin{align*}
\left(  \otimes f\right)  \left(  R_{n}\left(  V\right)  \right)   &  =\left(
\otimes f\right)  \left(  \sum_{i=1}^{n-1}V^{\otimes\left(  i-1\right)  }%
\cdot\left(  R_{2}\left(  V\right)  \right)  \cdot V^{\otimes\left(
n-1-i\right)  }\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.R(f).2}%
)}\right) \\
&  =\sum_{i=1}^{n-1}\underbrace{\left(  \otimes f\right)  \left(
V^{\otimes\left(  i-1\right)  }\right)  }_{=W^{\otimes\left(  i-1\right)
}\text{ (by (\ref{pf.R(f).9}))}}\cdot\underbrace{\left(  \otimes f\right)
\left(  R_{2}\left(  V\right)  \right)  }_{=R_{2}\left(  W\right)  \text{ (by
(\ref{pf.R(f).8}))}}\cdot\underbrace{\left(  \otimes f\right)  \left(
V^{\otimes\left(  n-1-i\right)  }\right)  }_{=W^{\otimes\left(  n-1-i\right)
}\text{ (by (\ref{pf.R(f).9}))}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\otimes f\text{ is a
}k\text{-algebra homomorphism}\right) \\
&  =\sum_{i=1}^{n-1}W^{\otimes\left(  i-1\right)  }\cdot\left(  R_{2}\left(
W\right)  \right)  \cdot W^{\otimes\left(  n-1-i\right)  }=R_{n}\left(
W\right)  .
\end{align*}
Since $\left(  \otimes f\right)  \left(  R_{n}\left(  V\right)  \right)
=\underbrace{\left(  \left(  \otimes f\right)  \mid_{V^{\otimes n}}\right)
}_{=f^{\otimes n}}\left(  R_{n}\left(  V\right)  \right)  =f^{\otimes
n}\left(  R_{n}\left(  V\right)  \right)  $, this rewrites as $f^{\otimes
n}\left(  R_{n}\left(  V\right)  \right)  =R_{n}\left(  W\right)  $.

We have $R\left(  V\right)  =\bigoplus\limits_{n\in\mathbb{N}}R_{n}\left(
V\right)  =\sum\limits_{n\in\mathbb{N}}R_{n}\left(  V\right)  $ (because
direct sums are sums) and $R\left(  W\right)  =\sum\limits_{n\in\mathbb{N}%
}R_{n}\left(  W\right)  $ (similarly). Since $R\left(  V\right)
=\sum\limits_{n\in\mathbb{N}}R_{n}\left(  V\right)  $, we have%
\begin{align*}
\left(  \otimes f\right)  \left(  R\left(  V\right)  \right)   &  =\left(
\otimes f\right)  \left(  \sum\limits_{n\in\mathbb{N}}R_{n}\left(  V\right)
\right)  =\sum\limits_{n\in\mathbb{N}}\underbrace{\left(  \otimes f\right)
\left(  R_{n}\left(  V\right)  \right)  }_{=R_{n}\left(  W\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\otimes f\text{ is }k\text{-linear}%
\right) \\
&  =\sum\limits_{n\in\mathbb{N}}R_{n}\left(  W\right)  =R\left(  W\right)  .
\end{align*}


This completes the proof of Lemma \ref{lem.R(f)} \textbf{(b)}.
\end{proof}

The following definition mirrors Definition \ref{def.ext-functor}:

\begin{definition}
\label{def.wedge-functor}Let $k$ be a commutative ring. Let $V$ and $W$ be two
$k$-modules. Let $f:V\rightarrow W$ be a $k$-module homomorphism. Then, the
$k$-algebra homomorphism $\otimes f:\otimes V\rightarrow\otimes W$ satisfies
$\left(  \otimes f\right)  \left(  R\left(  V\right)  \right)  \subseteq
R\left(  W\right)  $ (by Lemma \ref{lem.R(f)} \textbf{(a)}), and thus gives
rise to a $k$-algebra homomorphism $\left(  \otimes V\right)  \diagup R\left(
V\right)  \rightarrow\left(  \otimes W\right)  \diagup R\left(  W\right)  $.
This latter $k$-algebra homomorphism will be denoted by $\wedge f$. Since
$\left(  \otimes V\right)  \diagup R\left(  V\right)  =\wedge V$ and $\left(
\otimes W\right)  \diagup R\left(  W\right)  =\wedge W$, this homomorphism
$\wedge f:\left(  \otimes V\right)  \diagup R\left(  V\right)  \rightarrow
\left(  \otimes W\right)  \diagup R\left(  W\right)  $ is actually a
homomorphism from $\wedge V$ to $\wedge W$.\newline By the construction of
$\wedge f$, the diagram%
\begin{equation}%
%TCIMACRO{\TeXButton{wedge f from otimes f}{\xymatrix{
%\otimes V \ar[r]^{\otimes f}\ar[d]_{\operatorname*{wedge}_V} & \otimes
%W \ar[d]^{\operatorname*{wedge}_W} \\
%\wedge V \ar[r]_{\wedge f} & \wedge W
%}} }%
%BeginExpansion
\xymatrix{
\otimes V \ar[r]^{\otimes f}\ar[d]_{\operatorname*{wedge}_V} & \otimes
W \ar[d]^{\operatorname*{wedge}_W} \\
\wedge V \ar[r]_{\wedge f} & \wedge W
}
%EndExpansion
\label{def.wedge-functor.diag}%
\end{equation}
commutes (since $\operatorname*{wedge}\nolimits_{V}$ is the canonical
projection $\otimes V\rightarrow\wedge V$ and since $\operatorname*{wedge}%
\nolimits_{W}$ is the canonical projection $\otimes W\rightarrow\wedge W$).
\end{definition}

Needless to say, the notion $\wedge f$ introduced in this definition is an
analogue of the notion $\operatorname*{Exter}f$ introduced in Definition
\ref{def.ext-functor}.

Here is the analogue of Proposition \ref{prop.ext-surj}:

\begin{proposition}
\label{prop.wedge-surj}Let $k$ be a commutative ring. Let $V$ and $W$ be two
$k$-modules. Let $f:V\rightarrow W$ be a surjective $k$-module homomorphism.
Then:\newline\textbf{(a)} The $k$-module homomorphism $f^{\otimes
n}:V^{\otimes n}\rightarrow W^{\otimes n}$ is surjective for every
$n\in\mathbb{N}$.\newline\textbf{(b)} The $k$-algebra homomorphism $\otimes
f:\otimes V\rightarrow\otimes W$ is surjective.\newline\textbf{(c)} The
$k$-algebra homomorphism $\wedge f:\wedge V\rightarrow\wedge W$ is surjective.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.wedge-surj}.]The proof of this Proposition
\ref{prop.wedge-surj} is completely analogous to the proof of Proposition
\ref{prop.ext-surj} (and parts \textbf{(a)} and \textbf{(b)} are even the same).
\end{proof}

So much for analogues of the results of Subsection \ref{subsect.exter}. Now
let us formulate the analogues of the results of Subsection
\ref{subsect.Ker(exter)}. First, the analogue of Theorem \ref{thm.Ker(exter)}:

\begin{theorem}
\label{thm.Ker(wedge)}Let $k$ be a commutative ring. Let $V$ and $V^{\prime} $
be two $k$-modules, and let $f:V\rightarrow V^{\prime}$ be a surjective $k
$-module homomorphism. Then, the kernel of the map $\wedge f:\wedge
V\rightarrow\wedge V^{\prime}$ is%
\[
\operatorname*{Ker}\left(  \wedge f\right)  =\left(  \wedge V\right)
\cdot\operatorname*{wedge}\nolimits_{V}\left(  \operatorname*{Ker}f\right)
\cdot\left(  \wedge V\right)  =\left(  \wedge V\right)  \cdot
\operatorname*{wedge}\nolimits_{V}\left(  \operatorname*{Ker}f\right)
=\operatorname*{wedge}\nolimits_{V}\left(  \operatorname*{Ker}f\right)
\cdot\left(  \wedge V\right)  .
\]
Here, $\operatorname*{Ker}f$ is considered a $k$-submodule of $\otimes V$ by
means of the inclusion $\operatorname*{Ker}f\subseteq V=V^{\otimes1}%
\subseteq\otimes V$.
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.Ker(wedge)}.]The proof of this Theorem
\ref{thm.Ker(wedge)} is completely analogous to that of Theorem
\ref{thm.Ker(exter)}.
\end{proof}

The analogue of Corollary \ref{cor.exter.W} comes next:

\begin{corollary}
\label{cor.wedge.W}Let $k$ be a commutative ring. Let $V$ be a $k$-module, and
let $W$ be a $k$-submodule of $V$. Then,%
\[
\left(  \wedge V\right)  \cdot\operatorname*{wedge}\nolimits_{V}\left(
W\right)  \cdot\left(  \wedge V\right)  =\left(  \wedge V\right)
\cdot\operatorname*{wedge}\nolimits_{V}\left(  W\right)
=\operatorname*{wedge}\nolimits_{V}\left(  W\right)  \cdot\left(  \wedge
V\right)  .
\]
Here, $W$ is considered a $k$-submodule of $\otimes V$ by means of the
inclusion $W\subseteq V=V^{\otimes1}\subseteq\otimes V$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.wedge.W}.]Expectedly, the proof of Corollary
\ref{cor.wedge.W} is analogous to the proof of Corollary \ref{cor.exter.W}.
\end{proof}

Finally, the analogue of Corollary \ref{coro.Ker(exter)}:

\begin{corollary}
\label{coro.Ker(wedge)}Let $k$ be a commutative ring. Let $V$ be a $k$-module.
Let $W$ be a $k$-submodule of $V$, and let $f:V\rightarrow V\diagup W$ be the
canonical projection.\newline\textbf{(a)} Then, the kernel of the map $\wedge
f:\wedge V\rightarrow\wedge\left(  V\diagup W\right)  $ is%
\[
\operatorname*{Ker}\left(  \wedge f\right)  =\left(  \wedge V\right)
\cdot\operatorname*{wedge}\nolimits_{V}\left(  W\right)  \cdot\left(  \wedge
V\right)  =\left(  \wedge V\right)  \cdot\operatorname*{wedge}\nolimits_{V}%
\left(  W\right)  =\operatorname*{wedge}\nolimits_{V}\left(  W\right)
\cdot\left(  \wedge V\right)  .
\]
Here, $W$ is considered a $k$-submodule of $\otimes V$ by means of the
inclusion $W\subseteq V=V^{\otimes1}\subseteq\otimes V$.\newline\textbf{(b)}
We have%
\[
\left(  \wedge V\right)  \diagup\left(  \left(  \wedge V\right)
\cdot\operatorname*{wedge}\nolimits_{V}\left(  W\right)  \right)  \cong%
\wedge\left(  V\diagup W\right)  \ \ \ \ \ \ \ \ \ \ \text{as }%
k\text{-modules.}%
\]

\end{corollary}

\begin{proof}
[Proof of Corollary \ref{coro.Ker(wedge)}.]The proof of Corollary
\ref{coro.Ker(wedge)} is analogous to the proof of Corollary
\ref{coro.Ker(exter)}.
\end{proof}

\subsection{\label{subsect.ExtWedge}The relation between the exterior and
pseudoexterior algebras}

The name ``pseudoexterior'' for the algebra $\operatorname*{Exter}V$
introduced in Definition \ref{defs.ext-alg} already suggests a close relation
to the exterior algebra $\wedge V$. Indeed such a relation is given by the
following two theorems:

\begin{theorem}
\label{thm.QinR}Let $k$ be a commutative ring. Let $V$ be a $k$%
-module.\newline\textbf{(a)} We have $Q_{n}\left(  V\right)  \subseteq
R_{n}\left(  V\right)  $ for all $n\in\mathbb{N}$.\newline\textbf{(b)} We have
$Q\left(  V\right)  \subseteq R\left(  V\right)  $.\newline\textbf{(c)} For
every $n\in\mathbb{N}$, the projection $\operatorname*{wedge}\nolimits_{V,n}%
:V^{\otimes n}\rightarrow\wedge^{n}V$ factors through the projection
$\operatorname*{exter}\nolimits_{V,n}:V^{\otimes n}\rightarrow
\operatorname*{Exter}\nolimits^{n}V$.\newline\textbf{(d)} The projection
$\operatorname*{wedge}\nolimits_{V}:\otimes V\rightarrow\wedge V$ factors
through the projection $\operatorname*{exter}\nolimits_{V}:\otimes
V\rightarrow\operatorname*{Exter}V$.
\end{theorem}

\begin{theorem}
\label{thm.Q=R}Let $k$ be a commutative ring in which $2$ is invertible. Let
$V$ be a $k$-module.\newline\textbf{(a)} We have $Q_{n}\left(  V\right)
=R_{n}\left(  V\right)  $ for all $n\in\mathbb{N}$.\newline\textbf{(b)} We
have $Q\left(  V\right)  =R\left(  V\right)  $.\newline\textbf{(c)} For every
$n\in\mathbb{N}$, we have $\wedge^{n}V=\operatorname*{Exter}\nolimits^{n}V$
and $\operatorname*{wedge}\nolimits_{V,n}=\operatorname*{exter}\nolimits_{V,n}%
$.\newline\textbf{(d)} We have $\wedge V=\operatorname*{Exter}V$ and
$\operatorname*{wedge}\nolimits_{V}=\operatorname*{exter}\nolimits_{V}$.
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.QinR}.]\textbf{(a)} Let us use the notations of
Lemma \ref{lem.QinR-1}. For every $n\in\mathbb{N}$, we have%
\begin{align*}
Q_{n}\left(  V\right)   &  \subseteq\widetilde{R}_{n}\left(  V\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.QinR-1}}\right) \\
&  \subseteq R_{n}\left(  V\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.QinR-2}}\right)  .
\end{align*}
This proves Theorem \ref{thm.QinR} \textbf{(a)}.

\textbf{(b)} We have
\[
Q\left(  V\right)  =\bigoplus\limits_{n\in\mathbb{N}}\underbrace{Q_{n}\left(
V\right)  }_{\substack{\subseteq R_{n}\left(  V\right)  \\\text{(by Theorem
\ref{thm.QinR} \textbf{(a)})}}}\subseteq\bigoplus\limits_{n\in\mathbb{N}}%
R_{n}\left(  V\right)  =R\left(  V\right)  .
\]
This proves Theorem \ref{thm.QinR} \textbf{(b)}.

\textbf{(c)} Let $n\in\mathbb{N}$. The canonical projection $V^{\otimes
n}\rightarrow V^{\otimes n}\diagup R_{n}\left(  V\right)  $ factors through
the canonical projection $V^{\otimes n}\rightarrow V^{\otimes n}\diagup
Q_{n}\left(  V\right)  $ (because $Q_{n}\left(  V\right)  \subseteq
R_{n}\left(  V\right)  $ by Theorem \ref{thm.QinR} \textbf{(a)}). Since the
canonical projection $V^{\otimes n}\rightarrow V^{\otimes n}\diagup
R_{n}\left(  V\right)  $ is the map $\operatorname*{wedge}\nolimits_{V,n}%
:V^{\otimes n}\rightarrow\wedge^{n}V$, and since the canonical projection
$V^{\otimes n}\rightarrow V^{\otimes n}\diagup Q_{n}\left(  V\right)  $ is the
map $\operatorname*{exter}\nolimits_{V,n}:V^{\otimes n}\rightarrow
\operatorname*{Exter}\nolimits^{n}V$, this rewrites as follows: The map
$\operatorname*{wedge}\nolimits_{V,n}:V^{\otimes n}\rightarrow\wedge^{n}V$
factors through the map $\operatorname*{exter}\nolimits_{V,n}:V^{\otimes
n}\rightarrow\operatorname*{Exter}\nolimits^{n}V$. This proves Theorem
\ref{thm.QinR} \textbf{(c)}.

\textbf{(d)} The canonical projection $\otimes V\rightarrow\left(  \otimes
V\right)  \diagup R\left(  V\right)  $ factors through the canonical
projection $\otimes V\rightarrow\left(  \otimes V\right)  \diagup Q\left(
V\right)  $ (because $Q\left(  V\right)  \subseteq R\left(  V\right)  $ by
Theorem \ref{thm.QinR} \textbf{(b)}). Since the canonical projection $\otimes
V\rightarrow\left(  \otimes V\right)  \diagup R\left(  V\right)  $ is the map
$\operatorname*{wedge}\nolimits_{V}:\otimes V\rightarrow\wedge V$, and since
the canonical projection $\otimes V\rightarrow\left(  \otimes V\right)
\diagup Q\left(  V\right)  $ is the map $\operatorname*{exter}\nolimits_{V}%
:\otimes V\rightarrow\operatorname*{Exter}V$, this rewrites as follows: The
map $\operatorname*{wedge}\nolimits_{V}:\otimes V\rightarrow\wedge V$ factors
through the map $\operatorname*{exter}\nolimits_{V}:\otimes V\rightarrow
\operatorname*{Exter}V$. This proves Theorem \ref{thm.QinR} \textbf{(d)}.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.Q=R}.]The main step is to prove that $Q_{2}\left(
V\right)  =R_{2}\left(  V\right)  $. Let us do this now:

Corollary \ref{coro.Q_2} yields%
\[
Q_{2}\left(  V\right)  =\left\langle v_{1}\otimes v_{2}+v_{2}\otimes
v_{1}\ \mid\ \left(  v_{1},v_{2}\right)  \in V^{2}\right\rangle .
\]


Every $v\in V$ satisfies%
\begin{align*}
v\otimes v  &  =\dfrac{1}{2}\left(  v\otimes v+v\otimes v\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }2\text{ is invertible in }k\right) \\
&  =\dfrac{1}{2}v\otimes v+v\otimes\dfrac{1}{2}v\in\left\{  v_{1}\otimes
v_{2}+v_{2}\otimes v_{1}\ \mid\ \left(  v_{1},v_{2}\right)  \in V^{2}\right\}
\end{align*}
(since the tensor $\dfrac{1}{2}v\otimes v+v\otimes\dfrac{1}{2}v$ has the form
$v_{1}\otimes v_{2}+v_{2}\otimes v_{1}$ for $\left(  v_{1},v_{2}\right)
=\left(  \dfrac{1}{2}v,v\right)  $). In other words,%
\[
\left\{  v\otimes v\ \mid\ v\in V\right\}  \subseteq\left\{  v_{1}\otimes
v_{2}+v_{2}\otimes v_{1}\ \mid\ \left(  v_{1},v_{2}\right)  \in V^{2}\right\}
.
\]
Now, Corollary \ref{coro.R_2} yields%
\begin{align*}
R_{2}\left(  V\right)   &  =\left\langle v\otimes v\ \mid\ v\in V\right\rangle
=\left\langle \underbrace{\left\{  v\otimes v\ \mid\ v\in V\right\}
}_{\subseteq\left\{  v_{1}\otimes v_{2}+v_{2}\otimes v_{1}\ \mid\ \left(
v_{1},v_{2}\right)  \in V^{2}\right\}  }\right\rangle \subseteq\left\langle
\left\{  v_{1}\otimes v_{2}+v_{2}\otimes v_{1}\ \mid\ \left(  v_{1}%
,v_{2}\right)  \in V^{2}\right\}  \right\rangle \\
&  =\left\langle v_{1}\otimes v_{2}+v_{2}\otimes v_{1}\ \mid\ \left(
v_{1},v_{2}\right)  \in V^{2}\right\rangle =Q_{2}\left(  V\right)  .
\end{align*}
Combined with $Q_{2}\left(  V\right)  \subseteq R_{2}\left(  V\right)  $
(which follows from Theorem \ref{thm.QinR} \textbf{(a)}, applied to $n=2$),
this yields $Q_{2}\left(  V\right)  =R_{2}\left(  V\right)  $.

\textbf{(a)} Let $n\in\mathbb{N}$. Both $Q_{n}\left(  V\right)  $ and
$R_{n}\left(  V\right)  $ are $k$-submodules of $V^{\otimes n}$ and thus
$k$-submodules of $\otimes V$ (since $V^{\otimes n}\subseteq\otimes V$). Using
the multiplication on the $k$-algebra $\otimes V$, we have%
\begin{align*}
Q_{n}\left(  V\right)   &  =\sum_{i=1}^{n-1}V^{\otimes\left(  i-1\right)
}\cdot\underbrace{\left(  Q_{2}\left(  V\right)  \right)  }_{=R_{2}\left(
V\right)  }\cdot V^{\otimes\left(  n-1-i\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{by Corollary \ref{cor.Q_n}}\right) \\
&  =\sum_{i=1}^{n-1}V^{\otimes\left(  i-1\right)  }\cdot\left(  R_{2}\left(
V\right)  \right)  \cdot V^{\otimes\left(  n-1-i\right)  }=R_{n}\left(
V\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Corollary \ref{cor.R_n}%
}\right)  .
\end{align*}
This proves Theorem \ref{thm.Q=R} \textbf{(a)}.

\textbf{(b)} We have
\[
Q\left(  V\right)  =\bigoplus\limits_{n\in\mathbb{N}}\underbrace{Q_{n}\left(
V\right)  }_{\substack{=R_{n}\left(  V\right)  \\\text{(by Theorem
\ref{thm.Q=R} \textbf{(a)})}}}=\bigoplus\limits_{n\in\mathbb{N}}R_{n}\left(
V\right)  =R\left(  V\right)  .
\]
This proves Theorem \ref{thm.Q=R} \textbf{(b)}.

\textbf{(c)} Let $n\in\mathbb{N}$. Then, $V^{\otimes n}\diagup R_{n}\left(
V\right)  =V^{\otimes n}\diagup Q_{n}\left(  V\right)  $ (because
$R_{n}\left(  V\right)  =Q_{n}\left(  V\right)  $ by Theorem \ref{thm.Q=R}
\textbf{(a)}). Thus, $\wedge^{n}V=V^{\otimes n}\diagup R_{n}\left(  V\right)
=V^{\otimes n}\diagup Q_{n}\left(  V\right)  =\operatorname*{Exten}%
\nolimits^{n}V$.

Since the canonical projection $V^{\otimes n}\rightarrow V^{\otimes n}\diagup
R_{n}\left(  V\right)  $ is the map $\operatorname*{wedge}\nolimits_{V,n}%
:V^{\otimes n}\rightarrow\wedge^{n}V$, and since the canonical projection
$V^{\otimes n}\rightarrow V^{\otimes n}\diagup Q_{n}\left(  V\right)  $ is the
map $\operatorname*{exter}\nolimits_{V,n}:V^{\otimes n}\rightarrow
\operatorname*{Exter}\nolimits^{n}V$, we have $\operatorname*{wedge}%
\nolimits_{V,n}=\operatorname*{exter}\nolimits_{V,n}$ (because $R_{n}\left(
V\right)  =Q_{n}\left(  V\right)  $). This proves Theorem \ref{thm.Q=R}
\textbf{(c)}.

\textbf{(d)} We have $\left(  \otimes V\right)  \diagup R\left(  V\right)
=\left(  \otimes V\right)  \diagup Q\left(  V\right)  $ (because $R\left(
V\right)  =Q\left(  V\right)  $ by Theorem \ref{thm.Q=R} \textbf{(b)}). Since
the canonical projection $\otimes V\rightarrow\left(  \otimes V\right)
\diagup R\left(  V\right)  $ is the map $\operatorname*{wedge}\nolimits_{V}%
:\otimes V\rightarrow\wedge V$, and since the canonical projection $\otimes
V\rightarrow\left(  \otimes V\right)  \diagup Q\left(  V\right)  $ is the map
$\operatorname*{exter}\nolimits_{V}:\otimes V\rightarrow\operatorname*{Exter}%
V$, we have $\operatorname*{wedge}\nolimits_{V}=\operatorname*{exter}%
\nolimits_{V}$ (because $R\left(  V\right)  =Q\left(  V\right)  $). This
proves Theorem \ref{thm.Q=R} \textbf{(d)}.
\end{proof}

\subsection{\label{subsect.Sym.comm}The symmetric algebra is commutative}

In this section we are going to continue the study of the symmetric algebra
that we started in Section \ref{subsect.Sym} and prove some results which
don't have direct analogues for $\operatorname*{Exter}V$ and $\wedge V$
(although some analogues for $\wedge V$ can be found with a little more
effort, which we are not going to make).

The main result of this section will be:

\begin{theorem}
\label{thm.Sym.comm}Let $k$ be a commutative ring. Let $V$ be a $k$-module.
Then, the $k$-algebra $\operatorname*{Sym}V$ is commutative.
\end{theorem}

The standard proof of this theorem proceeds by double induction over the
degrees of the tensors that must be shown to commute. We are going to show a
slightly slicker version of this proof here, which replaces the double
induction by a double application of Lemma \ref{lem.ideal} (which, in its
proof, hides an induction). The intermediate step between these two
applications will be the following lemma:

\begin{lemma}
\label{lem.Symm.comm}Let $k$ be a commutative ring. Let $V$ be a $k$-module.
Every $v\in V$ and every $p\in\operatorname*{Sym}V$ satisfy
$\operatorname*{sym}\nolimits_{V}\left(  v\right)  \cdot p=p\cdot
\operatorname*{sym}\nolimits_{V}\left(  v\right)  $.
\end{lemma}

(Of course, the notations we are using here and everywhere throughout this
section are the notations of Section \ref{subsect.Sym}.)

\begin{proof}
[Proof of Lemma \ref{lem.Symm.comm}.]Let $v\in V$. Let $M$ be the subset%
\[
\left\{  q\in\operatorname*{Sym}V\ \mid\ \operatorname*{sym}\nolimits_{V}%
\left(  v\right)  \cdot q=q\cdot\operatorname*{sym}\nolimits_{V}\left(
v\right)  \right\}
\]
of $\operatorname*{Sym}V$. We are going to prove that $M$ is the whole
$\operatorname*{Sym}V$.

First of all, we have $0\in M$\ \ \ \ \footnote{\textit{Proof.} Clearly,
$\operatorname*{sym}\nolimits_{V}\left(  v\right)  \cdot0=0=0\cdot
\operatorname*{sym}\nolimits_{V}\left(  v\right)  $, so that $0\in\left\{
q\in\operatorname*{Sym}V\ \mid\ \operatorname*{sym}\nolimits_{V}\left(
v\right)  \cdot q=q\cdot\operatorname*{sym}\nolimits_{V}\left(  v\right)
\right\}  =M$.}. Furthermore, every $\alpha\in k$, $\beta\in k$, $p\in M$ and
$r\in M$ satisfy $\alpha p+\beta r\in M$.\ \ \ \ \footnote{\textit{Proof.} Let
$\alpha\in k$, $\beta\in k$, $p\in M$ and $r\in M$.
\par
Since $p\in M=\left\{  q\in\operatorname*{Sym}V\ \mid\ \operatorname*{sym}%
\nolimits_{V}\left(  v\right)  \cdot q=q\cdot\operatorname*{sym}%
\nolimits_{V}\left(  v\right)  \right\}  $, we have $\operatorname*{sym}%
\nolimits_{V}\left(  v\right)  \cdot p=p\cdot\operatorname*{sym}%
\nolimits_{V}\left(  v\right)  $. Similarly, $\operatorname*{sym}%
\nolimits_{V}\left(  v\right)  \cdot r=r\cdot\operatorname*{sym}%
\nolimits_{V}\left(  v\right)  $. Now,%
\begin{align*}
\operatorname*{sym}\nolimits_{V}\left(  v\right)  \cdot\left(  \alpha p+\beta
r\right)   &  =\alpha\underbrace{\operatorname*{sym}\nolimits_{V}\left(
v\right)  \cdot p}_{=p\cdot\operatorname*{sym}\nolimits_{V}\left(  v\right)
}+\beta\underbrace{\operatorname*{sym}\nolimits_{V}\left(  v\right)  \cdot
r}_{=r\cdot\operatorname*{sym}\nolimits_{V}\left(  v\right)  }=\alpha
p\cdot\operatorname*{sym}\nolimits_{V}\left(  v\right)  +\beta r\cdot
\operatorname*{sym}\nolimits_{V}\left(  v\right) \\
&  =\left(  \alpha p+\beta r\right)  \cdot\operatorname*{sym}\nolimits_{V}%
\left(  v\right)  .
\end{align*}
In other words, $\alpha p+\beta r\in\left\{  q\in\operatorname*{Sym}%
V\ \mid\ \operatorname*{sym}\nolimits_{V}\left(  v\right)  \cdot
q=q\cdot\operatorname*{sym}\nolimits_{V}\left(  v\right)  \right\}  =M$, qed.}
In other words, $M$ is a $k$-submodule of $\operatorname*{Sym}V$.

Second, $1\in M$ (with $1$ denoting the unity of the $k$-algebra
$\operatorname*{Sym}V$)\ \ \ \ \footnote{\textit{Proof.} Clearly,
$\operatorname*{sym}\nolimits_{V}\left(  v\right)  \cdot1=\operatorname*{sym}%
\nolimits_{V}\left(  v\right)  =1\cdot\operatorname*{sym}\nolimits_{V}\left(
v\right)  $, so that $1\in\left\{  q\in\operatorname*{Sym}V\ \mid
\ \operatorname*{sym}\nolimits_{V}\left(  v\right)  \cdot q=q\cdot
\operatorname*{sym}\nolimits_{V}\left(  v\right)  \right\}  =M$.}.

On the other hand, every $\left(  p,s\right)  \in M\times\operatorname*{sym}%
\nolimits_{V}\left(  V\right)  $ satisfy $p\cdot s\in M$%
\ \ \ \ \footnote{\textit{Proof.} Let $\left(  p,s\right)  \in M\times
\operatorname*{sym}\nolimits_{V}\left(  V\right)  $. Then, $p\in M$ and
$s\in\operatorname*{sym}\nolimits_{V}\left(  V\right)  $. Since $s\in
\operatorname*{sym}\nolimits_{V}\left(  V\right)  $, there exists some $w\in
V$ such that $s=\operatorname*{sym}\nolimits_{V}\left(  w\right)  $. Consider
this $w$.
\par
Since $p\in M=\left\{  q\in\operatorname*{Sym}V\ \mid\ \operatorname*{sym}%
\nolimits_{V}\left(  v\right)  \cdot q=q\cdot\operatorname*{sym}%
\nolimits_{V}\left(  v\right)  \right\}  $, we have $\operatorname*{sym}%
\nolimits_{V}\left(  v\right)  \cdot p=p\cdot\operatorname*{sym}%
\nolimits_{V}\left(  v\right)  $.
\par
We have $v\in V=V^{\otimes1}$ and $w\in V=V^{\otimes1}$. Hence, $v\cdot
w=v\otimes w$ (by (\ref{*=(X)}), applied to $v$, $w$, $1$ and $1$ instead of
$a$, $b$, $n$ and $m$) and similarly $w\cdot v=w\otimes v$. Thus,%
\begin{align*}
\underbrace{v\cdot w}_{=v\otimes w}-\underbrace{w\cdot v}_{=w\otimes v}  &
=v\otimes w-w\otimes v\\
&  \in\left\{  v_{1}\otimes v_{2}-v_{2}\otimes v_{1}\ \mid\ \left(
v_{1},v_{2}\right)  \in V^{2}\right\} \\
&  \subseteq\left\langle \left\{  v_{1}\otimes v_{2}-v_{2}\otimes v_{1}%
\ \mid\ \left(  v_{1},v_{2}\right)  \in V^{2}\right\}  \right\rangle \\
&  =\left\langle v_{1}\otimes v_{2}-v_{2}\otimes v_{1}\ \mid\ \left(
v_{1},v_{2}\right)  \in V^{2}\right\rangle =K_{2}\left(  V\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Corollary \ref{coro.K_2}}\right) \\
&  \subseteq\bigoplus\limits_{n\in\mathbb{N}}K_{n}\left(  V\right)  =K\left(
V\right)  .
\end{align*}
In other words, $v\cdot w\equiv w\cdot v\operatorname{mod}K\left(  V\right)
$. Since $\operatorname*{sym}\nolimits_{V}$ is the projection $\otimes
V\rightarrow\left(  \otimes V\right)  \diagup K\left(  V\right)  $, this
rewrites as $\operatorname*{sym}\nolimits_{V}\left(  v\cdot w\right)
=\operatorname*{sym}\nolimits_{V}\left(  w\cdot v\right)  $. Since
$\operatorname*{sym}\nolimits_{V}$ is a $k$-algebra homomorphism, we have
$\operatorname*{sym}\nolimits_{V}\left(  v\cdot w\right)  =\operatorname*{sym}%
\nolimits_{V}\left(  v\right)  \cdot\operatorname*{sym}\nolimits_{V}\left(
w\right)  $ and $\operatorname*{sym}\nolimits_{V}\left(  w\cdot v\right)
=\operatorname*{sym}\nolimits_{V}\left(  w\right)  \cdot\operatorname*{sym}%
\nolimits_{V}\left(  v\right)  $.
\par
Now,%
\begin{align*}
\underbrace{\operatorname*{sym}\nolimits_{V}\left(  v\right)  \cdot
p}_{=p\cdot\operatorname*{sym}\nolimits_{V}\left(  v\right)  }\cdot
\underbrace{s}_{=\operatorname*{sym}\nolimits_{V}\left(  w\right)  }  &
=p\cdot\underbrace{\operatorname*{sym}\nolimits_{V}\left(  v\right)
\cdot\operatorname*{sym}\nolimits_{V}\left(  w\right)  }%
_{\substack{=\operatorname*{sym}\nolimits_{V}\left(  v\cdot w\right)
=\operatorname*{sym}\nolimits_{V}\left(  w\cdot v\right)
\\=\operatorname*{sym}\nolimits_{V}\left(  w\right)  \cdot\operatorname*{sym}%
\nolimits_{V}\left(  v\right)  }}=p\cdot\underbrace{\operatorname*{sym}%
\nolimits_{V}\left(  w\right)  }_{=s}\cdot\operatorname*{sym}\nolimits_{V}%
\left(  v\right) \\
&  =p\cdot s\cdot\operatorname*{sym}\nolimits_{V}\left(  v\right)  .
\end{align*}
In other words, $p\cdot s\in\left\{  q\in\operatorname*{Sym}V\ \mid
\ \operatorname*{sym}\nolimits_{V}\left(  v\right)  \cdot q=q\cdot
\operatorname*{sym}\nolimits_{V}\left(  v\right)  \right\}  =M$, qed.}. In
other words, $\left\{  p\cdot s\ \mid\ \left(  p,s\right)  \in M\times
\operatorname*{sym}\nolimits_{V}\left(  V\right)  \right\}  \subseteq M$. By
Proposition \ref{prop.<>} \textbf{(a)} (applied to $\operatorname*{Sym}V$,
$\left\{  p\cdot s\ \mid\ \left(  p,s\right)  \in M\times\operatorname*{sym}%
\nolimits_{V}\left(  V\right)  \right\}  $ and $M$ instead of $M$, $S$ and
$Q$), this yields%
\[
\left\langle \left\{  p\cdot s\ \mid\ \left(  p,s\right)  \in M\times
\operatorname*{sym}\nolimits_{V}\left(  V\right)  \right\}  \right\rangle
\subseteq M.
\]
Now,%
\[
M\cdot\operatorname*{sym}\nolimits_{V}\left(  V\right)  =\left\langle p\cdot
s\ \mid\ \left(  p,s\right)  \in M\times\operatorname*{sym}\nolimits_{V}%
\left(  V\right)  \right\rangle =\left\langle \left\{  p\cdot s\ \mid\ \left(
p,s\right)  \in M\times\operatorname*{sym}\nolimits_{V}\left(  V\right)
\right\}  \right\rangle \subseteq M.
\]
By Lemma \ref{lem.ideal} (applied to $A=\operatorname*{Sym}V$ and
$\pi=\operatorname*{sym}\nolimits_{V}$), this yields that $M$ is a right ideal
of $\operatorname*{Sym}V$. Thus, $M\cdot\operatorname*{Sym}V\subseteq M$. But
since $1\in M$, we have $1\cdot\operatorname*{Sym}V\subseteq M\cdot
\operatorname*{Sym}V\subseteq M$, so that $\operatorname*{Sym}V=1\cdot
\operatorname*{Sym}V\subseteq M$. Combined with $M\subseteq\operatorname*{Sym}%
V$, this yields $M=\operatorname*{Sym}V$. Hence, every $p\in
\operatorname*{Sym}V$ satisfies $p\in M=\left\{  q\in\operatorname*{Sym}%
V\ \mid\ \operatorname*{sym}\nolimits_{V}\left(  v\right)  \cdot
q=q\cdot\operatorname*{sym}\nolimits_{V}\left(  v\right)  \right\}  $, so that
$\operatorname*{sym}\nolimits_{V}\left(  v\right)  \cdot p=p\cdot
\operatorname*{sym}\nolimits_{V}\left(  v\right)  $. This proves Lemma
\ref{lem.Symm.comm}.
\end{proof}

Using Lemma \ref{lem.Symm.comm}, we will now prove Theorem \ref{thm.Sym.comm}:

\begin{proof}
[Proof of Theorem \ref{thm.Sym.comm}.]\textbf{a)} Let $t\in\operatorname*{Sym}%
V$. We are going to prove that $tp=pt$ for every $p\in\operatorname*{Sym}V$.

\textit{Proof.} Let $M$ be the subset%
\[
\left\{  q\in\operatorname*{Sym}V\ \mid\ tq=qt\right\}
\]
of $\operatorname*{Sym}V$. We are going to prove that $M$ is the whole
$\operatorname*{Sym}V$.

First of all, we have $0\in M$\ \ \ \ \footnote{\textit{Proof.} Clearly,
$t0=0=0t$, so that $0\in\left\{  q\in\operatorname*{Sym}V\ \mid
\ tq=qt\right\}  =M$.}. Also, every $\alpha\in k$, $\beta\in k$, $p\in M$ and
$r\in M$ satisfy $\alpha p+\beta r\in M$.\ \ \ \ \footnote{\textit{Proof.} Let
$\alpha\in k$, $\beta\in k$, $p\in M$ and $r\in M$.
\par
Since $p\in M=\left\{  q\in\operatorname*{Sym}V\ \mid\ tq=qt\right\}  $, we
have $tp=pt$. Similarly, $tr=rt$. Now,%
\[
t\left(  \alpha p+\beta r\right)  =\alpha\underbrace{tp}_{=pt}+\beta
\underbrace{tr}_{=rt}=\alpha pt+\beta rt=\left(  \alpha p+\beta r\right)  t.
\]
In other words, $\alpha p+\beta r\in\left\{  q\in\operatorname*{Sym}%
V\ \mid\ tq=qt\right\}  =M$, qed.} In other words, $M$ is a $k$-submodule of
$\operatorname*{Sym}V$.

Second, $1\in M$ (with $1$ denoting the unity of the $k$-algebra
$\operatorname*{Sym}V$)\ \ \ \ \footnote{\textit{Proof.} Clearly,
$t\cdot1=t=1t$, so that $1\in\left\{  q\in\operatorname*{Sym}V\ \mid
\ tq=qt\right\}  =M$.}.

On the other hand, every $\left(  p,s\right)  \in M\times\operatorname*{sym}%
\nolimits_{V}\left(  V\right)  $ satisfy $p\cdot s\in M$%
\ \ \ \ \footnote{\textit{Proof.} Let $\left(  p,s\right)  \in M\times
\operatorname*{sym}\nolimits_{V}\left(  V\right)  $. Then, $p\in M$ and
$s\in\operatorname*{sym}\nolimits_{V}\left(  V\right)  $. Since $s\in
\operatorname*{sym}\nolimits_{V}\left(  V\right)  $, there exists some $v\in
V$ such that $s=\operatorname*{sym}\nolimits_{V}\left(  v\right)  $. Consider
this $v$. Lemma \ref{lem.Symm.comm} (applied to $t$ instead of $p$) yields
$\operatorname*{sym}\nolimits_{V}\left(  v\right)  \cdot t=t\cdot
\operatorname*{sym}\nolimits_{V}\left(  v\right)  $. Since
$\operatorname*{sym}\nolimits_{V}\left(  v\right)  =s$, this becomes $s\cdot
t=t\cdot s$.
\par
Since $p\in M=\left\{  q\in\operatorname*{Sym}V\ \mid\ tq=qt\right\}  $, we
have $tp=pt$. Now, $\underbrace{tp}_{=pt}s=p\underbrace{t\cdot s}_{=s\cdot
t=st}=pst$. In other words, $ps\in\left\{  q\in\operatorname*{Sym}%
V\ \mid\ tq=qt\right\}  =M$, so that $p\cdot s=ps\in M$, qed.}. In other
words, $\left\{  p\cdot s\ \mid\ \left(  p,s\right)  \in M\times
\operatorname*{sym}\nolimits_{V}\left(  V\right)  \right\}  \subseteq M$. By
Proposition \ref{prop.<>} \textbf{(a)} (applied to $\operatorname*{Sym}V$,
$\left\{  p\cdot s\ \mid\ \left(  p,s\right)  \in M\times\operatorname*{sym}%
\nolimits_{V}\left(  V\right)  \right\}  $ and $M$ instead of $M$, $S$ and
$Q$), this yields%
\[
\left\langle \left\{  p\cdot s\ \mid\ \left(  p,s\right)  \in M\times
\operatorname*{sym}\nolimits_{V}\left(  V\right)  \right\}  \right\rangle
\subseteq M.
\]
Now,%
\[
M\cdot\operatorname*{sym}\nolimits_{V}\left(  V\right)  =\left\langle p\cdot
s\ \mid\ \left(  p,s\right)  \in M\times\operatorname*{sym}\nolimits_{V}%
\left(  V\right)  \right\rangle =\left\langle \left\{  p\cdot s\ \mid\ \left(
p,s\right)  \in M\times\operatorname*{sym}\nolimits_{V}\left(  V\right)
\right\}  \right\rangle \subseteq M.
\]
By Lemma \ref{lem.ideal} (applied to $A=\operatorname*{Sym}V$ and
$\pi=\operatorname*{sym}\nolimits_{V}$), this yields that $M$ is a right ideal
of $\operatorname*{Sym}V$. Thus, $M\cdot\operatorname*{Sym}V\subseteq M$. But
since $1\in M$, we have $1\cdot\operatorname*{Sym}V\subseteq M\cdot
\operatorname*{Sym}V\subseteq M$, so that $\operatorname*{Sym}V=1\cdot
\operatorname*{Sym}V\subseteq M$. Combined with $M\subseteq\operatorname*{Sym}%
V$, this yields $M=\operatorname*{Sym}V$. Hence, every $p\in
\operatorname*{Sym}V$ satisfies $p\in M=\left\{  q\in\operatorname*{Sym}%
V\ \mid\ tq=qt\right\}  $, so that $tp=pt$. This proves \textbf{a)}.

\textbf{b)} Forget that we fixed $t$. We have proven that every $t\in
\operatorname*{Sym}V$ and every $p\in\operatorname*{Sym}V$ satisfy $tp=pt$ (by
part \textbf{a)}). In other words, the $k$-algebra $\operatorname*{Sym}V$ is
commutative. Theorem \ref{thm.Sym.comm} is proven.
\end{proof}

Theorem \ref{thm.Sym.comm} is a result on the nature of the factor algebra
$\operatorname*{Sym}V=\left(  \otimes V\right)  \diagup K\left(  V\right)  $,
so unsurprisingly it gives us an insight about the ideal $K\left(  V\right)  $
itself - namely, a new characterization of this ideal:

\begin{corollary}
\label{cor.Sym.symm}Let $k$ be a commutative ring. Let $V$ be a $k$-module.
Then,%
\[
K\left(  V\right)  =\left(  \otimes V\right)  \cdot\left\langle pq-qp\ \mid
\ \left(  p,q\right)  \in\left(  \otimes V\right)  ^{2}\right\rangle
\cdot\left(  \otimes V\right)  .
\]

\end{corollary}

This Corollary is usually formulated as follows: The ideal $K\left(  V\right)
$ is the commutator ideal of the $k$-algebra $\otimes V$. This is actually
often used as an alternative definition of $K\left(  V\right)  $.

\begin{proof}
[Proof of Corollary \ref{cor.Sym.symm}.]\textit{ }\textbf{a)} Let us first
show that $K\left(  V\right)  \subseteq\left(  \otimes V\right)
\cdot\left\langle pq-qp\ \mid\ \left(  p,q\right)  \in\left(  \otimes
V\right)  ^{2}\right\rangle \cdot\left(  \otimes V\right)  $.

\textit{Proof.} Every $\left(  v_{1},v_{2}\right)  \in V^{2}$ satisfies
$v_{1}\otimes v_{2}-v_{2}\otimes v_{1}=v_{1}v_{2}-v_{2}v_{1}$%
.\ \ \ \ \footnote{\textit{Proof.} Let $\left(  v_{1},v_{2}\right)  \in V^{2}%
$. Then, $v_{1}\in V=V^{\otimes1}$ and $v_{2}\in V=V^{\otimes1}$. Hence,
$v_{1}\cdot v_{2}=v_{1}\otimes v_{2}$ (by (\ref{*=(X)}), applied to $v_{1}$,
$v_{2}$, $1$ and $1$ instead of $a$, $b$, $n$ and $m$) and similarly
$v_{2}\cdot v_{1}=v_{2}\otimes v_{1}$. Hence,%
\[
\underbrace{v_{1}\otimes v_{2}}_{=v_{1}\cdot v_{2}=v_{1}v_{2}}%
-\underbrace{v_{2}\otimes v_{1}}_{=v_{2}\cdot v_{1}=v_{2}v_{1}}=v_{1}%
v_{2}-v_{2}v_{1},
\]
qed.} Hence,%
\begin{align*}
&  \left\{  v_{1}\otimes v_{2}-v_{2}\otimes v_{1}\ \mid\ \left(  v_{1}%
,v_{2}\right)  \in V^{2}\right\} \\
&  =\left\{  v_{1}v_{2}-v_{2}v_{1}\ \mid\ \left(  v_{1},v_{2}\right)  \in
V^{2}\right\}  =\left\{  pq-qp\ \mid\ \left(  p,q\right)  \in V^{2}\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed }\left(  v_{1}%
,v_{2}\right)  \text{ as }\left(  p,q\right)  \right) \\
&  \subseteq\left\{  pq-qp\ \mid\ \left(  p,q\right)  \in\left(  \otimes
V\right)  ^{2}\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
V^{2}\subseteq\left(  \otimes V\right)  ^{2}\right)  .
\end{align*}
Thus,%
\[
\left\langle \left\{  v_{1}\otimes v_{2}-v_{2}\otimes v_{1}\ \mid\ \left(
v_{1},v_{2}\right)  \in V^{2}\right\}  \right\rangle \subseteq\left\langle
\left\{  pq-qp\ \mid\ \left(  p,q\right)  \in\left(  \otimes V\right)
^{2}\right\}  \right\rangle .
\]
But by Corollary \ref{coro.K_2}, we have%
\begin{align*}
K_{2}\left(  V\right)   &  =\left\langle v_{1}\otimes v_{2}-v_{2}\otimes
v_{1}\ \mid\ \left(  v_{1},v_{2}\right)  \in V^{2}\right\rangle =\left\langle
\left\{  v_{1}\otimes v_{2}-v_{2}\otimes v_{1}\ \mid\ \left(  v_{1}%
,v_{2}\right)  \in V^{2}\right\}  \right\rangle \\
&  \subseteq\left\langle \left\{  pq-qp\ \mid\ \left(  p,q\right)  \in\left(
\otimes V\right)  ^{2}\right\}  \right\rangle =\left\langle pq-qp\ \mid
\ \left(  p,q\right)  \in\left(  \otimes V\right)  ^{2}\right\rangle .
\end{align*}
Hence,%
\[
\left(  \otimes V\right)  \cdot\left(  K_{2}\left(  V\right)  \right)
\cdot\left(  \otimes V\right)  \subseteq\left(  \otimes V\right)
\cdot\left\langle pq-qp\ \mid\ \left(  p,q\right)  \in\left(  \otimes
V\right)  ^{2}\right\rangle \cdot\left(  \otimes V\right)  .
\]
Since $\left(  \otimes V\right)  \cdot\left(  K_{2}\left(  V\right)  \right)
\cdot\left(  \otimes V\right)  =K\left(  V\right)  $, this rewrites as%
\[
K\left(  V\right)  \subseteq\left(  \otimes V\right)  \cdot\left\langle
pq-qp\ \mid\ \left(  p,q\right)  \in\left(  \otimes V\right)  ^{2}%
\right\rangle \cdot\left(  \otimes V\right)  .
\]
This proves part \textbf{a)}.

\textbf{b)} Now we will prove that $\left(  \otimes V\right)  \cdot
\left\langle pq-qp\ \mid\ \left(  p,q\right)  \in\left(  \otimes V\right)
^{2}\right\rangle \cdot\left(  \otimes V\right)  \subseteq K\left(  V\right)
$.

\textit{Proof.} Every $\left(  p,q\right)  \in\left(  \otimes V\right)  ^{2}$
satisfy $pq-qp\in K\left(  V\right)  $\ \ \ \ \footnote{\textit{Proof.} Let
$\left(  p,q\right)  \in\left(  \otimes V\right)  ^{2}$. Then,%
\begin{align*}
\operatorname*{sym}\nolimits_{V}\left(  pq\right)   &  =\operatorname*{sym}%
\nolimits_{V}\left(  p\right)  \cdot\operatorname*{sym}\nolimits_{V}\left(
q\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname*{sym}%
\nolimits_{V}\text{ is a }k\text{-algebra homomorphism}\right) \\
&  =\operatorname*{sym}\nolimits_{V}\left(  q\right)  \cdot\operatorname*{sym}%
\nolimits_{V}\left(  p\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\operatorname*{Sym}V\text{ is commutative by Theorem \ref{thm.Sym.comm}%
}\right) \\
&  =\operatorname*{sym}\nolimits_{V}\left(  qp\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname*{sym}\nolimits_{V}%
\text{ is a }k\text{-algebra homomorphism}\right)  .
\end{align*}
In other words, $pq\equiv qp\operatorname{mod}K\left(  V\right)  $ (since
$\operatorname*{sym}\nolimits_{V}$ is the projection $\otimes V\rightarrow
\left(  \otimes V\right)  \diagup K\left(  V\right)  $). In other words,
$pq-qp\in K\left(  V\right)  $, qed.}. In other words, $\left\{
pq-qp\ \mid\ \left(  p,q\right)  \in\left(  \otimes V\right)  ^{2}\right\}
\subseteq K\left(  V\right)  $. Hence, Proposition \ref{prop.<>} \textbf{(a)}
(applied to $\otimes V$, $\left\{  pq-qp\ \mid\ \left(  p,q\right)  \in\left(
\otimes V\right)  ^{2}\right\}  $ and $K\left(  V\right)  $ instead of $M$,
$S$ and $Q$) yields
\[
\left\langle \left\{  pq-qp\ \mid\ \left(  p,q\right)  \in\left(  \otimes
V\right)  ^{2}\right\}  \right\rangle \subseteq K\left(  V\right)  .
\]
Thus,%
\[
\left\langle pq-qp\ \mid\ \left(  p,q\right)  \in\left(  \otimes V\right)
^{2}\right\rangle =\left\langle \left\{  pq-qp\ \mid\ \left(  p,q\right)
\in\left(  \otimes V\right)  ^{2}\right\}  \right\rangle \subseteq K\left(
V\right)  .
\]
Hence,%
\[
\left(  \otimes V\right)  \cdot\left\langle pq-qp\ \mid\ \left(  p,q\right)
\in\left(  \otimes V\right)  ^{2}\right\rangle \cdot\left(  \otimes V\right)
\subseteq\left(  \otimes V\right)  \cdot\left(  K\left(  V\right)  \right)
\cdot\left(  \otimes V\right)  \subseteq K\left(  V\right)
\]
(since $K\left(  V\right)  $ is a two-sided ideal of $\otimes V$). This proves
part \textbf{b)}.

\textbf{c)} Combining $K\left(  V\right)  \subseteq\left(  \otimes V\right)
\cdot\left\langle pq-qp\ \mid\ \left(  p,q\right)  \in\left(  \otimes
V\right)  ^{2}\right\rangle \cdot\left(  \otimes V\right)  $ (which we know
from part \textbf{a)}) with $\left(  \otimes V\right)  \cdot\left\langle
pq-qp\ \mid\ \left(  p,q\right)  \in\left(  \otimes V\right)  ^{2}%
\right\rangle \cdot\left(  \otimes V\right)  \subseteq K\left(  V\right)  $
(which we know from part \textbf{b)}), we obtain $K\left(  V\right)  =\left(
\otimes V\right)  \cdot\left\langle pq-qp\ \mid\ \left(  p,q\right)
\in\left(  \otimes V\right)  ^{2}\right\rangle \cdot\left(  \otimes V\right)
$. This proves Corollary \ref{cor.Sym.symm}.
\end{proof}

\subsection{\label{subsect.uniprops}Some universal properties}

We shall next discuss some universal properties for the pseudoexterior powers
$\operatorname*{Exter}^{n}V$, the symmetric powers $\operatorname*{Sym}^{n}V$
and the exterior powers $\wedge^{n}V$.

Let us first recall the definition of a multilinear map:

\begin{definition}
Let $k$ be a commutative ring. Let $n\in\mathbb{N}$. Let $V_{1},V_{2}%
,\ldots,V_{n}$ be $k$-modules.

Let $W$ be any $k$-module, and let $f:V_{1}\times V_{2}\times\cdots\times
V_{n}\rightarrow W$ be a map. We say that the map $f$ is \textit{multilinear}
if and only if for each $i\in\left\{  1,2,\ldots,n\right\}  $ and each
\[
\left(  v_{1},v_{2},\ldots,v_{i-1},v_{i+1},v_{i+2},\ldots,v_{n}\right)  \in
V_{1}\times V_{2}\times\cdots\times V_{i-1}\times V_{i+1}\times V_{i+2}%
\times\cdots\times V_{n},
\]
the map%
\begin{align*}
V_{i}  &  \rightarrow W,\\
v  &  \mapsto f\left(  v_{1},v_{2},\ldots,v_{i-1},v,v_{i+1},v_{i+2}%
,\ldots,v_{n}\right)
\end{align*}
is $k$-linear.
\end{definition}

Now, we can state the classical universal property of a tensor product:

\begin{proposition}
\label{prop.uniprop.tensor-gen}Let $k$ be a commutative ring. Let
$n\in\mathbb{N}$. Let $V_{1},V_{2},\ldots,V_{n}$ be $k$-modules.

Let $W$ be any $k$-module, and let $f:V_{1}\times V_{2}\times\cdots\times
V_{n}\rightarrow W$ be a multilinear map. Then, there exists a unique
$k$-linear map $f_{\otimes}:V_{1}\otimes V_{2}\otimes\cdots\otimes
V_{n}\rightarrow W$ such that every $\left(  v_{1},v_{2},\ldots,v_{n}\right)
\in V_{1}\times V_{2}\times\cdots\times V_{n}$ satisfies $f_{\otimes}\left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  =f\left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  $.
\end{proposition}

Proposition \ref{prop.uniprop.tensor-gen} is the classical result that allows
one to construct maps from a tensor product comfortably.

The particular case of Proposition \ref{prop.uniprop.tensor-gen} when all of
$V_{1},V_{2},\ldots,V_{n}$ are identical will be the most useful to us:

\begin{corollary}
\label{cor.uniprop.tensor}Let $k$ be a commutative ring. Let $n\in\mathbb{N}$.
Let $V$ be a $k$-module.

Let $W$ be any $k$-module, and let $f:V^{n}\rightarrow W$ be a multilinear
map. Then, there exists a unique $k$-linear map $f_{\otimes}:V^{\otimes
n}\rightarrow W$ such that every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{n}$ satisfies $f_{\otimes}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\right)  =f\left(  v_{1},v_{2},\ldots,v_{n}\right)  $.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.uniprop.tensor}.]The map $f$ is a multilinear map
$V^{n}\rightarrow W$. In other words, the map $f$ is a multilinear map
$\underbrace{V\times V\times\cdots\times V}_{n\text{ times}}\rightarrow W$
(since $V^{n}=\underbrace{V\times V\times\cdots\times V}_{n\text{ times}}$).
Thus, Proposition \ref{prop.uniprop.tensor-gen} (applied to $V_{i}=V$) shows
that there exists a unique $k$-linear map $f_{\otimes}:\underbrace{V\otimes
V\otimes\cdots\otimes V}_{n\text{ times}}\rightarrow W$ such that every
$\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in\underbrace{V\times V\times
\cdots\times V}_{n\text{ times}}$ satisfies $f_{\otimes}\left(  v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}\right)  =f\left(  v_{1},v_{2},\ldots
,v_{n}\right)  $. Since $\underbrace{V\otimes V\otimes\cdots\otimes
V}_{n\text{ times}}=V^{\otimes n}$ and $\underbrace{V\times V\times
\cdots\times V}_{n\text{ times}}=V^{n}$, this rewrites as follows: There
exists a unique $k$-linear map $f_{\otimes}:V^{\otimes n}\rightarrow W$ such
that every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$ satisfies
$f_{\otimes}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)
=f\left(  v_{1},v_{2},\ldots,v_{n}\right)  $. This proves Corollary
\ref{cor.uniprop.tensor}.
\end{proof}

We shall now use Corollary \ref{cor.uniprop.tensor} to derive a universal
property for the pseudoexterior powers $\operatorname*{Exter}^{n}V$. We first
state an almost obvious fact:

\begin{lemma}
\label{lem.uniprop.ext-pow.uni}Let $k$ be a commutative ring. Let
$n\in\mathbb{N}$. Let $V$ be a $k$-module. Let $W$ be any $k$-module, and let
$f:V^{n}\rightarrow W$ be any map. Then, there exists \textbf{at most one}
$k$-linear map $f_{\operatorname*{Exter}}:\operatorname*{Exter}\nolimits^{n}%
V\rightarrow W$ such that every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{n}$ satisfies $f_{\operatorname*{Exter}}\left(  \operatorname*{exter}%
\nolimits_{V,n}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)
\right)  =f\left(  v_{1},v_{2},\ldots,v_{n}\right)  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.uniprop.ext-pow.uni}.]Let $\alpha$ and $\beta$ be two
$k$-linear maps $f_{\operatorname*{Exter}}:\operatorname*{Exter}%
\nolimits^{n}V\rightarrow W$ such that every $\left(  v_{1},v_{2},\ldots
,v_{n}\right)  \in V^{n}$ satisfies $f_{\operatorname*{Exter}}\left(
\operatorname*{exter}\nolimits_{V,n}\left(  v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}\right)  \right)  =f\left(  v_{1},v_{2},\ldots
,v_{n}\right)  $. We shall show that $\alpha=\beta$.

We know that $\alpha$ is a $k$-linear map $f_{\operatorname*{Exter}%
}:\operatorname*{Exter}\nolimits^{n}V\rightarrow W$ such that every $\left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$ satisfies
$f_{\operatorname*{Exter}}\left(  \operatorname*{exter}\nolimits_{V,n}\left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  \right)  =f\left(
v_{1},v_{2},\ldots,v_{n}\right)  $. In other words, $\alpha$ is a $k$-linear
map $\operatorname*{Exter}\nolimits^{n}V\rightarrow W$ and has the property
that every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$ satisfies%
\begin{equation}
\alpha\left(  \operatorname*{exter}\nolimits_{V,n}\left(  v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}\right)  \right)  =f\left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  . \label{pf.lem.uniprop.ext-pow.uni.alpha}%
\end{equation}
The same argument (applied to $\beta$ instead of $\alpha$) shows that $\beta$
is a $k$-linear map $\operatorname*{Exter}\nolimits^{n}V\rightarrow W$ and has
the property that every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$
satisfies%
\begin{equation}
\beta\left(  \operatorname*{exter}\nolimits_{V,n}\left(  v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}\right)  \right)  =f\left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  . \label{pf.lem.uniprop.ext-pow.uni.beta}%
\end{equation}


Now, the map $\alpha-\beta$ is $k$-linear (since the mps $\alpha$ and $\beta$
are $k$-linear). Hence, $\operatorname*{Ker}\left(  \alpha-\beta\right)  $ is
a $k$-submodule of $\operatorname*{Exter}\nolimits^{n}V$.

Define a subset $S$ of $\operatorname*{Exter}\nolimits^{n}V$ by%
\begin{equation}
S=\left\{  \operatorname*{exter}\nolimits_{V,n}\left(  v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}\right)  \ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{n}\right\}  .
\label{pf.lem.uniprop.ext-pow.uni.S=}%
\end{equation}
Then, $S\subseteq\operatorname*{Ker}\left(  \alpha-\beta\right)
$\ \ \ \ \footnote{\textit{Proof.} Let $s\in S$. Thus, $s\in S=\left\{
\operatorname*{exter}\nolimits_{V,n}\left(  v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}\right)  \ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)
\in V^{n}\right\}  $. In other words, $s=\operatorname*{exter}\nolimits_{V,n}%
\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  $ for some
$\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$. Consider this $\left(
v_{1},v_{2},\ldots,v_{n}\right)  $.
\par
Applying the map $\alpha$ to the equality $s=\operatorname*{exter}%
\nolimits_{V,n}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  $,
we obtain%
\[
\alpha\left(  s\right)  =\alpha\left(  \operatorname*{exter}\nolimits_{V,n}%
\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  \right)
=f\left(  v_{1},v_{2},\ldots,v_{n}\right)
\]
(by (\ref{pf.lem.uniprop.ext-pow.uni.alpha})). The same argument (applied to
$\beta$ instead of $\alpha$) shows that $\beta\left(  s\right)  =f\left(
v_{1},v_{2},\ldots,v_{n}\right)  $. Thus, $\alpha\left(  s\right)  =f\left(
v_{1},v_{2},\ldots,v_{n}\right)  =\beta\left(  s\right)  $. Now, $\left(
\alpha-\beta\right)  \left(  s\right)  =\underbrace{\alpha\left(  s\right)
}_{=\beta\left(  s\right)  }-\beta\left(  s\right)  =\beta\left(  s\right)
-\beta\left(  s\right)  =0$, so that $s\in\operatorname*{Ker}\left(
\alpha-\beta\right)  $.
\par
Now, let us forget that we fixed $s$. We thus have shown that $s\in
\operatorname*{Ker}\left(  \alpha-\beta\right)  $ for each $s\in S$. In other
words, $S\subseteq\operatorname*{Ker}\left(  \alpha-\beta\right)  $.}. Hence,
Proposition \ref{prop.<>} \textbf{(a)} (applied to $M=\operatorname*{Exter}%
\nolimits^{n}V$ and $Q=\operatorname*{Ker}\left(  \alpha-\beta\right)  $)
shows that $\left\langle S\right\rangle \subseteq\operatorname*{Ker}\left(
\alpha-\beta\right)  $.

On the other hand, define a subset $S^{\prime}$ of $V^{\otimes n}$ by
\begin{equation}
S^{\prime}=\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\}  .
\label{pf.lem.uniprop.ext-pow.uni.S'=}%
\end{equation}
Then,%
\begin{align}
\operatorname*{exter}\nolimits_{V,n}\left(  S^{\prime}\right)   &
=\operatorname*{exter}\nolimits_{V,n}\left(  \left\{  v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots
,v_{n}\right)  \in V^{n}\right\}  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.lem.uniprop.ext-pow.uni.S'=}%
)}\right) \nonumber\\
&  =\left\{  \operatorname*{exter}\nolimits_{V,n}\left(  v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}\right)  \ \mid\ \left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{n}\right\} \nonumber\\
&  =S\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.lem.uniprop.ext-pow.uni.S=}%
)}\right)  . \label{pf.lem.uniprop.ext-pow.1}%
\end{align}


However, the tensor product $V^{\otimes n}$ is generated (as a $k$-module) by
its pure tensors. In other words,%
\begin{align*}
V^{\otimes n}  &  =\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\rangle
\\
&  =\left\langle \underbrace{\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}\right\}
}_{=S^{\prime}}\right\rangle =\left\langle S^{\prime}\right\rangle .
\end{align*}
Applying the map $\operatorname*{exter}\nolimits_{V,n}$ to both sides of this
equality, we obtain%
\begin{align*}
\operatorname*{exter}\nolimits_{V,n}\left(  V^{\otimes n}\right)   &
=\operatorname*{exter}\nolimits_{V,n}\left(  \left\langle S^{\prime
}\right\rangle \right)  =\left\langle \underbrace{\operatorname*{exter}%
\nolimits_{V,n}\left(  S^{\prime}\right)  }_{\substack{=S\\\text{(by
(\ref{pf.lem.uniprop.ext-pow.1}))}}}\right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Proposition \ref{prop.<>} \textbf{(b)} (applied to}\\
V^{\otimes n}\text{, }S^{\prime}\text{, }\operatorname*{Exter}\nolimits^{n}%
V\text{ and }\operatorname*{exter}\nolimits_{V,n}\text{ instead of }M\text{,
}S\text{, }R\text{ and }f\text{)}%
\end{array}
\right) \\
&  =\left\langle S\right\rangle .
\end{align*}


But the map $\operatorname*{exter}\nolimits_{V,n}$ is surjective. Hence,
$\operatorname*{Exter}\nolimits^{n}V=\operatorname*{exter}\nolimits_{V,n}%
\left(  V^{\otimes n}\right)  =\left\langle S\right\rangle \subseteq
\operatorname*{Ker}\left(  \alpha-\beta\right)  $. In other words,
$\alpha-\beta=0$. Hence, $\alpha=\beta$.

Now, forget that we fixed $\alpha$ and $\beta$. We thus have shown that if
$\alpha$ and $\beta$ are two $k$-linear maps $f_{\operatorname*{Exter}%
}:\operatorname*{Exter}\nolimits^{n}V\rightarrow W$ such that every $\left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$ satisfies
$f_{\operatorname*{Exter}}\left(  \operatorname*{exter}\nolimits_{V,n}\left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  \right)  =f\left(
v_{1},v_{2},\ldots,v_{n}\right)  $, then $\alpha=\beta$. In other words, there
exists \textbf{at most one} $k$-linear map $f_{\operatorname*{Exter}%
}:\operatorname*{Exter}\nolimits^{n}V\rightarrow W$ such that every $\left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$ satisfies
$f_{\operatorname*{Exter}}\left(  \operatorname*{exter}\nolimits_{V,n}\left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  \right)  =f\left(
v_{1},v_{2},\ldots,v_{n}\right)  $. This proves Lemma
\ref{lem.uniprop.ext-pow.uni}.
\end{proof}

We shall furthermore need a definition:

\begin{definition}
\label{def.uniprop.antisym}Let $n\in\mathbb{N}$. Let $V$ be a set.

Let $W$ be a $\mathbb{Z}$-module. Let $f:V^{n}\rightarrow W$ be a map. We say
that the map $f$ is \textit{antisymmetric} if and only if each $\left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$ and $\gamma\in S_{n}$ satisfy%
\[
f\left(  v_{\gamma\left(  1\right)  },v_{\gamma\left(  2\right)  }%
,\ldots,v_{\gamma\left(  n\right)  }\right)  =\left(  -1\right)  ^{\gamma
}f\left(  v_{1},v_{2},\ldots,v_{n}\right)  .
\]

\end{definition}

Now, we can state a universal property for the pseudoexterior powers
$\operatorname*{Exter}^{n}V$:

\begin{corollary}
\label{cor.uniprop.ext-pow}Let $k$ be a commutative ring. Let $n\in\mathbb{N}%
$. Let $V$ be a $k$-module.

Let $W$ be any $k$-module, and let $f:V^{n}\rightarrow W$ be an antisymmetric
multilinear map. (The notion of \textquotedblleft
antisymmetric\textquotedblright\ makes sense here because the $k$-module $W$
is clearly a $\mathbb{Z}$-module.) Then, there exists a unique $k$-linear map
$f_{\operatorname*{Exter}}:\operatorname*{Exter}^{n}V\rightarrow W$ such that
every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$ satisfies
$f_{\operatorname*{Exter}}\left(  \operatorname*{exter}\nolimits_{V,n}\left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  \right)  =f\left(
v_{1},v_{2},\ldots,v_{n}\right)  $.
\end{corollary}

Before we prove this, let us recall a classical fact from abstract algebra --
viz. the universal property of quotient modules (also known as the
homomorphism theorem for $k$-modules):

\begin{proposition}
\label{prop.uniprop.quotient}Let $k$ be a commutative ring. Let $V$ be a
$k$-module. Let $I$ be a $k$-submodule of $V$. Let $\pi_{I}$ be the canonical
projection $V\rightarrow V\diagup I$.

Let $W$ be any $k$-module, and let $f:V\rightarrow W$ be a $k$-linear map
satisfying $f\left(  I\right)  =0$. Then, there exists a unique $k$-linear map
$f^{\prime}:V\diagup I\rightarrow W$ satisfying $f=f^{\prime}\circ\pi_{I}$.
\end{proposition}

\begin{proof}
[Proof of Corollary \ref{cor.uniprop.ext-pow}.]Corollary
\ref{cor.uniprop.tensor} shows that there exists a unique $k$-linear map
$f_{\otimes}:V^{\otimes n}\rightarrow W$ such that every $\left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  \in V^{n}$ satisfies $f_{\otimes}\left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  =f\left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  $. Consider this $f_{\otimes}$. The map
$f_{\otimes}$ is $k$-linear; thus, $\operatorname*{Ker}\left(  f_{\otimes
}\right)  $ is a $k$-submodule of $V^{\otimes n}$.

We know that every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$
satisfies
\begin{equation}
f_{\otimes}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)
=f\left(  v_{1},v_{2},\ldots,v_{n}\right)  .
\label{pf.cor.uniprop.ext-pow.ftens}%
\end{equation}


The map $f$ is antisymmetric. In other words, each $\left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  \in V^{n}$ and $\gamma\in S_{n}$ satisfy%
\begin{equation}
f\left(  v_{\gamma\left(  1\right)  },v_{\gamma\left(  2\right)  }%
,\ldots,v_{\gamma\left(  n\right)  }\right)  =\left(  -1\right)  ^{\gamma
}f\left(  v_{1},v_{2},\ldots,v_{n}\right)
\label{pf.cor.uniprop.ext-pow.symmetry}%
\end{equation}
(by the definition of \textquotedblleft antisymmetric\textquotedblright).

Define a subset $T$ of $V^{\otimes n}$ by%
\[
T=\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-\left(  -1\right)
^{\sigma}v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(  2\right)
}\otimes\cdots\otimes v_{\sigma\left(  n\right)  }\ \mid\ \left(  \left(
v_{1},v_{2},\ldots,v_{n}\right)  ,\sigma\right)  \in V^{n}\times
S_{n}\right\}  .
\]
Thus,%
\begin{align*}
&  \left\langle T\right\rangle \\
&  =\left\langle \left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-\left(
-1\right)  ^{\sigma}v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(
2\right)  }\otimes\cdots\otimes v_{\sigma\left(  n\right)  }\ \mid\ \left(
\left(  v_{1},v_{2},\ldots,v_{n}\right)  ,\sigma\right)  \in V^{n}\times
S_{n}\right\}  \right\rangle \\
&  =\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-\left(
-1\right)  ^{\sigma}v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(
2\right)  }\otimes\cdots\otimes v_{\sigma\left(  n\right)  }\ \mid\ \left(
\left(  v_{1},v_{2},\ldots,v_{n}\right)  ,\sigma\right)  \in V^{n}\times
S_{n}\right\rangle \\
&  =Q_{n}\left(  V\right)
\end{align*}
(since $Q_{n}\left(  V\right)  $ is defined to be \newline$\left\langle
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-\left(  -1\right)  ^{\sigma
}v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(  2\right)  }\otimes
\cdots\otimes v_{\sigma\left(  n\right)  }\ \mid\ \left(  \left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  ,\sigma\right)  \in V^{n}\times S_{n}\right\rangle
$).

Recall that $\operatorname*{exter}\nolimits_{V,n}$ is the canonical projection
$V^{\otimes n}\rightarrow V^{\otimes n}\diagup Q_{n}\left(  V\right)  $.

Now, $T\subseteq\operatorname*{Ker}\left(  f_{\otimes}\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $t\in T$. Then,%
\[
t\in T=\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-\left(
-1\right)  ^{\sigma}v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(
2\right)  }\otimes\cdots\otimes v_{\sigma\left(  n\right)  }\ \mid\ \left(
\left(  v_{1},v_{2},\ldots,v_{n}\right)  ,\sigma\right)  \in V^{n}\times
S_{n}\right\}  .
\]
In other words, $t$ has the form $t=v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}-\left(  -1\right)  ^{\sigma}v_{\sigma\left(  1\right)  }\otimes
v_{\sigma\left(  2\right)  }\otimes\cdots\otimes v_{\sigma\left(  n\right)  }$
for some $\left(  \left(  v_{1},v_{2},\ldots,v_{n}\right)  ,\sigma\right)  \in
V^{n}\times S_{n}$. Consider this $\left(  \left(  v_{1},v_{2},\ldots
,v_{n}\right)  ,\sigma\right)  $.
\par
It is known that $\left(  -1\right)  ^{\sigma}\in\left\{  1,-1\right\}  $. But
each $g\in\left\{  1,-1\right\}  $ satisfies $g^{2}=1$. Applying this to
$g=\left(  -1\right)  ^{\sigma}$, we obtain $\left(  \left(  -1\right)
^{\sigma}\right)  ^{2}=1$.
\par
From $\left(  \left(  v_{1},v_{2},\ldots,v_{n}\right)  ,\sigma\right)  \in
V^{n}\times S_{n}$, we obtain $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{n}$ and $\sigma\in S_{n}$. From (\ref{pf.cor.uniprop.ext-pow.ftens}), we
obtain $f_{\otimes}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\right)  =f\left(  v_{1},v_{2},\ldots,v_{n}\right)  $. From
(\ref{pf.cor.uniprop.ext-pow.ftens}) (applied to $\left(  v_{\sigma\left(
1\right)  },v_{\sigma\left(  2\right)  },\ldots,v_{\sigma\left(  n\right)
}\right)  $ instead of $\left(  v_{1},v_{2},\ldots,v_{n}\right)  $), we
obtain
\begin{align*}
f_{\otimes}\left(  v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(
2\right)  }\otimes\cdots\otimes v_{\sigma\left(  n\right)  }\right)   &
=f\left(  v_{\sigma\left(  1\right)  },v_{\sigma\left(  2\right)  }%
,\ldots,v_{\sigma\left(  n\right)  }\right) \\
&  =\left(  -1\right)  ^{\sigma}f\left(  v_{1},v_{2},\ldots,v_{n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.cor.uniprop.ext-pow.symmetry})
(applied to }\gamma=\sigma\text{)}\right)  .
\end{align*}
Multiplying both sides of this equality by $\left(  -1\right)  ^{\sigma}$, we
obtain%
\[
\left(  -1\right)  ^{\sigma}f_{\otimes}\left(  v_{\sigma\left(  1\right)
}\otimes v_{\sigma\left(  2\right)  }\otimes\cdots\otimes v_{\sigma\left(
n\right)  }\right)  =\underbrace{\left(  -1\right)  ^{\sigma}\left(
-1\right)  ^{\sigma}}_{=\left(  \left(  -1\right)  ^{\sigma}\right)  ^{2}%
=1}f\left(  v_{1},v_{2},\ldots,v_{n}\right)  =f\left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  .
\]
\par
Now, applying the map $f_{\otimes}$ to both sides of the equality
$t=v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}-\left(  -1\right)  ^{\sigma
}v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(  2\right)  }\otimes
\cdots\otimes v_{\sigma\left(  n\right)  }$, we find%
\begin{align*}
f_{\otimes}\left(  t\right)   &  =f_{\otimes}\left(  v_{1}\otimes v_{2}%
\otimes\cdots\otimes v_{n}-\left(  -1\right)  ^{\sigma}v_{\sigma\left(
1\right)  }\otimes v_{\sigma\left(  2\right)  }\otimes\cdots\otimes
v_{\sigma\left(  n\right)  }\right) \\
&  =\underbrace{f_{\otimes}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\right)  }_{=f\left(  v_{1},v_{2},\ldots,v_{n}\right)  }%
-\underbrace{\left(  -1\right)  ^{\sigma}f_{\otimes}\left(  v_{\sigma\left(
1\right)  }\otimes v_{\sigma\left(  2\right)  }\otimes\cdots\otimes
v_{\sigma\left(  n\right)  }\right)  }_{=f\left(  v_{1},v_{2},\ldots
,v_{n}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f_{\otimes}\text{ is
}k\text{-linear}\right) \\
&  =f\left(  v_{1},v_{2},\ldots,v_{n}\right)  -f\left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  =0.
\end{align*}
In other words, $t\in\operatorname*{Ker}\left(  f_{\otimes}\right)  $.
\par
Now, forget that we fixed $t$. We thus have proven that $t\in
\operatorname*{Ker}\left(  f_{\otimes}\right)  $ for each $t\in T$. In other
words, $T\subseteq\operatorname*{Ker}\left(  f_{\otimes}\right)  $.}. Thus,
$f_{\otimes}\left(  \underbrace{T}_{\subseteq\operatorname*{Ker}\left(
f_{\otimes}\right)  }\right)  \subseteq f_{\otimes}\left(  \operatorname*{Ker}%
\left(  f_{\otimes}\right)  \right)  =0$, so that $f_{\otimes}\left(
T\right)  =0$. But Proposition \ref{prop.<>} \textbf{(b)} (applied to
$V^{\otimes n}$, $T$, $W$ and $f_{\otimes}$ instead of $M$, $S$, $R$ and $f$)
yields $f_{\otimes}\left(  \left\langle T\right\rangle \right)  =\left\langle
\underbrace{f_{\otimes}\left(  T\right)  }_{=0}\right\rangle =\left\langle
0\right\rangle =0$. Since $\left\langle T\right\rangle =Q_{n}\left(  V\right)
$, this rewrites as $f_{\otimes}\left(  Q_{n}\left(  V\right)  \right)  =0$.

Hence, Proposition \ref{prop.uniprop.quotient} (applied to $V^{\otimes n}$,
$Q_{n}\left(  V\right)  $, $\operatorname*{exter}\nolimits_{V,n}$, $W$ and
$f_{\otimes}$ instead of $V$, $I$, $\pi_{I}$, $W$ and $f$) yields that there
exists a unique $k$-linear map $f^{\prime}:V^{\otimes n}\diagup Q_{n}\left(
V\right)  \rightarrow W$ satisfying $f_{\otimes}=f^{\prime}\circ
\operatorname*{exter}\nolimits_{V,n}$. Consider this $f^{\prime}$.

The map $f^{\prime}$ is a $k$-linear map $V^{\otimes n}\diagup Q_{n}\left(
V\right)  \rightarrow W$. In other words, the map $f^{\prime}$ is a $k$-linear
map $\operatorname*{Exter}^{n}V\rightarrow W$ (since $V^{\otimes n}\diagup
Q_{n}\left(  V\right)  =\operatorname*{Exter}^{n}V$). Every $\left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$ satisfies%
\begin{align*}
& f^{\prime}\left(  \operatorname*{exter}\nolimits_{V,n}\left(  v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}\right)  \right)  \\
& =\underbrace{\left(  f^{\prime}\circ\operatorname*{exter}\nolimits_{V,n}%
\right)  }_{=f_{\otimes}}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\right)  =f_{\otimes}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\right)  \\
& =f\left(  v_{1},v_{2},\ldots,v_{n}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{pf.cor.uniprop.ext-pow.ftens})}\right)  .
\end{align*}
Thus, $f^{\prime}$ is a $k$-linear map $\operatorname*{Exter}^{n}V\rightarrow
W$ and has the property that every $\left(  v_{1},v_{2},\ldots,v_{n}\right)
\in V^{n}$ satisfies $f^{\prime}\left(  \operatorname*{exter}\nolimits_{V,n}%
\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  \right)
=f\left(  v_{1},v_{2},\ldots,v_{n}\right)  $. Hence, there exists \textbf{at
least one} $k$-linear map $f_{\operatorname*{Exter}}:\operatorname*{Exter}%
^{n}V\rightarrow W$ such that every $\left(  v_{1},v_{2},\ldots,v_{n}\right)
\in V^{n}$ satisfies $f_{\operatorname*{Exter}}\left(  \operatorname*{exter}%
\nolimits_{V,n}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)
\right)  =f\left(  v_{1},v_{2},\ldots,v_{n}\right)  $ (namely,
$f_{\operatorname*{Exter}}=f^{\prime}$). Since we also know that there exists
\textbf{at most one} such map (in fact, this follows from Lemma
\ref{lem.uniprop.ext-pow.uni}), we can therefore conclude that there exists a
\textbf{unique} $k$-linear map $f_{\operatorname*{Exter}}%
:\operatorname*{Exter}^{n}V\rightarrow W$ such that every $\left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  \in V^{n}$ satisfies $f_{\operatorname*{Exter}%
}\left(  \operatorname*{exter}\nolimits_{V,n}\left(  v_{1}\otimes v_{2}%
\otimes\cdots\otimes v_{n}\right)  \right)  =f\left(  v_{1},v_{2},\ldots
,v_{n}\right)  $. This proves Corollary \ref{cor.uniprop.ext-pow}.
\end{proof}

We can similarly deal with symmetric powers. First, we state an analogue to
Lemma \ref{lem.uniprop.ext-pow.uni}:

\begin{lemma}
\label{lem.uniprop.sym-pow.uni}Let $k$ be a commutative ring. Let
$n\in\mathbb{N}$. Let $V$ be a $k$-module. Let $W$ be any $k$-module, and let
$f:V^{n}\rightarrow W$ be any map. Then, there exists \textbf{at most one}
$k$-linear map $f_{\operatorname*{Sym}}:\operatorname*{Sym}\nolimits^{n}%
V\rightarrow W$ such that every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{n}$ satisfies $f_{\operatorname*{Sym}}\left(  \operatorname*{sym}%
\nolimits_{V,n}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)
\right)  =f\left(  v_{1},v_{2},\ldots,v_{n}\right)  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.uniprop.sym-pow.uni}.] The proof of Lemma
\ref{lem.uniprop.sym-pow.uni} is completely analogous to the proof of Lemma
\ref{lem.uniprop.ext-pow.uni}, and thus is omitted.
\end{proof}

Next, we state a definition (which is analogous to Definition
\ref{def.uniprop.antisym}, but works in a greater generality, since $W$ no
longer needs to be a $\mathbb{Z}$-module):

\begin{definition}
\label{def.uniprop.sym}Let $n\in\mathbb{N}$. Let $V$ be a set.

Let $W$ be a set. Let $f:V^{n}\rightarrow W$ be a map. We say that the map $f$
is \textit{symmetric} if and only if each $\left(  v_{1},v_{2},\ldots
,v_{n}\right)  \in V^{n}$ and $\gamma\in S_{n}$ satisfy%
\[
f\left(  v_{\gamma\left(  1\right)  },v_{\gamma\left(  2\right)  }%
,\ldots,v_{\gamma\left(  n\right)  }\right)  =f\left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  .
\]

\end{definition}

Now, we can state a universal property for the symmetric powers
$\operatorname*{Sym}\nolimits^{n}V$:

\begin{corollary}
\label{cor.uniprop.sym-pow}Let $k$ be a commutative ring. Let $n\in\mathbb{N}%
$. Let $V$ be a $k$-module.

Let $W$ be any $k$-module, and let $f:V^{n}\rightarrow W$ be a symmetric
multilinear map. Then, there exists a unique $k$-linear map
$f_{\operatorname*{Sym}}:\operatorname*{Sym}\nolimits^{n}V\rightarrow W$ such
that every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$ satisfies
$f_{\operatorname*{Sym}}\left(  \operatorname*{sym}\nolimits_{V,n}\left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  \right)  =f\left(
v_{1},v_{2},\ldots,v_{n}\right)  $.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.uniprop.sym-pow}.] The proof of Corollary
\ref{cor.uniprop.sym-pow} is completely analogous to the proof of Corollary
\ref{cor.uniprop.ext-pow} (up to some replacing of $+$ signs by $-$ signs and
some removal of powers of $-1$), and thus is omitted.
\end{proof}

We shall next derive similar results for exterior powers. First of all, we can
again easily obtain an analogue to Lemma \ref{lem.uniprop.ext-pow.uni}:

\begin{lemma}
\label{lem.uniprop.wedge-pow.uni}Let $k$ be a commutative ring. Let
$n\in\mathbb{N}$. Let $V$ be a $k$-module. Let $W$ be any $k$-module, and let
$f:V^{n}\rightarrow W$ be any map. Then, there exists \textbf{at most one}
$k$-linear map $f_{\wedge}:\wedge^{n}V\rightarrow W$ such that every $\left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$ satisfies $f_{\wedge}\left(
\operatorname*{wedge}\nolimits_{V,n}\left(  v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}\right)  \right)  =f\left(  v_{1},v_{2},\ldots
,v_{n}\right)  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.uniprop.wedge-pow.uni}.] The proof of Lemma
\ref{lem.uniprop.wedge-pow.uni} is completely analogous to the proof of Lemma
\ref{lem.uniprop.ext-pow.uni}, and thus is omitted.
\end{proof}

Next, we define a notion of \textquotedblleft weakly
alternating\textquotedblright\ which is (in some weak sense) similar to
Definition \ref{def.uniprop.antisym} (but at this point, there is no direct
analogy any more):

\begin{definition}
\label{def.uniprop.weakly-alt}Let $n\in\mathbb{N}$. Let $V$ be a set.

Let $W$ be a $\mathbb{Z}$-module. Let $f:V^{n}\rightarrow W$ be a map. We say
that the map $f$ is \textit{weakly alternating} if and only if each
$i\in\left\{  1,2,\ldots,n-1\right\}  $ and $\left(  v_{1},v_{2},\ldots
,v_{n}\right)  \in V^{n}$ satisfying $v_{i}=v_{i+1}$ satisfy%
\[
f\left(  v_{1},v_{2},\ldots,v_{n}\right)  =0.
\]

\end{definition}

Now, we can state a universal property for the pseudoexterior powers
$\wedge^{n}V$:

\begin{corollary}
\label{cor.uniprop.wedge-pow}Let $k$ be a commutative ring. Let $n\in
\mathbb{N}$. Let $V$ be a $k$-module.

Let $W$ be any $k$-module, and let $f:V^{n}\rightarrow W$ be a weakly
alternating multilinear map. (The notion of \textquotedblleft weakly
alternating\textquotedblright\ makes sense here because the $k$-module $W$ is
clearly a $\mathbb{Z}$-module.) Then, there exists a unique $k$-linear map
$f_{\wedge}:\wedge^{n}V\rightarrow W$ such that every $\left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  \in V^{n}$ satisfies $f_{\wedge}\left(
\operatorname*{wedge}\nolimits_{V,n}\left(  v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}\right)  \right)  =f\left(  v_{1},v_{2},\ldots
,v_{n}\right)  $.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.uniprop.wedge-pow}.]Corollary
\ref{cor.uniprop.tensor} shows that there exists a unique $k$-linear map
$f_{\otimes}:V^{\otimes n}\rightarrow W$ such that every $\left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  \in V^{n}$ satisfies $f_{\otimes}\left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  =f\left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  $. Consider this $f_{\otimes}$. The map
$f_{\otimes}$ is $k$-linear; thus, $\operatorname*{Ker}\left(  f_{\otimes
}\right)  $ is a $k$-submodule of $V^{\otimes n}$.

We know that every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$
satisfies
\begin{equation}
f_{\otimes}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)
=f\left(  v_{1},v_{2},\ldots,v_{n}\right)
.\label{pf.cor.uniprop.wedge-pow.ftens}%
\end{equation}


The map $f$ is weakly alternating. In other words, each $i\in\left\{
1,2,\ldots,n-1\right\}  $ and $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{n}$ satisfying $v_{i}=v_{i+1}$ satisfy%
\begin{equation}
f\left(  v_{1},v_{2},\ldots,v_{n}\right)
=0.\label{pf.cor.uniprop.wedge-pow.symmetry}%
\end{equation}
(by the definition of \textquotedblleft weakly alternating\textquotedblright).

Fix $i\in\left\{  1,2,\ldots,n-1\right\}  $. Define a subset $T$ of
$V^{\otimes n}$ by%
\[
T=\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n};\ v_{i}=v_{i+1}\right\}  .
\]
Thus,%
\begin{align}
&  \left\langle T\right\rangle \nonumber\\
&  =\left\langle \left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}%
\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n};\ v_{i}%
=v_{i+1}\right\}  \right\rangle \nonumber\\
&  =\left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n};\ v_{i}=v_{i+1}\right\rangle
.\label{pf.cor.uniprop.wedge-pow.spanT}%
\end{align}


Recall that $\operatorname*{wedge}\nolimits_{V,n}$ is the canonical projection
$V^{\otimes n}\rightarrow V^{\otimes n}\diagup R_{n}\left(  V\right)  $.

Now, $T\subseteq\operatorname*{Ker}\left(  f_{\otimes}\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $t\in T$. Then,%
\[
t\in T=\left\{  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n};\ v_{i}=v_{i+1}\right\}  .
\]
In other words, $t$ has the form $t=v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}$ for some $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$
satisfying $v_{i}=v_{i+1}$. Consider this $\left(  v_{1},v_{2},\ldots
,v_{n}\right)  $.
\par
From (\ref{pf.cor.uniprop.wedge-pow.ftens}), we obtain $f_{\otimes}\left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  =f\left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  =0$ (by (\ref{pf.cor.uniprop.wedge-pow.symmetry}%
)).
\par
Now, applying the map $f_{\otimes}$ to both sides of the equality
$t=v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}$, we find $f_{\otimes}\left(
t\right)  =f_{\otimes}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\right)  =0$. In other words, $t\in\operatorname*{Ker}\left(  f_{\otimes
}\right)  $.
\par
Now, forget that we fixed $t$. We thus have proven that $t\in
\operatorname*{Ker}\left(  f_{\otimes}\right)  $ for each $t\in T$. In other
words, $T\subseteq\operatorname*{Ker}\left(  f_{\otimes}\right)  $.}. Thus,
$f_{\otimes}\left(  \underbrace{T}_{\subseteq\operatorname*{Ker}\left(
f_{\otimes}\right)  }\right)  \subseteq f_{\otimes}\left(  \operatorname*{Ker}%
\left(  f_{\otimes}\right)  \right)  =0$, so that $f_{\otimes}\left(
T\right)  =0$. But Proposition \ref{prop.<>} \textbf{(b)} (applied to
$V^{\otimes n}$, $T$, $W$ and $f_{\otimes}$ instead of $M$, $S$, $R$ and $f$)
yields $f_{\otimes}\left(  \left\langle T\right\rangle \right)  =\left\langle
\underbrace{f_{\otimes}\left(  T\right)  }_{=0}\right\rangle =\left\langle
0\right\rangle =0$. In view of (\ref{pf.cor.uniprop.wedge-pow.spanT}), this
rewrites as%
\begin{equation}
f_{\otimes}\left(  \left\langle v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n};\ v_{i}%
=v_{i+1}\right\rangle \right)  =0.\label{pf.cor.uniprop.wedge-pow.3}%
\end{equation}


Now, forget that we fixed $i$. We thus have proven
(\ref{pf.cor.uniprop.wedge-pow.3}) for each $i\in\left\{  1,2,\ldots
,n-1\right\}  $.

But Proposition \ref{prop.R_n} yields%
\[
R_{n}\left(  V\right)  =\sum_{i=1}^{n-1}\left\langle v_{1}\otimes v_{2}%
\otimes\cdots\otimes v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{n};\ v_{i}=v_{i+1}\right\rangle .
\]
Applying the map $f_{\otimes}$ to both sides of this equality, we obtain%
\begin{align*}
& f_{\otimes}\left(  R_{n}\left(  V\right)  \right)  \\
& =f_{\otimes}\left(  \sum_{i=1}^{n-1}\left\langle v_{1}\otimes v_{2}%
\otimes\cdots\otimes v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{n};\ v_{i}=v_{i+1}\right\rangle \right)  \\
& =\sum_{i=1}^{n-1}\underbrace{f_{\otimes}\left(  \left\langle v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}\ \mid\ \left(  v_{1},v_{2},\ldots
,v_{n}\right)  \in V^{n};\ v_{i}=v_{i+1}\right\rangle \right)  }%
_{\substack{=0\\\text{(by (\ref{pf.cor.uniprop.wedge-pow.3}))}}}\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f_{\otimes}\text{ is
}k\text{-linear}\right)  \\
& =\sum_{i=1}^{n-1}0=0.
\end{align*}


Hence, Proposition \ref{prop.uniprop.quotient} (applied to $V^{\otimes n}$,
$R_{n}\left(  V\right)  $, $\operatorname*{wedge}\nolimits_{V,n}$, $W$ and
$f_{\otimes}$ instead of $V$, $I$, $\pi_{I}$, $W$ and $f$) yields that there
exists a unique $k$-linear map $f^{\prime}:V^{\otimes n}\diagup R_{n}\left(
V\right)  \rightarrow W$ satisfying $f_{\otimes}=f^{\prime}\circ
\operatorname*{wedge}\nolimits_{V,n}$. Consider this $f^{\prime}$.

The map $f^{\prime}$ is a $k$-linear map $V^{\otimes n}\diagup R_{n}\left(
V\right)  \rightarrow W$. In other words, the map $f^{\prime}$ is a $k$-linear
map $\wedge^{n}V\rightarrow W$ (since $V^{\otimes n}\diagup R_{n}\left(
V\right)  =\wedge^{n}V$). Every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in
V^{n}$ satisfies%
\begin{align*}
& f^{\prime}\left(  \operatorname*{wedge}\nolimits_{V,n}\left(  v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}\right)  \right)  \\
& =\underbrace{\left(  f^{\prime}\circ\operatorname*{wedge}\nolimits_{V,n}%
\right)  }_{=f_{\otimes}}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\right)  =f_{\otimes}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\right)  \\
& =f\left(  v_{1},v_{2},\ldots,v_{n}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{pf.cor.uniprop.wedge-pow.ftens})}\right)  .
\end{align*}
Thus, $f^{\prime}$ is a $k$-linear map $\wedge^{n}V\rightarrow W$ and has the
property that every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$
satisfies $f^{\prime}\left(  \operatorname*{wedge}\nolimits_{V,n}\left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)  \right)  =f\left(
v_{1},v_{2},\ldots,v_{n}\right)  $. Hence, there exists \textbf{at least one}
$k$-linear map $f_{\wedge}:\wedge^{n}V\rightarrow W$ such that every $\left(
v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$ satisfies $f_{\wedge}\left(
\operatorname*{wedge}\nolimits_{V,n}\left(  v_{1}\otimes v_{2}\otimes
\cdots\otimes v_{n}\right)  \right)  =f\left(  v_{1},v_{2},\ldots
,v_{n}\right)  $ (namely, $f_{\wedge}=f^{\prime}$). Since we also know that
there exists \textbf{at most one} such map (in fact, this follows from Lemma
\ref{lem.uniprop.wedge-pow.uni}), we can therefore conclude that there exists
a \textbf{unique} $k$-linear map $f_{\wedge}:\wedge^{n}V\rightarrow W$ such
that every $\left(  v_{1},v_{2},\ldots,v_{n}\right)  \in V^{n}$ satisfies
$f_{\wedge}\left(  \operatorname*{wedge}\nolimits_{V,n}\left(  v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}\right)  \right)  =f\left(  v_{1},v_{2}%
,\ldots,v_{n}\right)  $. This proves Corollary \ref{cor.uniprop.wedge-pow}.
\end{proof}



\begin{noncompile}
Bibliography follows:
\end{noncompile}

\begin{thebibliography}{9}                                                                                                %


\bibitem {goodwillie-MO}Tom (Thomas) Goodwillie, \textit{MathOverflow post
\#65716 (answer to \textquotedblleft Commutator tensors and
submodules\textquotedblright)}.\newline\url{http://mathoverflow.net/questions/65716}

\bibitem {dg1}Darij Grinberg, \textit{The Clifford algebra and the Chevalley
map - a computational approach} (summary version).\newline%
\url{http://www.cip.ifi.lmu.de/~grinberg/algebra/chevalleys.pdf} \newline
Darij Grinberg, \textit{The Clifford algebra and the Chevalley map - a
computational approach} (detailed version).\newline\url{http://www.cip.ifi.lmu.de/~grinberg/algebra/chevalley.pdf}

\bibitem {dg0}Darij Grinberg, \textit{Poincar\'{e}-Birkhoff-Witt type results
for inclusions of Lie algebras}, 2011.\newline\url{http://www.cip.ifi.lmu.de/~grinberg/algebra/algebra.html#pbw}

\bibitem {dg-hopf}Vorlesung \textit{Hopfalgebren} von Hans-J\"{u}rgen
Schneider, mitgeschrieben von Darij Grinberg.\newline\url{https://sites.google.com/site/darijgrinberg/hopfalgebren}
\end{thebibliography}


\end{document}