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\begin{document}
\section*{Math 332 Winter 2023, Lecture 22: Modules}

\textbf{website:}
\texttt{\href{https://www.cip.ifi.lmu.de/~grinberg/t/23wa}{\texttt{https://www.cip.ifi.lmu.de/\symbol{126}%
grinberg/t/23wa}}}

\setcounter{section}{1}

\section{Modules}

\setcounter{subsection}{6}

\subsection{Bilinear maps}

Let $R$ be a commutative ring.

The addition map%
\begin{align*}
\operatorname*{add}:R\times R  &  \rightarrow R,\\
\left(  a,b\right)   &  \mapsto a+b
\end{align*}
is $R$-linear (where the domain is the direct product of two copies of $R$),
as you can easily check\footnote{Indeed:
\par
\begin{itemize}
\item This map $\operatorname*{add}$ respects addition, since any two elements
$\left(  a,b\right)  $ and $\left(  c,d\right)  $ of $R\times R$ satisfy%
\begin{align*}
\operatorname*{add}\left(  \left(  a,b\right)  +\left(  c,d\right)  \right)
&  =\operatorname*{add}\left(  a+c,\ b+d\right)  =\left(  a+c\right)  +\left(
b+d\right) \\
&  =\underbrace{a+b}_{=\operatorname*{add}\left(  a,b\right)  }%
+\underbrace{c+d}_{=\operatorname*{add}\left(  c,d\right)  }%
=\operatorname*{add}\left(  a,b\right)  +\operatorname*{add}\left(
c,d\right)  .
\end{align*}
\par
\item This map $\operatorname*{add}$ respects scaling, since any $r\in R$ and
$\left(  a,b\right)  \in R\times R$ satisfy%
\[
\operatorname*{add}\left(  r\left(  a,b\right)  \right)  =\operatorname*{add}%
\left(  ra,rb\right)  =ra+rb=r\left(  \underbrace{a+b}_{=\operatorname*{add}%
\left(  a,b\right)  }\right)  =r\operatorname*{add}\left(  a,b\right)  .
\]
\par
\item This map $\operatorname*{add}$ respects zero, since $\operatorname*{add}%
\left(  0,0\right)  =0+0=0$.
\end{itemize}
}. However, the multiplication map%
\begin{align*}
\operatorname*{mul}:R\times R  &  \rightarrow R,\\
\left(  a,b\right)   &  \mapsto ab
\end{align*}
is not. But there is \textquotedblleft some linearity\textquotedblright\ in
$\operatorname*{mul}$: Namely, if we fix one argument, then
$\operatorname*{mul}$ is linear in the other. That is:

\begin{itemize}
\item For any $a\in R$, the map%
\begin{align*}
R  &  \rightarrow R,\\
b  &  \mapsto ab
\end{align*}
is $R$-linear.

\item For any $b\in R$, the map%
\begin{align*}
R  &  \rightarrow R,\\
a  &  \mapsto ab
\end{align*}
is $R$-linear.
\end{itemize}

Maps with these properties are called \textbf{bilinear}:

\begin{definition}
Let $R$ be a commutative ring. Let $M$, $N$ and $P$ be three $R$-modules. A
map $f:M\times N\rightarrow P$ is said to be $R$\textbf{-bilinear} (or just
\textbf{bilinear}) if it satisfies the following two conditions:

\begin{enumerate}
\item For any $n\in N$, the map%
\begin{align*}
M  &  \rightarrow P,\\
m  &  \mapsto f\left(  m,n\right)
\end{align*}
is $R$-linear. Explicitly, this is saying that for any $n\in N$, we have%
\begin{align*}
f\left(  m_{1}+m_{2},\ n\right)   &  =f\left(  m_{1},n\right)  +f\left(
m_{2},n\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }m_{1},m_{2}\in M;\\
f\left(  rm,n\right)   &  =rf\left(  m,n\right)  \ \ \ \ \ \ \ \ \ \ \text{for
all }r\in R\text{ and }m\in M;\\
f\left(  0,n\right)   &  =0.
\end{align*}
This is called \textquotedblleft$f$ is \textbf{linear in its first
argument}\textquotedblright.

\item For any $m\in M$, the map%
\begin{align*}
N  &  \rightarrow P,\\
n  &  \mapsto f\left(  m,n\right)
\end{align*}
is $R$-linear. Explicitly, this is saying that for any $m\in M$, we have%
\begin{align*}
f\left(  m,\ n_{1}+n_{2}\right)   &  =f\left(  m,n_{1}\right)  +f\left(
m,n_{2}\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }n_{1},n_{2}\in N;\\
f\left(  m,rn\right)   &  =rf\left(  m,n\right)  \ \ \ \ \ \ \ \ \ \ \text{for
all }r\in R\text{ and }n\in N;\\
f\left(  m,0\right)   &  =0.
\end{align*}
This is called \textquotedblleft$f$ is \textbf{linear in its second
argument}\textquotedblright.
\end{enumerate}
\end{definition}

Here are some examples of bilinear maps (where $R$ denotes a commutative ring throughout):

\begin{itemize}
\item As we just teased, the multiplication map $\operatorname*{mul}:R\times
R\rightarrow R$ is bilinear.

\item For any $n\in\mathbb{N}$, the map%
\begin{align*}
R^{n}\times R^{n}  &  \rightarrow R,\\
\left(  \left(  a_{1},a_{2},\ldots,a_{n}\right)  ,\ \left(  b_{1},b_{2}%
,\ldots,b_{n}\right)  \right)   &  \mapsto a_{1}b_{1}+a_{2}b_{2}+\cdots
+a_{n}b_{n}%
\end{align*}
is $R$-bilinear. This map is called the \textbf{dot product} or the
\textbf{standard scalar product}. At least in the case $R=\mathbb{R}$, it
should be familiar from basic linear algebra.

\item Consider the field $\mathbb{C}$ of complex numbers. For any
$n\in\mathbb{N}$, the \textbf{standard inner product}
\begin{align*}
\mathbb{C}^{n}\times\mathbb{C}^{n}  &  \rightarrow\mathbb{C},\\
\left(  \left(  a_{1},a_{2},\ldots,a_{n}\right)  ,\ \left(  b_{1},b_{2}%
,\ldots,b_{n}\right)  \right)   &  \mapsto a_{1}\overline{b_{1}}%
+a_{2}\overline{b_{2}}+\cdots+a_{n}\overline{b_{n}}%
\end{align*}
(where $\overline{z}$ denotes the complex conjugate of a complex number $z$)
is $\mathbb{R}$-bilinear but not $\mathbb{C}$-bilinear (since it is antilinear
rather than linear in the second argument). However, it becomes $\mathbb{C}%
$-bilinear if you view it as a map $\mathbb{C}^{n}\times\overline{\mathbb{C}%
}^{n}\rightarrow\mathbb{C}$ (with $\overline{\mathbb{C}}$ being the
\textquotedblleft twisted\textquotedblright\ $\mathbb{C}$-module $\mathbb{C}$
constructed in \S 2.4.3 in Lecture 19).

\item The determinant map%
\begin{align*}
\det:R^{2}\times R^{2}  &  \rightarrow R,\\
\left(  \left(  a,b\right)  ,\ \left(  c,d\right)  \right)   &  \mapsto ad-bc
\end{align*}
is $R$-bilinear. (I call it the determinant map because it sends $\left(
\left(  a,b\right)  ,\ \left(  c,d\right)  \right)  $ to $\det\left(
\begin{array}
[c]{cc}%
a & b\\
c & d
\end{array}
\right)  $.)

\item Matrix multiplication is $R$-bilinear. That is: For any $m,n,p\in
\mathbb{N}$, the map%
\begin{align*}
R^{m\times n}\times R^{n\times p}  &  \rightarrow R^{m\times p},\\
\left(  A,B\right)   &  \mapsto AB
\end{align*}
is $R$-bilinear. (Indeed, this boils down to standard properties of matrix
multiplication, such as $\left(  A_{1}+A_{2}\right)  B=A_{1}B+A_{2}B$ and
$A\left(  \lambda B\right)  =\lambda AB$ for any matrices $A_{1},A_{2},A,B$
and any scalar $\lambda\in R$.)

\item The \textbf{cross product} map%
\begin{align*}
R^{3}\times R^{3}  &  \rightarrow R^{3},\\
\left(  \left(  a,b,c\right)  ,\ \left(  a^{\prime},b^{\prime},c^{\prime
}\right)  \right)   &  \mapsto\left(  bc^{\prime}-cb^{\prime},\ ca^{\prime
}-ac^{\prime},\ ab^{\prime}-ba^{\prime}\right)
\end{align*}
is $R$-bilinear.

\item For any $R$-module $M$, the action%
\begin{align*}
R\times M  &  \rightarrow M,\\
\left(  r,m\right)   &  \mapsto rm
\end{align*}
is a $R$-bilinear map. It is a nice exercise to see what the conditions in the
definition of \textquotedblleft bilinear\textquotedblright\ translate to for
this action (you can find the answer in \S 3.9 of the text). Note that we
really need $R$ to be commutative here.
\end{itemize}

\bigskip

For free modules, we have proved a universal property (Theorem 2.6.2 in
Lecture 21) that lets us define a linear map out of a free module just by
specifying its values on a given basis. The same can be done for bilinear maps:

\begin{theorem}
[Universal property of free modules wrt bilinear maps]%
\label{thm.mods.uniprop-free-2}Let $R$ be a commutative ring. Let $M$ be a
free $R$-module with basis $\left(  m_{i}\right)  _{i\in I}$. Let $N$ be a
free $R$-module with basis $\left(  n_{j}\right)  _{j\in J}$. Let $P$ be a
further $R$-module (free or not). Let $p_{i,j}$ be a vector in $P$ for each
pair $\left(  i,j\right)  \in I\times J$. Then, there exists a \textbf{unique}
$R$-bilinear map $f:M\times N\rightarrow P$ such that%
\[
\text{each }\left(  i,j\right)  \in I\times J\text{ satisfies }f\left(
m_{i},n_{j}\right)  =p_{i,j}.
\]

\end{theorem}

\begin{proof}
Similar to the proof of Theorem 2.6.2 in Lecture 21. (Details can be found in
the proof of Theorem 3.9.2 in the text.)
\end{proof}

\subsection{Multilinear maps}

\textbf{Multilinear maps} are a generalization of linear and bilinear maps.
Linear maps have one argument; bilinear maps have two. Multilinear maps are
\textquotedblleft the same thing\textquotedblright\ but with $n$ arguments:

\begin{definition}
Let $R$ be a commutative ring. Let $M_{1},M_{2},\ldots,M_{n}$ be finitely many
$R$-modules. Let $P$ be any $R$-module. A map $f:M_{1}\times M_{2}\times
\cdots\times M_{n}\rightarrow P$ is said to be $R$\textbf{-multilinear} (or
just \textbf{multilinear}) if it satisfies the following condition:

\begin{itemize}
\item For any $i\in\left\{  1,2,\ldots,n\right\}  $ and any $m_{1}%
,m_{2},\ldots,m_{i-1},m_{i+1},\ldots,m_{n}$ in the respective modules (meaning
that $m_{k}\in M_{k}$ for each $k\neq i$), the map%
\begin{align*}
M_{i}  &  \rightarrow P,\\
m_{i}  &  \mapsto f\left(  m_{1},m_{2},\ldots,m_{n}\right)
\end{align*}
is $R$-linear. In other words, if we fix all arguments of $f$ other than the
$i$-th argument, then $f$ is $R$-linear as a function of this $i$-th argument.
This is called \textquotedblleft$f$ is \textbf{linear in its }$i$\textbf{-th
argument}\textquotedblright.
\end{itemize}
\end{definition}

Thus, \textquotedblleft bilinear\textquotedblright\ means \textquotedblleft
multilinear for $n=2$\textquotedblright, whereas \textquotedblleft
linear\textquotedblright\ means \textquotedblleft multilinear for
$n=1$\textquotedblright.

The simplest example of a multilinear map is the map%
\begin{align*}
\operatorname*{prod}\nolimits_{n}:R^{n}  &  \rightarrow R,\\
\left(  a_{1},a_{2},\ldots,a_{n}\right)   &  \mapsto a_{1}a_{2}\cdots a_{n}.
\end{align*}
Another famous example of a multilinear map is
\begin{align*}
\det:\underbrace{R^{n}\times R^{n}\times\cdots\times R^{n}}_{n\text{ times}}
&  \rightarrow R,\\
\left(  v_{1},v_{2},\ldots,v_{n}\right)   &  \mapsto\det\left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  ,
\end{align*}
where $\det\left(  v_{1},v_{2},\ldots,v_{n}\right)  $ is the determinant of
the matrix whose rows are $v_{1},v_{2},\ldots,v_{n}$.

There is a universal property of free modules wrt multilinear maps, but you
can state and prove it at home. (It involves no new ideas compared to the
bilinear one, but a lot more subscripts.)

\subsection{Algebras over commutative rings}

\begin{convention}
In this section, we fix a \textbf{commutative} ring $R$.
\end{convention}

\subsubsection{Definition}

We know rings and we know $R$-modules. The former have addition and
multiplication; the latter have addition and scaling. What happens if we
combine these features, to obtain an object that has addition, multiplication
and scaling?

That kind of object turns out to be very useful. Here is its precise
definition (we do impose an extra axiom to keep the multiplication and scaling
in harmony):

\begin{definition}
An $R$\textbf{-algebra} is a set $A$ that is endowed with

\begin{itemize}
\item two binary operations (i.e., maps from $A\times A$ to $A$) that are
called \textbf{addition} and \textbf{multiplication} and denoted by $+$ and
$\cdot$,

\item a map $\cdot$ from $R\times A$ to $A$ that is called \textbf{action} of
$R$ on $A$ (and should not be confused with the multiplication map, which is
also denoted by $\cdot$), and

\item two elements of $A$ that are called \textbf{zero} and \textbf{unity} and
are denoted by $0$ and $1$,
\end{itemize}

\noindent such that the following properties (the \textquotedblleft%
\textbf{algebra axioms}\textquotedblright) hold:

\begin{itemize}
\item The addition, the multiplication, the zero and the unity satisfy all the
ring axioms (so that $A$ becomes a ring).

\item The addition, the action and the zero satisfy all the module axioms (so
that $A$ becomes an $R$-module).

\item \textbf{Scale-invariance of multiplication:} We have%
\[
r\left(  ab\right)  =\left(  ra\right)  b=a\left(  rb\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }r\in R\text{ and }a,b\in A.
\]
Here (and in the following), we omit the $\cdot$ signs.
\end{itemize}
\end{definition}

Thus, an $R$-algebra is an $R$-module that is also a ring at the same time,
with the same addition, and satisfying the \textquotedblleft
scale-invariance\textquotedblright\ axiom so that the structures work together nicely.

The \textquotedblleft scale-invariance\textquotedblright\ axiom can be
replaced by requiring that the multiplication map%
\begin{align*}
A\times A  &  \rightarrow A,\\
\left(  a,b\right)   &  \mapsto ab
\end{align*}
be $R$-bilinear. (This $R$-bilinearity also includes the distributive laws
$\left(  a_{1}+a_{2}\right)  b=a_{1}b+a_{2}b$ and $a\left(  b_{1}%
+b_{2}\right)  =ab_{1}+ab_{2}$ as well as the axioms $a0=0$ and $0b=0$, but
these are also part of the ring axioms.)

A simple-to-memorize way to restate the \textquotedblleft
scale-invariance\textquotedblright\ axiom is the following: \textquotedblleft
Scalars commute with vectors\textquotedblright. Here, \textquotedblleft
scalars\textquotedblright\ mean elements of $R$ (as expected, since $A$ is an
$R$-module), while \textquotedblleft vectors\textquotedblright\ mean elements
of $A$, and the word \textquotedblleft commute\textquotedblright\ is used in a
slightly broader sense as usual (as we are dealing with actions, not just products).

Examples of $R$-algebras include the following:

\begin{itemize}
\item The commutative ring $R$ itself is an $R$-algebra. (Here, the
multiplication of $R$ plays both the role of multiplication and the role of action.)

\item The zero ring $\left\{  0\right\}  $ is an $R$-algebra.

\item The matrix ring $R^{n\times n}$ is an $R$-algebra for any $n\in
\mathbb{N}$ (since it is both a ring and an $R$-module, and \textquotedblleft
scale-invariance\textquotedblright\ is easily seen to hold). (Note that
$R^{n\times n}$ is usually not commutative!)

\item The ring $\mathbb{C}$ is an $\mathbb{R}$-algebra (since it is both a
ring and an $\mathbb{R}$-module, and \textquotedblleft
scale-invariance\textquotedblright\ is easily seen to hold).

\item The ring $\mathbb{R}$ is a $\mathbb{Q}$-algebra (similarly).

\item More generally: If a commutative ring $R$ is a subring of a
\textbf{commutative} ring $S$, then $S$ becomes an $R$-module in a natural way
(by the method we have learnt in \S 2.1.5 in Lecture 18: the action of $R$ on
$S$ is just the multiplication of $S$, restricted to $R\times S$), and thus
becomes an $R$-algebra (because it is also a ring and satisfies the
\textquotedblleft scale-invariance\textquotedblright\ axiom).

\item Even more generally: If $R$ and $S$ are two \textbf{commutative} rings,
and if $f:R\rightarrow S$ is a ring morphism, then $S$ becomes an $R$-module
(by the method we have learnt in \S 2.1.5 in Lecture 18: the action of $R$ on
$S$ is given by%
\[
\underbrace{rs}_{\text{action}}=\underbrace{f\left(  r\right)  \cdot
s}_{\substack{\text{multiplication}\\\text{inside }S}%
}\ \ \ \ \ \ \ \ \ \ \text{for all }r\in R\text{ and }s\in S
\]
), and thus becomes an $R$-algebra (since it is also a ring and satisfies the
\textquotedblleft scale-invariance axiom\textquotedblright). This $R$-algebra
structure on $S$ is said to be \textbf{induced} by the morphism $f$.

\item Even more generally: If $R$ and $S$ are two commutative rings, and if
$f:R\rightarrow S$ is a ring morphism, then any $S$-algebra $A$ becomes an
$R$-algebra via the \textquotedblleft restriction of scalars\textquotedblright%
\ rule%
\[
r\cdot a=f\left(  r\right)  \cdot a\ \ \ \ \ \ \ \ \ \ \text{for all }r\in
R\text{ and }a\in A.
\]
(Again, this is the same rule that we have seen in \S 2.1.5 in Lecture 18, but
we now additionally have a ring structure on $A$, so that $A$ becomes not just
an $R$-module but also an $R$-algebra.)

\item The quaternion ring $\mathbb{H}$ is an $\mathbb{R}$-algebra, but
\textbf{not} a $\mathbb{C}$-algebra (despite $\mathbb{C}$ being a subring of
$\mathbb{H}$). Why not? Because it violates the scale-invariance axiom, which
in this case says that%
\[
r\left(  ab\right)  =\left(  ra\right)  b=a\left(  rb\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }r\in\mathbb{C}\text{ and }a,b\in
\mathbb{H}.
\]
For example, for $r=i$ and $a=j$ and $b=1$, we have%
\begin{align*}
\left(  ra\right)  b  &  =\left(  ij\right)
1=k1=k\ \ \ \ \ \ \ \ \ \ \text{but}\\
a\left(  rb\right)   &  =j\left(  i1\right)  =ji=-k\neq k.
\end{align*}
In a nutshell, this axiom is failing because quaternions don't commute with
complex numbers. As we said, the vectors in an $R$-algebra must
\textquotedblleft commute\textquotedblright\ with the scalars (meaning
$ra=a\left(  r1_{A}\right)  $ for all $r\in R$ and $a\in A$).

\item The polynomial ring $R\left[  x\right]  $ (to be defined soon) is an $R$-algebra.

\item More examples can be found in \S 3.11.2 of the text.
\end{itemize}

\subsubsection{Rings as $\mathbb{Z}$-algebras}

Proposition 2.3.1 \textbf{(a)} in Lecture 19 shows that every abelian group
(written additively) automatically becomes a $\mathbb{Z}$-module. Likewise
(and in fact extending this construction), any ring automatically becomes a
$\mathbb{Z}$-algebra:

\begin{proposition}
Let $A$ be any ring. Then, $A$ is an abelian group (with respect to addition),
and thus becomes a $\mathbb{Z}$-module (by Proposition 2.3.1 \textbf{(a)} in
Lecture 19). Combining this $\mathbb{Z}$-module structure with the given ring
structure on $A$, we obtain a $\mathbb{Z}$-algebra. Thus, $A$ becomes a
$\mathbb{Z}$-algebra.
\end{proposition}

\begin{proof}
Easy.
\end{proof}

\subsubsection{The underlying structures}

Every $R$-algebra $A$ has an underlying ring (i.e., the ring that you are left
with if you forget the action of $R$ on $A$) and an underlying $R$-module
(i.e., the $R$-module that you are left with if you forget the multiplication
and the unity of $A$). This ring and this $R$-module are inherent in $A$; we
will refer to them simply as \textquotedblleft the ring $A$\textquotedblright%
\ and \textquotedblleft the $R$-module $A$\textquotedblright.

Thus, when $A$ and $B$ are two $R$-algebras, a \textquotedblleft ring morphism
from $A$ to $B$\textquotedblright\ means a ring morphism from the underlying
ring of $A$ to the underlying ring of $B$, whereas an \textquotedblleft%
$R$-module morphism from $A$ to $B$\textquotedblright\ means an $R$-module
morphism from the underlying $R$-module of $A$ to the underlying $R$-module of
$B$.

\begin{fineprint}
\textbf{Warning:} The notion of an \textquotedblleft underlying
ring\textquotedblright\ has nothing to do with the notion of a
\textquotedblleft base ring\textquotedblright! The \textbf{base ring} of an
$R$-algebra $A$ is defined to be $R$ (not $A$).
\end{fineprint}

\subsubsection{Commutative $R$-algebras}

\begin{definition}
An $R$-algebra is said to be \textbf{commutative} if its underlying ring is
commutative (i.e., its multiplication is commutative).
\end{definition}

\subsubsection{Subalgebras}

Subalgebras are to algebras what subrings are to rings (and what submodules
are to modules):

\begin{definition}
Let $A$ be an $R$-algebra. An $R$\textbf{-subalgebra} of $A$ means a subset of
$A$ that is simultaneously a subring and an $R$-submodule of $A$ (that is,
that is closed under addition, multiplication and scaling and contains $0$ and
$1$).
\end{definition}

Every $R$-subalgebra of an algebra $A$ becomes an algebra in its own right
automatically (by inheriting the operations from $A$).

\subsubsection{$R$-algebra morphisms}

As you could expect, $R$-algebras (just like rings and modules) have their own
kind of morphisms and isomorphisms:

\begin{definition}
Let $A$ and $B$ be two $R$-algebras.

\textbf{(a)} An $R$\textbf{-algebra morphism} (or, short, \textbf{algebra
morphism}) from $A$ to $B$ means a map $f:A\rightarrow B$ that is both a ring
morphism and an $R$-module morphism.

\textbf{(b)} An $R$\textbf{-algebra isomorphism} from $A$ to $B$ means an
invertible $R$-algebra morphism $f:A\rightarrow B$ whose inverse
$f^{-1}:B\rightarrow A$ is also an $R$-algebra morphism.

\textbf{(c)} The $R$-algebras $A$ and $B$ are said to be \textbf{isomorphic}
(this is written $A\cong B$) if there exists an $R$-algebra isomorphism from
$A$ to $B$.
\end{definition}

All the fundamental properties of ring morphisms and of ring isomorphisms (as
discussed in \S 1.7 of Lecture 6) have analogues for algebras instead of
rings. For example:

\begin{itemize}
\item Any invertible $R$-algebra morphism is an isomorphism. (This is an
analogue of Proposition 1.7.6 in Lecture 6.)

\item The image of an $R$-algebra morphism $f:A\rightarrow B$ is an
$R$-subalgebra of $B$. (This is an analogue of Proposition 1.7.5 in Lecture 6.)
\end{itemize}

The proofs of these analogues are analogous to the proofs of the original results.

Furthermore, if $A$ and $B$ are two $\mathbb{Z}$-algebras, then the
$\mathbb{Z}$-algebra morphisms from $A$ to $B$ are precisely the ring
morphisms from $A$ to $B$. (In fact, this follows easily from Proposition
2.4.2 in Lecture 19.)

\subsubsection{Direct products}

\begin{definition}
Direct products of $R$-algebras are defined just as for rings and $R$-modules:
Addition, multiplication and action are entrywise.
\end{definition}

For details, see \S 3.11.8 in the text.
\end{document}