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\ihead{Collected trivialities on algebra derivations}
\ohead{\today}
\begin{document}

\title{Collected trivialities on algebra derivations}
\author{Darij Grinberg}
\date{\today}
\maketitle
\tableofcontents

\subsection*{***}

The purpose of this note is to state and prove in detail some folklore
properties of derivations on $\mathbf{k}$-algebras. It contains no deep result
or complicated proofs; its length is chiefly due to the level of detail and
slow pacing.

\section{Derivations in general}

\subsection{Definitions and conventions}

Let us first recall some definitions and set up some notations that will be
used for the rest of this note.

\begin{condition}
\label{conv.k-alg}In the following, $\mathbb{N}$ denotes the set $\left\{
0,1,2,\ldots\right\}  $.

The word \textquotedblleft ring\textquotedblright\ shall always mean an
associative ring with unity.

We fix a commutative ring $\mathbf{k}$ (once and for all). If $M$ and $N$ are
two $\mathbf{k}$-modules, then we let $\operatorname*{Hom}\left(  M,N\right)
$ denote the $\mathbf{k}$-module consisting of all $\mathbf{k}$-module
homomorphisms $M\rightarrow N$. (We use this notation even if $M$ and $N$ are
equipped with some other structure; e.g., even if $M$ and $N$ are $\mathbf{k}%
$-algebras, we write $\operatorname*{Hom}\left(  M,N\right)  $ for the
$\mathbf{k}$-module homomorphisms $M\rightarrow N$, not the $\mathbf{k}%
$-algebra homomorphisms $M\rightarrow N$.) All tensor products are understood
to be tensor products over $\mathbf{k}$ unless said otherwise.

A \textit{magmatic }$\mathbf{k}$\textit{-algebra} means a $\mathbf{k}$-module
$A$ equipped with a $\mathbf{k}$-bilinear map $m:A\times A\rightarrow A$. This
map $m$ is called the \textit{multiplication} of the magmatic $\mathbf{k}%
$-algebra $A$. The multiplication of a magmatic $\mathbf{k}$-algebra $A$ is
often \textquotedblleft written as multiplication\textquotedblright; i.e., one
writes $a\cdot b$ (or, even shorter, $ab$) for $m\left(  \left(  a, b\right)
\right)  $ whenever $a$ and $b$ are two elements of $A$. (Often, a magmatic
$\mathbf{k}$-algebra is called a \textit{nonassociative }$\mathbf{k}%
$\textit{-algebra}. However, the word \textquotedblleft
nonassociative\textquotedblright\ is slightly confusing here, since it does
not mean that associativity must be violated; it only means that associativity
is not required.)

Notice that any magmatic $\mathbf{k}$-algebra $A$ automatically satisfies%
\begin{align}
\lambda\left(  ab\right)   &  =\left(  \lambda a\right)  b=a\left(  \lambda
b\right) \label{eq.conv.k-alg.centrality}\\
&  \ \ \ \ \ \ \ \ \ \ \text{for every }\lambda\in\mathbf{k}\text{ and every
}a\in A\text{ and }b\in A\nonumber
\end{align}
(because its multiplication is $\mathbf{k}$-bilinear). Some authors use a
different concept of $\mathbf{k}$-algebras, which is more general and does not
always satisfy (\ref{eq.conv.k-alg.centrality}). In their notations, what we
call \textquotedblleft magmatic $\mathbf{k}$-algebra\textquotedblright\ is
called \textquotedblleft central magmatic $\mathbf{k}$%
-algebra\textquotedblright.

A \textit{unital magmatic }$\mathbf{k}$\textit{-algebra} is defined to be a
magmatic $\mathbf{k}$-algebra $A$ equipped with an element $e\in A$ which
satisfies%
\[
\left(  ea=ae=a\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\right)  .
\]
This element $e$ is unique when it exists (this is easy to prove), and is
called the \textit{unity} of $A$. (Some authors use the word \textquotedblleft
identity\textquotedblright, \textquotedblleft unit\textquotedblright\ or
\textquotedblleft one\textquotedblright\ instead of \textquotedblleft
unity\textquotedblright, but these words are sometimes ambiguous.) Since the
unity of $A$ is unique, we can afford not specifying it when defining a unital
magmatic $\mathbf{k}$-algebra (as long as we make sure that it exists); we
thus can say that some magmatic $\mathbf{k}$-algebra $A$ \textquotedblleft is
unital\textquotedblright, when we really mean that there exists an $e\in A$
such that $A$ equipped with this $e$ is a unital magmatic $\mathbf{k}$-algebra.

The unity of a unital magmatic $\mathbf{k}$-algebra $A$ is denoted by $1_{A}$,
or by $1$ when no confusion can arise.

Notice that any unital magmatic $\mathbf{k}$-algebra $A$ automatically
satisfies%
\[
\lambda a=\left(  \lambda\cdot1_{A}\right)  a=a\left(  \lambda\cdot
1_{A}\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }\lambda\in\mathbf{k}\text{
and every }a\in A.
\]
(Indeed, this follows easily from (\ref{eq.conv.k-alg.centrality}).)

A magmatic $\mathbf{k}$-algebra $A$ is said to be \textit{associative} if it
satisfies%
\[
\left(  a\left(  bc\right)  =\left(  ab\right)  c\ \ \ \ \ \ \ \ \ \ \text{for
every }a\in A\text{, }b\in A\text{ and }c\in A\right)  .
\]


We use the notation \textquotedblleft$\mathbf{k}$\textit{-algebra}%
\textquotedblright\ for \textquotedblleft associative unital magmatic
$\mathbf{k}$-algebra\textquotedblright. Thus, of course, a magmatic
$\mathbf{k}$-algebra is not always a $\mathbf{k}$-algebra.

The base ring $\mathbf{k}$ itself becomes a $\mathbf{k}$-algebra (equipped
with its multiplication and its unity).

If $A$ and $B$ are two $\mathbf{k}$-algebras, then a $\mathbf{k}%
$\textit{-algebra homomorphism from }$A$ \textit{to }$B$ means a $\mathbf{k}%
$-module homomorphism $f:A\rightarrow B$ satisfying $f\left(  1_{A}\right)
=1_{B}$ and%
\begin{equation}
\left(  f\left(  ab\right)  =f\left(  a\right)  f\left(  b\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{ and }b\in A\right)  .
\label{eq.conv.k-alg.morphism}%
\end{equation}
We notice that the condition $f\left(  1_{A}\right)  =1_{B}$ does not follow
from (\ref{eq.conv.k-alg.morphism}).

Every $\mathbf{k}$-algebra is a ring (when equipped with its multiplication
and its unity), and every $\mathbf{k}$-algebra homomorphism is a ring
homomorphism. Actually, if $A$ and $B$ are two $\mathbf{k}$-algebras, then a
$\mathbf{k}$-algebra homomorphism $A\rightarrow B$ is the same as a
$\mathbf{k}$-linear ring homomorphism $A\rightarrow B$. If $A$ is a
$\mathbf{k}$-algebra, then the map%
\[
\mathbf{k}\rightarrow A,\ \ \ \ \ \ \ \ \ \ \lambda\mapsto\lambda\cdot1_{A}%
\]
is a ring homomorphism and a $\mathbf{k}$-algebra homomorphism. Thus, any left
$A$-module $M$ canonically becomes a $\mathbf{k}$-module (via this
homomorphism); explicitly, its $\mathbf{k}$-module structure is given by%
\[
\left(  \lambda m=\left(  \lambda\cdot1_{A}\right)
m\ \ \ \ \ \ \ \ \ \ \text{for every }\lambda\in\mathbf{k}\text{ and }m\in
M\right)  .
\]
Similarly, any right $A$-module $M$ canonically becomes a $\mathbf{k}$-module.
\end{condition}

The terminology we have introduced in Convention \ref{conv.k-alg} is fairly
standard in some parts of the mathematical world, and nonstandard in others.
For instance, we use the word \textquotedblleft$\mathbf{k}$%
-algebra\textquotedblright\ as an abbreviation for \textquotedblleft
associative unital magmatic $\mathbf{k}$-algebra\textquotedblright; some
others use it as an abbreviation for \textquotedblleft associative magmatic
$\mathbf{k}$-algebra\textquotedblright, whereas others use it for
\textquotedblleft magmatic $\mathbf{k}$-algebra\textquotedblright. Some
authors also say \textquotedblleft unitary\textquotedblright\ instead of
\textquotedblleft unital\textquotedblright.

\begin{definition}
\label{def.bimod}Let $A$ and $B$ be $\mathbf{k}$-algebras. An $\left(
A,B\right)  _{\mathbf{k}}$\textit{-bimodule} means a $\mathbf{k}$-module $M$
equipped with a left $A$-module structure and a right $B$-module structure
satisfying the following properties:

\begin{enumerate}
\item We have%
\begin{equation}
\left(  am\right)  b=a\left(  mb\right)  \ \ \ \ \ \ \ \ \ \ \text{for any
}a\in A\text{, }m\in M\text{ and }b\in B. \label{eq.def.bimod.1}%
\end{equation}


\item We have
\begin{equation}
\left(  \lambda1_{A}\right)  m=m\left(  \lambda1_{B}\right)  =\lambda
m\ \ \ \ \ \ \ \ \ \ \text{for any }\lambda\in\mathbf{k}\text{ and }m\in M.
\label{eq.def.bimod.2}%
\end{equation}
(In other words, the $\mathbf{k}$-module structure on $M$ obtained from the
left $A$-module structure on $M$ is identical to the $\mathbf{k}$-module
structure on $M$ obtained from the right $B$-module structure on $M$, and also
identical to the $\mathbf{k}$-module structure that was given on $M$.)
\end{enumerate}

We shall abbreviate \textquotedblleft$\left(  A,B\right)  _{\mathbf{k}}%
$-bimodule\textquotedblright\ as \textquotedblleft$\left(  A,B\right)
$-bimodule\textquotedblright, since the $\mathbf{k}$ is fixed.

The condition (\ref{eq.def.bimod.1}) in the definition of an $\left(
A,B\right)  $-bimodule allows us to write $amb$ (without bracketing) for both
$\left(  am\right)  b$ and $a\left(  mb\right)  $ (where $M$ is an $\left(
A,B\right)  $-bimodule, and where $a\in A$, $m\in M$ and $b\in B$) without
having to worry about ambiguity.
\end{definition}

Notice that the notion of an $\left(  A,B\right)  $-bimodule depends on
$\mathbf{k}$ (even if we suppress $\mathbf{k}$ from the notation), because
$\mathbf{k}$ plays a role in the condition (\ref{eq.def.bimod.2}%
).\footnote{For example, if $\mathbb{H}$ denotes the ring of quaternions, then
$\mathbb{H}$ becomes a $\left(  \mathbb{C},\mathbb{C}\right)  $-bimodule when
$\mathbf{k}=\mathbb{R}$ (by multiplication from the left and from the right),
but not when $\mathbf{k}=\mathbb{C}$ (because (\ref{eq.def.bimod.2}) does not
hold for $\lambda\in\mathbb{C}$ and $m\in\mathbb{H}$).} Various authors use a
slightly different notion of an $\left(  A,B\right)  $-bimodule, in whose
definition the $\mathbf{k}$ does not occur\footnote{More precisely: They
require $A$ and $B$ to be rings (rather than $\mathbf{k}$-algebras); they
require $M$ to be an abelian group (rather than a $\mathbf{k}$-module); and
they omit the condition (\ref{eq.def.bimod.2}).}; their notion of an $\left(
A,B\right)  $-bimodule is equivalent to our notion of an $\left(  A,B\right)
_{\mathbb{Z}}$-bimodule. Our definition is thus more general. My impression is
that several authors use our general notion of an $\left(  A,B\right)
$-bimodule while only defining their less general one; this has led to some
confusion (\cite{CSA14}), and is the reason why I have done Definition
\ref{def.bimod} in this much detail.

\begin{proposition}
\label{prop.bimod.A}Let $A$ be a $\mathbf{k}$-algebra. Then, equipping $A$
with the natural left $A$-module structure on $A$ (which is given by
$am=a\cdot m$ for every $a\in A$ and $m\in A$) and with the natural right
$A$-module structure on $A$ (which is given by $mb=m\cdot b$ for every $b\in
A$ and $m\in A$) yields an $\left(  A,A\right)  $-bimodule. This $\left(
A,A\right)  $-bimodule will be called the $\left(  A,A\right)  $%
\textit{-bimodule }$A$. We consider $A$ equipped with this $\left(
A,A\right)  $-bimodule structure by default.

(Let us point out that the axiom (\ref{eq.def.bimod.1}) is satisfied for the
$\left(  A,A\right)  $-bimodule $A$ because the $\mathbf{k}$-algebra $A$ is associative.)
\end{proposition}

\begin{remark}
Let $A$ and $B$ be $\mathbf{k}$-algebras. Let $M$ be an $\left(  A,B\right)
$-bimodule. Then, the maps%
\[
A\times M\rightarrow M,\ \ \ \ \ \ \ \ \ \ \left(  a,m\right)  \mapsto am
\]
and%
\[
M\times B\rightarrow M,\ \ \ \ \ \ \ \ \ \ \left(  m,b\right)  \mapsto mb
\]
are $\mathbf{k}$-bilinear. (This follows easily from (\ref{eq.def.bimod.2}).)
\end{remark}

We can now define the notion of a derivation:

\begin{definition}
\label{def.deriv}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be an $\left(
A,A\right)  $-bimodule. A $\mathbf{k}$-linear map $f:A\rightarrow M$ is said
to be a $\mathbf{k}$\textit{-derivation} if it satisfies%
\begin{equation}
\left(  f\left(  ab\right)  =af\left(  b\right)  +f\left(  a\right)
b\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{ and }b\in A\right)  .
\label{eq.def.deriv.axiom}%
\end{equation}


We shall often abbreviate \textquotedblleft$\mathbf{k}$%
-derivation\textquotedblright\ as \textquotedblleft
derivation\textquotedblright, since the $\mathbf{k}$ is fixed.

The equality (\ref{eq.def.deriv.axiom}) is usually called the \textit{Leibniz
law}.
\end{definition}

\begin{remark}
When authors speak of a \textquotedblleft derivation\textquotedblright, they
usually mean a $\mathbf{k}$-derivation, where $\mathbf{k}$ is some base ring
which is \textbf{usually} clear from the context. Sometimes, it is
$\mathbb{Z}$; sometimes, it is whatever base ring their modules are defined
over; sometimes it is $A$. Unfortunately, in some situations (e.g., field
extensions), the base ring is ambiguous; forgetting to specify it is then a
common source of mistakes.
\end{remark}

We notice that the product $ab$ on the left hand side of
(\ref{eq.def.deriv.axiom}) is a product inside the $\mathbf{k}$-algebra $A$,
whereas the \textquotedblleft products\textquotedblright\ $af\left(  b\right)
$ and $f\left(  a\right)  b$ on the right hand side of
(\ref{eq.def.deriv.axiom}) are defined using the left and right $A$-module
structures on $M$.

\begin{verlong}
\begin{corollary}
\label{cor.deriv.verlong}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be an
$\left(  A,A\right)  $-bimodule. Let $f:A\rightarrow M$ be a derivation. Then,
$f$ is a $\mathbf{k}$-linear map and satisfies (\ref{eq.def.deriv.axiom}).
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.deriv.verlong}.]The map $f:A\rightarrow M$ is a
derivation if and only if it is a $\mathbf{k}$-linear map and satisfies
(\ref{eq.def.deriv.axiom}) (by the definition of a \textquotedblleft
derivation\textquotedblright). Since the map $f:A\rightarrow M$ is a
derivation, we thus conclude that $f$ is a $\mathbf{k}$-linear map and
satisfies (\ref{eq.def.deriv.axiom}). This proves Corollary
\ref{cor.deriv.verlong}.
\end{proof}

\begin{corollary}
\label{cor.deriv.verlong.converse}Let $A$ be a $\mathbf{k}$-algebra. Let $M$
be an $\left(  A,A\right)  $-bimodule. Let $f:A\rightarrow M$ be a
$\mathbf{k}$-linear map. Assume that%
\begin{equation}
f\left(  ab\right)  =af\left(  b\right)  +f\left(  a\right)
b\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{ and }b\in A.
\label{eq.cor.deriv.verlong.converse.cond}%
\end{equation}
Then, $f$ is a derivation.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.deriv.verlong}.]The map $f$ satisfies
(\ref{eq.def.deriv.axiom}) (since (\ref{eq.cor.deriv.verlong.converse.cond})
holds). Now, the map $f:A\rightarrow M$ is a derivation if and only if it is a
$\mathbf{k}$-linear map and satisfies (\ref{eq.def.deriv.axiom}) (by the
definition of a \textquotedblleft derivation\textquotedblright). We thus
conclude that $f$ is a derivation (since the map $f:A\rightarrow M$ is a
$\mathbf{k}$-linear map and satisfies (\ref{eq.def.deriv.axiom})). This proves
Corollary \ref{cor.deriv.verlong.converse}.
\end{proof}
\end{verlong}

\begin{example}
\label{exa.diff}Let $\mathbf{k}\left[  x\right]  $ denote the polynomial ring
in an indeterminate $x$ over $\mathbf{k}$. Then, the $\mathbf{k}$-linear map
$\partial_{x}:\mathbf{k}\left[  x\right]  \rightarrow\mathbf{k}\left[
x\right]  $ defined by%
\[
\left(  \partial_{x}\left(  x^{n}\right)  =nx^{n-1}%
\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}\right)
\]
(where $nx^{n-1}$ is to be interpreted as $0$ in the case when $n=0$) is a
derivation. So is the map $p\cdot\partial_{x}$ (which sends every
$g\in\mathbf{k}\left[  x\right]  $ to $p\cdot\partial_{x}\left(  g\right)  $)
for every $p\in\mathbf{k}\left[  x\right]  $. More generally, if $M$ is any
$\left(  \mathbf{k}\left[  x\right]  ,\mathbf{k}\left[  x\right]  \right)
$-bimodule, and if $p$ is a \textit{central} element of $M$ (that is, an
element of $M$ satisfying $fp=pf$ for every $f\in\mathbf{k}\left[  x\right]
$), then the map $p\cdot\partial_{x}:\mathbf{k}\left[  x\right]  \rightarrow
M$ (which sends every $g\in\mathbf{k}\left[  x\right]  $ to $p\cdot
\partial_{x}\left(  g\right)  $) is a derivation.

The map $\operatorname*{id}:\mathbf{k}\left[  x\right]  \rightarrow
\mathbf{k}\left[  x\right]  $ is not a derivation (unless the ring
$\mathbf{k}$ is trivial), and the map $\partial_{x}^{2}:\mathbf{k}\left[
x\right]  \rightarrow\mathbf{k}\left[  x\right]  $ is not a derivation (unless
$2=0$ in $\mathbf{k}$, in which case $\partial_{x}^{2}=0$).
\end{example}

\begin{example}
Let $A$ be a $\mathbb{Z}$-graded $\mathbf{k}$-algebra. Let $\mathcal{E}%
:A\rightarrow A$ be the $\mathbf{k}$-linear map defined by the following
property: If $n\in\mathbb{Z}$, and if $a\in A$ is a homogeneous element of
degree $n$, then $\mathcal{E}\left(  a\right)  =na$. It is easy to see that
$\mathcal{E}$ is a derivation.
\end{example}

\begin{example}
\label{exa.ad}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be an $\left(
A,A\right)  $-bimodule. Let $p\in M$. Define a map $\operatorname*{ad}%
\nolimits_{p}:A\rightarrow M$ by%
\begin{equation}
\left(  \operatorname*{ad}\nolimits_{p}\left(  a\right)
=pa-ap\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\right)  .
\label{eq.exa.ad.def}%
\end{equation}
Then, the map $\operatorname*{ad}\nolimits_{p}$ is a derivation. It is called
an \textit{inner derivation}.
\end{example}

\begin{verlong}
\begin{proof}
[Proof of Example \ref{exa.ad}.]The element $pa-ap$ depends $\mathbf{k}%
$-linearly on $a\in A$. Thus, the map $\operatorname*{ad}\nolimits_{p}$ is
$\mathbf{k}$-linear. Moreover, for every $a\in A$ and $b\in A$, we have%
\begin{align*}
\operatorname*{ad}\nolimits_{p}\left(  ab\right)   &
=pab-abp\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\operatorname*{ad}\nolimits_{p}\right) \\
&  =\underbrace{\left(  pab-apb\right)  }_{=\left(  pa-ap\right)
b}+\underbrace{\left(  apb-abp\right)  }_{=a\left(  pb-bp\right)  }\\
&  =\underbrace{\left(  pa-ap\right)  }_{\substack{=\operatorname*{ad}%
\nolimits_{p}a\\\text{(by (\ref{eq.exa.ad.def}))}}}b+a\underbrace{\left(
pb-bp\right)  }_{\substack{=\operatorname*{ad}\nolimits_{p}b\\\text{(since
}\operatorname*{ad}\nolimits_{p}b=pb-bp\\\text{(by (\ref{eq.exa.ad.def}),
applied}\\\text{to }b\text{ instead of }a\text{))}}}\\
&  =\left(  \operatorname*{ad}\nolimits_{p}a\right)  b+a\left(
\operatorname*{ad}\nolimits_{p}b\right)  =a\left(  \operatorname*{ad}%
\nolimits_{p}b\right)  +\left(  \operatorname*{ad}\nolimits_{p}a\right)  b.
\end{align*}
In other words, the map $\operatorname*{ad}\nolimits_{p}$ satisfies%
\begin{equation}
\left(  \operatorname*{ad}\nolimits_{p}\left(  ab\right)  =a\left(
\operatorname*{ad}\nolimits_{p}b\right)  +\left(  \operatorname*{ad}%
\nolimits_{p}a\right)  b\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{ and
}b\in A\right)  \label{pf.exa.ad.3}%
\end{equation}


But the map $\operatorname*{ad}\nolimits_{p}$ is a derivation if and only if
$\operatorname*{ad}\nolimits_{p}$ is $\mathbf{k}$-linear and satisfies
(\ref{pf.exa.ad.3}) (by the definition of a \textquotedblleft
derivation\textquotedblright). Hence, the map $\operatorname*{ad}%
\nolimits_{p}$ is a derivation (because we know that $\operatorname*{ad}%
\nolimits_{p}$ is $\mathbf{k}$-linear and satisfies (\ref{pf.exa.ad.3})). This
proves Example \ref{exa.ad}.
\end{proof}
\end{verlong}

\begin{remark}
\label{rmk.deriv.AA}Many authors use a notion of \textquotedblleft
derivation\textquotedblright\ that is less general than ours. Namely, given a
$\mathbf{k}$-algebra $A$, they define a \textit{derivation of }$A$ to be a
$\mathbf{k}$-linear map $f:A\rightarrow A$ satisfying
(\ref{eq.def.deriv.axiom}). This concept of a \textquotedblleft
derivation\textquotedblright\ is a particular case of our concept defined
above. Namely, it is precisely what we call a derivation from the $\mathbf{k}%
$-algebra $A$ to the $\left(  A,A\right)  $-bimodule $A$.
\end{remark}

\begin{remark}
\label{rmk.deriv.rhotau}Let $A$ and $B$ be two $\mathbf{k}$-algebras. Let
$\rho:A\rightarrow B$ and $\tau:A\rightarrow B$ be two $\mathbf{k}$-algebra
homomorphisms. Some authors define a $\left(  \rho,\tau\right)  $%
\textit{-derivation} to be a $\mathbf{k}$-linear map $f:A\rightarrow B$
satisfying%
\[
\left(  f\left(  ab\right)  =\rho\left(  a\right)  f\left(  b\right)
+f\left(  a\right)  \tau\left(  b\right)  \ \ \ \ \ \ \ \ \ \ \text{for every
}a\in A\text{ and }b\in A\right)  .
\]
This concept of a \textquotedblleft$\left(  \rho,\tau\right)  $%
-derivation\textquotedblright\ is, too, a particular case of the concept of a
\textquotedblleft derivation\textquotedblright\ defined in Definition
\ref{def.deriv}. Namely, the left $B$-module $B$ becomes a left $A$-module via
the $\mathbf{k}$-algebra homomorphism $\rho:A\rightarrow B$. Furthermore, the
right $B$-module $B$ becomes a right $A$-module via the $\mathbf{k}$-algebra
homomorphism $\tau:A\rightarrow B$. Thus we have defined a left $A$-module
structure on $B$ and a right $A$-module structure on $B$. These two
structures, combined, make $B$ into an $\left(  A,A\right)  $-bimodule. Now, a
$\left(  \rho,\tau\right)  $-derivation from $A$ to $B$ is the same as a
derivation from $A$ to this $\left(  A,A\right)  $-bimodule $B$.
\end{remark}

\subsection{Basic properties}

The following fact is probably the simplest property of derivations:

\begin{theorem}
\label{thm.deriv.0}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be an $\left(
A,A\right)  $-bimodule. Let $f:A\rightarrow M$ be a derivation. Then,
$f\left(  1\right)  =0$. (Keep in mind that $1$ denotes the unity $1_{A}$ of
$A$ here.)
\end{theorem}

\begin{vershort}
\begin{proof}
[Proof of Theorem \ref{thm.deriv.0}.]The map $f:A\rightarrow M$ is a
derivation. In other words, $f$ is a $\mathbf{k}$-linear map and satisfies
(\ref{eq.def.deriv.axiom}) (by the definition of a \textquotedblleft
derivation\textquotedblright). Applying (\ref{eq.def.deriv.axiom}) to $a=1$
and $b=1$, we obtain $f\left(  1\cdot1\right)  =\underbrace{1f\left(
1\right)  }_{=f\left(  1\right)  }+\underbrace{f\left(  1\right)
1}_{=f\left(  1\right)  }=f\left(  1\right)  +f\left(  1\right)  $. Since
$1\cdot1=1$, this rewrites as $f\left(  1\right)  =f\left(  1\right)
+f\left(  1\right)  $. Subtracting $f\left(  1\right)  $ from this equality,
we obtain $0=f\left(  1\right)  $. This proves Theorem \ref{thm.deriv.0}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Theorem \ref{thm.deriv.0}.]Corollary \ref{cor.deriv.verlong} shows
that $f$ is a $\mathbf{k}$-linear map and satisfies (\ref{eq.def.deriv.axiom}%
). Applying (\ref{eq.def.deriv.axiom}) to $a=1$ and $b=1$, we obtain%
\[
f\left(  1\cdot1\right)  =\underbrace{1f\left(  1\right)  }_{=f\left(
1\right)  }+\underbrace{f\left(  1\right)  1}_{=f\left(  1\right)  }=f\left(
1\right)  +f\left(  1\right)  .
\]
Since $1\cdot1=1$, this rewrites as $f\left(  1\right)  =f\left(  1\right)
+f\left(  1\right)  $. Subtracting $f\left(  1\right)  $ from this equality,
we obtain $0=f\left(  1\right)  $. This proves Theorem \ref{thm.deriv.0}.
\end{proof}
\end{verlong}

\begin{remark}
Theorem \ref{thm.deriv.0} is extremely basic; it holds even more generally
when $A$ is a unital magmatic $\mathbf{k}$-algebra rather than a $\mathbf{k}%
$-algebra. (Of course, in this case, the definition of a derivation should be
generalized appropriately.)

Notice, however, that Theorem \ref{thm.deriv.0} does not generalize to
algebras over semirings (because we cannot subtract over semirings). When
defining derivations in this generality, it thus is advisable to include the
equality $f\left(  1\right)  =0$ as an axiom in the definition of a derivation.
\end{remark}

The following fact (sometimes called the \textit{generalized Leibniz law}, or
simply the \textit{Leibniz law} again) generalizes (\ref{eq.def.deriv.axiom}):

\begin{theorem}
\label{thm.deriv.leibniz}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be an
$\left(  A,A\right)  $-bimodule. Let $f:A\rightarrow M$ be a derivation. Let
$n\in\mathbb{N}$, and let $a_{1},a_{2},\ldots,a_{n}\in A$. Then,%
\[
f\left(  a_{1}a_{2}\cdots a_{n}\right)  =\sum_{i=1}^{n}a_{1}a_{2}\cdots
a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{n}.
\]

\end{theorem}

\begin{vershort}
\begin{proof}
[Proof of Theorem \ref{thm.deriv.leibniz}.]We shall prove Theorem
\ref{thm.deriv.leibniz} by induction over $n$:

\textit{Induction base:} Theorem \ref{thm.deriv.leibniz} holds for $n=0$
\ \ \ \ \footnote{\textit{Proof.} Let $a_{1},a_{2},\ldots,a_{0}$ be $0$
elements of $A$. Then, $a_{1}a_{2}\cdots a_{0}=\left(  \text{empty
product}\right)  =1$, and thus $f\left(  \underbrace{a_{1}a_{2}\cdots a_{0}%
}_{=1}\right)  =f\left(  1\right)  =0$ (by Theorem \ref{thm.deriv.0}).
Comparing this with%
\[
\sum_{i=1}^{0}a_{1}a_{2}\cdots a_{i-1}f\left(  a_{i}\right)  a_{i+1}%
a_{i+2}\cdots a_{0}=\left(  \text{empty sum}\right)  =0,
\]
we obtain $f\left(  a_{1}a_{2}\cdots a_{0}\right)  =\sum_{i=1}^{0}a_{1}%
a_{2}\cdots a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{0}$. Thus,
we have shown that Theorem \ref{thm.deriv.leibniz} holds for $n=0$.}. This
completes the induction base.

\textit{Induction step:} Let $N\in\mathbb{N}$. Assume that Theorem
\ref{thm.deriv.leibniz} holds for $n=N$. We must prove that Theorem
\ref{thm.deriv.leibniz} holds for $n=N+1$.

Let $a_{1},a_{2},\ldots,a_{N+1}$ be $N+1$ elements of $A$. We assumed that
Theorem \ref{thm.deriv.leibniz} holds for $n=N$. Hence, we can apply Theorem
\ref{thm.deriv.leibniz} to $N$ and $a_{1},a_{2},\ldots,a_{N}$ instead of $n$
and $a_{1},a_{2},\ldots,a_{n}$. As a result, we obtain%
\begin{equation}
f\left(  a_{1}a_{2}\cdots a_{N}\right)  =\sum_{i=1}^{N}a_{1}a_{2}\cdots
a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{N}.
\label{pf.thm.deriv.leibniz.short.indhyp}%
\end{equation}


But $f$ is a derivation. In other words, $f$ is a $\mathbf{k}$-linear map and
satisfies (\ref{eq.def.deriv.axiom}). Hence, we can apply
(\ref{eq.def.deriv.axiom}) to $a_{1}a_{2}\cdots a_{N}$ and $a_{N+1}$ instead
of $a$ and $b$. As a result, we obtain%
\begin{align*}
&  f\left(  \left(  a_{1}a_{2}\cdots a_{N}\right)  a_{N+1}\right) \\
&  =\underbrace{\left(  a_{1}a_{2}\cdots a_{N}\right)  f\left(  a_{N+1}%
\right)  }_{=a_{1}a_{2}\cdots a_{N}f\left(  a_{N+1}\right)  }%
+\underbrace{f\left(  a_{1}a_{2}\cdots a_{N}\right)  }_{\substack{=\sum
_{i=1}^{N}a_{1}a_{2}\cdots a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots
a_{N}\\\text{(by (\ref{pf.thm.deriv.leibniz.short.indhyp}))}}}a_{N+1}\\
&  =a_{1}a_{2}\cdots a_{N}f\left(  a_{N+1}\right)  +\underbrace{\left(
\sum_{i=1}^{N}a_{1}a_{2}\cdots a_{i-1}f\left(  a_{i}\right)  a_{i+1}%
a_{i+2}\cdots a_{N}\right)  a_{N+1}}_{=\sum_{i=1}^{N}\left(  a_{1}a_{2}\cdots
a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{N}\right)  a_{N+1}}\\
&  =a_{1}a_{2}\cdots a_{N}f\left(  a_{N+1}\right)  +\sum_{i=1}^{N}%
\underbrace{\left(  a_{1}a_{2}\cdots a_{i-1}f\left(  a_{i}\right)
a_{i+1}a_{i+2}\cdots a_{N}\right)  a_{N+1}}_{\substack{=a_{1}a_{2}\cdots
a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{N}a_{N+1}\\=a_{1}%
a_{2}\cdots a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{N+1}}}\\
&  =a_{1}a_{2}\cdots a_{N}f\left(  a_{N+1}\right)  +\sum_{i=1}^{N}a_{1}%
a_{2}\cdots a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{N+1}.
\end{align*}
Compared with%
\begin{align*}
&  \sum_{i=1}^{N+1}a_{1}a_{2}\cdots a_{i-1}f\left(  a_{i}\right)
a_{i+1}a_{i+2}\cdots a_{N+1}\\
&  =\underbrace{a_{1}a_{2}\cdots a_{\left(  N+1\right)  -1}}_{=a_{1}%
a_{2}\cdots a_{N}}f\left(  a_{N+1}\right)  \underbrace{a_{\left(  N+1\right)
+1}a_{\left(  N+1\right)  +2}\cdots a_{N+1}}_{=\left(  \text{empty
product}\right)  =1}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{i=1}^{N}a_{1}a_{2}\cdots a_{i-1}f\left(
a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{N+1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}i=N+1\text{ from the sum}\right) \\
&  =a_{1}a_{2}\cdots a_{N}\underbrace{f\left(  a_{N+1}\right)  1}_{=f\left(
a_{N+1}\right)  }+\sum_{i=1}^{N}a_{1}a_{2}\cdots a_{i-1}f\left(  a_{i}\right)
a_{i+1}a_{i+2}\cdots a_{N+1}\\
&  =a_{1}a_{2}\cdots a_{N}f\left(  a_{N+1}\right)  +\sum_{i=1}^{N}a_{1}%
a_{2}\cdots a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{N+1},
\end{align*}
this shows that%
\begin{equation}
f\left(  a_{1}a_{2}\cdots a_{N+1}\right)  =\sum_{i=1}^{N+1}a_{1}a_{2}\cdots
a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{N+1}.
\label{pf.thm.deriv.leibniz.short.indres}%
\end{equation}


Now, let us forget that we fixed $a_{1},a_{2},\ldots,a_{N+1}$. We thus have
shown that (\ref{pf.thm.deriv.leibniz.short.indres}) holds for any
$a_{1},a_{2},\ldots,a_{N+1}\in A$. In other words, Theorem
\ref{thm.deriv.leibniz} holds for $n=N+1$. This completes the induction step.
Thus, the proof of Theorem \ref{thm.deriv.leibniz} is complete.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Theorem \ref{thm.deriv.leibniz}.]Corollary \ref{cor.deriv.verlong}
shows that $f$ is a $\mathbf{k}$-linear map and satisfies
(\ref{eq.def.deriv.axiom}).

We shall prove Theorem \ref{thm.deriv.leibniz} by induction over $n$:

\textit{Induction base:} If $a_{1},a_{2},\ldots,a_{0}$ are $0$ elements of
$A$, then
\[
f\left(  a_{1}a_{2}\cdots a_{0}\right)  =\sum_{i=1}^{0}a_{1}a_{2}\cdots
a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{0}%
\]
\footnote{\textit{Proof.} Let $a_{1},a_{2},\ldots,a_{0}$ be $0$ elements of
$A$. Then, $a_{1}a_{2}\cdots a_{0}=\left(  \text{empty product}\right)  =1$,
and thus $f\left(  \underbrace{a_{1}a_{2}\cdots a_{0}}_{=1}\right)  =f\left(
1\right)  =0$ (by Theorem \ref{thm.deriv.0}). Comparing this with%
\[
\sum_{i=1}^{0}a_{1}a_{2}\cdots a_{i-1}f\left(  a_{i}\right)  a_{i+1}%
a_{i+2}\cdots a_{0}=\left(  \text{empty sum}\right)  =0,
\]
we obtain $f\left(  a_{1}a_{2}\cdots a_{0}\right)  =\sum_{i=1}^{0}a_{1}%
a_{2}\cdots a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{0}$. Qed.}.
In other words, Theorem \ref{thm.deriv.leibniz} holds for $n=0$. This
completes the induction base.

\textit{Induction step:} Let $N\in\mathbb{N}$. Assume that Theorem
\ref{thm.deriv.leibniz} holds for $n=N$. We must prove that Theorem
\ref{thm.deriv.leibniz} holds for $n=N+1$.

Let $a_{1},a_{2},\ldots,a_{N+1}$ be $N+1$ elements of $A$. We assumed that
Theorem \ref{thm.deriv.leibniz} holds for $n=N$. Hence, we can apply Theorem
\ref{thm.deriv.leibniz} to $N$ and $a_{1},a_{2},\ldots,a_{N}$ instead of $n$
and $a_{1},a_{2},\ldots,a_{n}$. As a result, we obtain%
\begin{equation}
f\left(  a_{1}a_{2}\cdots a_{N}\right)  =\sum_{i=1}^{N}a_{1}a_{2}\cdots
a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{N}.
\label{pf.thm.deriv.leibniz.indhyp}%
\end{equation}


But Corollary \ref{cor.deriv.verlong} shows that $f$ is a $\mathbf{k}$-linear
map and satisfies (\ref{eq.def.deriv.axiom}). Hence, we can apply
(\ref{eq.def.deriv.axiom}) to $a_{1}a_{2}\cdots a_{N}$ and $a_{N+1}$ instead
of $a$ and $b$. As a result, we obtain%
\begin{align*}
&  f\left(  \left(  a_{1}a_{2}\cdots a_{N}\right)  a_{N+1}\right) \\
&  =\underbrace{\left(  a_{1}a_{2}\cdots a_{N}\right)  f\left(  a_{N+1}%
\right)  }_{=a_{1}a_{2}\cdots a_{N}f\left(  a_{N+1}\right)  }%
+\underbrace{f\left(  a_{1}a_{2}\cdots a_{N}\right)  }_{\substack{=\sum
_{i=1}^{N}a_{1}a_{2}\cdots a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots
a_{N}\\\text{(by (\ref{pf.thm.deriv.leibniz.indhyp}))}}}a_{N+1}\\
&  =a_{1}a_{2}\cdots a_{N}f\left(  a_{N+1}\right)  +\underbrace{\left(
\sum_{i=1}^{N}a_{1}a_{2}\cdots a_{i-1}f\left(  a_{i}\right)  a_{i+1}%
a_{i+2}\cdots a_{N}\right)  a_{N+1}}_{=\sum_{i=1}^{N}\left(  a_{1}a_{2}\cdots
a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{N}\right)  a_{N+1}}\\
&  =a_{1}a_{2}\cdots a_{N}f\left(  a_{N+1}\right)  +\sum_{i=1}^{N}%
\underbrace{\left(  a_{1}a_{2}\cdots a_{i-1}f\left(  a_{i}\right)
a_{i+1}a_{i+2}\cdots a_{N}\right)  a_{N+1}}_{\substack{=a_{1}a_{2}\cdots
a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{N}a_{N+1}\\=a_{1}%
a_{2}\cdots a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{N+1}}}\\
&  =a_{1}a_{2}\cdots a_{N}f\left(  a_{N+1}\right)  +\sum_{i=1}^{N}a_{1}%
a_{2}\cdots a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{N+1}.
\end{align*}
Compared with%
\begin{align*}
&  \sum_{i=1}^{N+1}a_{1}a_{2}\cdots a_{i-1}f\left(  a_{i}\right)
a_{i+1}a_{i+2}\cdots a_{N+1}\\
&  =\underbrace{a_{1}a_{2}\cdots a_{\left(  N+1\right)  -1}}_{=a_{1}%
a_{2}\cdots a_{N}}f\left(  a_{N+1}\right)  \underbrace{a_{\left(  N+1\right)
+1}a_{\left(  N+1\right)  +2}\cdots a_{N+1}}_{=\left(  \text{empty
product}\right)  =1}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{i=1}^{N}a_{1}a_{2}\cdots a_{i-1}f\left(
a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{N+1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}i=N+1\text{ from the sum}\right) \\
&  =a_{1}a_{2}\cdots a_{N}\underbrace{f\left(  a_{N+1}\right)  1}_{=f\left(
a_{N+1}\right)  }+\sum_{i=1}^{N}a_{1}a_{2}\cdots a_{i-1}f\left(  a_{i}\right)
a_{i+1}a_{i+2}\cdots a_{N+1}\\
&  =a_{1}a_{2}\cdots a_{N}f\left(  a_{N+1}\right)  +\sum_{i=1}^{N}a_{1}%
a_{2}\cdots a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{N+1},
\end{align*}
this shows that%
\begin{equation}
f\left(  a_{1}a_{2}\cdots a_{N+1}\right)  =\sum_{i=1}^{N+1}a_{1}a_{2}\cdots
a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{N+1}.
\label{pf.thm.deriv.leibniz.indres}%
\end{equation}


Now, let us forget that we fixed $a_{1},a_{2},\ldots,a_{N+1}$. We thus have
shown that (\ref{pf.thm.deriv.leibniz.indres}) holds for any $a_{1}%
,a_{2},\ldots,a_{N+1}\in A$. In other words, Theorem \ref{thm.deriv.leibniz}
holds for $n=N+1$. This completes the induction step. Thus, the induction
proof of Theorem \ref{thm.deriv.leibniz} is complete.
\end{proof}
\end{verlong}

Let us gather some consequences of Theorem \ref{thm.deriv.leibniz}:

\begin{corollary}
\label{cor.deriv.leibniz.power1}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be
an $\left(  A,A\right)  $-bimodule. Let $f:A\rightarrow M$ be a derivation.
Let $n\in\mathbb{N}$, and let $a\in A$. Then,%
\[
f\left(  a^{n}\right)  =\sum_{i=1}^{n}a^{i-1}f\left(  a\right)  a^{n-i}.
\]

\end{corollary}

\begin{vershort}
\begin{proof}
[Proof of Corollary \ref{cor.deriv.leibniz.power1}.]Set $a_{k}=a$ in Theorem
\ref{thm.deriv.leibniz}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Corollary \ref{cor.deriv.leibniz.power1}.]Theorem
\ref{thm.deriv.leibniz} (applied to $a_{k}=a$) shows that%
\[
f\left(  \underbrace{aa\cdots a}_{n\text{ factors}}\right)  =\sum_{i=1}%
^{n}\underbrace{\underbrace{aa\cdots a}_{i-1\text{ factors}}}_{=a^{i-1}%
}f\left(  a_{i}\right)  \underbrace{\underbrace{aa\cdots a}_{n-i\text{
factors}}}_{=a^{n-i}}=\sum_{i=1}^{n}a^{i-1}f\left(  a\right)  a^{n-i}.
\]
Since $\underbrace{aa\cdots a}_{n\text{ factors}}=a^{n}$, this rewrites as
$f\left(  a^{n}\right)  =\sum_{i=1}^{n}a^{i-1}f\left(  a\right)  a^{n-i}$.
This proves Corollary \ref{cor.deriv.leibniz.power1}.
\end{proof}
\end{verlong}

For the next corollary, we need another definition:

\begin{definition}
\label{def.bimod.symm}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be an
$\left(  A,A\right)  $-bimodule. The $\left(  A,A\right)  $-bimodule $M$ is
said to be \textit{symmetric} if it satisfies%
\begin{equation}
\left(  am=ma\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{ and }m\in
M\right)  . \label{eq.def.bimod.symm.axiom}%
\end{equation}

\end{definition}

\begin{remark}
If $A$ is a commutative ring, then the symmetric $\left(  A,A\right)
$-bimodules can be identified with the $A$-modules. Namely, a symmetric
$\left(  A,A\right)  $-bimodule structure is the same as an $A$-module
structure, and conversely, any $A$-module structure can be interpreted both as
a left $A$-module structure and as a right $A$-module structure, and the
latter two structures combined yield a symmetric $\left(  A,A\right)
$-bimodule structure.
\end{remark}

\begin{corollary}
\label{cor.deriv.leibniz.power2}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be
a symmetric $\left(  A,A\right)  $-bimodule. Let $f:A\rightarrow M$ be a
derivation. Let $n\in\mathbb{N}$, and let $a\in A$. Then,%
\[
f\left(  a^{n}\right)  =nf\left(  a\right)  a^{n-1}.
\]
Here, $nf\left(  a\right)  a^{n-1}$ is to be understood as $0$ if $n=0$.
\end{corollary}

\begin{vershort}
\begin{proof}
[Proof of Corollary \ref{cor.deriv.leibniz.power2}.]Corollary
\ref{cor.deriv.leibniz.power1} yields%
\begin{align*}
f\left(  a^{n}\right)   &  =\sum_{i=1}^{n}\underbrace{a^{i-1}f\left(
a\right)  }_{\substack{=f\left(  a\right)  a^{i-1}\\\text{(since }M\text{ is
symmetric)}}}a^{n-i}=\sum_{i=1}^{n}f\left(  a\right)  \underbrace{a^{i-1}%
a^{n-i}}_{=a^{\left(  i-1\right)  +\left(  n-i\right)  }=a^{n-1}}\\
&  =\sum_{i=1}^{n}f\left(  a\right)  a^{n-1}=nf\left(  a\right)  a^{n-1}.
\end{align*}

\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Corollary \ref{cor.deriv.leibniz.power2}.]If $n=0$, then Corollary
\ref{cor.deriv.leibniz.power2} holds\footnote{\textit{Proof.} Assume that
$n=0$. Then, $a^{n}=a^{0}=1$, so that $f\left(  \underbrace{a^{n}}%
_{=1}\right)  =f\left(  1\right)  =0$ (by Theorem \ref{thm.deriv.0}). Compared
with $nf\left(  a\right)  a^{n-1}=0$ (since $n=0$), this yields $f\left(
a^{n}\right)  =nf\left(  a\right)  a^{n-1}$. Thus, Corollary
\ref{cor.deriv.leibniz.power2} holds, qed.}. Hence, for the rest of this
proof, we can WLOG assume that we don't have $n=0$. Assume this. Thus,
$n\geq1$ (since we don't have $n=0$). Hence, $a^{n-1}$ is well-defined.

The $\left(  A,A\right)  $-bimodule $M$ is symmetric if and only if it
satisfies (\ref{eq.def.bimod.symm.axiom}) (by the definition of a
\textquotedblleft symmetric $\left(  A,A\right)  $-bimodule\textquotedblright%
). Hence, $M$ satisfies (\ref{eq.def.bimod.symm.axiom}) (since $M$ is
symmetric). Now, Corollary \ref{cor.deriv.leibniz.power1} yields%
\begin{align*}
f\left(  a^{n}\right)   &  =\sum_{i=1}^{n}\underbrace{a^{i-1}f\left(
a\right)  }_{\substack{=f\left(  a\right)  a^{i-1}\\\text{(by
(\ref{eq.def.bimod.symm.axiom}), applied to }a^{i-1}\text{ and }f\left(
a\right)  \\\text{instead of }a\text{ and }m\text{)}}}a^{n-i}=\sum_{i=1}%
^{n}f\left(  a\right)  \underbrace{a^{i-1}a^{n-i}}_{=a^{\left(  i-1\right)
+\left(  n-i\right)  }=a^{n-1}}\\
&  =\sum_{i=1}^{n}f\left(  a\right)  a^{n-1}=nf\left(  a\right)  a^{n-1}.
\end{align*}
This proves Corollary \ref{cor.deriv.leibniz.power2}.
\end{proof}
\end{verlong}

The following corollary can be regarded as a \textquotedblleft chain
rule\textquotedblright\ for derivations:

\begin{corollary}
\label{cor.deriv.leibniz.chainrule}Let $A$ be a $\mathbf{k}$-algebra. Let $M$
be a symmetric $\left(  A,A\right)  $-bimodule. Let $f:A\rightarrow M$ be a
derivation. Let $p\in\mathbf{k}\left[  x\right]  $ be a polynomial. Let $a\in
A$. Then,%
\[
f\left(  p\left(  a\right)  \right)  =p^{\prime}\left(  a\right)  \cdot
f\left(  a\right)  .
\]
Here, $p^{\prime}$ denotes the (usual) derivative of $p$ in $\mathbf{k}\left[
x\right]  $.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.deriv.leibniz.chainrule}.]Write the polynomial
$p$ in the form $p=\sum_{n=0}^{N}p_{n}x^{n}$ for some $N\in\mathbb{N}$ and
some $\left(  p_{0},p_{1},\ldots,p_{N}\right)  \in\mathbf{k}^{N+1}$. Then, the
definition of $p^{\prime}$ shows that $p^{\prime}=\sum_{n=1}^{N}np_{n}x^{n-1}%
$. Substituting $a$ for $x$ in this equality, we obtain $p^{\prime}\left(
a\right)  =\sum_{n=1}^{N}np_{n}a^{n-1}$.

We have $f\left(  \underbrace{a^{0}}_{=1}\right)  =f\left(  1\right)  =0$ (by
Theorem \ref{thm.deriv.0}), and thus $p_{0}\underbrace{f\left(  a^{0}\right)
}_{=0}=0$.

\begin{verlong}
Corollary \ref{cor.deriv.verlong} shows that $f$ is a $\mathbf{k}$-linear map
and satisfies (\ref{eq.def.deriv.axiom}).
\end{verlong}

\begin{verlong}
The $\left(  A,A\right)  $-bimodule $M$ is symmetric if and only if it
satisfies (\ref{eq.def.bimod.symm.axiom}) (by the definition of a
\textquotedblleft symmetric $\left(  A,A\right)  $-bimodule\textquotedblright%
). Hence, $M$ satisfies (\ref{eq.def.bimod.symm.axiom}) (since $M$ is symmetric).
\end{verlong}

On the other hand, substituting $a$ for $x$ in the equality $p=\sum_{n=0}%
^{N}p_{n}x^{n}$, we obtain $p\left(  a\right)  =\sum_{n=0}^{N}p_{n}a^{n}$.
Applying the map $f$ to both sides of this equality, we obtain%
\begin{align*}
f\left(  p\left(  a\right)  \right)   &  =f\left(  \sum_{n=0}^{N}p_{n}%
a^{n}\right)  =\sum_{n=0}^{N}p_{n}f\left(  a^{n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f\text{ is }\mathbf{k}%
\text{-linear}\right) \\
&  =\underbrace{p_{0}f\left(  a^{0}\right)  }_{=0}+\sum_{n=1}^{N}p_{n}f\left(
a^{n}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}n=0\text{ from the sum}\right) \\
&  =\sum_{n=1}^{N}p_{n}\underbrace{f\left(  a^{n}\right)  }%
_{\substack{=nf\left(  a\right)  a^{n-1}\\\text{(by Corollary
\ref{cor.deriv.leibniz.power2})}}}=\sum_{n=1}^{N}\underbrace{p_{n}n}_{=np_{n}%
}\underbrace{f\left(  a\right)  a^{n-1}}_{\substack{=a^{n-1}f\left(  a\right)
\\\text{(because (\ref{eq.def.bimod.symm.axiom}) (applied to }a^{n-1}\text{
and }f\left(  a\right)  \\\text{instead of }a\text{ and }m\text{)
shows}\\\text{that }a^{n-1}f\left(  a\right)  =f\left(  a\right)
a^{n-1}\text{)}}}\\
&  =\sum_{n=1}^{N}np_{n}a^{n-1}f\left(  a\right)  =\underbrace{\left(
\sum_{n=1}^{N}np_{n}a^{n-1}\right)  }_{=p^{\prime}\left(  a\right)  }\cdot
f\left(  a\right)  =p^{\prime}\left(  a\right)  \cdot f\left(  a\right)  .
\end{align*}
This proves Corollary \ref{cor.deriv.leibniz.chainrule}.
\end{proof}

\begin{definition}
\label{def.prod-of-ideals}Let $A$ be a $\mathbf{k}$-algebra.

\textbf{(a)} If $I_{1},I_{2},\ldots,I_{n}$ are some two-sided ideals of $A$,
then $I_{1}I_{2}\cdots I_{n}$ denotes the two-sided ideal of $A$ generated by%
\[
\left\{  u_{1}u_{2}\cdots u_{n}\ \mid\ \left(  u_{1},u_{2},\ldots
,u_{n}\right)  \in I_{1}\times I_{2}\times\cdots\times I_{n}\right\}  .
\]
(In particular, for $n=0$, this means that $I_{1}I_{2}\cdots I_{n}$ is the
two-sided ideal of $A$ generated by the empty product. Since the empty product
is defined to be $1_{A}$, this ideal is therefore $A$.)

\textbf{(b)} If $I$ is a two-sided ideal of $A$, and if $n\in\mathbb{N}$, then
$I^{n}$ is defined to be $\underbrace{II\cdots I}_{n\text{ times}}$.

[\textbf{Note:} This notation is not really standard, particularly as far as
the meaning of an empty product of ideals is concerned.]

\textbf{(c)} If $S$ is any set, and if $n\in\mathbb{N}$, then we use the
notation $S^{\times n}$ for the Cartesian product $\underbrace{S\times
S\times\cdots\times S}_{n\text{ times}}$. (This is commonly denoted by $S^{n}%
$, but this notation would conflict with the one in Definition
\ref{def.prod-of-ideals} \textbf{(b)}.)
\end{definition}

\begin{verlong}
\begin{proposition}
\label{prop.prod-of-two-ideals}Let $A$ be a $\mathbf{k}$-algebra. Let $U$ and
$V$ be two two-sided ideals of $A$. Let $x\in UV$. Then, there exists an
$m\in\mathbb{N}$ and two $m$-tuples $\left(  u_{1},u_{2},\ldots,u_{m}\right)
\in U^{\times m}$ and $\left(  v_{1},v_{2},\ldots,v_{m}\right)  \in V^{\times
m}$ such that $x=\sum_{i=1}^{m}u_{i}v_{i}$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.prod-of-two-ideals}.]Let $Z$ denote the subset
$\left\{  uv\ \mid\ \left(  u,v\right)  \in U\times V\right\}  $ of $A$.

We know that $UV$ is the two-sided ideal generated by $\left\{  uv\ \mid
\ \left(  u,v\right)  \in U\times V\right\}  $\ \ \ \ \footnote{Indeed, this
is a particular case of Definition \ref{def.prod-of-ideals} \textbf{(a)}.}. In
other words, $UV$ is the two-sided ideal generated by $Z$ (since
\newline$Z=\left\{  uv\ \mid\ \left(  u,v\right)  \in U\times V\right\}  $).
In other words, $UV$ is the set
\[
\left\{  \sum_{i=1}^{N}a_{i}z_{i}b_{i}\ \mid\ \left(
\begin{array}
[c]{c}%
N\in\mathbb{N},\ \left(  a_{1},a_{2},\ldots,a_{N}\right)  \in A^{\times
N},\ \left(  b_{1},b_{2},\ldots,b_{N}\right)  \in A^{\times N},\\
\text{and }\left(  z_{1},z_{2},\ldots,z_{N}\right)  \in Z^{\times N}%
\end{array}
\right)  \right\}
\]
(because the two-sided ideal generated by $Z$ is the set \newline$\left\{
\sum_{i=1}^{N}a_{i}z_{i}b_{i}\ \mid\ \left(
\begin{array}
[c]{c}%
N\in\mathbb{N},\ \left(  a_{1},a_{2},\ldots,a_{N}\right)  \in A^{\times
N},\ \left(  b_{1},b_{2},\ldots,b_{N}\right)  \in A^{\times N},\\
\text{and }\left(  z_{1},z_{2},\ldots,z_{N}\right)  \in Z^{\times N}%
\end{array}
\right)  \right\}  $). Hence,%
\begin{align*}
&  UV\\
&  =\left\{  \sum_{i=1}^{N}a_{i}z_{i}b_{i}\ \mid\ \left(
\begin{array}
[c]{c}%
N\in\mathbb{N},\ \left(  a_{1},a_{2},\ldots,a_{N}\right)  \in A^{\times
N},\ \left(  b_{1},b_{2},\ldots,b_{N}\right)  \in A^{\times N},\\
\text{and }\left(  z_{1},z_{2},\ldots,z_{N}\right)  \in Z^{\times N}%
\end{array}
\right)  \right\}  .
\end{align*}
Thus,%
\begin{align*}
x  &  \in UV\\
&  =\left\{  \sum_{i=1}^{N}a_{i}z_{i}b_{i}\ \mid\ \left(
\begin{array}
[c]{c}%
N\in\mathbb{N},\ \left(  a_{1},a_{2},\ldots,a_{N}\right)  \in A^{\times
N},\ \left(  b_{1},b_{2},\ldots,b_{N}\right)  \in A^{\times N},\\
\text{and }\left(  z_{1},z_{2},\ldots,z_{N}\right)  \in Z^{\times N}%
\end{array}
\right)  \right\}  .
\end{align*}
In other words, $x$ has the form $x=\sum_{i=1}^{N}a_{i}z_{i}b_{i}$ for some
$N\in\mathbb{N}$, some $\left(  a_{1},a_{2},\ldots,a_{N}\right)  \in A^{\times
N}$, some $\left(  b_{1},b_{2},\ldots,b_{N}\right)  \in A^{\times N}$ and some
$\left(  z_{1},z_{2},\ldots,z_{N}\right)  \in Z^{\times N}$. Consider this
$N$, this $\left(  a_{1},a_{2},\ldots,a_{N}\right)  $, this $\left(
b_{1},b_{2},\ldots,b_{N}\right)  $ and this $\left(  z_{1},z_{2},\ldots
,z_{N}\right)  $.

For every $i\in\left\{  1,2,\ldots,N\right\}  $, we define two elements
$u_{i}^{\prime}\in U$ and $v_{i}^{\prime}\in V$ as follows: We have $z_{i}\in
Z=\left\{  uv\ \mid\ \left(  u,v\right)  \in U\times V\right\}  $. Thus, we
can write the element $z_{i}$ of $Z$ in the form $z_{i}=uv$ for some $\left(
u,v\right)  \in U\times V$. Let $\left(  u_{i}^{\prime},v_{i}^{\prime}\right)
$ be this $\left(  u,v\right)  $. Thus, $\left(  u_{i}^{\prime},v_{i}^{\prime
}\right)  \in U\times V$ satisfies $z_{i}=u_{i}^{\prime}v_{i}^{\prime}$. From
$\left(  u_{i}^{\prime},v_{i}^{\prime}\right)  \in U\times V$, we obtain
$u_{i}^{\prime}\in U$ and $v_{i}^{\prime}\in V$.

Thus, we have defined two elements $u_{i}^{\prime}\in U$ and $v_{i}^{\prime
}\in V$ for each $i\in\left\{  1,2,\ldots,N\right\}  $. These two elements
$u_{i}^{\prime}$ and $v_{i}^{\prime}$ satisfy $z_{i}=u_{i}^{\prime}%
v_{i}^{\prime}$ (by their definition).

We have $x=\sum_{i=1}^{N}a_{i}\underbrace{z_{i}}_{=u_{i}^{\prime}v_{i}%
^{\prime}}b_{i}=\sum_{i=1}^{N}\underbrace{a_{i}u_{i}^{\prime}v_{i}^{\prime
}b_{i}}_{=\left(  a_{i}u_{i}^{\prime}\right)  \left(  v_{i}^{\prime}%
b_{i}\right)  }=\sum_{i=1}^{N}\left(  a_{i}u_{i}^{\prime}\right)  \left(
v_{i}^{\prime}b_{i}\right)  $. But for each $i\in\left\{  1,2,\ldots
,N\right\}  $, we have $a_{i}u_{i}^{\prime}\in U$ (since $u_{i}^{\prime}\in
U$, and since $U$ is a two-sided ideal of $A$). Hence, $\left(  a_{1}%
u_{1}^{\prime},a_{2}u_{2}^{\prime},\ldots,a_{N}u_{N}^{\prime}\right)  \in
U^{\times N}$. Also, for each $i\in\left\{  1,2,\ldots,N\right\}  $, we have
$v_{i}^{\prime}b_{i}\in V$ (since $v_{i}^{\prime}\in V$, and since $V$ is a
two-sided ideal). Hence, $\left(  v_{1}^{\prime}b_{1},v_{2}^{\prime}%
b_{2},\ldots,v_{N}^{\prime}b_{N}\right)  \in V^{\times N}$. Therefore, there
exists an $m\in\mathbb{N}$ and two $m$-tuples $\left(  u_{1},u_{2}%
,\ldots,u_{m}\right)  \in U^{\times m}$ and $\left(  v_{1},v_{2},\ldots
,v_{m}\right)  \in V^{\times m}$ such that $x=\sum_{i=1}^{m}u_{i}v_{i}$
(namely, $m=N$, $u_{i}=a_{i}u_{i}^{\prime}$ and $v_{i}=v_{i}^{\prime}b_{i}$).
This proves Proposition \ref{prop.prod-of-two-ideals}.
\end{proof}
\end{verlong}

\begin{proposition}
\label{prop.deriv.leibniz.ideal}Let $A$ be a $\mathbf{k}$-algebra. Let $I$ be
a two-sided ideal of $A$. Let $f:A\rightarrow A$ be a derivation. Then,
$f\left(  I^{n+1}\right)  \subseteq I^{n}$ for every $n\in\mathbb{N}$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.deriv.leibniz.ideal}.]It is not hard to derive
Proposition \ref{prop.deriv.leibniz.ideal} from Theorem
\ref{thm.deriv.leibniz}. We shall proceed differently, however.

We shall prove Proposition \ref{prop.deriv.leibniz.ideal} by induction over
$n$:

\textit{Induction base:} We have $I^{0}=A$. Now, $f\left(  I^{0+1}\right)
\subseteq A=I^{0}$. In other words, Proposition \ref{prop.deriv.leibniz.ideal}
holds for $n=0$. This completes the induction base.

\textit{Induction step:} Let $N\in\mathbb{N}$. Assume that Proposition
\ref{prop.deriv.leibniz.ideal} holds for $n=N$. We must prove that Proposition
\ref{prop.deriv.leibniz.ideal} holds for $n=N+1$.

\begin{vershort}
Proposition \ref{prop.deriv.leibniz.ideal} holds for $n=N$. In other words, we
have $f\left(  I^{N+1}\right)  \subseteq I^{N}$.

Now, let $r\in f\left(  I^{\left(  N+1\right)  +1}\right)  $. Thus, there
exists a $q\in I^{\left(  N+1\right)  +1}$ such that $r=f\left(  q\right)  $.
Consider this $q$. We have $q\in I^{\left(  N+1\right)  +1}=I^{N+1}%
\underbrace{I^{1}}_{=I}=I^{N+1}I$. But it is easy to see the following general
fact: If $U$ and $V$ are two two-sided ideals of $A$, then every element of
$UV$ has the form $\sum_{i=1}^{m}u_{i}v_{i}$ for some $m\in\mathbb{N}$ and
some two $m$-tuples $\left(  u_{1},u_{2},\ldots,u_{m}\right)  \in U^{\times
m}$ and $\left(  v_{1},v_{2},\ldots,v_{m}\right)  \in V^{\times m}%
$\ \ \ \ \footnote{\textit{Proof.} Let $x$ be an element of $UV$. We need to
prove that $x$ has the form $\sum_{i=1}^{m}u_{i}v_{i}$ for some $m\in
\mathbb{N}$ and some two $m$-tuples $\left(  u_{1},u_{2},\ldots,u_{m}\right)
\in U^{\times m}$ and $\left(  v_{1},v_{2},\ldots,v_{m}\right)  \in V^{\times
m}$.
\par
The ideal $UV$ is the two-sided ideal generated by the set $\left\{
uv\ \mid\ \left(  u,v\right)  \in U\times V\right\}  $. Thus, the ideal $UV$
is the set of all sums of the form $\sum_{i=1}^{p}a_{i}u_{i}^{\prime}%
v_{i}^{\prime}b_{i}$ with $p\in\mathbb{N}$, $\left(  a_{1},a_{2},\ldots
,a_{p}\right)  \in A^{\times p}$, $\left(  u_{1}^{\prime},u_{2}^{\prime
},\ldots,u_{p}^{\prime}\right)  \in U^{\times p}$, $\left(  v_{1}^{\prime
},v_{2}^{\prime},\ldots,v_{p}^{\prime}\right)  \in V^{\times p}$ and $\left(
b_{1},b_{2},\ldots,b_{p}\right)  \in A^{\times p}$. In particular, $x$
therefore has this form (since $x\in UV$). So let us write $x$ in the form
$x=\sum_{i=1}^{p}a_{i}u_{i}^{\prime}v_{i}^{\prime}b_{i}$ with $p\in\mathbb{N}%
$, $\left(  a_{1},a_{2},\ldots,a_{p}\right)  \in A^{\times p}$, $\left(
u_{1}^{\prime},u_{2}^{\prime},\ldots,u_{p}^{\prime}\right)  \in U^{\times p}$,
$\left(  v_{1}^{\prime},v_{2}^{\prime},\ldots,v_{p}^{\prime}\right)  \in
V^{\times p}$ and $\left(  b_{1},b_{2},\ldots,b_{p}\right)  \in A^{\times p}$.
For every $i\in\left\{  1,2,\ldots,p\right\}  $, we have $a_{i}u_{i}^{\prime
}\in U$ (since $u_{i}^{\prime}\in U$ and since $U$ is an ideal) and
$v_{i}^{\prime}b_{i}\in V$ (since $v_{i}^{\prime}\in V$ and since $V$ is an
ideal). Hence, $x$ has the form $\sum_{i=1}^{m}u_{i}v_{i}$ for some
$m\in\mathbb{N}$ and some two $m$-tuples $\left(  u_{1},u_{2},\ldots
,u_{m}\right)  \in U^{\times m}$ and $\left(  v_{1},v_{2},\ldots,v_{m}\right)
\in V^{\times m}$ (namely, for $m=p$, $u_{i}=a_{i}u_{i}^{\prime}$ and
$v_{i}=v_{i}^{\prime}b_{i}$). This is what we wanted to prove.}. Applying this
to $U=I^{N+1}$ and $V=I$ and to the element $q$ of $I^{N+1}I$, we conclude
that the element $q$ has the form $\sum_{i=1}^{m}u_{i}v_{i}$ for some
$m\in\mathbb{N}$ and some two $m$-tuples $\left(  u_{1},u_{2},\ldots
,u_{m}\right)  \in\left(  I^{N+1}\right)  ^{\times m}$ and $\left(
v_{1},v_{2},\ldots,v_{m}\right)  \in I^{\times m}$. Consider this $m$ and
these $\left(  u_{1},u_{2},\ldots,u_{m}\right)  $ and $\left(  v_{1}%
,v_{2},\ldots,v_{m}\right)  $. Clearly, $u_{i}\in I^{N+1}$ and $v_{i}\in I$
for every $i\in\left\{  1,2,\ldots,m\right\}  $. Hence, for every
$i\in\left\{  1,2,\ldots,m\right\}  $, we have $f\left(  \underbrace{u_{i}%
}_{\in I^{N+1}}\right)  \in f\left(  I^{N+1}\right)  \subseteq I^{N}$ and
thus
\begin{equation}
\underbrace{f\left(  u_{i}\right)  }_{\in I^{N}}\underbrace{v_{i}}_{\in I}\in
I^{N}I=I^{N+1}. \label{pf.prop.deriv.leibniz.ideal.short.fuivi}%
\end{equation}

\end{vershort}

\begin{verlong}
Proposition \ref{prop.deriv.leibniz.ideal} holds for $n=N$. In other words, we
have $f\left(  I^{N+1}\right)  \subseteq I^{N}$. Now, recall that $I^{a}%
I^{b}=I^{a+b}$ for any $a\in\mathbb{N}$ and $b\in\mathbb{N}$. Applying this to
$a=N+1$ and $b=1$, we obtain $I^{N+1}I^{1}=I^{\left(  N+1\right)  +1}$. Also,
applying $I^{a}I^{b}=I^{a+b}$ to $a=N$ and $b=1$, we obtain $I^{N}%
I^{1}=I^{N+1}$.

Let $r\in f\left(  I^{\left(  N+1\right)  +1}\right)  $. Then, there exists a
$q\in I^{\left(  N+1\right)  +1}$ such that $r=f\left(  q\right)  $ (since
$r\in f\left(  I^{\left(  N+1\right)  +1}\right)  $). Consider this $q$. We
have $q\in I^{\left(  N+1\right)  +1}=I^{N+1}\underbrace{I^{1}}_{=I}=I^{N+1}%
I$. Proposition \ref{prop.prod-of-two-ideals} (applied to $U=I^{N+1}$, $V=I$
and $x=q$) thus shows that there exists an $m\in\mathbb{N}$ and two $m$-tuples
$\left(  u_{1},u_{2},\ldots,u_{m}\right)  \in\left(  I^{N+1}\right)  ^{\times
m}$ and $\left(  v_{1},v_{2},\ldots,v_{m}\right)  \in I^{\times m}$ such that
$q=\sum_{i=1}^{m}u_{i}v_{i}$. Consider this $m$ and these $\left(  u_{1}%
,u_{2},\ldots,u_{m}\right)  $ and $\left(  v_{1},v_{2},\ldots,v_{m}\right)  $.
\end{verlong}

\begin{vershort}
The map $f$ is a derivation. Thus, it is $\mathbf{k}$-linear and satisfies
(\ref{eq.def.deriv.axiom}). Applying the map $f$ to both sides of the equality
$q=\sum_{i=1}^{m}u_{i}v_{i}$, we obtain%
\begin{align*}
f\left(  q\right)   &  =f\left(  \sum_{i=1}^{m}u_{i}v_{i}\right)  =\sum
_{i=1}^{m}\underbrace{f\left(  u_{i}v_{i}\right)  }_{\substack{=u_{i}f\left(
v_{i}\right)  +f\left(  u_{i}\right)  v_{i}\\\text{(by
(\ref{eq.def.deriv.axiom}), applied to }a=u_{i}\text{ and }b=v_{i}\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f\text{ is }\mathbf{k}%
\text{-linear}\right) \\
&  =\sum_{i=1}^{m}\left(  \underbrace{u_{i}f\left(  v_{i}\right)
}_{\substack{\in I^{N+1}\\\text{(since }u_{i}\in I^{N+1}\text{, and}%
\\\text{since }I^{N+1}\text{ is an ideal)}}}+\underbrace{f\left(
u_{i}\right)  v_{i}}_{\substack{\in I^{N+1}\\\text{(by
(\ref{pf.prop.deriv.leibniz.ideal.short.fuivi}))}}}\right)  \in\sum_{i=1}%
^{m}\left(  I^{N+1}+I^{N+1}\right) \\
&  \subseteq I^{N+1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }I^{N+1}\text{ is
an ideal of }A\right)  .
\end{align*}
Thus, $r=f\left(  q\right)  \in I^{N+1}$.
\end{vershort}

\begin{verlong}
We have%
\begin{equation}
u_{i}f\left(  v_{i}\right)  \in I^{N+1}\ \ \ \ \ \ \ \ \ \ \text{for every
}i\in\left\{  1,2,\ldots,m\right\}  \label{pf.prop.deriv.leibniz.ideal.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.deriv.leibniz.ideal.1}):} Let
$i\in\left\{  1,2,\ldots,m\right\}  $. Then, $u_{i}\in I^{N+1}$ (since
$\left(  u_{1},u_{2},\ldots,u_{m}\right)  \in\left(  I^{N+1}\right)  ^{\times
m}$). Thus, $u_{i}f\left(  v_{i}\right)  \in I^{N+1}$ (since $I^{N+1}$ is a
two-sided ideal). This proves (\ref{pf.prop.deriv.leibniz.ideal.1}).}. Also,%
\begin{equation}
f\left(  u_{i}\right)  v_{i}\in I^{N+1}\ \ \ \ \ \ \ \ \ \ \text{for every
}i\in\left\{  1,2,\ldots,m\right\}  \label{pf.prop.deriv.leibniz.ideal.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.deriv.leibniz.ideal.2}):} Let
$i\in\left\{  1,2,\ldots,m\right\}  $. Then, $u_{i}\in I^{N+1}$ (since
$\left(  u_{1},u_{2},\ldots,u_{m}\right)  \in\left(  I^{N+1}\right)  ^{\times
m}$). Hence, $f\left(  \underbrace{u_{i}}_{\in I^{N+1}}\right)  \in f\left(
I^{N+1}\right)  \subseteq I^{N}$. But $v_{i}\in I$ (since $\left(  v_{1}%
,v_{2},\ldots,v_{m}\right)  \in I^{\times m}$). Hence, $\underbrace{f\left(
u_{i}\right)  }_{\in I^{N}}\underbrace{v_{i}}_{\in I=I^{1}}\in I^{N}%
I^{1}=I^{N+1}$. This proves (\ref{pf.prop.deriv.leibniz.ideal.2}).}.

Corollary \ref{cor.deriv.verlong} shows that $f$ is a $\mathbf{k}$-linear map
and satisfies (\ref{eq.def.deriv.axiom}). Applying the map $f$ to both sides
of the equality $q=\sum_{i=1}^{m}u_{i}v_{i}$, we obtain%
\begin{align*}
f\left(  q\right)   &  =f\left(  \sum_{i=1}^{m}u_{i}v_{i}\right)  =\sum
_{i=1}^{m}\underbrace{f\left(  u_{i}v_{i}\right)  }_{\substack{=u_{i}f\left(
v_{i}\right)  +f\left(  u_{i}\right)  v_{i}\\\text{(by
(\ref{eq.def.deriv.axiom}), applied to }a=u_{i}\text{ and }b=v_{i}\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f\text{ is }\mathbf{k}%
\text{-linear}\right) \\
&  =\sum_{i=1}^{m}\left(  \underbrace{u_{i}f\left(  v_{i}\right)
}_{\substack{\in I^{N+1}\\\text{(by (\ref{pf.prop.deriv.leibniz.ideal.1}))}%
}}+\underbrace{f\left(  u_{i}\right)  v_{i}}_{\substack{\in I^{N+1}\\\text{(by
(\ref{pf.prop.deriv.leibniz.ideal.2}))}}}\right)  \in\sum_{i=1}^{m}%
\underbrace{\left(  I^{N+1}+I^{N+1}\right)  }_{\substack{\subseteq
I^{N+1}\\\text{(since }I^{N+1}\text{ is an ideal of }A\text{)}}}\\
&  \subseteq\sum_{i=1}^{m}I^{N+1}\subseteq I^{N+1}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }I^{N+1}\text{ is an ideal of }A\right)  .
\end{align*}
Thus, $r=f\left(  q\right)  \in I^{N+1}$.
\end{verlong}

Now let us forget that we fixed $r$. We thus have proven that $r\in I^{N+1}$
for every $r\in f\left(  I^{\left(  N+1\right)  +1}\right)  $. In other words,
$f\left(  I^{\left(  N+1\right)  +1}\right)  \subseteq I^{N+1}$. In other
words, Proposition \ref{prop.deriv.leibniz.ideal} holds for $n=N+1$. This
completes the induction step. Proposition \ref{prop.deriv.leibniz.ideal} is
thus proven by induction.
\end{proof}

\subsection{The module of derivations}

We have so far studied properties of a single derivation. Let us next consider
the set of all derivations.

\begin{definition}
\label{def.Der-mod}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be an $\left(
A,A\right)  $-bimodule. We let $\operatorname*{Der}\left(  A,M\right)  $
denote the set of all derivations from $A$ to $M$.
\end{definition}

\begin{proposition}
\label{prop.Der-mod}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be an $\left(
A,A\right)  $-bimodule. Then, $\operatorname*{Der}\left(  A,M\right)  $ is a
$\mathbf{k}$-submodule of $\operatorname*{Hom}\left(  A,M\right)  $.
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.Der-mod}.]All derivations are $\mathbf{k}%
$-linear; hence, $\operatorname*{Der}\left(  A,M\right)  \subseteq
\operatorname*{Hom}\left(  A,M\right)  $. Also, the equation
(\ref{eq.def.deriv.axiom}) is $\mathbf{k}$-linear in $d$. Thus, any
$\mathbf{k}$-linear combination of derivations from $A$ to $M$ is a derivation
from $A$ to $M$. In other words, $\operatorname*{Der}\left(  A,M\right)  $ is
a $\mathbf{k}$-submodule of $\operatorname*{Hom}\left(  A,M\right)  $.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.Der-mod}.]We have $\operatorname*{Der}\left(
A,M\right)  \subseteq\operatorname*{Hom}\left(  A,M\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $f\in\operatorname*{Der}\left(
A,M\right)  $. Thus, $f$ is an element of $\operatorname*{Der}\left(
A,M\right)  $. In other words, $f$ is an element of the set of all derivations
from $A$ to $M$ (since $\operatorname*{Der}\left(  A,M\right)  $ is the set of
all derivations from $A$ to $M$). In other words, $f$ is a derivation from $A$
to $M$. Thus, Corollary \ref{cor.deriv.verlong} shows that $f$ is a
$\mathbf{k}$-linear map and satisfies (\ref{eq.def.deriv.axiom}). In
particular, $f$ is a $\mathbf{k}$-linear map from $A$ to $M$. In other words,
$f\in\operatorname*{Hom}\left(  A,M\right)  $.
\par
Now, let us forget that we fixed $f$. We thus have proven that $f\in
\operatorname*{Hom}\left(  A,M\right)  $ for every $f\in\operatorname*{Der}%
\left(  A,M\right)  $. In other words, $\operatorname*{Der}\left(  A,M\right)
\subseteq\operatorname*{Hom}\left(  A,M\right)  $, qed.}. Furthermore, the
following three statements hold:

\begin{enumerate}
\item We have $f+g\in\operatorname*{Der}\left(  A,M\right)  $ for every
$f\in\operatorname*{Der}\left(  A,M\right)  $ and $g\in\operatorname*{Der}%
\left(  A,M\right)  $\ \ \ \ \footnote{\textit{Proof.} Let $f\in
\operatorname*{Der}\left(  A,M\right)  $ and $g\in\operatorname*{Der}\left(
A,M\right)  $.
\par
We have $f\in\operatorname*{Der}\left(  A,M\right)  $. Thus, $f$ is an element
of $\operatorname*{Der}\left(  A,M\right)  $. In other words, $f$ is an
element of the set of all derivations from $A$ to $M$ (since
$\operatorname*{Der}\left(  A,M\right)  $ is the set of all derivations from
$A$ to $M$). In other words, $f$ is a derivation from $A$ to $M$. Thus,
Corollary \ref{cor.deriv.verlong} shows that $f$ is a $\mathbf{k}$-linear map
and satisfies (\ref{eq.def.deriv.axiom}). The same argument (but made for $g$
instead of $f$) shows that $g$ is a $\mathbf{k}$-linear map and satisfies
(\ref{eq.def.deriv.axiom}) (with $f$ replaced by $g$).
\par
We know that $g$ satisfies (\ref{eq.def.deriv.axiom}) (with $g$ replaced by
$f$). In other words, we have%
\begin{equation}
\left(  g\left(  ab\right)  =ag\left(  b\right)  +g\left(  a\right)
b\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{ and }b\in A\right)  .
\label{pf.prop.Der-mod.fn2.1}%
\end{equation}
\par
Now, for every $a\in A$ and $b\in A$, we have%
\begin{align*}
\left(  f+g\right)  \left(  ab\right)   &  =\underbrace{f\left(  ab\right)
}_{\substack{=af\left(  b\right)  +f\left(  a\right)  b\\\text{(by
(\ref{eq.def.deriv.axiom}))}}}+\underbrace{g\left(  ab\right)  }%
_{\substack{=ag\left(  b\right)  +g\left(  a\right)  b\\\text{(by
(\ref{pf.prop.Der-mod.fn2.1}))}}}\\
&  =\left(  af\left(  b\right)  +f\left(  a\right)  b\right)  +\left(
ag\left(  b\right)  +g\left(  a\right)  b\right) \\
&  =\underbrace{\left(  af\left(  b\right)  +ag\left(  b\right)  \right)
}_{=a\left(  f\left(  b\right)  +g\left(  b\right)  \right)  }%
+\underbrace{\left(  f\left(  a\right)  b+g\left(  a\right)  b\right)
}_{=\left(  f\left(  a\right)  +g\left(  a\right)  \right)  b}\\
&  =a\underbrace{\left(  f\left(  b\right)  +g\left(  b\right)  \right)
}_{=\left(  f+g\right)  \left(  b\right)  }+\underbrace{\left(  f\left(
a\right)  +g\left(  a\right)  \right)  }_{=\left(  f+g\right)  \left(
a\right)  }b=a\left(  f+g\right)  \left(  b\right)  +\left(  f+g\right)
\left(  a\right)  b.
\end{align*}
In other words, we have%
\[
\left(  \left(  f+g\right)  \left(  ab\right)  =a\left(  f+g\right)  \left(
b\right)  +\left(  f+g\right)  \left(  a\right)
b\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{ and }b\in A\right)  .
\]
In other words, the map $f+g$ satisfies (\ref{eq.def.deriv.axiom}) (with $f$
replaced by $f+g$).
\par
Also, the map $f+g$ is $\mathbf{k}$-linear (since the maps $f$ and $g$ are
$\mathbf{k}$-linear).
\par
But $f+g$ is a derivation from $A$ to $M$ if and only if it is a $\mathbf{k}%
$-linear map and satisfies (\ref{eq.def.deriv.axiom}) (with $f$ replaced by
$f+g$) (by the definition of a \textquotedblleft derivation\textquotedblright%
). This shows that $f+g$ is a derivation from $A$ to $M$ (since the map $f+g$
is a $\mathbf{k}$-linear map and satisfies (\ref{eq.def.deriv.axiom}) (with
$f$ replaced by $f+g$)). In other words, $f+g$ is an element of the set of all
derivations from $A$ to $M$. In other words, $f+g$ is an element of
$\operatorname*{Der}\left(  A,M\right)  $ (since $\operatorname*{Der}\left(
A,M\right)  $ is the set of all derivations from $A$ to $M$). In other words,
$f+g\in\operatorname*{Der}\left(  A,M\right)  $, qed.}.

\item We have $\lambda f\in\operatorname*{Der}\left(  A,M\right)  $ for every
$f\in\operatorname*{Der}\left(  A,M\right)  $ and $\lambda\in\mathbf{k}%
$\ \ \ \ \footnote{\textit{Proof.} Let $f\in\operatorname*{Der}\left(
A,M\right)  $ and $\lambda\in\mathbf{k}$.
\par
We have $f\in\operatorname*{Der}\left(  A,M\right)  $. Thus, $f$ is an element
of $\operatorname*{Der}\left(  A,M\right)  $. In other words, $f$ is an
element of the set of all derivations from $A$ to $M$ (since
$\operatorname*{Der}\left(  A,M\right)  $ is the set of all derivations from
$A$ to $M$). In other words, $f$ is a derivation from $A$ to $M$. Thus,
Corollary \ref{cor.deriv.verlong} shows that $f$ is a $\mathbf{k}$-linear map
and satisfies (\ref{eq.def.deriv.axiom}).
\par
Now, for every $a\in A$ and $b\in A$, we have%
\begin{align*}
\left(  \lambda f\right)  \left(  ab\right)   &  =\lambda\underbrace{f\left(
ab\right)  }_{\substack{=af\left(  b\right)  +f\left(  a\right)  b\\\text{(by
(\ref{eq.def.deriv.axiom}))}}}\\
&  =\lambda\left(  af\left(  b\right)  +f\left(  a\right)  b\right)
=\underbrace{\lambda a}_{=a\lambda}f\left(  b\right)  +\lambda f\left(
a\right)  b=a\underbrace{\lambda f\left(  b\right)  }_{=\left(  \lambda
f\right)  \left(  b\right)  }+\underbrace{\lambda f\left(  a\right)
}_{=\left(  \lambda f\right)  \left(  a\right)  }b\\
&  =a\left(  \lambda f\right)  \left(  b\right)  +\left(  \lambda f\right)
\left(  a\right)  b.
\end{align*}
In other words, we have%
\[
\left(  \left(  \lambda f\right)  \left(  ab\right)  =a\left(  \lambda
f\right)  \left(  b\right)  +\left(  \lambda f\right)  \left(  a\right)
b\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{ and }b\in A\right)  .
\]
In other words, the map $\lambda f$ satisfies (\ref{eq.def.deriv.axiom}) (with
$f$ replaced by $\lambda f$).
\par
Also, the map $\lambda f$ is $\mathbf{k}$-linear (since the map $f$ is
$\mathbf{k}$-linear).
\par
But $\lambda f$ is a derivation from $A$ to $M$ if and only if it is a
$\mathbf{k}$-linear map and satisfies (\ref{eq.def.deriv.axiom}) (with $f$
replaced by $\lambda f$) (by the definition of a \textquotedblleft
derivation\textquotedblright). This shows that $\lambda f$ is a derivation
from $A$ to $M$ (since the map $\lambda f$ is a $\mathbf{k}$-linear map and
satisfies (\ref{eq.def.deriv.axiom}) (with $f$ replaced by $\lambda f$)). In
other words, $\lambda f$ is an element of the set of all derivations from $A$
to $M$. In other words, $\lambda f$ is an element of $\operatorname*{Der}%
\left(  A,M\right)  $ (since $\operatorname*{Der}\left(  A,M\right)  $ is the
set of all derivations from $A$ to $M$). In other words, $\lambda
f\in\operatorname*{Der}\left(  A,M\right)  $, qed.}.

\item We have $0\in\operatorname*{Der}\left(  A,M\right)  $, where $0$ denotes
the zero of $\operatorname*{Hom}\left(  A,M\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Consider the zero $0$ of
$\operatorname*{Hom}\left(  A,M\right)  $.
\par
For every $a\in A$ and $b\in A$, we have
\[
a\underbrace{0\left(  b\right)  }_{=0}+\underbrace{0\left(  a\right)  }%
_{=0}b=0+0=0=0\left(  ab\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}0\left(  ab\right)  =0\right)  .
\]
In other words, for every $a\in A$ and $b\in A$, we have $0\left(  ab\right)
=a0\left(  b\right)  +0\left(  a\right)  b$. In other words, we have%
\[
\left(  0\left(  ab\right)  =a0\left(  b\right)  +0\left(  a\right)
b\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{ and }b\in A\right)  .
\]
In other words, the map $0$ satisfies (\ref{eq.def.deriv.axiom}) (with $f$
replaced by $0$).
\par
Also, the map $0$ is $\mathbf{k}$-linear (since the map $f$ is $\mathbf{k}%
$-linear).
\par
But $0$ is a derivation from $A$ to $M$ if and only if it is a $\mathbf{k}%
$-linear map and satisfies (\ref{eq.def.deriv.axiom}) (with $f$ replaced by
$0$) (by the definition of a \textquotedblleft derivation\textquotedblright).
This shows that $0$ is a derivation from $A$ to $M$ (since the map $0$ is a
$\mathbf{k}$-linear map and satisfies (\ref{eq.def.deriv.axiom}) (with $f$
replaced by $0$)). In other words, $0$ is an element of the set of all
derivations from $A$ to $M$. In other words, $0$ is an element of
$\operatorname*{Der}\left(  A,M\right)  $ (since $\operatorname*{Der}\left(
A,M\right)  $ is the set of all derivations from $A$ to $M$). In other words,
$0\in\operatorname*{Der}\left(  A,M\right)  $, qed.}.
\end{enumerate}

Combining these three statements, we conclude that $\operatorname*{Der}\left(
A,M\right)  $ is a $\mathbf{k}$-submodule of $\operatorname*{Hom}\left(
A,M\right)  $. This proves Proposition \ref{prop.Der-mod}.
\end{proof}
\end{verlong}

A more interesting property of $\operatorname*{Der}\left(  A,M\right)  $ holds
in the case when $M=A$:

\begin{theorem}
\label{thm.Der-mod.lie}Let $A$ be a $\mathbf{k}$-algebra. Let $f\in
\operatorname*{Der}\left(  A,A\right)  $ and $g\in\operatorname*{Der}\left(
A,A\right)  $. (See Proposition \ref{prop.bimod.A} for the definition of the
$\left(  A,A\right)  $-bimodule $A$ that is used here.) Then, $f\circ g-g\circ
f\in\operatorname*{Der}\left(  A,A\right)  $.
\end{theorem}

\begin{example}
Let $A$ be a $\mathbf{k}$-algebra. Let $x\in A$ and $y\in A$. As we know from
Example \ref{exa.ad}, for every $p\in A$, we can define a map
$\operatorname*{ad}\nolimits_{p}:A\rightarrow A$ by
\[
\left(  \operatorname*{ad}\nolimits_{p}\left(  a\right)
=pa-ap\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\right)  .
\]
Thus, we have two maps $\operatorname*{ad}\nolimits_{x}:A\rightarrow A$ and
$\operatorname*{ad}\nolimits_{y}:A\rightarrow A$. Example \ref{exa.ad} shows
that these two maps are derivations, i.e., belong to $\operatorname*{Der}%
\left(  A,A\right)  $. Hence, Theorem \ref{thm.Der-mod.lie} (applied to
$f=\operatorname*{ad}\nolimits_{x}$ and $g=\operatorname*{ad}\nolimits_{y}$)
yields that $\operatorname*{ad}\nolimits_{x}\circ\operatorname*{ad}%
\nolimits_{y}-\operatorname*{ad}\nolimits_{y}\circ\operatorname*{ad}%
\nolimits_{x}\in\operatorname*{Der}\left(  A,A\right)  $. This can also be
proven in a simpler way: A short calculation confirms that $\operatorname*{ad}%
\nolimits_{x}\circ\operatorname*{ad}\nolimits_{y}-\operatorname*{ad}%
\nolimits_{y}\circ\operatorname*{ad}\nolimits_{x}=\operatorname*{ad}%
\nolimits_{xy-yx}\in\operatorname*{Der}\left(  A,A\right)  $ (again by Example
\ref{exa.ad}).
\end{example}

\begin{remark}
Theorem \ref{thm.Der-mod.lie} is usually worded in the language of Lie
algebras. Namely, let $A$ be a $\mathbf{k}$-algebra. We shall use the standard
notation $\operatorname*{End}A$ for the $\mathbf{k}$-algebra
$\operatorname*{Hom}\left(  A,A\right)  $ (where the multiplication is given
by the composition of maps, and where the unity is the identity map
$\operatorname*{id}\nolimits_{A}$). Then, we can define a Lie algebra
structure on the $\mathbf{k}$-module $\operatorname*{End}A$ by setting%
\[
\left(  \left[  f,g\right]  =f\circ g-g\circ f\ \ \ \ \ \ \ \ \ \ \text{for
every }f\in\operatorname*{End}A\text{ and }g\in\operatorname*{End}A\right)  .
\]
This Lie algebra is denoted by $\left(  \operatorname*{End}A\right)  ^{-}$ or
by $\mathfrak{gl}\left(  A\right)  $. Now, Theorem \ref{thm.Der-mod.lie}
states that $\operatorname*{Der}\left(  A,A\right)  $ is a Lie subalgebra of
this Lie algebra $\left(  \operatorname*{End}A\right)  ^{-}$. When $f$ and $g$
are two elements of $\operatorname*{End}A$, the element $\left[  f,g\right]  $
is called the \textit{commutator} of $f$ and $g$.
\end{remark}

Rather than proving Theorem \ref{thm.Der-mod.lie}, we shall prove a more
general result:

\begin{theorem}
\label{thm.Der-mod.lie.gen1}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be an
$\left(  A,A\right)  $-bimodule. Let $f:A\rightarrow M$ and $h:A\rightarrow A$
be two derivations. Let $g:M\rightarrow M$ be a $\mathbf{k}$-linear map.
Assume that%
\begin{equation}
\left(  g\left(  am\right)  =ag\left(  m\right)  +h\left(  a\right)
m\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{ and }m\in M\right)
\label{eq.thm.Der-mod.lie.gen1.cond}%
\end{equation}
and%
\begin{equation}
\left(  g\left(  mb\right)  =mh\left(  b\right)  +g\left(  m\right)
b\ \ \ \ \ \ \ \ \ \ \text{for every }b\in A\text{ and }m\in M\right)  .
\label{eq.thm.Der-mod.lie.gen1.cond2}%
\end{equation}


Then, $f\circ h-g\circ f\in\operatorname*{Der}\left(  A,M\right)  $.
\end{theorem}

\begin{vershort}
\begin{proof}
[Proof of Theorem \ref{thm.Der-mod.lie.gen1}.]Let $a\in A$ and $b\in A$. Then,%
\begin{align}
\left(  f\circ h\right)  \left(  ab\right)   &  =f\left(  \underbrace{h\left(
ab\right)  }_{\substack{=ah\left(  b\right)  +h\left(  a\right)
b\\\text{(since }h\text{ is a derivation)}}}\right)  =f\left(  ah\left(
b\right)  +h\left(  a\right)  b\right) \nonumber\\
&  =\underbrace{f\left(  ah\left(  b\right)  \right)  }_{\substack{=af\left(
h\left(  b\right)  \right)  +f\left(  a\right)  h\left(  b\right)
\\\text{(since }f\text{ is a derivation)}}}+\underbrace{f\left(  h\left(
a\right)  b\right)  }_{\substack{=h\left(  a\right)  f\left(  b\right)
+f\left(  h\left(  a\right)  \right)  b\\\text{(since }f\text{ is a
derivation)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f\text{ is
}\mathbf{k}\text{-linear}\right) \nonumber\\
&  =\left(  af\left(  h\left(  b\right)  \right)  +f\left(  a\right)  h\left(
b\right)  \right)  +\left(  h\left(  a\right)  f\left(  b\right)  +f\left(
h\left(  a\right)  \right)  b\right)  .
\label{pf.thm.Der-mod.lie.gen1.short.4}%
\end{align}
On the other hand,%
\begin{align*}
\left(  g\circ f\right)  \left(  ab\right)   &  =g\left(  \underbrace{f\left(
ab\right)  }_{\substack{=af\left(  b\right)  +f\left(  a\right)
b\\\text{(since }f\text{ is a derivation)}}}\right)  =g\left(  af\left(
b\right)  +f\left(  a\right)  b\right) \\
&  =\underbrace{g\left(  af\left(  b\right)  \right)  }_{\substack{=ag\left(
f\left(  b\right)  \right)  +h\left(  a\right)  f\left(  b\right)  \\\text{(by
(\ref{eq.thm.Der-mod.lie.gen1.cond}), applied to}\\f\left(  b\right)  \text{
instead of }m\text{)}}}+\underbrace{g\left(  f\left(  a\right)  b\right)
}_{\substack{=f\left(  a\right)  h\left(  b\right)  +g\left(  f\left(
a\right)  \right)  b\\\text{(by (\ref{eq.thm.Der-mod.lie.gen1.cond2}), applied
to}\\f\left(  a\right)  \text{ instead of }m\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }g\text{ is }\mathbf{k}%
\text{-linear}\right) \\
&  =\left(  ag\left(  f\left(  b\right)  \right)  +h\left(  a\right)  f\left(
b\right)  \right)  +\left(  f\left(  a\right)  h\left(  b\right)  +g\left(
f\left(  a\right)  \right)  b\right)  .
\end{align*}
Subtracting this equality from (\ref{pf.thm.Der-mod.lie.gen1.short.4}), we
obtain%
\begin{align*}
&  \left(  f\circ h\right)  \left(  ab\right)  -\left(  g\circ f\right)
\left(  ab\right) \\
&  =\left(  \left(  af\left(  h\left(  b\right)  \right)  +f\left(  a\right)
h\left(  b\right)  \right)  +\left(  h\left(  a\right)  f\left(  b\right)
+f\left(  h\left(  a\right)  \right)  b\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  \left(  ag\left(  f\left(  b\right)  \right)
+h\left(  a\right)  f\left(  b\right)  \right)  +\left(  f\left(  a\right)
h\left(  b\right)  +g\left(  f\left(  a\right)  \right)  b\right)  \right) \\
&  =af\left(  h\left(  b\right)  \right)  +f\left(  h\left(  a\right)
\right)  b-ag\left(  f\left(  b\right)  \right)  -g\left(  f\left(  a\right)
\right)  b\\
&  =\underbrace{af\left(  h\left(  b\right)  \right)  -ag\left(  f\left(
b\right)  \right)  }_{=a\left(  f\left(  h\left(  b\right)  \right)  -g\left(
f\left(  b\right)  \right)  \right)  }+\underbrace{f\left(  h\left(  a\right)
\right)  b-g\left(  f\left(  a\right)  \right)  b}_{=\left(  f\left(  h\left(
a\right)  \right)  -g\left(  f\left(  a\right)  \right)  \right)  b}\\
&  =a\underbrace{\left(  f\left(  h\left(  b\right)  \right)  -g\left(
f\left(  b\right)  \right)  \right)  }_{=\left(  f\circ h-g\circ f\right)
\left(  b\right)  }+\underbrace{\left(  f\left(  h\left(  a\right)  \right)
-g\left(  f\left(  a\right)  \right)  \right)  }_{=\left(  f\circ h-g\circ
f\right)  \left(  a\right)  }b\\
&  =a\left(  f\circ h-g\circ f\right)  \left(  b\right)  +\left(  f\circ
h-g\circ f\right)  \left(  a\right)  b.
\end{align*}
Thus,%
\begin{align*}
\left(  f\circ h-g\circ f\right)  \left(  ab\right)   &  =\left(  f\circ
h\right)  \left(  ab\right)  -\left(  g\circ f\right)  \left(  ab\right) \\
&  =a\left(  f\circ h-g\circ f\right)  \left(  b\right)  +\left(  f\circ
h-g\circ f\right)  \left(  a\right)  b.
\end{align*}


Let us now forget that we fixed $a$ and $b$. We thus have proven that
\[
\left(  f\circ h-g\circ f\right)  \left(  ab\right)  =a\left(  f\circ h-g\circ
f\right)  \left(  b\right)  +\left(  f\circ h-g\circ f\right)  \left(
a\right)  b
\]
for every $a\in A$ and $b\in A$. Thus, $f\circ h-g\circ f$ is a derivation
from $A$ to $M$ (since $f\circ h-g\circ f$ is $\mathbf{k}$-linear). In other
words, $f\circ h-g\circ f\in\operatorname*{Der}\left(  A,M\right)  $. This
proves Theorem \ref{thm.Der-mod.lie.gen1}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Theorem \ref{thm.Der-mod.lie.gen1}.]The map $f:A\rightarrow M$ is a
derivation. Corollary \ref{cor.deriv.verlong} thus shows that $f$ is a
$\mathbf{k}$-linear map and satisfies (\ref{eq.def.deriv.axiom}).

The map $h:A\rightarrow A$ is a derivation. Corollary \ref{cor.deriv.verlong}
(applied to $A$ and $h$ instead of $M$ and $f$) thus shows that $h$ is a
$\mathbf{k}$-linear map and satisfies (\ref{eq.def.deriv.axiom}) (with $M$ and
$f$ replaced by $A$ and $h$).

We know that $h$ satisfies (\ref{eq.def.deriv.axiom}) (with $M$ and $f$
replaced by $A$ and $h$). In other words, we have%
\begin{equation}
\left(  h\left(  ab\right)  =ah\left(  b\right)  +h\left(  a\right)
b\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{ and }b\in A\right)  .
\label{pf.thm.Der-mod.lie.gen1.1}%
\end{equation}


The maps $f$, $h$, $g$ and $f$ are $\mathbf{k}$-linear. Hence, the map $f\circ
h-g\circ f$ is also $\mathbf{k}$-linear.

Let $a\in A$ and $b\in A$. Then,%
\begin{align}
\left(  f\circ h\right)  \left(  ab\right)   &  =f\left(  \underbrace{h\left(
ab\right)  }_{\substack{=ah\left(  b\right)  +h\left(  a\right)  b\\\text{(by
(\ref{pf.thm.Der-mod.lie.gen1.1}))}}}\right)  =f\left(  ah\left(  b\right)
+h\left(  a\right)  b\right) \nonumber\\
&  =\underbrace{f\left(  ah\left(  b\right)  \right)  }_{\substack{=af\left(
h\left(  b\right)  \right)  +f\left(  a\right)  h\left(  b\right)  \\\text{(by
(\ref{eq.def.deriv.axiom}), applied to}\\h\left(  b\right)  \text{ instead of
}b\text{)}}}+\underbrace{f\left(  h\left(  a\right)  b\right)  }%
_{\substack{=h\left(  a\right)  f\left(  b\right)  +f\left(  h\left(
a\right)  \right)  b\\\text{(by (\ref{eq.def.deriv.axiom}), applied
to}\\h\left(  a\right)  \text{ instead of }a\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f\text{ is }\mathbf{k}%
\text{-linear}\right) \nonumber\\
&  =\left(  af\left(  h\left(  b\right)  \right)  +f\left(  a\right)  h\left(
b\right)  \right)  +\left(  h\left(  a\right)  f\left(  b\right)  +f\left(
h\left(  a\right)  \right)  b\right) \nonumber\\
&  =af\left(  h\left(  b\right)  \right)  +f\left(  a\right)  h\left(
b\right)  +h\left(  a\right)  f\left(  b\right)  +f\left(  h\left(  a\right)
\right)  b. \label{pf.thm.Der-mod.lie.gen1.4}%
\end{align}
Also,%
\begin{align*}
\left(  g\circ f\right)  \left(  ab\right)   &  =g\left(  \underbrace{f\left(
ab\right)  }_{\substack{=af\left(  b\right)  +f\left(  a\right)  b\\\text{(by
(\ref{eq.def.deriv.axiom}))}}}\right)  =g\left(  af\left(  b\right)  +f\left(
a\right)  b\right) \\
&  =\underbrace{g\left(  af\left(  b\right)  \right)  }_{\substack{=ag\left(
f\left(  b\right)  \right)  +h\left(  a\right)  f\left(  b\right)  \\\text{(by
(\ref{eq.thm.Der-mod.lie.gen1.cond}), applied to}\\f\left(  b\right)  \text{
instead of }m\text{)}}}+\underbrace{g\left(  f\left(  a\right)  b\right)
}_{\substack{=f\left(  a\right)  h\left(  b\right)  +g\left(  f\left(
a\right)  \right)  b\\\text{(by (\ref{eq.thm.Der-mod.lie.gen1.cond2}), applied
to}\\f\left(  a\right)  \text{ instead of }m\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }g\text{ is }\mathbf{k}%
\text{-linear}\right) \\
&  =\left(  ag\left(  f\left(  b\right)  \right)  +h\left(  a\right)  f\left(
b\right)  \right)  +\left(  f\left(  a\right)  h\left(  b\right)  +g\left(
f\left(  a\right)  \right)  b\right) \\
&  =ag\left(  f\left(  b\right)  \right)  +h\left(  a\right)  f\left(
b\right)  +f\left(  a\right)  h\left(  b\right)  +g\left(  f\left(  a\right)
\right)  b.
\end{align*}
Subtracting this equality from (\ref{pf.thm.Der-mod.lie.gen1.4}), we obtain%
\begin{align*}
&  \left(  f\circ h\right)  \left(  ab\right)  -\left(  g\circ f\right)
\left(  ab\right) \\
&  =\left(  af\left(  h\left(  b\right)  \right)  +f\left(  a\right)  h\left(
b\right)  +h\left(  a\right)  f\left(  b\right)  +f\left(  h\left(  a\right)
\right)  b\right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  ag\left(  f\left(  b\right)  \right)
+h\left(  a\right)  f\left(  b\right)  +f\left(  a\right)  h\left(  b\right)
+g\left(  f\left(  a\right)  \right)  b\right) \\
&  =af\left(  h\left(  b\right)  \right)  +f\left(  h\left(  a\right)
\right)  b-ag\left(  f\left(  b\right)  \right)  -g\left(  f\left(  a\right)
\right)  b\\
&  =\underbrace{af\left(  h\left(  b\right)  \right)  -ag\left(  f\left(
b\right)  \right)  }_{=a\left(  f\left(  h\left(  b\right)  \right)  -g\left(
f\left(  b\right)  \right)  \right)  }+\underbrace{f\left(  h\left(  a\right)
\right)  b-g\left(  f\left(  a\right)  \right)  b}_{=\left(  f\left(  h\left(
a\right)  \right)  -g\left(  f\left(  a\right)  \right)  \right)  b}\\
&  =a\left(  \underbrace{f\left(  h\left(  b\right)  \right)  }_{=\left(
f\circ h\right)  \left(  b\right)  }-\underbrace{g\left(  f\left(  b\right)
\right)  }_{=\left(  g\circ f\right)  \left(  b\right)  }\right)  +\left(
\underbrace{f\left(  h\left(  a\right)  \right)  }_{=\left(  f\circ h\right)
\left(  a\right)  }-\underbrace{g\left(  f\left(  a\right)  \right)
}_{=\left(  g\circ f\right)  \left(  a\right)  }\right)  b\\
&  =a\underbrace{\left(  \left(  f\circ h\right)  \left(  b\right)  -\left(
g\circ f\right)  \left(  b\right)  \right)  }_{=\left(  f\circ h-g\circ
f\right)  \left(  b\right)  }+\underbrace{\left(  \left(  f\circ h\right)
\left(  a\right)  -\left(  g\circ f\right)  \left(  a\right)  \right)
}_{=\left(  f\circ h-g\circ f\right)  \left(  a\right)  }b\\
&  =a\left(  f\circ h-g\circ f\right)  \left(  b\right)  +\left(  f\circ
h-g\circ f\right)  \left(  a\right)  b.
\end{align*}
Thus,%
\begin{align*}
\left(  f\circ h-g\circ f\right)  \left(  ab\right)   &  =\left(  f\circ
h\right)  \left(  ab\right)  -\left(  g\circ f\right)  \left(  ab\right) \\
&  =a\left(  f\circ h-g\circ f\right)  \left(  b\right)  +\left(  f\circ
h-g\circ f\right)  \left(  a\right)  b.
\end{align*}


Let us now forget that we fixed $a$ and $b$. We thus have proven that
\[
\left(  f\circ h-g\circ f\right)  \left(  ab\right)  =a\left(  f\circ h-g\circ
f\right)  \left(  b\right)  +\left(  f\circ h-g\circ f\right)  \left(
a\right)  b
\]
for every $a\in A$ and $b\in A$. Thus, Corollary
\ref{cor.deriv.verlong.converse} (applied to $f\circ h-g\circ f$ instead of
$f$) shows that $f\circ h-g\circ f$ is a derivation. Thus, $f\circ h-g\circ f$
is a derivation from $A$ to $M$. In other words, $f\circ h-g\circ f$ is an
element of the set of all derivations from $A$ to $M$. In other words, $f\circ
h-g\circ f$ is an element of $\operatorname*{Der}\left(  A,M\right)  $ (since
$\operatorname*{Der}\left(  A,M\right)  $ is the set of all derivations from
$A$ to $M$). In other words, $f\circ h-g\circ f\in\operatorname*{Der}\left(
A,M\right)  $. This proves Theorem \ref{thm.Der-mod.lie.gen1}.
\end{proof}
\end{verlong}

\begin{vershort}
\begin{proof}
[Proof of Theorem \ref{thm.Der-mod.lie}.]Set $M=A$ and $h=g$. Recall that $g$
is a derivation; thus, $g\left(  ab\right)  =ag\left(  b\right)  +g\left(
a\right)  b$ for any $a\in A$ and $b\in A$. Hence, the equalities
(\ref{eq.thm.Der-mod.lie.gen1.cond}) and (\ref{eq.thm.Der-mod.lie.gen1.cond2})
hold (because $h=g$). Theorem \ref{thm.Der-mod.lie.gen1} thus shows that
$f\circ h-g\circ f\in\operatorname*{Der}\left(  A,M\right)  $. Since $h=g$ and
$M=A$, this rewrites as $f\circ g-g\circ f\in\operatorname*{Der}\left(
A,A\right)  $.
\end{proof}

\begin{verlong}
\begin{proof}
[Proof of Theorem \ref{thm.Der-mod.lie}.]We have $f\in\operatorname*{Der}%
\left(  A,A\right)  $. In other words, $f$ belongs to $\operatorname*{Der}%
\left(  A,A\right)  $. In other words, $f$ belongs to the set of all
derivations from $A$ to $A$ (since $\operatorname*{Der}\left(  A,A\right)  $
is the set of all derivations from $A$ to $A$ (by the definition of
$\operatorname*{Der}\left(  A,A\right)  $)). In other words, $f$ is a
derivation from $A$ to $A$. The same argument (with $f$ replaced by $g$) shows
that $g$ is a derivation from $A$ to $A$.

The map $g:A\rightarrow A$ is a derivation. Corollary \ref{cor.deriv.verlong}
(applied to $A$ and $g$ instead of $M$ and $f$) thus shows that $g$ is a
$\mathbf{k}$-linear map and satisfies (\ref{eq.def.deriv.axiom}) (with $M$ and
$f$ replaced by $A$ and $g$).

We know that $g$ satisfies (\ref{eq.def.deriv.axiom}) (with $M$ and $f$
replaced by $A$ and $g$). In other words, we have%
\begin{equation}
\left(  g\left(  ab\right)  =ag\left(  b\right)  +g\left(  a\right)
b\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{ and }b\in A\right)  .
\label{pf.thm.Der-mod.lie.3}%
\end{equation}


Now, for every $a\in A$ and $m\in A$, we have $g\left(  am\right)  =ag\left(
m\right)  +g\left(  a\right)  m$ (by (\ref{pf.thm.Der-mod.lie.3}), applied to
$b=m$). Furthermore, for every $b\in A$ and $m\in M$, we have $g\left(
mb\right)  =mg\left(  b\right)  +gm\left(  b\right)  $ (by
(\ref{pf.thm.Der-mod.lie.3}), applied to $a=m$). Thus, Theorem
\ref{thm.Der-mod.lie.gen1} (applied to $M=A$ and $h=g$) shows that $f\circ
g-g\circ f\in\operatorname*{Der}\left(  A,A\right)  $. This proves Theorem
\ref{thm.Der-mod.lie}.
\end{proof}
\end{verlong}
\end{vershort}

Let us state another particular case of Theorem \ref{thm.Der-mod.lie.gen1}.

\begin{corollary}
\label{cor.Der-mod.lie.gen2}Let $B$ be a $\mathbf{k}$-algebra. Let $A$ be a
$\mathbf{k}$-subalgebra of $B$. Thus, $B$ becomes an $\left(  A,A\right)
$-bimodule. (The left and the right $A$-module structures on $B$ are given by
multiplication inside $B$.) Let $f:A\rightarrow B$ and $g:B\rightarrow B$ be
two derivations such that $g\left(  A\right)  \subseteq A$. Then,
$f\circ\left(  g\mid_{A}\right)  -g\circ f:A\rightarrow B$ is a derivation.
\end{corollary}

This corollary is sometimes useful (e.g., it appears in \cite[Proposition
4.6.22]{etilie}\footnote{More precisely, the particular case of Corollary
\ref{cor.Der-mod.lie.gen2} when $\mathbf{k}=\mathbb{C}$ appears in
\cite[Proposition 4.6.22]{etilie}. But the general case is proven in the same
way as this particular case.
\par
Notice that Theorem \ref{thm.Der-mod.lie} (in the particular case when
$\mathbf{k}=\mathbb{C}$) is \cite[Proposition 4.6.21]{etilie}.}, and is used there).

\begin{vershort}
\begin{proof}
[Proof of Corollary \ref{cor.Der-mod.lie.gen2}.]The map $g\mid_{A}%
:A\rightarrow A$ is a derivation (since $g:B\rightarrow B$ is a derivation).
Moreover, the equalities (\ref{eq.thm.Der-mod.lie.gen1.cond}) and
(\ref{eq.thm.Der-mod.lie.gen1.cond2}) hold for $M=B$ and $h=g\mid_{A}$ (again
because $g:B\rightarrow B$ is a derivation). Thus, Theorem
\ref{thm.Der-mod.lie.gen1} (applied to $M=B$ and $h=g\mid_{A}$) shows that
$f\circ\left(  g\mid_{A}\right)  -g\circ f\in\operatorname*{Der}\left(
A,B\right)  $.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Corollary \ref{cor.Der-mod.lie.gen2}.]The map $g\mid_{A}%
:A\rightarrow A$ is well-defined (since $g\left(  A\right)  \subseteq A$) and
$\mathbf{k}$-linear (since the map $g$ is $\mathbf{k}$-linear).

The map $g:B\rightarrow B$ is a derivation. Corollary \ref{cor.deriv.verlong}
(applied to $B$, $B$ and $g$ instead of $A$, $M$ and $f$) thus shows that $g$
is a $\mathbf{k}$-linear map and satisfies (\ref{eq.def.deriv.axiom}) (with
$A$, $M$ and $f$ replaced by $B$, $B$ and $g$).

We know that $g$ satisfies (\ref{eq.def.deriv.axiom}) (with $A$, $M$ and $f$
replaced by $B$, $B$ and $g$). In other words, we have%
\begin{equation}
\left(  g\left(  ab\right)  =ag\left(  b\right)  +g\left(  a\right)
b\ \ \ \ \ \ \ \ \ \ \text{for every }a\in B\text{ and }b\in B\right)  .
\label{pf.cor.Der-mod.lie.gen2.1}%
\end{equation}


The map $g\mid_{A}:A\rightarrow A$ is a derivation\footnote{\textit{Proof.}
Let $a\in A$ and $b\in A$. Then, $\left(  g\mid_{A}\right)  \left(  a\right)
=g\left(  a\right)  $ and $\left(  g\mid_{A}\right)  \left(  b\right)
=g\left(  b\right)  $. Now,
\begin{align*}
\left(  g\mid_{A}\right)  \left(  ab\right)   &  =g\left(  ab\right)
=a\underbrace{g\left(  b\right)  }_{=\left(  g\mid_{A}\right)  \left(
b\right)  }+\underbrace{g\left(  a\right)  }_{=\left(  g\mid_{A}\right)
\left(  a\right)  }b\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.cor.Der-mod.lie.gen2.1})}\right) \\
&  =a\left(  g\mid_{A}\right)  \left(  b\right)  +\left(  g\mid_{A}\right)
\left(  a\right)  b.
\end{align*}
\par
Let us now forget that we fixed $a$ and $b$. We thus have proven that $\left(
g\mid_{A}\right)  \left(  ab\right)  =a\left(  g\mid_{A}\right)  \left(
b\right)  +\left(  g\mid_{A}\right)  \left(  a\right)  b$ for every $a\in A$
and $b\in A$. Hence, Corollary \ref{cor.deriv.verlong.converse} (applied to
$A$ and $g\mid_{A}$ instead of $M$ and $f$) shows that $g\mid_{A}$ is a
derivation. Qed.}. Moreover, we have%
\begin{equation}
\left(  g\left(  am\right)  =ag\left(  m\right)  +\left(  g\mid_{A}\right)
\left(  a\right)  m\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{ and }m\in
B\right)  \label{pf.cor.Der-mod.lie.gen2.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.cor.Der-mod.lie.gen2.2}):} Let $a\in A$
and $m\in B$. Then, $\left(  g\mid_{A}\right)  \left(  a\right)  =g\left(
a\right)  $. Applying (\ref{pf.cor.Der-mod.lie.gen2.1}) to $b=m$, we obtain
\[
g\left(  am\right)  =ag\left(  m\right)  +\underbrace{g\left(  a\right)
}_{=\left(  g\mid_{A}\right)  \left(  a\right)  }m=ag\left(  m\right)
+\left(  g\mid_{A}\right)  \left(  a\right)  m.
\]
This proves (\ref{pf.cor.Der-mod.lie.gen2.2}).} and%
\begin{equation}
\left(  g\left(  mb\right)  =m\left(  g\mid_{A}\right)  \left(  b\right)
+g\left(  m\right)  b\ \ \ \ \ \ \ \ \ \ \text{for every }b\in A\text{ and
}m\in B\right)  \label{pf.cor.Der-mod.lie.gen2.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.cor.Der-mod.lie.gen2.3}):} Let $b\in A$
and $m\in B$. Then, $\left(  g\mid_{A}\right)  \left(  b\right)  =g\left(
b\right)  $. Applying (\ref{pf.cor.Der-mod.lie.gen2.1}) to $a=m$, we obtain
\[
g\left(  mb\right)  =m\underbrace{g\left(  b\right)  }_{=\left(  g\mid
_{A}\right)  \left(  b\right)  }+g\left(  m\right)  b=m\left(  g\mid
_{A}\right)  \left(  b\right)  +g\left(  m\right)  b.
\]
This proves (\ref{pf.cor.Der-mod.lie.gen2.3}).}. Hence, we can apply Theorem
\ref{thm.Der-mod.lie.gen1} to $M=B$ and $h=g\mid_{A}$. As a result, we obtain
$f\circ\left(  g\mid_{A}\right)  -g\circ f\in\operatorname*{Der}\left(
A,B\right)  $. In other words, $f\circ\left(  g\mid_{A}\right)  -g\circ f$ is
an element of $\operatorname*{Der}\left(  A,B\right)  $. In other words,
$f\circ\left(  g\mid_{A}\right)  -g\circ f$ is an element of the set of all
derivations from $A$ to $B$ (since $\operatorname*{Der}\left(  A,B\right)  $
is the set of all derivations from $A$ to $B$ (by the definition of
$\operatorname*{Der}\left(  A,B\right)  $)). In other words, $f\circ\left(
g\mid_{A}\right)  -g\circ f$ is a derivation from $A$ to $B$. This proves
Corollary \ref{cor.Der-mod.lie.gen2}.
\end{proof}
\end{verlong}

Let us now prove another fact about derivations.\footnote{Proposition
\ref{prop.derivation.unique} appears in \cite[Proposition 4.6.13]{etilie}.}

\begin{proposition}
\label{prop.derivation.unique}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be an
$\left(  A,A\right)  $-bimodule. Let $d:A\rightarrow M$ and $e:A\rightarrow M$
be two derivations. Let $S$ be a subset of $A$ which generates $A$ as a
$\mathbf{k}$-algebra. Assume that $d\mid_{S}=e\mid_{S}$. Then, $d=e$.
\end{proposition}

Proposition \ref{prop.derivation.unique} shows that a derivation $A\rightarrow
M$ is uniquely determined by its restriction to a generating set of the
$\mathbf{k}$-algebra $A$. Of course, this does not yield that any map from
such a generating set can be extended to a derivation $A\rightarrow M$ (though
we will see some situations when such extensions are possible
later\footnote{in Proposition \ref{prop.T.deruniv} and Proposition
\ref{prop.Sym.deruniv}}).

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.derivation.unique}.]We have $d\mid_{S}%
=e\mid_{S}$. In other words, $d\left(  s\right)  =e\left(  s\right)  $ for
every $s\in S$.

Let $z$ be the map $d-e:A\rightarrow M$. Then, $z\left(  s\right)  =0$ for
every $s\in S$ (since $d\left(  s\right)  =e\left(  s\right)  $ for every
$s\in S$). In other words, $S\subseteq\operatorname*{Ker}z$.

The maps $d$ and $e$ are two derivations, and thus belong to
$\operatorname*{Der}\left(  A,M\right)  $. Hence, $d-e\in\operatorname*{Der}%
\left(  A,M\right)  $ as well (since Proposition \ref{prop.Der-mod} shows that
$\operatorname*{Der}\left(  A,M\right)  $ is a $\mathbf{k}$-submodule of
$\operatorname*{Hom}\left(  A,M\right)  $). Hence, $z=d-e\in
\operatorname*{Der}\left(  A,M\right)  $. In other words, $z$ is a derivation
from $A$ to $M$. Hence, $z\left(  1\right)  =0$ (by Theorem \ref{thm.deriv.0},
applied to $f=z$).

Also, $z$ is a derivation and thus $\mathbf{k}$-linear. Hence,
$\operatorname*{Ker}z$ is a $\mathbf{k}$-submodule of $A$. This $\mathbf{k}%
$-submodule $\operatorname*{Ker}z$ furthermore satisfies $1\in
\operatorname*{Ker}z$ (since $z\left(  1\right)  =0$) and%
\[
\left(  ab\in\operatorname*{Ker}z\text{ for every }a\in\operatorname*{Ker}%
z\text{ and }b\in\operatorname*{Ker}z\right)
\]
\footnote{\textit{Proof.} Let $a\in\operatorname*{Ker}z$ and $b\in
\operatorname*{Ker}z$. We have $z\left(  a\right)  =0$ (since $a\in
\operatorname*{Ker}z$) and $z\left(  b\right)  =0$ (since $b\in
\operatorname*{Ker}z$). But $z$ is a derivation. Hence,%
\[
z\left(  ab\right)  =a\underbrace{z\left(  b\right)  }_{=0}%
+\underbrace{z\left(  a\right)  }_{=0}b=0+0=0.
\]
In other words, $ab\in\operatorname*{Ker}z$, qed.}; therefore, this
$\mathbf{k}$-submodule $\operatorname*{Ker}z$ is a $\mathbf{k}$-subalgebra of
$A$. More precisely, it is a $\mathbf{k}$-subalgebra of $A$ which contains $S$
as a subset (since $S\subseteq\operatorname*{Ker}z$).

But recall that the subset $S$ generates $A$ as a $\mathbf{k}$-algebra. Hence,
the smallest $\mathbf{k}$-subalgebra of $A$ which contains $S$ as a subset
must be $A$ itself. Hence, if $B$ is any $\mathbf{k}$-subalgebra of $A$ which
contains $S$ as a subset, then $A\subseteq B$. We can apply this to
$B=\operatorname*{Ker}z$ (since $\operatorname*{Ker}z$ is a $\mathbf{k}%
$-subalgebra of $A$ which contains $S$ as a subset). Thus, we obtain
$A\subseteq\operatorname*{Ker}z$. Hence, $z=0$. Since $z=d-e$, this rewrites
as $d-e=0$. Hence, $d=e$. Proposition \ref{prop.derivation.unique} is thus proven.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.derivation.unique}.]Let $z$ be the map
$d-e:A\rightarrow M$. Then, $z=d-e$. Hence,%
\begin{equation}
z\left(  s\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }s\in S
\label{pf.prop.derivation.unique.zs}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.derivation.unique.zs}):} Let $s\in
S$. Then, $\left(  d\mid_{S}\right)  \left(  s\right)  =d\left(  s\right)  $
and $\left(  e\mid_{S}\right)  =e\left(  s\right)  $. Hence, $d\left(
s\right)  =\underbrace{\left(  d\mid_{S}\right)  }_{=\left(  e\mid_{S}\right)
}\left(  s\right)  =\left(  e\mid_{S}\right)  =e\left(  s\right)  $. Now,
\[
\underbrace{z}_{=d-e}\left(  s\right)  =\left(  d-e\right)  \left(  s\right)
=\underbrace{d\left(  s\right)  }_{=e\left(  s\right)  }-e\left(  s\right)
=e\left(  s\right)  -e\left(  s\right)  =0.
\]
This proves (\ref{pf.prop.derivation.unique.zs}).}.

But $z:A\rightarrow M$ is a derivation\footnote{\textit{Proof.} Proposition
\ref{prop.Der-mod} shows that $\operatorname*{Der}\left(  A,M\right)  $ is a
$\mathbf{k}$-submodule of $\operatorname*{Hom}\left(  A,M\right)  $.
\par
The map $d$ is a derivation from $A$ to $M$. In other words, $d$ belongs to
the set of all derivations from $A$ to $M$. In other words, $d$ belongs to
$\operatorname*{Der}\left(  A,M\right)  $ (since $\operatorname*{Der}\left(
A,M\right)  $ is the set of all derivations from $A$ to $M$). In other words,
$d\in\operatorname*{Der}\left(  A,M\right)  $. The same reasoning (but with
$d$ replaced by $e$) shows that $e\in\operatorname*{Der}\left(  A,M\right)  $.
\par
Now,%
\[
z=\underbrace{d}_{\in\operatorname*{Der}\left(  A,M\right)  }-\underbrace{e}%
_{\in\operatorname*{Der}\left(  A,M\right)  }\in\operatorname*{Der}\left(
A,M\right)  -\operatorname*{Der}\left(  A,M\right)  \subseteq
\operatorname*{Der}\left(  A,M\right)
\]
(since $\operatorname*{Der}\left(  A,M\right)  $ is a $\mathbf{k}$-submodule
of $\operatorname*{Hom}\left(  A,M\right)  $). In other words, $z$ belongs to
$\operatorname*{Der}\left(  A,M\right)  $. In other words, $z$ belongs to the
set of all derivations from $A$ to $M$ (since $\operatorname*{Der}\left(
A,M\right)  $ is the set of all derivations from $A$ to $M$). In other words,
$z$ is a derivation from $A$ to $M$. Qed.}. Corollary \ref{cor.deriv.verlong}
(applied to $f=z$) thus shows that $z$ is a $\mathbf{k}$-linear map and
satisfies (\ref{eq.def.deriv.axiom}) (with $f$ replaced by $z$).

We know that $z$ satisfies (\ref{eq.def.deriv.axiom}) (with $f$ replaced by
$z$). In other words, we have%
\begin{equation}
\left(  z\left(  ab\right)  =az\left(  b\right)  +z\left(  a\right)
b\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{ and }b\in A\right)  .
\label{pf.prop.derivation.unique.zderi}%
\end{equation}


The set $\operatorname*{Ker}z$ is a $\mathbf{k}$-submodule of $A$ (since $z$
is a $\mathbf{k}$-linear map). Moreover, $z\left(  1\right)  =0$ (by Theorem
\ref{thm.deriv.0}, applied to $f=z$). In other words, $1\in\operatorname*{Ker}%
z$. Furthermore, every $a\in\operatorname*{Ker}z$ and $b\in\operatorname*{Ker}%
z$ satisfy $ab\in\operatorname*{Ker}z$\ \ \ \ \footnote{\textit{Proof.} Let
$a\in\operatorname*{Ker}z$ and $b\in\operatorname*{Ker}z$. We have $z\left(
a\right)  =0$ (since $a\in\operatorname*{Ker}z$) and $z\left(  b\right)  =0$
(since $b\in\operatorname*{Ker}z$). Now,
(\ref{pf.prop.derivation.unique.zderi}) shows that%
\[
z\left(  ab\right)  =a\underbrace{z\left(  b\right)  }_{=0}%
+\underbrace{z\left(  a\right)  }_{=0}b=0+0=0.
\]
In other words, $ab\in\operatorname*{Ker}z$, qed.}.

We thus have proven the following three statements:

\begin{itemize}
\item The set $\operatorname*{Ker}z$ is a $\mathbf{k}$-submodule of $A$.

\item We have $1\in\operatorname*{Ker}z$.

\item Every $a\in\operatorname*{Ker}z$ and $b\in\operatorname*{Ker}z$ satisfy
$ab\in\operatorname*{Ker}z$.
\end{itemize}

Combining these three statements, we conclude that $\operatorname*{Ker}z$ is a
$\mathbf{k}$-subalgebra of $A$.

Moreover, $S\subseteq\operatorname*{Ker}z$\ \ \ \ \footnote{\textit{Proof.}
Let $s\in S$. Then, $z\left(  s\right)  =0$ (by
(\ref{pf.prop.derivation.unique.zs})). In other words, $s\in
\operatorname*{Ker}z$.
\par
Now, let us forget that we fixed $s$. We thus have shown that $s\in
\operatorname*{Ker}z$ for every $s\in S$. In other words, $S\subseteq
\operatorname*{Ker}z$, qed.}. In other words, $\operatorname*{Ker}z$ contains
$S$ as a subset.

The subset $S$ generates $A$ as a $\mathbf{k}$-algebra. In other words, the
$\mathbf{k}$-subalgebra of $A$ generated by $S$ is $A$ itself.

Now, let us recall that the $\mathbf{k}$-subalgebra of $A$ generated by $S$ is
the smallest $\mathbf{k}$-subalgebra of $A$ which contains $S$ as a subset. In
other words, $A$ is the smallest $\mathbf{k}$-subalgebra of $A$ which contains
$S$ as a subset (because the $\mathbf{k}$-subalgebra of $A$ generated by $S$
is $A$). Hence, if $B$ is any $\mathbf{k}$-subalgebra of $A$ which contains
$S$ as a subset, then $A\subseteq B$. We can apply this to
$B=\operatorname*{Ker}z$ (since $\operatorname*{Ker}z$ is a $\mathbf{k}%
$-subalgebra of $A$ which contains $S$ as a subset). Thus, we obtain
$A\subseteq\operatorname*{Ker}z$. Combined with $\operatorname*{Ker}z\subseteq
A$, this yields $A=\operatorname*{Ker}z$. Hence, $z=0$. Since $z=d-e$, this
rewrites as $d-e=0$. Hence, $d=e$. Proposition \ref{prop.derivation.unique} is
thus proven.
\end{proof}
\end{verlong}

\begin{remark}
Again, much of what we have proven (e.g., Proposition \ref{prop.Der-mod},
Theorem \ref{thm.Der-mod.lie}, Theorem \ref{thm.Der-mod.lie.gen1} and
Proposition \ref{prop.derivation.unique}) can be generalized to the situation
when $A$ is a magmatic $\mathbf{k}$-algebra (instead of being a $\mathbf{k}%
$-algebra), provided that the definition of a derivation is properly adjusted
to this generality.
\end{remark}

\subsection{The $\mathbf{k}$-algebra $\mathcal{R}_{A}\left(  M\right)  $}

We shall now introduce a construction that will allow us to reduce questions
about derivations to questions about $\mathbf{k}$-algebra homomorphisms.

\begin{theorem}
\label{thm.RAM}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be an $\left(
A,A\right)  $-bimodule. Consider the $\mathbf{k}$-module $A\oplus M$. Define a
map $m:\left(  A\oplus M\right)  \times\left(  A\oplus M\right)  \rightarrow
A\oplus M$ by%
\[
\left(  m\left(  \left(  a,p\right)  ,\left(  b,q\right)  \right)  =\left(
ab,aq+pb\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }\left(  a,p\right)  \in
A\oplus M\text{ and }\left(  b,q\right)  \in A\oplus M\right)  .
\]
Then, the map $m$ is $\mathbf{k}$-bilinear. Furthermore, the $\mathbf{k}%
$-module $A\oplus M$, equipped with the $\mathbf{k}$-bilinear map $m$, becomes
a $\mathbf{k}$-algebra with unity $\left(  1,0\right)  $.
\end{theorem}

\begin{definition}
\label{def.RAM}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be an $\left(
A,A\right)  $-bimodule. Consider the $\mathbf{k}$-module $A\oplus M$, and the
map $m$ defined in Theorem \ref{thm.RAM}. Theorem \ref{thm.RAM} shows that the
$\mathbf{k}$-module $A\oplus M$, equipped with the $\mathbf{k}$-bilinear map
$m$, becomes a $\mathbf{k}$-algebra with unity $\left(  1,0\right)  $. This
$\mathbf{k}$-algebra will be denoted by $\mathcal{R}_{A}\left(  M\right)  $.
Thus, its multiplication satisfies%
\begin{align}
\left(  a,p\right)  \cdot\left(  b,q\right)   &  =m\left(  \left(  a,p\right)
,\left(  b,q\right)  \right)  =\left(  ab,aq+pb\right)
\label{eq.def.RAM.product}\\
&  \ \ \ \ \ \ \ \ \ \ \text{for all }\left(  a,p\right)  \in A\oplus M\text{
and }\left(  b,q\right)  \in A\oplus M.\nonumber
\end{align}

\end{definition}

\begin{vershort}
\begin{proof}
[Proof of Theorem \ref{thm.RAM}.]The map $m$ is $\mathbf{k}$%
-bilinear\footnote{The proof of this is just a straightforward application of
the facts that
\par
\begin{itemize}
\item the multiplication of the $\mathbf{k}$-algebra $A$ (that is, the map
$A\times A\rightarrow A,\ \left(  a,b\right)  \mapsto ab$) is $\mathbf{k}%
$-bilinear;
\par
\item the map $A\times M\rightarrow M,\ \left(  a,n\right)  \mapsto an$ is
$\mathbf{k}$-bilinear;
\par
\item the map $M\times A\rightarrow M,\ \left(  n,a\right)  \mapsto na$ is
$\mathbf{k}$-bilinear.
\end{itemize}
}. Thus, the $\mathbf{k}$-module $A\oplus M$, equipped with the $\mathbf{k}%
$-bilinear map $m$, is a magmatic $\mathbf{k}$-algebra. We denote this
magmatic $\mathbf{k}$-algebra by $\mathbf{R}$. Thus, $\mathbf{R}=A\oplus M$ as
a $\mathbf{k}$-module, and the multiplication on $\mathbf{R}$ is the map $m$.
Thus, the multiplication on $\mathbf{R}$ is given by%
\begin{align}
\left(  a,p\right)  \left(  b,q\right)   &  =m\left(  \left(  a,p\right)
,\left(  b,q\right)  \right)  =\left(  ab,aq+pb\right)
\label{pf.thm.RAM.short.product}\\
&  \ \ \ \ \ \ \ \ \ \ \text{for all }\left(  a,p\right)  \in A\oplus M\text{
and }\left(  b,q\right)  \in A\oplus M.\nonumber
\end{align}


The magmatic $\mathbf{k}$-algebra $\mathbf{R}$ is
associative\footnote{\textit{Proof.} We want to show that the magmatic
$\mathbf{k}$-algebra $\mathbf{R}$ is associative. In order to do so, it is
clearly enough to show that $x\left(  yz\right)  =\left(  xy\right)  z$ for
all $x\in\mathbf{R}$, $y\in\mathbf{R}$ and $z\in\mathbf{R}$. So let us do
this.
\par
Fix $x\in\mathbf{R}$, $y\in\mathbf{R}$ and $z\in\mathbf{R}$. The elements $x$,
$y$ and $z$ belong to $A\oplus M$, and therefore can be written in the forms
$x=\left(  a,p\right)  $, $y=\left(  b,q\right)  $ and $z=\left(  c,r\right)
$ for some elements $a$, $b$ and $c$ of $A$ and some elements $p$, $q$ and $r$
of $M$. Consider these $a$, $b$ and $c$ and these $p$, $q$ and $r$.
\par
We have%
\[
\underbrace{y}_{=\left(  b,q\right)  }\underbrace{z}_{=\left(  c,r\right)
}=\left(  b,q\right)  \left(  c,r\right)  =\left(  bc,br+qc\right)
\]
(by (\ref{pf.thm.RAM.short.product})) and thus%
\begin{align}
\underbrace{x}_{=\left(  a,p\right)  }\underbrace{\left(  yz\right)
}_{=\left(  bc,br+qc\right)  }  &  =\left(  a,p\right)  \left(
bc,br+qc\right)  =\left(  a\left(  bc\right)  ,\underbrace{a\left(
br+qc\right)  }_{=a\left(  br\right)  +a\left(  qc\right)  }+p\left(
bc\right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.RAM.short.product}),
applied to }\left(  bc,br+qc\right)  \text{ instead of }\left(  b,q\right)
\right) \nonumber\\
&  =\left(  a\left(  bc\right)  ,a\left(  br\right)  +a\left(  qc\right)
+p\left(  bc\right)  \right)  . \label{pf.thm.RAM.short.assoc.pf.1}%
\end{align}
A similar computation shows that%
\begin{equation}
\left(  xy\right)  z=\left(  \left(  ab\right)  c,\left(  ab\right)  r+\left(
aq\right)  c+\left(  pb\right)  c\right)  .
\label{pf.thm.RAM.short.assoc.pf.2}%
\end{equation}
But $\left(  ab\right)  c=a\left(  bc\right)  $ (since $A$ is associative),
and $\left(  ab\right)  r=a\left(  br\right)  $ (since $M$ is a left
$A$-module), and $\left(  aq\right)  c=a\left(  qc\right)  $ (by
(\ref{eq.def.bimod.1}), applied to $A$, $q$ and $c$ instead of $B$, $m$ and
$b$), and $\left(  pb\right)  c=p\left(  bc\right)  $ (since $M$ is a right
$A$-module). Hence, (\ref{pf.thm.RAM.short.assoc.pf.2}) becomes%
\begin{align*}
\left(  xy\right)  z  &  =\left(  \underbrace{\left(  ab\right)  c}_{=a\left(
bc\right)  },\underbrace{\left(  ab\right)  r}_{=a\left(  br\right)
}+\underbrace{\left(  aq\right)  c}_{=a\left(  qc\right)  }%
+\underbrace{\left(  pb\right)  c}_{=p\left(  bc\right)  }\right)  =\left(
a\left(  bc\right)  ,a\left(  br\right)  +a\left(  qc\right)  +p\left(
bc\right)  \right) \\
&  =x\left(  yz\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.thm.RAM.short.assoc.pf.1})}\right)  .
\end{align*}
Thus, $x\left(  yz\right)  =\left(  xy\right)  z$ is proven, qed.}.
Furthermore, straightforward computations show that%
\[
\left(  1,0\right)  a=a\left(  1,0\right)  =a\ \ \ \ \ \ \ \ \ \ \text{for
every }a\in\mathbf{R}.
\]
Thus, the magmatic $\mathbf{k}$-algebra $\mathbf{R}$, equipped with the
element $\left(  1,0\right)  $, becomes a unital magmatic $\mathbf{k}%
$-algebra. Since it is also associative, it is therefore a $\mathbf{k}%
$-algebra with unity $\left(  1,0\right)  $.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Theorem \ref{thm.RAM}.]We have%
\begin{equation}
m\left(  \lambda x,y\right)  =\lambda m\left(  x,y\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\lambda\in\mathbf{k}\text{, }x\in A\oplus
M\text{ and }y\in A\oplus M \label{pf.thm.RAM.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.RAM.1}):} Let $\lambda\in\mathbf{k}$,
$x\in A\oplus M$ and $y\in A\oplus M$.
\par
We have $x\in A\oplus M$. Thus, we can write $x$ in the form $x=\left(
a,p\right)  $ for some $a\in A$ and $p\in M$. Consider these $a$ and $p$.
\par
We have $y\in A\oplus M$. Thus, we can write $y$ in the form $y=\left(
b,q\right)  $ for some $b\in A$ and $q\in M$. Consider these $b$ and $q$.
\par
We have%
\begin{align*}
m\left(  \lambda\underbrace{x}_{=\left(  a,p\right)  },\underbrace{y}%
_{=\left(  b,q\right)  }\right)   &  =m\left(  \underbrace{\lambda\left(
a,p\right)  }_{=\left(  \lambda a,\lambda p\right)  },\left(  b,q\right)
\right)  =m\left(  \left(  \lambda a,\lambda p\right)  ,\left(  b,q\right)
\right)  =\left(  \lambda ab,\underbrace{\lambda aq+\lambda pb}_{=\lambda
\left(  aq+pb\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.def.RAM.product}), applied to
}\left(  \lambda a,\lambda p\right)  \text{ instead of }\left(  a,p\right)
\right) \\
&  =\left(  \lambda ab,\lambda\left(  aq+pb\right)  \right)  .
\end{align*}
Compared with%
\[
\lambda m\left(  \underbrace{x}_{=\left(  a,p\right)  },\underbrace{y}%
_{=\left(  b,q\right)  }\right)  =\lambda\underbrace{m\left(  \left(
a,p\right)  ,\left(  b,q\right)  \right)  }_{\substack{=\left(
ab,aq+pb\right)  \\\text{(by (\ref{eq.def.RAM.product}))}}}=\lambda\left(
ab,aq+pb\right)  =\left(  \lambda ab,\lambda\left(  aq+pb\right)  \right)  ,
\]
this yields $m\left(  \lambda x,y\right)  =\lambda m\left(  x,y\right)  $.
This proves (\ref{pf.thm.RAM.1}).}. In other words, the map $m$ is
$\mathbf{k}$-linear in its first argument.

We have
\begin{equation}
m\left(  x,\lambda y\right)  =\lambda m\left(  x,y\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\lambda\in\mathbf{k}\text{, }x\in A\oplus
M\text{ and }y\in A\oplus M \label{pf.thm.RAM.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.RAM.2}):} Let $\lambda\in\mathbf{k}$,
$x\in A\oplus M$ and $y\in A\oplus M$.
\par
We have $x\in A\oplus M$. Thus, we can write $x$ in the form $x=\left(
a,p\right)  $ for some $a\in A$ and $p\in M$. Consider these $a$ and $p$.
\par
We have $y\in A\oplus M$. Thus, we can write $y$ in the form $y=\left(
b,q\right)  $ for some $b\in A$ and $q\in M$. Consider these $b$ and $q$.
\par
We have%
\begin{align*}
m\left(  \underbrace{x}_{=\left(  a,p\right)  },\lambda\underbrace{y}%
_{=\left(  b,q\right)  }\right)   &  =m\left(  \left(  a,p\right)
,\underbrace{\lambda\left(  b,q\right)  }_{=\left(  \lambda b,\lambda
q\right)  }\right)  =m\left(  \left(  a,p\right)  ,\left(  \lambda b,\lambda
q\right)  \right) \\
&  =\left(  \underbrace{a\left(  \lambda b\right)  }_{\substack{=\lambda
ab\\\text{(since the multiplication}\\\text{on }A\text{ is }\mathbf{k}%
\text{-bilinear)}}},\underbrace{a\left(  \lambda q\right)  }%
_{\substack{=\lambda aq\\\text{(since the map}\\A\times M\rightarrow
M,\ \left(  c,n\right)  \mapsto cn\\\text{is }\mathbf{k}\text{-bilinear)}%
}}+\underbrace{p\left(  \lambda b\right)  }_{\substack{=\lambda
pb\\\text{(since the map}\\M\times A\rightarrow M,\ \left(  n,c\right)
\mapsto nc\\\text{is }\mathbf{k}\text{-bilinear)}}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.def.RAM.product}), applied to
}\left(  \lambda a,\lambda p\right)  \text{ instead of }\left(  a,p\right)
\right) \\
&  =\left(  \lambda ab,\underbrace{\lambda aq+\lambda pb}_{=\lambda\left(
aq+pb\right)  }\right)  =\left(  \lambda ab,\lambda\left(  aq+pb\right)
\right)  .
\end{align*}
Compared with%
\[
\lambda m\left(  \underbrace{x}_{=\left(  a,p\right)  },\underbrace{y}%
_{=\left(  b,q\right)  }\right)  =\lambda\underbrace{m\left(  \left(
a,p\right)  ,\left(  b,q\right)  \right)  }_{\substack{=\left(
ab,aq+pb\right)  \\\text{(by (\ref{eq.def.RAM.product}))}}}=\lambda\left(
ab,aq+pb\right)  =\left(  \lambda ab,\lambda\left(  aq+pb\right)  \right)  ,
\]
this yields $m\left(  x,\lambda y\right)  =\lambda m\left(  x,y\right)  $.
This proves (\ref{pf.thm.RAM.2}).}. In other words, the map $m$ is
$\mathbf{k}$-linear in its second argument.

The map $m$ is $\mathbf{k}$-bilinear (because it is $\mathbf{k}$-linear in its
first argument and $\mathbf{k}$-linear in its second argument). Hence, the
$\mathbf{k}$-module $A\oplus M$, equipped with the $\mathbf{k}$-bilinear map
$m$, is a magmatic $\mathbf{k}$-algebra. We denote this magmatic $\mathbf{k}%
$-algebra by $\mathbf{R}$. Thus, $\mathbf{R}=A\oplus M$ as a $\mathbf{k}%
$-module, and the multiplication on $\mathbf{R}$ is the map $m$. Thus, the
multiplication on $\mathbf{R}$ is given by%
\begin{align}
\left(  a,p\right)  \cdot\left(  b,q\right)   &  =m\left(  \left(  a,p\right)
,\left(  b,q\right)  \right)  =\left(  ab,aq+pb\right)
\label{pf.thm.RAM.product}\\
&  \ \ \ \ \ \ \ \ \ \ \text{for all }\left(  a,p\right)  \in A\oplus M\text{
and }\left(  b,q\right)  \in A\oplus M.\nonumber
\end{align}


The element $\left(  1,0\right)  $ of $A\oplus M=\mathbf{R}$ satisfies%
\begin{equation}
\left(  1,0\right)  x=x\left(  1,0\right)  =x\ \ \ \ \ \ \ \ \ \ \text{for
every }x\in\mathbf{R} \label{pf.thm.RAM.unity}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.RAM.unity}):} Let $x\in\mathbf{R}$. We
must then prove that $\left(  1,0\right)  x=x\left(  1,0\right)  =x$.
\par
We have $x\in A\oplus M=\mathbf{R}$. Thus, we can write $x$ in the form
$x=\left(  a,p\right)  $ for some $a\in A$ and $p\in M$. Consider these $a$
and $p$. Now,%
\begin{align*}
\left(  1,0\right)  \underbrace{x}_{=\left(  a,p\right)  }  &  =\left(
1,0\right)  \left(  a,p\right)  =\left(  1,0\right)  \cdot\left(  a,p\right)
=\left(  \underbrace{1a}_{=a},\underbrace{1p}_{=p}+\underbrace{0\cdot a}%
_{=0}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.RAM.product}), applied to
}\left(  1,0\right)  \text{ and }\left(  a,p\right)  \text{ instead of
}\left(  a,p\right)  \text{ and }\left(  b,q\right)  \right) \\
&  =\left(  a,\underbrace{p+0}_{=p}\right)  =\left(  a,p\right)  =x
\end{align*}
and
\begin{align*}
\underbrace{x}_{=\left(  a,p\right)  }\left(  1,0\right)   &  =\left(
a,p\right)  \left(  1,0\right)  =\left(  a,p\right)  \cdot\left(  1,0\right)
=\left(  \underbrace{a\cdot1}_{=a},\underbrace{a\cdot0}_{=0}%
+\underbrace{p\cdot1}_{=p}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.RAM.product}), applied to
}\left(  1,0\right)  \text{ instead of }\left(  b,q\right)  \right) \\
&  =\left(  a,\underbrace{0+p}_{=p}\right)  =\left(  a,p\right)  =x.
\end{align*}
Thus, $\left(  1,0\right)  x=x=x\left(  1,0\right)  =x$. This proves
(\ref{pf.thm.RAM.unity}).}. Renaming $x$ as $a$ in this statement, we obtain
that%
\[
\left(  1,0\right)  a=a\left(  1,0\right)  =a\ \ \ \ \ \ \ \ \ \ \text{for
every }a\in\mathbf{R}.
\]


But if $B$ is any magmatic $\mathbf{k}$-algebra, and if $e\in B$ is any
element which satisfies%
\[
\left(  ea=ae=a\ \ \ \ \ \ \ \ \ \ \text{for every }a\in B\right)  ,
\]
then the magmatic $\mathbf{k}$-algebra $B$, equipped with the element $e$, is
a unital magmatic $\mathbf{k}$-algebra (by the definition of a
\textquotedblleft unital magmatic $\mathbf{k}$-algebra\textquotedblright). We
can apply this to $B=\mathbf{R}$ and $e=\left(  1,0\right)  $ (because we have
$\left(  1,0\right)  a=a\left(  1,0\right)  =a$ for every $a\in\mathbf{R}$).
As a consequence, we conclude that the magmatic $\mathbf{k}$-algebra
$\mathbf{R}$, equipped with the element $\left(  1,0\right)  \in\mathbf{R}$,
is a unital magmatic $\mathbf{k}$-algebra. As usual, we denote this unital
magmatic $\mathbf{k}$-algebra again by $\mathbf{R}$. Thus, strictly speaking,
$\mathbf{R}$ is the $\mathbf{k}$-module $A\oplus M$, equipped with the
$\mathbf{k}$-bilinear map $m$ and the unity $\left(  1,0\right)  $.

We have%
\begin{equation}
x\left(  yz\right)  =\left(  xy\right)  z\ \ \ \ \ \ \ \ \ \ \text{for all
}x\in\mathbf{R}\text{, }y\in\mathbf{R}\text{ and }z\in\mathbf{R}
\label{pf.thm.RAM.assoc}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.RAM.assoc}):} Let $x\in\mathbf{R}$,
$y\in\mathbf{R}$ and $z\in\mathbf{R}$.
\par
We have $x\in A\oplus M=\mathbf{R}$. Thus, we can write $x$ in the form
$x=\left(  a,p\right)  $ for some $a\in A$ and $p\in M$. Consider these $a$
and $p$.
\par
We have $y\in A\oplus M=\mathbf{R}$. Thus, we can write $y$ in the form
$y=\left(  b,q\right)  $ for some $b\in A$ and $q\in M$. Consider these $b$
and $q$.
\par
We have $z\in A\oplus M=\mathbf{R}$. Thus, we can write $z$ in the form
$z=\left(  c,r\right)  $ for some $c\in A$ and $r\in M$. Consider these $c$
and $r$.
\par
We have%
\begin{align*}
&  \underbrace{y}_{=\left(  b,q\right)  }\underbrace{z}_{=\left(  c,r\right)
}=\left(  b,q\right)  \left(  c,r\right)  =\left(  b,q\right)  \cdot\left(
c,r\right)  =\left(  bc,br+qc\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.RAM.product}), applied to
}\left(  b,q\right)  \text{ and }\left(  c,r\right)  \text{ instead of
}\left(  a,p\right)  \text{ and }\left(  b,q\right)  \right)
\end{align*}
and%
\[
\underbrace{x}_{=\left(  a,p\right)  }\underbrace{y}_{=\left(  b,q\right)
}=\left(  a,p\right)  \left(  b,q\right)  =\left(  a,p\right)  \cdot\left(
b,q\right)  =\left(  ab,aq+pb\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.thm.RAM.product})}\right)  .
\]
Now, comparing%
\begin{align*}
x\left(  yz\right)   &  =\underbrace{x}_{=\left(  a,p\right)  }\cdot
\underbrace{\left(  yz\right)  }_{=\left(  bc,br+qc\right)  }=\left(
a,p\right)  \cdot\left(  bc,br+qc\right) \\
&  =\left(  \underbrace{a\left(  bc\right)  }_{\substack{=\left(  ab\right)
c\\\text{(since }A\text{ is associative)}}},\underbrace{a\left(  br+qc\right)
}_{=a\left(  br\right)  +a\left(  qc\right)  }+\underbrace{p\left(  bc\right)
}_{\substack{=\left(  pb\right)  c\\\text{(since }M\text{ is a right}%
\\A\text{-module)}}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.RAM.product}), applied to
}\left(  bc,br+qc\right)  \text{ instead of }\left(  b,q\right)  \right) \\
&  =\left(  \left(  ab\right)  c,a\left(  br\right)  +a\left(  qc\right)
+\left(  pb\right)  c\right)
\end{align*}
with%
\begin{align*}
\left(  xy\right)  z  &  =\underbrace{\left(  xy\right)  }_{=\left(
ab,aq+pb\right)  }\cdot\underbrace{z}_{=\left(  c,r\right)  }=\left(
ab,aq+pb\right)  \cdot\left(  c,r\right) \\
&  =\left(  \left(  ab\right)  c,\underbrace{\left(  ab\right)  r}%
_{\substack{=a\left(  br\right)  \\\text{(since }M\text{ is a left}%
\\A\text{-module)}}}+\underbrace{\left(  aq+pb\right)  c}_{=\left(  aq\right)
c+\left(  pb\right)  c}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.RAM.product}), applied to
}\left(  ab,aq+pb\right)  \text{ and }\left(  c,r\right)  \text{ instead of
}\left(  a,p\right)  \text{ and }\left(  b,q\right)  \right) \\
&  =\left(  \left(  ab\right)  c,a\left(  br\right)  +\underbrace{\left(
aq\right)  c}_{\substack{=a\left(  qc\right)  \\\text{(by
(\ref{eq.def.bimod.1}), applied to }A\text{,}\\q\text{ and }c\text{ instead of
}B\text{, }m\text{ and }b\text{)}}}+\left(  pb\right)  c\right)  =\left(
\left(  ab\right)  c,a\left(  br\right)  +a\left(  qc\right)  +\left(
pb\right)  c\right)  ,
\end{align*}
we obtain $x\left(  yz\right)  =\left(  xy\right)  z$. This proves
(\ref{pf.thm.RAM.assoc}).}. In other words, the unital magmatic $\mathbf{k}%
$-algebra $\mathbf{R}$ is associative. Hence, $\mathbf{R}$ is an associative
unital magmatic $\mathbf{k}$-algebra. In other words, $\mathbf{R}$ is a
$\mathbf{k}$-algebra (because a $\mathbf{k}$-algebra is the same as an
associative unital magmatic $\mathbf{k}$-algebra). In other words, the
$\mathbf{k}$-module $A\oplus M$, equipped with the $\mathbf{k}$-bilinear map
$m$ and the unity $\left(  1,0\right)  $, is a $\mathbf{k}$-algebra (because
$\mathbf{R}$ is the $\mathbf{k}$-module $A\oplus M$, equipped with the
$\mathbf{k}$-bilinear map $m$ and the unity $\left(  1,0\right)  $). In other
words, the $\mathbf{k}$-module $A\oplus M$, equipped with the $\mathbf{k}%
$-bilinear map $m$, becomes a $\mathbf{k}$-algebra with unity $\left(
1,0\right)  $. This proves Theorem \ref{thm.RAM}.
\end{proof}
\end{verlong}

\begin{definition}
\label{def.RAM.projs}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be an $\left(
A,A\right)  $-bimodule. Recall that $\mathcal{R}_{A}\left(  M\right)  =A\oplus
M$ as a $\mathbf{k}$-module.

Let $\mathbf{projl}_{A,M}:A\oplus M\rightarrow A$ and $\mathbf{projr}%
_{A,M}:A\oplus M\rightarrow M$ be the canonical projections from the direct
sum $A\oplus M$ to its two addends. These two projections $\mathbf{projl}%
_{A,M}$ and $\mathbf{projr}_{A,M}$ are two surjective $\mathbf{k}$-linear
maps. Explicitly, they are given by%
\[
\left(  \mathbf{projl}_{A,M}\left(  \left(  a,m\right)  \right)
=a\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  a,m\right)  \in A\oplus
M\right)
\]
and%
\[
\left(  \mathbf{projr}_{A,M}\left(  \left(  a,m\right)  \right)
=m\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  a,m\right)  \in A\oplus
M\right)  .
\]


The map $\mathbf{projl}_{A,M}$ is a $\mathbf{k}$-linear map $A\oplus
M\rightarrow A$, therefore a $\mathbf{k}$-linear map $\mathcal{R}_{A}\left(
M\right)  \rightarrow A$ (since $\mathcal{R}_{A}\left(  M\right)  =A\oplus M$).

The map $\mathbf{projr}_{A,M}$ is a $\mathbf{k}$-linear map $A\oplus
M\rightarrow M$, therefore a $\mathbf{k}$-linear map $\mathcal{R}_{A}\left(
M\right)  \rightarrow M$ (since $\mathcal{R}_{A}\left(  M\right)  =A\oplus M$).

Let $\mathbf{inl}_{A,M}:A\rightarrow A\oplus M$ and $\mathbf{inr}%
_{A,M}:M\rightarrow A\oplus M$ be the canonical injections of the two
$\mathbf{k}$-modules $A$ and $M$ into their direct sum $A\oplus M$. These two
injections $\mathbf{inl}_{A,M}$ and $\mathbf{inr}_{A,M}$ are two injective
$\mathbf{k}$-linear maps. Explicitly, they are given by%
\[
\left(  \mathbf{inl}_{A,M}\left(  a\right)  =\left(  a,0\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\right)
\]
and%
\[
\left(  \mathbf{inr}_{A,M}\left(  m\right)  =\left(  0,m\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }m\in M\right)  .
\]


The map $\mathbf{inl}_{A,M}$ is a $\mathbf{k}$-linear map $A\rightarrow
A\oplus M$, therefore a $\mathbf{k}$-linear map $A\rightarrow\mathcal{R}%
_{A}\left(  M\right)  $ (since $\mathcal{R}_{A}\left(  M\right)  =A\oplus M$).

The map $\mathbf{inr}_{A,M}$ is a $\mathbf{k}$-linear map $M\rightarrow
A\oplus M$, therefore a $\mathbf{k}$-linear map $M\rightarrow\mathcal{R}%
_{A}\left(  M\right)  $ (since $\mathcal{R}_{A}\left(  M\right)  =A\oplus M$).
\end{definition}

The basic properties of direct sums of $\mathbf{k}$-modules (known from linear
algebra) now show:

\begin{proposition}
\label{prop.RAM.props.trivial}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be an
$\left(  A,A\right)  $-bimodule. Then,%
\begin{align}
\mathbf{projl}_{A,M}\circ\mathbf{inl}_{A,M}  &  =\operatorname*{id}%
\nolimits_{A};\label{eq.prop.RAM.props.trivial.1}\\
\mathbf{projl}_{A,M}\circ\mathbf{inr}_{A,M}  &
=0;\label{eq.prop.RAM.props.trivial.2}\\
\mathbf{projr}_{A,M}\circ\mathbf{inl}_{A,M}  &
=0;\label{eq.prop.RAM.props.trivial.3}\\
\mathbf{projr}_{A,M}\circ\mathbf{inr}_{A,M}  &  =\operatorname*{id}%
\nolimits_{M};\label{eq.prop.RAM.props.trivial.4}\\
\mathbf{inl}_{A,M}\circ\mathbf{projl}_{A,M}+\mathbf{inr}_{A,M}\circ
\mathbf{projr}_{A,M}  &  =\operatorname*{id}\nolimits_{\mathcal{R}_{A}\left(
M\right)  }. \label{eq.prop.RAM.props.trivial.5}%
\end{align}

\end{proposition}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.RAM.props.trivial}.]\textit{Proof of
(\ref{eq.prop.RAM.props.trivial.1}):} For every $a\in A$, we have%
\begin{align*}
\left(  \mathbf{projl}_{A,M}\circ\mathbf{inl}_{A,M}\right)  \left(  a\right)
&  =\mathbf{projl}_{A,M}\left(  \underbrace{\mathbf{inl}_{A,M}\left(
a\right)  }_{\substack{=\left(  a,0\right)  \\\text{(by the definition of
}\mathbf{inl}_{A,M}\text{)}}}\right)  =\mathbf{projl}_{A,M}\left(  \left(
a,0\right)  \right) \\
&  =a\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{projl}%
_{A,M}\right) \\
&  =\operatorname*{id}\nolimits_{A}\left(  a\right)  .
\end{align*}
In other words, we have $\mathbf{projl}_{A,M}\circ\mathbf{inl}_{A,M}%
=\operatorname*{id}\nolimits_{A}$. This proves
(\ref{eq.prop.RAM.props.trivial.1}).

\textit{Proof of (\ref{eq.prop.RAM.props.trivial.2}):} For every $m\in M$, we
have%
\begin{align*}
\left(  \mathbf{projl}_{A,M}\circ\mathbf{inr}_{A,M}\right)  \left(  M\right)
&  =\mathbf{projl}_{A,M}\left(  \underbrace{\mathbf{inr}_{A,M}\left(
m\right)  }_{\substack{=\left(  0,m\right)  \\\text{(by the definition of
}\mathbf{inr}_{A,M}\text{)}}}\right)  =\mathbf{projl}_{A,M}\left(  \left(
0,m\right)  \right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{projl}%
_{A,M}\right) \\
&  =0\left(  m\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }0\left(
m\right)  =0\right)  .
\end{align*}
In other words, we have $\mathbf{projl}_{A,M}\circ\mathbf{inr}_{A,M}=0$. This
proves (\ref{eq.prop.RAM.props.trivial.2}).

\textit{Proof of (\ref{eq.prop.RAM.props.trivial.3}):} For every $a\in A$, we
have%
\begin{align*}
\left(  \mathbf{projr}_{A,M}\circ\mathbf{inl}_{A,M}\right)  \left(  a\right)
&  =\mathbf{projr}_{A,M}\left(  \underbrace{\mathbf{inl}_{A,M}\left(
a\right)  }_{\substack{=\left(  a,0\right)  \\\text{(by the definition of
}\mathbf{inl}_{A,M}\text{)}}}\right)  =\mathbf{projr}_{A,M}\left(  \left(
a,0\right)  \right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{projr}%
_{A,M}\right) \\
&  =0\left(  a\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }0\left(
a\right)  =0\right)  .
\end{align*}
In other words, we have $\mathbf{projr}_{A,M}\circ\mathbf{inl}_{A,M}=0$. This
proves (\ref{eq.prop.RAM.props.trivial.3}).

\textit{Proof of (\ref{eq.prop.RAM.props.trivial.4}):} For every $m\in M$, we
have%
\begin{align*}
\left(  \mathbf{projr}_{A,M}\circ\mathbf{inr}_{A,M}\right)  \left(  m\right)
&  =\mathbf{projr}_{A,M}\left(  \underbrace{\mathbf{inr}_{A,M}\left(
m\right)  }_{\substack{=\left(  0,m\right)  \\\text{(by the definition of
}\mathbf{inl}_{A,M}\text{)}}}\right)  =\mathbf{projr}_{A,M}\left(  \left(
0,m\right)  \right) \\
&  =m\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{projr}%
_{A,M}\right) \\
&  =\operatorname*{id}\nolimits_{M}\left(  m\right)  .
\end{align*}
In other words, we have $\mathbf{projr}_{A,M}\circ\mathbf{inr}_{A,M}%
=\operatorname*{id}\nolimits_{M}$. This proves
(\ref{eq.prop.RAM.props.trivial.4}).

\textit{Proof of (\ref{eq.prop.RAM.props.trivial.5}):} Let $x\in
\mathcal{R}_{A}\left(  M\right)  $. Then, $x\in\mathcal{R}_{A}\left(
M\right)  =A\oplus M$. Thus, we can write $x$ in the form $x=\left(
a,p\right)  $ for some $a\in A$ and $p\in M$. Consider these $a$ and $p$. Now,%
\begin{align*}
&  \left(  \mathbf{inl}_{A,M}\circ\mathbf{projl}_{A,M}+\mathbf{inr}_{A,M}%
\circ\mathbf{projr}_{A,M}\right)  \left(  \underbrace{x}_{=\left(  a,p\right)
}\right) \\
&  =\left(  \mathbf{inl}_{A,M}\circ\mathbf{projl}_{A,M}+\mathbf{inr}%
_{A,M}\circ\mathbf{projr}_{A,M}\right)  \left(  \left(  a,p\right)  \right) \\
&  =\underbrace{\left(  \mathbf{inl}_{A,M}\circ\mathbf{projl}_{A,M}\right)
\left(  \left(  a,p\right)  \right)  }_{=\mathbf{inl}_{A,M}\left(
\mathbf{projl}_{A,M}\left(  \left(  a,p\right)  \right)  \right)
}+\underbrace{\left(  \mathbf{inr}_{A,M}\circ\mathbf{projr}_{A,M}\right)
\left(  \left(  a,p\right)  \right)  }_{=\mathbf{inr}_{A,M}\left(
\mathbf{projr}_{A,M}\left(  \left(  a,p\right)  \right)  \right)  }\\
&  =\mathbf{inl}_{A,M}\left(  \underbrace{\mathbf{projl}_{A,M}\left(  \left(
a,p\right)  \right)  }_{\substack{=a\\\text{(by the definition of
}\mathbf{projr}_{A,M}\text{)}}}\right)  +\mathbf{inr}_{A,M}\left(
\underbrace{\mathbf{projr}_{A,M}\left(  \left(  a,p\right)  \right)
}_{\substack{=p\\\text{(by the definition of }\mathbf{projr}_{A,M}\text{)}%
}}\right) \\
&  =\underbrace{\mathbf{inl}_{A,M}\left(  a\right)  }_{\substack{=\left(
a,0\right)  \\\text{(by the definition of }\mathbf{inl}_{A,M}\text{)}%
}}+\underbrace{\mathbf{inr}_{A,M}\left(  p\right)  }_{\substack{=\left(
0,p\right)  \\\text{(by the definition of }\mathbf{inr}_{A,M}\text{)}%
}}=\left(  a,0\right)  +\left(  0,p\right) \\
&  =\left(  \underbrace{a+0}_{=a},\underbrace{0+p}_{=p}\right)  =\left(
a,p\right)  =x=\operatorname*{id}\nolimits_{\mathcal{R}_{A}\left(  M\right)
}\left(  x\right)  .
\end{align*}


Now, let us forget that we fixed $x$. We thus have proven that
\[
\left(  \mathbf{inl}_{A,M}\circ\mathbf{projl}_{A,M}+\mathbf{inr}_{A,M}%
\circ\mathbf{projr}_{A,M}\right)  \left(  x\right)  =\operatorname*{id}%
\nolimits_{\mathcal{R}_{A}\left(  M\right)  }\left(  x\right)
\]
for every $x\in\mathcal{R}_{A}\left(  M\right)  $. In other words, we have
$\mathbf{inl}_{A,M}\circ\mathbf{projl}_{A,M}+\mathbf{inr}_{A,M}\circ
\mathbf{projr}_{A,M}=\operatorname*{id}\nolimits_{\mathcal{R}_{A}\left(
M\right)  }$. This proves (\ref{eq.prop.RAM.props.trivial.5}).

The proof of Proposition \ref{prop.RAM.props.trivial} is thus complete.
\end{proof}
\end{verlong}

\begin{proposition}
\label{prop.RAM.props.alg}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be an
$\left(  A,A\right)  $-bimodule.

\textbf{(a)} The map $\mathbf{projl}_{A,M}:\mathcal{R}_{A}\left(  M\right)
\rightarrow A$ is a $\mathbf{k}$-algebra homomorphism from $\mathcal{R}%
_{A}\left(  M\right)  $ to $A$.

\textbf{(b)} The map $\mathbf{inl}_{A,M}:A\rightarrow\mathcal{R}_{A}\left(
M\right)  $ is a $\mathbf{k}$-algebra homomorphism from $A$ to $\mathcal{R}%
_{A}\left(  M\right)  $.
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.RAM.props.alg}.]This is really straightforward.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.RAM.props.alg}.]Recall that the unity of the
$\mathbf{k}$-algebra $\mathcal{R}_{A}\left(  M\right)  $ is $\left(
1,0\right)  $. In other words, $1_{\mathcal{R}_{A}\left(  M\right)  }=\left(
1,0\right)  $.

\textbf{(a)} We have $\mathbf{projl}_{A,M}\left(  \underbrace{1_{\mathcal{R}%
_{A}\left(  M\right)  }}_{=\left(  1,0\right)  }\right)  =\mathbf{projl}%
_{A,M}\left(  \left(  1,0\right)  \right)  =1$ (by the definition of
$\mathbf{projl}_{A,M}$).

Every two elements $x\in\mathcal{R}_{A}\left(  M\right)  $ and $y\in
\mathcal{R}_{A}\left(  M\right)  $ satisfy%
\[
\mathbf{projl}_{A,M}\left(  xy\right)  =\mathbf{projl}_{A,M}\left(  x\right)
\cdot\mathbf{projl}_{A,M}\left(  y\right)
\]
\footnote{\textit{Proof.} Let $x\in\mathcal{R}_{A}\left(  M\right)  $ and
$y\in\mathcal{R}_{A}\left(  M\right)  $.
\par
We have $x\in\mathcal{R}_{A}\left(  M\right)  = A\oplus M$. Thus, we can write
$x$ in the form $x=\left(  a,p\right)  $ for some $a\in A$ and $p\in M$.
Consider these $a$ and $p$.
\par
We have $y\in\mathcal{R}_{A}\left(  M\right)  = A\oplus M$. Thus, we can write
$y$ in the form $y=\left(  b,q\right)  $ for some $b\in A$ and $q\in M$.
Consider these $b$ and $q$.
\par
Now, $\mathbf{projl}_{A,M}\left(  \underbrace{x}_{=\left(  a,p\right)
}\right)  =\mathbf{projl}_{A,M}\left(  \left(  a,p\right)  \right)  =a$ (by
the definition of $\mathbf{projl}_{A,M}$). Also, $\mathbf{projl}_{A,M}\left(
\underbrace{y}_{=\left(  b,q\right)  }\right)  =\mathbf{projl}_{A,M}\left(
\left(  b,q\right)  \right)  =b$ (by the definition of $\mathbf{projl}_{A,M}%
$). Furthermore,
\[
xy=\underbrace{x}_{=\left(  a,p\right)  }\cdot\underbrace{y}_{=\left(
b,q\right)  }=\left(  a,p\right)  \cdot\left(  b,q\right)  =\left(
ab,aq+pb\right)
\]
(by (\ref{eq.def.RAM.product})). Applying the map $\mathbf{projl}_{A,M}$ to
both sides of this equality, we obtain%
\[
\mathbf{projl}_{A,M}\left(  xy\right)  =\mathbf{projl}_{A,M}\left(  \left(
ab,aq+pb\right)  \right)  =ab\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\mathbf{projl}_{A,M}\right)  .
\]
Comparing this with%
\[
\underbrace{\mathbf{projl}_{A,M}\left(  x\right)  }_{=a}\cdot
\underbrace{\mathbf{projl}_{A,M}\left(  y\right)  }_{=b}=ab,
\]
we obtain $\mathbf{projl}_{A,M}\left(  xy\right)  =\mathbf{projl}_{A,M}\left(
x\right)  \cdot\mathbf{projl}_{A,M}\left(  y\right)  $, qed.}. This (combined
with the fact that the map $\mathbf{projl}_{A,M}$ is $\mathbf{k}$-linear, and
with the fact that $\mathbf{projl}_{A,M}\left(  1_{\mathcal{R}_{A}\left(
M\right)  }\right)  =1$) shows that $\mathbf{projl}_{A,M}$ is a $\mathbf{k}%
$-algebra homomorphism from $\mathcal{R}_{A}\left(  M\right)  $ to $A$. This
proves Proposition \ref{prop.RAM.props.alg} \textbf{(a)}.

\textbf{(b)} The definition of $\mathbf{inl}_{A,M}$ shows that $\mathbf{inl}%
_{A,M}\left(  1\right)  =\left(  1,0\right)  =1_{\mathcal{R}_{A}\left(
M\right)  }$.

Every $a\in A$ and $b\in A$ satisfy%
\[
\mathbf{inl}_{A,M}\left(  ab\right)  =\mathbf{inl}_{A,M}\left(  a\right)
\cdot\mathbf{inl}_{A,M}\left(  b\right)
\]
\footnote{\textit{Proof.} Let $a\in A$ and $b\in A$. Then, the definition of
$\mathbf{inl}_{A,M}$ shows that $\mathbf{inl}_{A,M}\left(  ab\right)  =\left(
ab,0\right)  $. On the other hand, the definition of $\mathbf{inl}_{A,M}$
shows that $\mathbf{inl}_{A,M}\left(  a\right)  =\left(  a,0\right)  $. Also,
the definition of $\mathbf{inl}_{A,M}$ shows that $\mathbf{inl}_{A,M}\left(
b\right)  =\left(  b,0\right)  $. Now,%
\begin{align*}
\underbrace{\mathbf{inl}_{A,M}\left(  a\right)  }_{=\left(  a,0\right)  }%
\cdot\underbrace{\mathbf{inl}_{A,M}\left(  b\right)  }_{=\left(  b,0\right)
}  &  =\left(  a,0\right)  \cdot\left(  b,0\right)  =\left(
ab,\underbrace{a\cdot0}_{=0}+\underbrace{0\cdot b}_{=0}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{eq.def.RAM.product}), applied to }\left(  a,0\right)  \text{
and }\left(  b,0\right) \\
\text{instead of }\left(  a,p\right)  \text{ and }\left(  b,q\right)
\end{array}
\right) \\
&  =\left(  ab,\underbrace{0+0}_{=0}\right)  =\left(  ab,0\right)  .
\end{align*}
Comparing this with $\mathbf{inl}_{A,M}\left(  ab\right)  =\left(
ab,0\right)  $, we obtain $\mathbf{inl}_{A,M}\left(  ab\right)  =\mathbf{inl}%
_{A,M}\left(  a\right)  \cdot\mathbf{inl}_{A,M}\left(  b\right)  $. Qed.}.
This (combined with the fact that the map $\mathbf{inl}_{A,M}$ is $\mathbf{k}%
$-linear, and with the fact that $\mathbf{inl}_{A,M}\left(  1\right)
=1_{\mathcal{R}_{A}\left(  M\right)  }$) shows that $\mathbf{inl}_{A,M}$ is a
$\mathbf{k}$-algebra homomorphism from $A$ to $\mathcal{R}_{A}\left(
M\right)  $. This proves Proposition \ref{prop.RAM.props.alg} \textbf{(b)}.
\end{proof}
\end{verlong}

\begin{remark}
\label{rmk.RAM.isomorphs}Let $A$ be a $\mathbf{k}$-algebra, and let $M$ be an
$\left(  A,A\right)  $-bimodule. An alternative realization of $\mathcal{R}%
_{A}\left(  M\right)  $ (i.e., a $\mathbf{k}$-algebra that is canonically
isomorphic to $\mathcal{R}_{A}\left(  M\right)  $) can be constructed as follows:

Let $\left(
\begin{array}
[c]{cc}%
A & M\\
0 & A
\end{array}
\right)  $ denote the $\mathbf{k}$-module of all $2\times2$-matrices of the
form $\left(
\begin{array}
[c]{cc}%
a & m\\
0 & b
\end{array}
\right)  $ with $a\in A$, $m\in M$ and $b\in A$. We can make this $\mathbf{k}%
$-module $\left(
\begin{array}
[c]{cc}%
A & M\\
0 & A
\end{array}
\right)  $ into a ring by defining the product of two matrices in the usual
way:%
\[
\left(
\begin{array}
[c]{cc}%
a & m\\
0 & b
\end{array}
\right)  \left(
\begin{array}
[c]{cc}%
a^{\prime} & m^{\prime}\\
0 & b^{\prime}%
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
aa^{\prime} & am^{\prime}+mb^{\prime}\\
0 & bb^{\prime}%
\end{array}
\right)  .
\]
The set of all $\left(
\begin{array}
[c]{cc}%
a & m\\
0 & b
\end{array}
\right)  \in\left(
\begin{array}
[c]{cc}%
A & M\\
0 & A
\end{array}
\right)  $ with $a=b$ is easily seen to be a $\mathbf{k}$-subalgebra of this
$\mathbf{k}$-algebra. This $\mathbf{k}$-subalgebra is canonically isomorphic
to $\mathcal{R}_{A}\left(  M\right)  $; the isomorphism sends the element
$\left(
\begin{array}
[c]{cc}%
a & m\\
0 & a
\end{array}
\right)  $ of this $\mathbf{k}$-subalgebra to the element $\left(  a,m\right)
$ of $\mathcal{R}_{A}\left(  M\right)  $.

This realization of $\mathcal{R}_{A}\left(  M\right)  $ is particularly suited
for generalization; but we shall not go further in this direction.
\end{remark}

\begin{definition}
\label{def.RAM.Homid}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be an $\left(
A,A\right)  $-bimodule. We let $\operatorname*{Hom}\nolimits_{\mathcal{R}%
}\left(  A,M\right)  $ denote the subset%
\[
\left\{  \varphi\in\operatorname*{Hom}\left(  A,\mathcal{R}_{A}\left(
M\right)  \right)  \ \mid\ \mathbf{projl}_{A,M}\circ\varphi=\operatorname*{id}%
\nolimits_{A}\right\}
\]
of $\operatorname*{Hom}\left(  A,\mathcal{R}_{A}\left(  M\right)  \right)  $.

(\textbf{Warning:} This subset $\operatorname*{Hom}\nolimits_{\mathcal{R}%
}\left(  A,M\right)  $ is not a $\mathbf{k}$-submodule.)
\end{definition}

\begin{proposition}
\label{prop.RAM.hombij1}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be an
$\left(  A,A\right)  $-bimodule.

\textbf{(a)} For every $f\in\operatorname*{Hom}\left(  A,M\right)  $, we have
$\mathbf{inl}_{A,M}+\mathbf{inr}_{A,M}\circ f\in\operatorname*{Hom}%
\nolimits_{\mathcal{R}}\left(  A,M\right)  $.

\textbf{(b)} For every $F\in\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(
A,M\right)  $, we have $\mathbf{projr}_{A,M}\circ F\in\operatorname*{Hom}%
\left(  A,M\right)  $.
\end{proposition}

We delay the (simple) proof of this proposition until later, as we would like
to first explain what its purpose is.

\begin{definition}
\label{def.RAM.hombij}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be an
$\left(  A,A\right)  $-bimodule.

We define a map $\operatorname*{dth}\nolimits_{A,M}:\operatorname*{Hom}\left(
A,M\right)  \rightarrow\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(
A,M\right)  $ by%
\[
\left(  \operatorname*{dth}\nolimits_{A,M}\left(  f\right)  =\mathbf{inl}%
_{A,M}+\mathbf{inr}_{A,M}\circ f\ \ \ \ \ \ \ \ \ \ \text{for every }%
f\in\operatorname*{Hom}\left(  A,M\right)  \right)  .
\]
This map is well-defined, because for every $f\in\operatorname*{Hom}\left(
A,M\right)  $, we have $\mathbf{inl}_{A,M}+\mathbf{inr}_{A,M}\circ
f\in\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  A,M\right)  $ (according
to Proposition \ref{prop.RAM.hombij1} \textbf{(a)}).

We define a map $\operatorname*{htd}\nolimits_{A,M}:\operatorname*{Hom}%
\nolimits_{\mathcal{R}}\left(  A,M\right)  \rightarrow\operatorname*{Hom}%
\left(  A,M\right)  $ by%
\[
\left(  \operatorname*{htd}\nolimits_{A,M}\left(  F\right)  =\mathbf{projr}%
_{A,M}\circ F\ \ \ \ \ \ \ \ \ \ \text{for every }F\in\operatorname*{Hom}%
\nolimits_{\mathcal{R}}\left(  A,M\right)  \right)  .
\]
This map is well-defined, because for every $F\in\operatorname*{Hom}%
\nolimits_{\mathcal{R}}\left(  A,M\right)  $, we have $\mathbf{projr}%
_{A,M}\circ F\in\operatorname*{Hom}\left(  A,M\right)  $ (according to
Proposition \ref{prop.RAM.hombij1} \textbf{(b)}).
\end{definition}

\begin{theorem}
\label{thm.RAM.dth-htd}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be an
$\left(  A,A\right)  $-bimodule.

\textbf{(a)} If $f\in\operatorname*{Hom}\left(  A,M\right)  $ and $a\in A$,
then $\left(  \operatorname*{dth}\nolimits_{A,M}\left(  f\right)  \right)
\left(  a\right)  =\left(  a,f\left(  a\right)  \right)  $.

\textbf{(b)} The maps $\operatorname*{dth}\nolimits_{A,M}$ and
$\operatorname*{htd}\nolimits_{A,M}$ are mutually inverse.

\textbf{(c)} Let $f\in\operatorname*{Hom}\left(  A,M\right)  $. Then, $f$ is a
derivation if and only if $\operatorname*{dth}\nolimits_{A,M}\left(  f\right)
$ is a $\mathbf{k}$-algebra homomorphism from $A$ to $\mathcal{R}_{A}\left(
M\right)  $.
\end{theorem}

Theorem \ref{thm.RAM.dth-htd} shows that the derivations from a $\mathbf{k}%
$-algebra $A$ to an $\left(  A,A\right)  $-bimodule $M$ are in 1-to-1
correspondence with $\mathbf{k}$-algebra homomorphisms from $A$ to
$\mathcal{R}_{A}\left(  M\right)  $ that lie in $\operatorname*{Hom}%
\nolimits_{\mathcal{R}}\left(  A,M\right)  $. (This correspondence is given by
the maps $\operatorname*{dth}\nolimits_{A,M}$ and $\operatorname*{htd}%
\nolimits_{A,M}$; it also explains the names that we gave these
maps\footnote{Namely, $\operatorname*{dth}\nolimits_{A,M}$ is short for
\textquotedblleft derivation to homomorphism\textquotedblright, and
$\operatorname*{htd}\nolimits_{A,M}$ for \textquotedblleft homomorphism to
derivation\textquotedblright. Of course, the map $\operatorname*{dth}%
\nolimits_{A,M}$ is defined not only on derivations, and $\operatorname*{htd}%
\nolimits_{A,M}$ not only on homomorphisms, but this should motivate the names
well enough.}.) This allows deriving properties of derivations from properties
of $\mathbf{k}$-algebra homomorphisms. For instance, we could have used this
tactic to derive Proposition \ref{prop.derivation.unique} from the analogous
property of $\mathbf{k}$-algebra homomorphisms.

As promised, let us now prove Proposition \ref{prop.RAM.hombij1} and Theorem
\ref{thm.RAM.dth-htd}. These proofs are again rather straightforward.

\begin{proof}
[Proof of Proposition \ref{prop.RAM.hombij1}.]\textbf{(a)} Let $f\in
\operatorname*{Hom}\left(  A,M\right)  $. We must prove that $\mathbf{inl}%
_{A,M}+\mathbf{inr}_{A,M}\circ f\in\operatorname*{Hom}\nolimits_{\mathcal{R}%
}\left(  A,M\right)  $.

\begin{vershort}
We have%
\begin{align*}
&  \mathbf{projl}_{A,M}\circ\left(  \mathbf{inl}_{A,M}+\mathbf{inr}_{A,M}\circ
f\right) \\
&  =\underbrace{\mathbf{projl}_{A,M}\circ\mathbf{inl}_{A,M}}%
_{\substack{=\operatorname*{id}\nolimits_{A}\\\text{(by
(\ref{eq.prop.RAM.props.trivial.1}))}}}+\underbrace{\mathbf{projl}_{A,M}%
\circ\mathbf{inr}_{A,M}}_{\substack{=0\\\text{(by
(\ref{eq.prop.RAM.props.trivial.2}))}}}\circ f=\operatorname*{id}%
\nolimits_{A}+\underbrace{0\circ f}_{=0}=\operatorname*{id}\nolimits_{A}.
\end{align*}
Thus, $\mathbf{inl}_{A,M}+\mathbf{inr}_{A,M}\circ f$ is an element of
$\operatorname*{Hom}\left(  A,\mathcal{R}_{A}\left(  M\right)  \right)  $
(because $\mathbf{inl}_{A,M}+\mathbf{inr}_{A,M}\circ f$ is $\mathbf{k}%
$-linear) and satisfies $\mathbf{projl}_{A,M}\circ\left(  \mathbf{inl}%
_{A,M}+\mathbf{inr}_{A,M}\circ f\right)  =\operatorname*{id}\nolimits_{A}$. In
other words, $\mathbf{inl}_{A,M}+\mathbf{inr}_{A,M}\circ f$ is an element
$\varphi$ of $\operatorname*{Hom}\left(  A,\mathcal{R}_{A}\left(  M\right)
\right)  $ satisfying $\mathbf{projl}_{A,M}\circ\varphi=\operatorname*{id}%
\nolimits_{A}$. In other words,
\[
\mathbf{inl}_{A,M}+\mathbf{inr}_{A,M}\circ f\in\left\{  \varphi\in
\operatorname*{Hom}\left(  A,\mathcal{R}_{A}\left(  M\right)  \right)
\ \mid\ \mathbf{projl}_{A,M}\circ\varphi=\operatorname*{id}\nolimits_{A}%
\right\}  =\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  A,M\right)
\]
(since $\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  A,M\right)  $ was
defined as $\left\{  \varphi\in\operatorname*{Hom}\left(  A,\mathcal{R}%
_{A}\left(  M\right)  \right)  \ \mid\ \mathbf{projl}_{A,M}\circ
\varphi=\operatorname*{id}\nolimits_{A}\right\}  $). This proves Proposition
\ref{prop.RAM.hombij1} \textbf{(a)}.
\end{vershort}

\begin{verlong}
The map $f$ belongs to $\operatorname*{Hom}\left(  A,M\right)  $. In other
words, $f$ is a $\mathbf{k}$-linear map from $A$ to $M$. The map
$\mathbf{inr}_{A,M}\circ f$ is $\mathbf{k}$-linear (since it is the
composition of the $\mathbf{k}$-linear maps $\mathbf{inr}_{A,M}$ and $f$). The
map $\mathbf{inl}_{A,M}+\mathbf{inr}_{A,M}\circ f$ is $\mathbf{k}$-linear
(since it is the sum of the $\mathbf{k}$-linear maps $\mathbf{inl}_{A,M}$ and
$\mathbf{inr}_{A,M}\circ f$). In other words, $\mathbf{inl}_{A,M}%
+\mathbf{inr}_{A,M}\circ f\in\operatorname*{Hom}\left(  A,\mathcal{R}%
_{A}\left(  M\right)  \right)  $.

Now, composition of $\mathbf{k}$-linear maps is $\mathbf{k}$-bilinear. Since
$\mathbf{projl}_{A,M}$ and $\mathbf{inl}_{A,M}+\mathbf{inr}_{A,M}\circ f$ are
$\mathbf{k}$-linear maps, we thus have
\begin{align*}
&  \mathbf{projl}_{A,M}\circ\left(  \mathbf{inl}_{A,M}+\mathbf{inr}_{A,M}\circ
f\right) \\
&  =\underbrace{\mathbf{projl}_{A,M}\circ\mathbf{inl}_{A,M}}%
_{\substack{=\operatorname*{id}\nolimits_{A}\\\text{(by
(\ref{eq.prop.RAM.props.trivial.1}))}}}+\underbrace{\mathbf{projl}_{A,M}%
\circ\mathbf{inr}_{A,M}}_{\substack{=0\\\text{(by
(\ref{eq.prop.RAM.props.trivial.2}))}}}\circ f=\operatorname*{id}%
\nolimits_{A}+\underbrace{0\circ f}_{=0}=\operatorname*{id}\nolimits_{A}.
\end{align*}
Thus, $\mathbf{inl}_{A,M}+\mathbf{inr}_{A,M}\circ f$ is an element of
$\operatorname*{Hom}\left(  A,\mathcal{R}_{A}\left(  M\right)  \right)  $ and
satisfies $\mathbf{projl}_{A,M}\circ\left(  \mathbf{inl}_{A,M}+\mathbf{inr}%
_{A,M}\circ f\right)  =\operatorname*{id}\nolimits_{A}$. In other words,
$\mathbf{inl}_{A,M}+\mathbf{inr}_{A,M}\circ f$ is an element $\varphi$ of
$\operatorname*{Hom}\left(  A,\mathcal{R}_{A}\left(  M\right)  \right)  $
satisfying $\mathbf{projl}_{A,M}\circ\varphi=\operatorname*{id}\nolimits_{A}$.
In other words,
\[
\mathbf{inl}_{A,M}+\mathbf{inr}_{A,M}\circ f\in\left\{  \varphi\in
\operatorname*{Hom}\left(  A,\mathcal{R}_{A}\left(  M\right)  \right)
\ \mid\ \mathbf{projl}_{A,M}\circ\varphi=\operatorname*{id}\nolimits_{A}%
\right\}  =\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  A,M\right)
\]
(since $\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  A,M\right)
=\left\{  \varphi\in\operatorname*{Hom}\left(  A,\mathcal{R}_{A}\left(
M\right)  \right)  \ \mid\ \mathbf{projl}_{A,M}\circ\varphi=\operatorname*{id}%
\nolimits_{A}\right\}  $ (by the definition of $\operatorname*{Hom}%
\nolimits_{\mathcal{R}}\left(  A,M\right)  $)). This proves Proposition
\ref{prop.RAM.hombij1} \textbf{(a)}.
\end{verlong}

\textbf{(b)} Let $F\in\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(
A,M\right)  $. We must prove that $\mathbf{projr}_{A,M}\circ F\in
\operatorname*{Hom}\left(  A,M\right)  $.

\begin{vershort}
But this follows immediately from $F\in\operatorname*{Hom}%
\nolimits_{\mathcal{R}}\left(  A,M\right)  \subseteq\operatorname*{Hom}\left(
A,\mathcal{R}_{A}\left(  M\right)  \right)  $. Thus, Proposition
\ref{prop.RAM.hombij1} \textbf{(b)} is proven.
\end{vershort}

\begin{verlong}
We have%
\[
F\in\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  A,M\right)  =\left\{
\varphi\in\operatorname*{Hom}\left(  A,\mathcal{R}_{A}\left(  M\right)
\right)  \ \mid\ \mathbf{projl}_{A,M}\circ\varphi=\operatorname*{id}%
\nolimits_{A}\right\}
\]
(by the definition of $\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(
A,M\right)  $). In other words, $F$ is an element $\varphi$ of
$\operatorname*{Hom}\left(  A,\mathcal{R}_{A}\left(  M\right)  \right)  $
satisfying $\mathbf{projl}_{A,M}\circ\varphi=\operatorname*{id}\nolimits_{A}$.
In other words, $F$ is an element of $\operatorname*{Hom}\left(
A,\mathcal{R}_{A}\left(  M\right)  \right)  $ and satisfies $\mathbf{projl}%
_{A,M}\circ F=\operatorname*{id}\nolimits_{A}$.

The map $F$ is $\mathbf{k}$-linear (since $F\in\operatorname*{Hom}\left(
A,\mathcal{R}_{A}\left(  M\right)  \right)  $). Hence, the map $\mathbf{projr}%
_{A,M}\circ F$ is $\mathbf{k}$-linear (since it is the composition of the
$\mathbf{k}$-linear maps $\mathbf{projr}_{A,M}$ and $F$). In other words,
$\mathbf{projr}_{A,M}\circ F\in\operatorname*{Hom}\left(  A,M\right)  $. This
proves Proposition \ref{prop.RAM.hombij1} \textbf{(b)}.
\end{verlong}
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.RAM.dth-htd}.]\textbf{(a)} Let $f\in
\operatorname*{Hom}\left(  A,M\right)  $ and $a\in A$. We must show that
$\left(  \operatorname*{dth}\nolimits_{A,M}\left(  f\right)  \right)  \left(
a\right)  =\left(  a,f\left(  a\right)  \right)  $.

The definition of $\operatorname*{dth}\nolimits_{A,M}$ shows that
$\operatorname*{dth}\nolimits_{A,M}\left(  f\right)  =\mathbf{inl}%
_{A,M}+\mathbf{inr}_{A,M}\circ f$. Hence,%
\begin{align*}
\underbrace{\left(  \operatorname*{dth}\nolimits_{A,M}\left(  f\right)
\right)  }_{=\mathbf{inl}_{A,M}+\mathbf{inr}_{A,M}\circ f}\left(  a\right)
&  =\left(  \mathbf{inl}_{A,M}+\mathbf{inr}_{A,M}\circ f\right)  \left(
a\right) \\
&  =\underbrace{\mathbf{inl}_{A,M}\left(  a\right)  }_{\substack{=\left(
a,0\right)  \\\text{(by the definition of }\mathbf{inl}_{A,M}\text{)}%
}}+\underbrace{\left(  \mathbf{inr}_{A,M}\circ f\right)  \left(  a\right)
}_{\substack{=\mathbf{inr}_{A,M}\left(  f\left(  a\right)  \right)  =\left(
0,f\left(  a\right)  \right)  \\\text{(by the definition of }\mathbf{inr}%
_{A,M}\text{)}}}\\
&  =\left(  a,0\right)  +\left(  0,f\left(  a\right)  \right)  =\left(
\underbrace{a+0}_{=a},\underbrace{0+f\left(  a\right)  }_{=f\left(  a\right)
}\right)  =\left(  a,f\left(  a\right)  \right)  .
\end{align*}
This proves Theorem \ref{thm.RAM.dth-htd} \textbf{(a)}.

\begin{vershort}
\textbf{(b)} Every $F\in\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(
A,M\right)  $ satisfies $\left(  \operatorname*{dth}\nolimits_{A,M}%
\circ\operatorname*{htd}\nolimits_{A,M}\right)  \left(  F\right)
=\operatorname*{id}\nolimits_{\operatorname*{Hom}\nolimits_{\mathcal{R}%
}\left(  A,M\right)  }\left(  F\right)  $\ \ \ \ \footnote{\textit{Proof.} Let
$F\in\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  A,M\right)  $. We must
prove that $\left(  \operatorname*{dth}\nolimits_{A,M}\circ\operatorname*{htd}%
\nolimits_{A,M}\right)  \left(  F\right)  =\operatorname*{id}%
\nolimits_{A,\mathcal{R}_{A}\left(  M\right)  }\left(  F\right)  $.
\par
We have
\[
F\in\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  A,M\right)  =\left\{
\varphi\in\operatorname*{Hom}\left(  A,\mathcal{R}_{A}\left(  M\right)
\right)  \ \mid\ \mathbf{projl}_{A,M}\circ\varphi=\operatorname*{id}%
\nolimits_{A}\right\}
\]
(by the definition of $\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(
A,M\right)  $). In other words, $F$ is an element of $\operatorname*{Hom}%
\left(  A,\mathcal{R}_{A}\left(  M\right)  \right)  $ and satisfies
$\mathbf{projl}_{A,M}\circ F=\operatorname*{id}\nolimits_{A}$.
\par
On the other hand, $\operatorname*{htd}\nolimits_{A,M}\left(  F\right)
=\mathbf{projr}_{A,M}\circ F$ (by the definition of $\operatorname*{htd}%
\nolimits_{A,M}$). Now,%
\begin{align*}
\left(  \operatorname*{dth}\nolimits_{A,M}\circ\operatorname*{htd}%
\nolimits_{A,M}\right)  \left(  F\right)   &  =\operatorname*{dth}%
\nolimits_{A,M}\left(  \underbrace{\operatorname*{htd}\nolimits_{A,M}\left(
F\right)  }_{=\mathbf{projr}_{A,M}\circ F}\right)  =\operatorname*{dth}%
\nolimits_{A,M}\left(  \mathbf{projr}_{A,M}\circ F\right) \\
&  =\underbrace{\mathbf{inl}_{A,M}}_{=\mathbf{inl}_{A,M}\circ
\operatorname*{id}\nolimits_{A}}+\underbrace{\mathbf{inr}_{A,M}\circ\left(
\mathbf{projr}_{A,M}\circ F\right)  }_{=\mathbf{inr}_{A,M}\circ\mathbf{projr}%
_{A,M}\circ F}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\operatorname*{dth}\nolimits_{A,M}\right) \\
&  =\mathbf{inl}_{A,M}\circ\underbrace{\operatorname*{id}\nolimits_{A}%
}_{=\mathbf{projl}_{A,M}\circ F}+\mathbf{inr}_{A,M}\circ\mathbf{projr}%
_{A,M}\circ F\\
&  =\mathbf{inl}_{A,M}\circ\mathbf{projl}_{A,M}\circ F+\mathbf{inr}_{A,M}%
\circ\mathbf{projr}_{A,M}\circ F\\
&  =\underbrace{\left(  \mathbf{inl}_{A,M}\circ\mathbf{projl}_{A,M}%
+\mathbf{inr}_{A,M}\circ\mathbf{projr}_{A,M}\right)  }%
_{\substack{=\operatorname*{id}\nolimits_{\mathcal{R}_{A}\left(  M\right)
}\\\text{(by (\ref{eq.prop.RAM.props.trivial.5}))}}}\circ F\\
&  =\operatorname*{id}\nolimits_{\mathcal{R}_{A}\left(  M\right)  }\circ
F=F=\operatorname*{id}\nolimits_{\operatorname*{Hom}\nolimits_{\mathcal{R}%
}\left(  A,M\right)  }\left(  F\right)  ,
\end{align*}
qed.}. In other words, we have $\operatorname*{dth}\nolimits_{A,M}%
\circ\operatorname*{htd}\nolimits_{A,M}=\operatorname*{id}%
\nolimits_{\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  A,M\right)  }$.

On the other hand, every $f\in\operatorname*{Hom}\left(  A,M\right)  $
satisfies $\left(  \operatorname*{htd}\nolimits_{A,M}\circ\operatorname*{dth}%
\nolimits_{A,M}\right)  \left(  f\right)  =\operatorname*{id}%
\nolimits_{\operatorname*{Hom}\left(  A,M\right)  }\left(  f\right)
$\ \ \ \ \footnote{\textit{Proof.} Let $f\in\operatorname*{Hom}\left(
A,M\right)  $. We must prove that $\left(  \operatorname*{htd}\nolimits_{A,M}%
\circ\operatorname*{dth}\nolimits_{A,M}\right)  \left(  f\right)
=\operatorname*{id}\nolimits_{\operatorname*{Hom}\left(  A,M\right)  }\left(
f\right)  $.
\par
The definition of $\operatorname*{dth}\nolimits_{A,M}$ shows that
$\operatorname*{dth}\nolimits_{A,M}\left(  f\right)  =\mathbf{inl}%
_{A,M}+\mathbf{inr}_{A,M}\circ f$. Now,%
\begin{align*}
\left(  \operatorname*{htd}\nolimits_{A,M}\circ\operatorname*{dth}%
\nolimits_{A,M}\right)  \left(  f\right)   &  =\operatorname*{htd}%
\nolimits_{A,M}\left(  \underbrace{\operatorname*{dth}\nolimits_{A,M}\left(
f\right)  }_{=\mathbf{inl}_{A,M}+\mathbf{inr}_{A,M}\circ f}\right)
=\operatorname*{htd}\nolimits_{A,M}\left(  \mathbf{inl}_{A,M}+\mathbf{inr}%
_{A,M}\circ f\right) \\
&  =\mathbf{projr}_{A,M}\circ\left(  \mathbf{inl}_{A,M}+\mathbf{inr}%
_{A,M}\circ f\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\operatorname*{htd}\nolimits_{A,M}\right) \\
&  =\mathbf{projr}_{A,M}\circ\mathbf{inl}_{A,M}+\underbrace{\mathbf{projr}%
_{A,M}\circ\left(  \mathbf{inr}_{A,M}\circ f\right)  }_{=\mathbf{projr}%
_{A,M}\circ\mathbf{inr}_{A,M}\circ f}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since composition of }\mathbf{k}\text{-linear maps is }\mathbf{k}%
\text{-bilinear (and since}\\
\text{the three maps }\mathbf{projr}_{A,M}\text{, }\mathbf{inl}_{A,M}\text{
and }\mathbf{inr}_{A,M}\circ f\text{ are }\mathbf{k}\text{-linear)}%
\end{array}
\right) \\
&  =\underbrace{\mathbf{projr}_{A,M}\circ\mathbf{inl}_{A,M}}%
_{\substack{=0\\\text{(by (\ref{eq.prop.RAM.props.trivial.3}))}}%
}+\underbrace{\mathbf{projr}_{A,M}\circ\mathbf{inr}_{A,M}}%
_{\substack{=\operatorname*{id}\nolimits_{M}\\\text{(by
(\ref{eq.prop.RAM.props.trivial.4}))}}}\circ f\\
&  =\operatorname*{id}\nolimits_{M}\circ f=f=\operatorname*{id}%
\nolimits_{\operatorname*{Hom}\left(  A,M\right)  }\left(  f\right)  ,
\end{align*}
qed.}. In other words, we have $\operatorname*{htd}\nolimits_{A,M}%
\circ\operatorname*{dth}\nolimits_{A,M}=\operatorname*{id}%
\nolimits_{\operatorname*{Hom}\left(  A,M\right)  }$.

Combining $\operatorname*{htd}\nolimits_{A,M}\circ\operatorname*{dth}%
\nolimits_{A,M}=\operatorname*{id}\nolimits_{\operatorname*{Hom}\left(
A,M\right)  }$ and $\operatorname*{htd}\nolimits_{A,M}\circ\operatorname*{dth}%
\nolimits_{A,M}=\operatorname*{id}\nolimits_{\operatorname*{Hom}\left(
A,M\right)  }$, we conclude that the maps $\operatorname*{dth}\nolimits_{A,M}$
and $\operatorname*{htd}\nolimits_{A,M}$ are mutually inverse. Theorem
\ref{thm.RAM.dth-htd} \textbf{(b)} is thus proven.
\end{vershort}

\begin{verlong}
\textbf{(b)} Every $F\in\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(
A,M\right)  $ satisfies $\left(  \operatorname*{dth}\nolimits_{A,M}%
\circ\operatorname*{htd}\nolimits_{A,M}\right)  \left(  F\right)
=\operatorname*{id}\nolimits_{\operatorname*{Hom}\nolimits_{\mathcal{R}%
}\left(  A,M\right)  }\left(  F\right)  $\ \ \ \ \footnote{\textit{Proof.} Let
$F\in\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  A,M\right)  $. We must
prove that $\left(  \operatorname*{dth}\nolimits_{A,M}\circ\operatorname*{htd}%
\nolimits_{A,M}\right)  \left(  F\right)  =\operatorname*{id}%
\nolimits_{A,\mathcal{R}_{A}\left(  M\right)  }\left(  F\right)  $.
\par
We have
\[
F\in\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  A,M\right)  =\left\{
\varphi\in\operatorname*{Hom}\left(  A,\mathcal{R}_{A}\left(  M\right)
\right)  \ \mid\ \mathbf{projl}_{A,M}\circ\varphi=\operatorname*{id}%
\nolimits_{A}\right\}
\]
(by the definition of $\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(
A,M\right)  $). In other words, $F$ is an element $\varphi$ of
$\operatorname*{Hom}\left(  A,\mathcal{R}_{A}\left(  M\right)  \right)  $
satisfying $\mathbf{projl}_{A,M}\circ\varphi=\operatorname*{id}\nolimits_{A}$.
In other words, $F$ is an element of $\operatorname*{Hom}\left(
A,\mathcal{R}_{A}\left(  M\right)  \right)  $ and satisfies $\mathbf{projl}%
_{A,M}\circ F=\operatorname*{id}\nolimits_{A}$. In other words,
$\operatorname*{id}\nolimits_{A}=\mathbf{projl}_{A,M}\circ F$.
\par
On the other hand, $\operatorname*{htd}\nolimits_{A,M}\left(  F\right)
=\mathbf{projr}_{A,M}\circ F$ (by the definition of $\operatorname*{htd}%
\nolimits_{A,M}$). The map $F$ is $\mathbf{k}$-linear (since $F\in
\operatorname*{Hom}\left(  A,\mathcal{R}_{A}\left(  M\right)  \right)  $).
\par
Now,%
\begin{align*}
\left(  \operatorname*{dth}\nolimits_{A,M}\circ\operatorname*{htd}%
\nolimits_{A,M}\right)  \left(  F\right)   &  =\operatorname*{dth}%
\nolimits_{A,M}\left(  \underbrace{\operatorname*{htd}\nolimits_{A,M}\left(
F\right)  }_{=\mathbf{projr}_{A,M}\circ F}\right)  =\operatorname*{dth}%
\nolimits_{A,M}\left(  \mathbf{projr}_{A,M}\circ F\right) \\
&  =\underbrace{\mathbf{inl}_{A,M}}_{=\mathbf{inl}_{A,M}\circ
\operatorname*{id}\nolimits_{A}}+\underbrace{\mathbf{inr}_{A,M}\circ\left(
\mathbf{projr}_{A,M}\circ F\right)  }_{=\mathbf{inr}_{A,M}\circ\mathbf{projr}%
_{A,M}\circ F}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\operatorname*{dth}\nolimits_{A,M}\right) \\
&  =\mathbf{inl}_{A,M}\circ\underbrace{\operatorname*{id}\nolimits_{A}%
}_{=\mathbf{projl}_{A,M}\circ F}+\mathbf{inr}_{A,M}\circ\mathbf{projr}%
_{A,M}\circ F\\
&  =\mathbf{inl}_{A,M}\circ\mathbf{projl}_{A,M}\circ F+\mathbf{inr}_{A,M}%
\circ\mathbf{projr}_{A,M}\circ F\\
&  =\left(  \mathbf{inl}_{A,M}\circ\mathbf{projl}_{A,M}\right)  \circ
F+\left(  \mathbf{inr}_{A,M}\circ\mathbf{projr}_{A,M}\right)  \circ F\\
&  =\underbrace{\left(  \mathbf{inl}_{A,M}\circ\mathbf{projl}_{A,M}%
+\mathbf{inr}_{A,M}\circ\mathbf{projr}_{A,M}\right)  }%
_{\substack{=\operatorname*{id}\nolimits_{\mathcal{R}_{A}\left(  M\right)
}\\\text{(by (\ref{eq.prop.RAM.props.trivial.5}))}}}\circ F\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }\mathbf{k}%
\text{-linear maps is }\mathbf{k}\text{-bilinear}\right) \\
&  =\operatorname*{id}\nolimits_{\mathcal{R}_{A}\left(  M\right)  }\circ
F=F=\operatorname*{id}\nolimits_{\operatorname*{Hom}\nolimits_{\mathcal{R}%
}\left(  A,M\right)  }\left(  F\right)  ,
\end{align*}
qed.}. In other words, we have $\operatorname*{dth}\nolimits_{A,M}%
\circ\operatorname*{htd}\nolimits_{A,M}=\operatorname*{id}%
\nolimits_{\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  A,M\right)  }$.

On the other hand, every $f\in\operatorname*{Hom}\left(  A,M\right)  $
satisfies $\left(  \operatorname*{htd}\nolimits_{A,M}\circ\operatorname*{dth}%
\nolimits_{A,M}\right)  \left(  f\right)  =\operatorname*{id}%
\nolimits_{\operatorname*{Hom}\left(  A,M\right)  }\left(  f\right)
$\ \ \ \ \footnote{\textit{Proof.} Let $f\in\operatorname*{Hom}\left(
A,M\right)  $. We must prove that $\left(  \operatorname*{htd}\nolimits_{A,M}%
\circ\operatorname*{dth}\nolimits_{A,M}\right)  \left(  f\right)
=\operatorname*{id}\nolimits_{\operatorname*{Hom}\left(  A,M\right)  }\left(
f\right)  $.
\par
The map $f$ belongs to $\operatorname*{Hom}\left(  A,M\right)  $. In other
words, $f$ is a $\mathbf{k}$-linear map from $A$ to $M$. The map
$\mathbf{inr}_{A,M}\circ f$ is $\mathbf{k}$-linear (since it is the
composition of the $\mathbf{k}$-linear maps $\mathbf{inr}_{A,M}$ and $f$).
\par
The definition of $\operatorname*{dth}\nolimits_{A,M}$ shows that
$\operatorname*{dth}\nolimits_{A,M}\left(  f\right)  =\mathbf{inl}%
_{A,M}+\mathbf{inr}_{A,M}\circ f$. Now,%
\begin{align*}
\left(  \operatorname*{htd}\nolimits_{A,M}\circ\operatorname*{dth}%
\nolimits_{A,M}\right)  \left(  f\right)   &  =\operatorname*{htd}%
\nolimits_{A,M}\left(  \underbrace{\operatorname*{dth}\nolimits_{A,M}\left(
f\right)  }_{=\mathbf{inl}_{A,M}+\mathbf{inr}_{A,M}\circ f}\right)
=\operatorname*{htd}\nolimits_{A,M}\left(  \mathbf{inl}_{A,M}+\mathbf{inr}%
_{A,M}\circ f\right) \\
&  =\mathbf{projr}_{A,M}\circ\left(  \mathbf{inl}_{A,M}+\mathbf{inr}%
_{A,M}\circ f\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\operatorname*{htd}\nolimits_{A,M}\right) \\
&  =\mathbf{projr}_{A,M}\circ\mathbf{inl}_{A,M}+\underbrace{\mathbf{projr}%
_{A,M}\circ\left(  \mathbf{inr}_{A,M}\circ f\right)  }_{=\mathbf{projr}%
_{A,M}\circ\mathbf{inr}_{A,M}\circ f}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since composition of }\mathbf{k}\text{-linear maps is }\mathbf{k}%
\text{-bilinear (and since the}\\
\text{three maps }\mathbf{projr}_{A,M}\text{, }\mathbf{inl}_{A,M}\text{ and
}\mathbf{inr}_{A,M}\circ f\text{ are }\mathbf{k}\text{-linear)}%
\end{array}
\right) \\
&  =\underbrace{\mathbf{projr}_{A,M}\circ\mathbf{inl}_{A,M}}%
_{\substack{=0\\\text{(by (\ref{eq.prop.RAM.props.trivial.3}))}}%
}+\underbrace{\mathbf{projr}_{A,M}\circ\mathbf{inr}_{A,M}}%
_{\substack{=\operatorname*{id}\nolimits_{M}\\\text{(by
(\ref{eq.prop.RAM.props.trivial.4}))}}}\circ f\\
&  =0+\operatorname*{id}\nolimits_{M}\circ f=\operatorname*{id}\nolimits_{M}%
\circ f=f=\operatorname*{id}\nolimits_{\operatorname*{Hom}\left(  A,M\right)
}\left(  f\right)  ,
\end{align*}
qed.}. In other words, we have $\operatorname*{htd}\nolimits_{A,M}%
\circ\operatorname*{dth}\nolimits_{A,M}=\operatorname*{id}%
\nolimits_{\operatorname*{Hom}\left(  A,M\right)  }$.

Combining $\operatorname*{htd}\nolimits_{A,M}\circ\operatorname*{dth}%
\nolimits_{A,M}=\operatorname*{id}\nolimits_{\operatorname*{Hom}\left(
A,M\right)  }$ and $\operatorname*{htd}\nolimits_{A,M}\circ\operatorname*{dth}%
\nolimits_{A,M}=\operatorname*{id}\nolimits_{\operatorname*{Hom}\left(
A,M\right)  }$, we conclude that the maps $\operatorname*{dth}\nolimits_{A,M}$
and $\operatorname*{htd}\nolimits_{A,M}$ are mutually inverse. Theorem
\ref{thm.RAM.dth-htd} \textbf{(b)} is thus proven.
\end{verlong}

\textbf{(c)} Set $\varphi=\operatorname*{dth}\nolimits_{A,M}\left(  f\right)
$. Thus, $\varphi=\operatorname*{dth}\nolimits_{A,M}\left(  f\right)
\in\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  A,M\right)  $. For every
$a\in A$, we have%
\begin{equation}
\underbrace{\varphi}_{=\operatorname*{dth}\nolimits_{A,M}\left(  f\right)
}\left(  a\right)  =\left(  \operatorname*{dth}\nolimits_{A,M}\left(
f\right)  \right)  \left(  a\right)  =\left(  a,f\left(  a\right)  \right)  .
\label{pf.thm.RAM.dth-htd.c.1}%
\end{equation}


We must prove that $f$ is a derivation if and only if $\operatorname*{dth}%
\nolimits_{A,M}\left(  f\right)  $ is a $\mathbf{k}$-algebra homomorphism from
$A$ to $\mathcal{R}_{A}\left(  M\right)  $. In other words, we must prove that
$f$ is a derivation if and only if $\varphi$ is a $\mathbf{k}$-algebra
homomorphism from $A$ to $\mathcal{R}_{A}\left(  M\right)  $ (because
$\varphi=\operatorname*{dth}\nolimits_{A,M}\left(  f\right)  $). In other
words, we must prove the following two statements:

\textit{Statement 1:} If $f$ is a derivation, then $\varphi$ is a $\mathbf{k}%
$-algebra homomorphism from $A$ to $\mathcal{R}_{A}\left(  M\right)  $.

\textit{Statement 2:} If $\varphi$ is a $\mathbf{k}$-algebra homomorphism from
$A$ to $\mathcal{R}_{A}\left(  M\right)  $, then $f$ is a derivation.

\textit{Proof of Statement 1:} Assume that $f$ is a derivation. We must prove
that $\varphi$ is a $\mathbf{k}$-algebra homomorphism from $A$ to
$\mathcal{R}_{A}\left(  M\right)  $.

Recall that $\left(  1,0\right)  $ is the unity of the $\mathbf{k}$-algebra
$\mathcal{R}_{A}\left(  M\right)  $. In other words, $\left(  1,0\right)
=1_{\mathcal{R}_{A}\left(  M\right)  }$.

\begin{vershort}
Theorem \ref{thm.deriv.0} shows that $f\left(  1\right)  =0$. Now,
(\ref{pf.thm.RAM.dth-htd.c.1}) (applied to $a=1$) shows that $\varphi\left(
1\right)  =\left(  1,\underbrace{f\left(  1\right)  }_{=0}\right)  =\left(
1,0\right)  =1_{\mathcal{R}_{A}\left(  M\right)  }$.

Also, $\varphi=\operatorname*{dth}\nolimits_{A,M}\left(  f\right)
\in\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  A,M\right)
\subseteq\operatorname*{Hom}\left(  A,\mathcal{R}_{A}\left(  M\right)
\right)  $. Thus, the map $\varphi$ is $\mathbf{k}$-linear.
\end{vershort}

\begin{verlong}
We know that $f$ is a derivation. Thus, Corollary \ref{cor.deriv.verlong}
shows that $f$ is a $\mathbf{k}$-linear map and satisfies
(\ref{eq.def.deriv.axiom}). Also, Theorem \ref{thm.deriv.0} shows that
$f\left(  1\right)  =0$. Now, (\ref{pf.thm.RAM.dth-htd.c.1}) (applied to
$a=1$) shows that $\varphi\left(  1\right)  =\left(  1,\underbrace{f\left(
1\right)  }_{=0}\right)  =\left(  1,0\right)  =1_{\mathcal{R}_{A}\left(
M\right)  }$.

We have $\varphi=\operatorname*{dth}\nolimits_{A,M}\left(  f\right)
\in\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  A,M\right)
\subseteq\operatorname*{Hom}\left(  A,\mathcal{R}_{A}\left(  M\right)
\right)  $ (since $\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(
A,M\right)  $ was defined as a certain subset of $\operatorname*{Hom}\left(
A,\mathcal{R}_{A}\left(  M\right)  \right)  $). In other words, $\varphi$ is a
$\mathbf{k}$-linear map from $A$ to $\mathcal{R}_{A}\left(  M\right)  $.
\end{verlong}

Furthermore, $\varphi\left(  ab\right)  =\varphi\left(  a\right)
\varphi\left(  b\right)  $ for every $a\in A$ and $b\in A$%
\ \ \ \ \footnote{\textit{Proof.} Let $a\in A$ and $b\in A$. Then,
$\varphi\left(  a\right)  =\left(  a,f\left(  a\right)  \right)  $ (by
(\ref{pf.thm.RAM.dth-htd.c.1})) and $\varphi\left(  b\right)  =\left(
b,f\left(  b\right)  \right)  $ (by (\ref{pf.thm.RAM.dth-htd.c.1}), applied to
$b$ instead of $a$). Furthermore, (\ref{pf.thm.RAM.dth-htd.c.1}) (applied to
$ab$ instead of $a$) shows that%
\[
\varphi\left(  ab\right)  =\left(  ab,\underbrace{f\left(  ab\right)
}_{\substack{=af\left(  b\right)  +f\left(  a\right)  b\\\text{(by
(\ref{eq.def.deriv.axiom}))}}}\right)  =\left(  ab,af\left(  b\right)
+f\left(  a\right)  b\right)  .
\]
Comparing this with%
\begin{align*}
\varphi\left(  a\right)  \varphi\left(  b\right)   &  =\underbrace{\varphi
\left(  a\right)  }_{=\left(  a,f\left(  a\right)  \right)  }\cdot
\underbrace{\varphi\left(  b\right)  }_{=\left(  b,f\left(  b\right)  \right)
}=\left(  a,f\left(  a\right)  \right)  \cdot\left(  b,f\left(  b\right)
\right) \\
&  =\left(  ab,af\left(  b\right)  +f\left(  a\right)  b\right)
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{eq.def.RAM.product}), applied to }\left(  a,f\left(  a\right)
\right)  \text{ and }\left(  b,f\left(  b\right)  \right) \\
\text{instead of }\left(  a,p\right)  \text{ and }\left(  b,q\right)
\end{array}
\right)  ,
\end{align*}
we obtain $\varphi\left(  ab\right)  =\varphi\left(  a\right)  \varphi\left(
b\right)  $. Qed.}. This (combined with the fact that $\varphi$ is a
$\mathbf{k}$-linear map and the fact that $\varphi\left(  1\right)
=1_{\mathcal{R}_{A}\left(  M\right)  }$) shows that $\varphi$ is a
$\mathbf{k}$-algebra homomorphism from $A$ to $\mathcal{R}_{A}\left(
M\right)  $. Statement 1 is thus proven.

\textit{Proof of Statement 2:} Assume that $\varphi$ is a $\mathbf{k}$-algebra
homomorphism from $A$ to $\mathcal{R}_{A}\left(  M\right)  $. We must prove
that $f$ is a derivation.

We have $f\in\operatorname*{Hom}\left(  A,M\right)  $. In other words, $f$ is
a $\mathbf{k}$-linear map from $A$ to $M$.

Now, let $a\in A$ and $b\in A$. We shall show that $f\left(  ab\right)
=af\left(  b\right)  +f\left(  a\right)  b$.

Indeed, we have $\varphi\left(  a\right)  =\left(  a,f\left(  a\right)
\right)  $ (by (\ref{pf.thm.RAM.dth-htd.c.1})) and $\varphi\left(  b\right)
=\left(  b,f\left(  b\right)  \right)  $ (by (\ref{pf.thm.RAM.dth-htd.c.1}),
applied to $b$ instead of $a$). Furthermore, (\ref{pf.thm.RAM.dth-htd.c.1})
(applied to $ab$ instead of $a$) shows that $\varphi\left(  ab\right)
=\left(  ab,f\left(  ab\right)  \right)  $. Thus,%
\begin{align*}
\left(  ab,f\left(  ab\right)  \right)   &  =\varphi\left(  ab\right)
=\underbrace{\varphi\left(  a\right)  }_{=\left(  a,f\left(  a\right)
\right)  }\cdot\underbrace{\varphi\left(  b\right)  }_{=\left(  b,f\left(
b\right)  \right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varphi\text{ is
a }\mathbf{k}\text{-algebra homomorphism}\right) \\
&  =\left(  a,f\left(  a\right)  \right)  \cdot\left(  b,f\left(  b\right)
\right) \\
&  =\left(  ab,af\left(  b\right)  +f\left(  a\right)  b\right)
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{eq.def.RAM.product}), applied to }\left(  a,f\left(  a\right)
\right)  \text{ and }\left(  b,f\left(  b\right)  \right) \\
\text{instead of }\left(  a,p\right)  \text{ and }\left(  b,q\right)
\end{array}
\right)  .
\end{align*}
In other words, $ab=ab$ and $f\left(  ab\right)  =af\left(  b\right)
+f\left(  a\right)  b$.

\begin{vershort}
Now, let us forget that we fixed $a$ and $b$. We thus have proven that
$f\left(  ab\right)  =af\left(  b\right)  +f\left(  a\right)  b$ for every
$a\in A$ and $b\in A$. In other words, $f$ is a derivation. This proves
Statement 2.
\end{vershort}

\begin{verlong}
Now, let us forget that we fixed $a$ and $b$. We thus have proven that
$f\left(  ab\right)  =af\left(  b\right)  +f\left(  a\right)  b$ for every
$a\in A$ and $b\in A$. Thus, Corollary \ref{cor.deriv.verlong.converse} shows
that $f$ is a derivation. This proves Statement 2.
\end{verlong}

Now, Statement 1 and Statement 2 are both proven. As we have said above, this
concludes the proof of Theorem \ref{thm.RAM.dth-htd} \textbf{(c)}.
\end{proof}

Theorem \ref{thm.RAM.dth-htd} is a classical folklore result. In the
particular case when $A$ is a commutative $\mathbf{k}$-algebra and $M$ is a
symmetric $\left(  A,A\right)  $-bimodule, it is explicitly stated in
\cite{Becker} (where $\mathcal{R}_{A}\left(  M\right)  $ is denoted by
$M\rtimes A$, and called a \textit{square zero extension}\footnote{This name
has a slightly more general meaning in other sources.}), but it was widely
used before.

The following is a sufficient criterion for $\mathcal{R}_{A}\left(  M\right)
$ to be commutative\footnote{Actually, it is a necessary criterion as well
(i.e., if $\mathcal{R}_{A}\left(  M\right)  $ is commutative, then $A$ is
commutative and $M$ is symmetric), but we will not have use for this fact.}:

\begin{proposition}
\label{prop.RAM.commut}Let $A$ be a commutative $\mathbf{k}$-algebra. Let $M$
be a symmetric $\left(  A,A\right)  $-bimodule. Then, the $\mathbf{k}$-algebra
$\mathcal{R}_{A}\left(  M\right)  $ is commutative.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.RAM.commut}.]Let $x\in\mathcal{R}_{A}\left(
M\right)  $ and $y\in\mathcal{R}_{A}\left(  M\right)  $.

\begin{verlong}
The $\left(  A,A\right)  $-bimodule $M$ is symmetric. Thus, it satisfies
(\ref{eq.def.bimod.symm.axiom}) (because the $\left(  A,A\right)  $-bimodule
$M$ is symmetric if and only if it satisfies (\ref{eq.def.bimod.symm.axiom})
(by the definition of a \textquotedblleft symmetric $\left(  A,A\right)
$-bimodule\textquotedblright)).
\end{verlong}

We have $x\in\mathcal{R}_{A}\left(  M\right)  $. Then, $x\in\mathcal{R}%
_{A}\left(  M\right)  =A\oplus M$. Thus, we can write $x$ in the form
$x=\left(  a,p\right)  $ for some $a\in A$ and $p\in M$. Consider these $a$
and $p$.

We have $y\in\mathcal{R}_{A}\left(  M\right)  $. Then, $y\in\mathcal{R}%
_{A}\left(  M\right)  =A\oplus M$. Thus, we can write $y$ in the form
$y=\left(  b,q\right)  $ for some $b\in A$ and $q\in M$. Consider these $b$
and $q$.

\begin{vershort}
Since the $\left(  A,A\right)  $-bimodule $M$ is symmetric, we have $aq=qa$
and $bp=pb$. Finally, $ab=ba$ (since $A$ is commutative).
\end{vershort}

\begin{verlong}
Applying (\ref{eq.def.bimod.symm.axiom}) to $m=q$, we obtain $aq=qa$. Also,
(\ref{eq.def.bimod.symm.axiom}) (applied to $b$ and $p$ instead of $a$ and
$m$) shows that $bp=pb$. Thus, $pb=bp$. Finally, $ab=ba$ (since $A$ is commutative).
\end{verlong}

Now,%
\begin{align*}
xy  &  =\underbrace{x}_{=\left(  a,p\right)  }\cdot\underbrace{y}_{=\left(
b,q\right)  }=\left(  a,p\right)  \cdot\left(  b,q\right)  =\left(
\underbrace{ab}_{=ba},\underbrace{aq}_{=qa}+\underbrace{pb}_{=bp}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.def.RAM.product})}\right) \\
&  =\left(  ba,\underbrace{qa+bp}_{=bp+qa}\right)  =\left(  ba,bp+qa\right)  .
\end{align*}
Comparing this with%
\begin{align*}
yx  &  =\underbrace{y}_{=\left(  b,q\right)  }\cdot\underbrace{x}_{=\left(
a,p\right)  }=\left(  b,q\right)  \cdot\left(  a,p\right)  =\left(
ba,bp+qa\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{eq.def.RAM.product}), applied to }\left(  b,q\right)  \text{
and }\left(  a,p\right) \\
\text{instead of }\left(  a,p\right)  \text{ and }\left(  b,q\right)
\end{array}
\right)  ,
\end{align*}
we obtain $xy=yx$.

Let us now forget that we fixed $x$ and $y$. We thus have proven that $xy=yx$
for every $x\in\mathcal{R}_{A}\left(  M\right)  $ and $y\in\mathcal{R}%
_{A}\left(  M\right)  $. In other words, the $\mathbf{k}$-algebra
$\mathcal{R}_{A}\left(  M\right)  $ is commutative.
\end{proof}

\subsection{Compositions and tensor products}

The following two (almost trivial) facts show how derivations can be obtained
by composing derivations with other maps:

\begin{proposition}
\label{prop.der.compos1}Let $A$ and $B$ be two $\mathbf{k}$-algebras. Let
$f:A\rightarrow B$ be a $\mathbf{k}$-algebra homomorphism. Let $M$ be a
$\left(  B,B\right)  $-bimodule. Let $d:B\rightarrow M$ be a derivation.

We consider the $\left(  B,B\right)  $-bimodule $M$ as an $\left(  A,A\right)
$-bimodule (via the $\mathbf{k}$-algebra homomorphism $f:A\rightarrow B$).
Then, $d\circ f:A\rightarrow M$ is a derivation.
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.der.compos1}.]We know that $d$ is a
derivation. In other words, we have%
\begin{equation}
\left(  d\left(  ab\right)  =ad\left(  b\right)  +d\left(  a\right)
b\ \ \ \ \ \ \ \ \ \ \text{for every }a\in B\text{ and }b\in B\right)  .
\label{pf.prop.der.compos1.short.1}%
\end{equation}


Now, let $a\in A$ and $b\in A$. Then, $f\left(  ab\right)  =f\left(  a\right)
f\left(  b\right)  $ (since $f$ is a $\mathbf{k}$-algebra homomorphism). Now,%
\begin{equation}
\left(  d\circ f\right)  \left(  ab\right)  =d\left(  \underbrace{f\left(
ab\right)  }_{=f\left(  a\right)  f\left(  b\right)  }\right)  =d\left(
f\left(  a\right)  f\left(  b\right)  \right)  =f\left(  a\right)  d\left(
f\left(  b\right)  \right)  +d\left(  f\left(  a\right)  \right)  f\left(
b\right)  \label{pf.prop.der.compos1.short.l}%
\end{equation}
(by (\ref{pf.prop.der.compos1.short.1}), applied to $f\left(  a\right)  $ and
$f\left(  b\right)  $ instead of $a$ and $b$).

On the other hand, recall that the $\left(  B,B\right)  $-module $M$ was made
into an $\left(  A,A\right)  $-bimodule via the $\mathbf{k}$-algebra
homomorphism $f:A\rightarrow B$. Hence, $xu=f\left(  x\right)  u$ for every
$x\in A$ and $u\in M$. Applying this to $x=a$ and $u=d\left(  f\left(
b\right)  \right)  $, we obtain $ad\left(  f\left(  b\right)  \right)
=f\left(  a\right)  d\left(  f\left(  b\right)  \right)  $. Similarly, we
obtain $d\left(  f\left(  a\right)  \right)  b=d\left(  f\left(  a\right)
\right)  f\left(  b\right)  $. Now,%
\begin{align*}
&  a\underbrace{\left(  d\circ f\right)  \left(  b\right)  }_{=d\left(
f\left(  b\right)  \right)  }+\underbrace{\left(  d\circ f\right)  \left(
a\right)  }_{=d\left(  f\left(  a\right)  \right)  }b\\
&  =\underbrace{ad\left(  f\left(  b\right)  \right)  }_{=f\left(  a\right)
d\left(  f\left(  b\right)  \right)  }+\underbrace{d\left(  f\left(  a\right)
\right)  b}_{=d\left(  f\left(  a\right)  \right)  f\left(  b\right)
}=f\left(  a\right)  d\left(  f\left(  b\right)  \right)  +d\left(  f\left(
a\right)  \right)  f\left(  b\right)  .
\end{align*}
Comparing this with (\ref{pf.prop.der.compos1.short.l}), we obtain $\left(
d\circ f\right)  \left(  ab\right)  =a\left(  d\circ f\right)  \left(
b\right)  +\left(  d\circ f\right)  \left(  a\right)  b$.

Let us now forget that we fixed $a$ and $b$. We thus have shown that $\left(
d\circ f\right)  \left(  ab\right)  =a\left(  d\circ f\right)  \left(
b\right)  +\left(  d\circ f\right)  \left(  a\right)  b$ for all $a\in A$ and
$b\in A$. In other words, $d\circ f$ is a derivation (by the definition of a
\textquotedblleft derivation\textquotedblright, since $d\circ f$ is a
$\mathbf{k}$-linear map). This proves Proposition \ref{prop.der.compos1}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.der.compos1}.]We know that $d:B\rightarrow M$
is a derivation. Hence, Corollary \ref{cor.deriv.verlong} (applied to $B$ and
$d$ instead of $A$ and $f$) shows that $d$ is a $\mathbf{k}$-linear map and
satisfies (\ref{eq.def.deriv.axiom}) (with $A$ and $f$ replaced by $B$ and $d$).

We know that $d$ satisfies (\ref{eq.def.deriv.axiom}) (with $A$ and $f$
replaced by $B$ and $d$). In other words, we have%
\begin{equation}
\left(  d\left(  ab\right)  =ad\left(  b\right)  +d\left(  a\right)
b\ \ \ \ \ \ \ \ \ \ \text{for every }a\in B\text{ and }b\in B\right)  .
\label{pf.prop.der.compos1.1}%
\end{equation}


Now, the map $d\circ f:A\rightarrow M$ is $\mathbf{k}$-linear (because it is
the composition of the $\mathbf{k}$-linear maps $d$ and $f$).

Let $a\in A$ and $b\in A$. Then, $f\left(  ab\right)  =f\left(  a\right)
f\left(  b\right)  $ (since $f$ is a $\mathbf{k}$-algebra homomorphism). Now,%
\begin{equation}
\left(  d\circ f\right)  \left(  ab\right)  =d\left(  \underbrace{f\left(
ab\right)  }_{=f\left(  a\right)  f\left(  b\right)  }\right)  =d\left(
f\left(  a\right)  f\left(  b\right)  \right)  =f\left(  a\right)  d\left(
f\left(  b\right)  \right)  +d\left(  f\left(  a\right)  \right)  f\left(
b\right)  \label{pf.prop.der.compos1.l}%
\end{equation}
(by (\ref{pf.prop.der.compos1.1}), applied to $f\left(  a\right)  $ and
$f\left(  b\right)  $ instead of $a$ and $b$).

On the other hand, recall that the $\left(  B,B\right)  $-module $M$ was made
into an $\left(  A,A\right)  $-bimodule via the $\mathbf{k}$-algebra
homomorphism $f:A\rightarrow B$. Hence,%
\begin{equation}
\left(  xu=f\left(  x\right)  u\ \ \ \ \ \ \ \ \ \ \text{for every }x\in
A\text{ and }u\in M\right)  \label{pf.prop.der.compos1.3}%
\end{equation}
and%
\begin{equation}
\left(  ux=uf\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }x\in
A\text{ and }u\in M\right)  . \label{pf.prop.der.compos1.4}%
\end{equation}


Applying (\ref{pf.prop.der.compos1.3}) to $x=a$ and $u=d\left(  f\left(
b\right)  \right)  $, we obtain $ad\left(  f\left(  b\right)  \right)
=f\left(  a\right)  d\left(  f\left(  b\right)  \right)  $. Applying
(\ref{pf.prop.der.compos1.4}) to $x=b$ and $u=d\left(  f\left(  a\right)
\right)  $, we obtain $d\left(  f\left(  a\right)  \right)  b=d\left(
f\left(  a\right)  \right)  f\left(  b\right)  $. Now,%
\begin{align*}
&  a\underbrace{\left(  d\circ f\right)  \left(  b\right)  }_{=d\left(
f\left(  b\right)  \right)  }+\underbrace{\left(  d\circ f\right)  \left(
a\right)  }_{=d\left(  f\left(  a\right)  \right)  }b\\
&  =\underbrace{ad\left(  f\left(  b\right)  \right)  }_{=f\left(  a\right)
d\left(  f\left(  b\right)  \right)  }+\underbrace{d\left(  f\left(  a\right)
\right)  b}_{=d\left(  f\left(  a\right)  \right)  f\left(  b\right)
}=f\left(  a\right)  d\left(  f\left(  b\right)  \right)  +d\left(  f\left(
a\right)  \right)  f\left(  b\right)  .
\end{align*}
Comparing this with (\ref{pf.prop.der.compos1.l}), we obtain $\left(  d\circ
f\right)  \left(  ab\right)  =a\left(  d\circ f\right)  \left(  b\right)
+\left(  d\circ f\right)  \left(  a\right)  b$.

Let us now forget that we fixed $a$ and $b$. We thus have shown that $\left(
d\circ f\right)  \left(  ab\right)  =a\left(  d\circ f\right)  \left(
b\right)  +\left(  d\circ f\right)  \left(  a\right)  b$ for all $a\in A$ and
$b\in A$. Thus, Corollary \ref{cor.deriv.verlong.converse} (applied to $d\circ
f$ instead of $f$) shows that $d\circ f$ is a derivation. This proves
Proposition \ref{prop.der.compos1}.
\end{proof}
\end{verlong}

\begin{proposition}
\label{prop.der.compos2}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ and $N$ be
two $\left(  A,A\right)  $-bimodules. Let $f:M\rightarrow N$ be an $\left(
A,A\right)  $-bimodule homomorphism. Let $d:A\rightarrow M$ be a derivation.
Then, $f\circ d:A\rightarrow N$ is a derivation.
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.der.compos2}.]This is even more
straightforward than the proof of Proposition \ref{prop.der.compos1}, and thus
left to the reader.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.der.compos2}.]We know that $d:A\rightarrow M$
is a derivation. Hence, Corollary \ref{cor.deriv.verlong} (applied to $d$
instead of $f$) shows that $d$ is a $\mathbf{k}$-linear map and satisfies
(\ref{eq.def.deriv.axiom}) (with $f$ replaced by $d$).

The map $f\circ d:A\rightarrow N$ is $\mathbf{k}$-linear (because it is the
composition of the $\mathbf{k}$-linear maps $f$ and $d$).

We know that $d$ satisfies (\ref{eq.def.deriv.axiom}) (with $f$ replaced by
$d$). In other words, we have%
\begin{equation}
\left(  d\left(  ab\right)  =ad\left(  b\right)  +d\left(  a\right)
b\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{ and }b\in A\right)  .
\label{pf.prop.der.compos2.1}%
\end{equation}


For every $a\in A$ and $b\in A$, we have%
\begin{align*}
\left(  f\circ d\right)  \left(  ab\right)   &  =f\left(  \underbrace{d\left(
ab\right)  }_{\substack{=ad\left(  b\right)  +d\left(  a\right)  b\\\text{(by
(\ref{pf.prop.der.compos2.1}))}}}\right)  =f\left(  ad\left(  b\right)
+d\left(  a\right)  b\right) \\
&  =\underbrace{f\left(  ad\left(  b\right)  \right)  }_{\substack{=af\left(
d\left(  b\right)  \right)  \\\text{(since }f\text{ is an }\left(  A,A\right)
\text{-bimodule}\\\text{homomorphism)}}}+\underbrace{f\left(  d\left(
a\right)  b\right)  }_{\substack{=f\left(  d\left(  a\right)  \right)
b\\\text{(since }f\text{ is an }\left(  A,A\right)  \text{-bimodule}%
\\\text{homomorphism)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is an }\left(  A,A\right)
\text{-bimodule homomorphism}\right) \\
&  =a\underbrace{f\left(  d\left(  b\right)  \right)  }_{=\left(  f\circ
d\right)  \left(  b\right)  }+\underbrace{f\left(  d\left(  a\right)  \right)
}_{=\left(  f\circ d\right)  \left(  a\right)  }b=a\left(  f\circ d\right)
\left(  b\right)  +\left(  f\circ d\right)  \left(  a\right)  b.
\end{align*}
Thus, Corollary \ref{cor.deriv.verlong.converse} (applied to $f\circ d$ and
$N$ instead of $f$ and $M$) shows that $f\circ d$ is a derivation. This proves
Proposition \ref{prop.der.compos2}.
\end{proof}
\end{verlong}

A composition of two derivations is usually not a derivation (although we will
see some properties of composition powers of derivations in the next section).

A tensor product of two derivations is usually not a derivation either.
However, here is an example of how derivations interact with tensor products:

\begin{proposition}
\label{prop.der.tensid}Let $A$ be a $\mathbf{k}$-algebra. Let $M$ be an
$\left(  A,A\right)  $-bimodule. Let $f:A\rightarrow M$ be a derivation. Let
$B$ be a $\mathbf{k}$-algebra. Clearly, the tensor product $M\otimes B$ of the
$\left(  A,A\right)  $-bimodule $M$ with the $\left(  B,B\right)  $-bimodule
$B$ is an $\left(  A\otimes B,A\otimes B\right)  $-bimodule. Similarly,
$B\otimes M$ is an $\left(  B\otimes A,B\otimes A\right)  $-bimodule.

\textbf{(a)} The map $f\otimes\operatorname*{id}\nolimits_{B}:A\otimes
B\rightarrow M\otimes B$ is a derivation.

\textbf{(b)} The map $\operatorname*{id}\nolimits_{B}\otimes f:B\otimes
A\rightarrow B\otimes M$ is a derivation.
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.der.tensid}.]\textbf{(a)} The map
$f\otimes\operatorname*{id}\nolimits_{B}$ is clearly $\mathbf{k}$-linear
(since it is the tensor product of the two $\mathbf{k}$-linear maps $f$ and
$\operatorname*{id}\nolimits_{B}$).

Now, we are going to prove that%
\begin{equation}
\left(  f\otimes\operatorname*{id}\nolimits_{B}\right)  \left(  ab\right)
=a\left(  f\otimes\operatorname*{id}\nolimits_{B}\right)  \left(  b\right)
+\left(  f\otimes\operatorname*{id}\nolimits_{B}\right)  \left(  a\right)  b
\label{pf.prop.der.tensid.a.short.goal}%
\end{equation}
for every $a\in A\otimes B$ and $b\in A\otimes B$.

\textit{Proof of (\ref{pf.prop.der.tensid.a.short.goal}):} Let $a\in A\otimes
B$ and $b\in A\otimes B$. We need to prove the equality
(\ref{pf.prop.der.tensid.a.short.goal}). But this equality is $\mathbf{k}%
$-linear in each of $a$ and $b$ (because the multiplication of the
$\mathbf{k}$-algebra $A\otimes B$ is $\mathbf{k}$-bilinear, because the left
and right actions of $A\otimes B$ on $M\otimes B$ are $\mathbf{k}$-bilinear,
and because the map $f\otimes\operatorname*{id}\nolimits_{B}$ is $\mathbf{k}%
$-linear). Hence, we can WLOG assume that $a$ and $b$ are pure tensors (since
the $\mathbf{k}$-module $A\otimes B$ is spanned by pure tensors). Assume this.

We know that $a$ is a pure tensor. In other words, $a=x\otimes y$ for some
$x\in A$ and $y\in B$. Consider these $x$ and $y$.

We know that $b$ is a pure tensor. In other words, $b=z\otimes w$ for some
$z\in A$ and $w\in B$. Consider these $z$ and $w$.

We have $\underbrace{a}_{=x\otimes y}\underbrace{b}_{=z\otimes w}=\left(
x\otimes y\right)  \left(  z\otimes w\right)  =xz\otimes yw$. Hence,%
\begin{align*}
\left(  f\otimes\operatorname*{id}\nolimits_{B}\right)  \left(
\underbrace{ab}_{=xz\otimes yw}\right)   &  =\left(  f\otimes
\operatorname*{id}\nolimits_{B}\right)  \left(  xz\otimes yw\right)
=\underbrace{f\left(  xz\right)  }_{\substack{=xf\left(  z\right)  +f\left(
x\right)  z\\\text{(since }f\text{ is a derivation)}}}\otimes
\underbrace{\operatorname*{id}\nolimits_{B}\left(  yw\right)  }_{=yw}\\
&  =\left(  xf\left(  z\right)  +f\left(  x\right)  z\right)  \otimes
yw=xf\left(  z\right)  \otimes yw+f\left(  x\right)  z\otimes yw.
\end{align*}
Compared with%
\begin{align*}
&  \underbrace{a}_{=x\otimes y}\left(  f\otimes\operatorname*{id}%
\nolimits_{B}\right)  \left(  \underbrace{b}_{=z\otimes w}\right)  +\left(
f\otimes\operatorname*{id}\nolimits_{B}\right)  \left(  \underbrace{a}%
_{=x\otimes y}\right)  \underbrace{b}_{=z\otimes w}\\
&  =\left(  x\otimes y\right)  \cdot\underbrace{\left(  f\otimes
\operatorname*{id}\nolimits_{B}\right)  \left(  z\otimes w\right)
}_{=f\left(  z\right)  \otimes\operatorname*{id}\nolimits_{B}\left(  w\right)
}+\underbrace{\left(  f\otimes\operatorname*{id}\nolimits_{B}\right)  \left(
x\otimes y\right)  }_{=f\left(  x\right)  \otimes\operatorname*{id}%
\nolimits_{B}\left(  y\right)  }\cdot\left(  z\otimes w\right) \\
&  =\left(  x\otimes y\right)  \cdot\left(  f\left(  z\right)  \otimes
\underbrace{\operatorname*{id}\nolimits_{B}\left(  w\right)  }_{=w}\right)
+\left(  f\left(  x\right)  \otimes\underbrace{\operatorname*{id}%
\nolimits_{B}\left(  y\right)  }_{=y}\right)  \cdot\left(  z\otimes w\right)
\\
&  =\underbrace{\left(  x\otimes y\right)  \cdot\left(  f\left(  z\right)
\otimes w\right)  }_{=xf\left(  z\right)  \otimes yw}+\underbrace{\left(
f\left(  x\right)  \otimes y\right)  \cdot\left(  z\otimes w\right)
}_{=f\left(  x\right)  z\otimes yw}=xf\left(  z\right)  \otimes yw+f\left(
x\right)  z\otimes yw,
\end{align*}
this shows that $\left(  f\otimes\operatorname*{id}\nolimits_{B}\right)
\left(  ab\right)  =a\left(  f\otimes\operatorname*{id}\nolimits_{B}\right)
\left(  b\right)  +\left(  f\otimes\operatorname*{id}\nolimits_{B}\right)
\left(  a\right)  b$. Thus, (\ref{pf.prop.der.tensid.a.short.goal}) is proven.

We thus have shown that $\left(  f\otimes\operatorname*{id}\nolimits_{B}%
\right)  \left(  ab\right)  =a\left(  f\otimes\operatorname*{id}%
\nolimits_{B}\right)  \left(  b\right)  +\left(  f\otimes\operatorname*{id}%
\nolimits_{B}\right)  \left(  a\right)  b$ for every $a\in A\otimes B$ and
$b\in A\otimes B$. Since $f\otimes\operatorname*{id}\nolimits_{B}$ is a
$\mathbf{k}$-linear map, this shows that $f\otimes\operatorname*{id}%
\nolimits_{B}$ is a derivation (by the definition of a \textquotedblleft
derivation\textquotedblright). This proves Proposition \ref{prop.der.tensid}
\textbf{(a)}.

\textbf{(b)} The proof of Proposition \ref{prop.der.tensid} \textbf{(b)} is
analogous to the proof of Proposition \ref{prop.der.tensid} \textbf{(a)}
(indeed, Proposition \ref{prop.der.tensid} \textbf{(b)} differs from
Proposition \ref{prop.der.tensid} \textbf{(a)} only in the order of the
tensorands), and thus is left to the reader. (Alternatively, Proposition
\ref{prop.der.tensid} \textbf{(b)} can be deduced from Proposition
\ref{prop.der.tensid} \textbf{(a)} via the isomorphisms $A\otimes B\cong
B\otimes A$ and $M\otimes B\cong B\otimes M$.)
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.der.tensid}.]\textbf{(a)} We know that
$f:A\rightarrow M$ is a derivation. Hence, Corollary \ref{cor.deriv.verlong}
shows that $f$ is a $\mathbf{k}$-linear map and satisfies
(\ref{eq.def.deriv.axiom}).

The map $f\otimes\operatorname*{id}\nolimits_{B}$ is clearly $\mathbf{k}%
$-linear (since it is the tensor product of the two $\mathbf{k}$-linear maps
$f$ and $\operatorname*{id}\nolimits_{B}$).

Now, we are going to prove that%
\begin{equation}
\left(  f\otimes\operatorname*{id}\nolimits_{B}\right)  \left(  ab\right)
=a\left(  f\otimes\operatorname*{id}\nolimits_{B}\right)  \left(  b\right)
+\left(  f\otimes\operatorname*{id}\nolimits_{B}\right)  \left(  a\right)  b
\label{pf.prop.der.tensid.a.goal}%
\end{equation}
for every $a\in A\otimes B$ and $b\in A\otimes B$.

\textit{Proof of (\ref{pf.prop.der.tensid.a.goal}):} Let $a\in A\otimes B$ and
$b\in A\otimes B$. We need to prove the equality
(\ref{pf.prop.der.tensid.a.goal}). But this equality is $\mathbf{k}$-linear in
each of $a$ and $b$ (because the multiplication of the $\mathbf{k}$-algebra
$A\otimes B$ is $\mathbf{k}$-bilinear, because the left and right actions of
$A\otimes B$ on $M\otimes B$ are $\mathbf{k}$-bilinear, and because the map
$f\otimes\operatorname*{id}\nolimits_{B}$ is $\mathbf{k}$-linear). Hence, we
can WLOG assume that $a$ and $b$ are pure tensors (since the $\mathbf{k}%
$-module $A\otimes B$ is spanned by pure tensors). Assume this.

We know that $a$ is a pure tensor. In other words, $a=x\otimes y$ for some
$x\in A$ and $y\in B$. Consider these $x$ and $y$.

We know that $b$ is a pure tensor. In other words, $b=z\otimes w$ for some
$z\in A$ and $w\in B$. Consider these $z$ and $w$.

We have $\underbrace{a}_{=x\otimes y}\underbrace{b}_{=z\otimes w}=\left(
x\otimes y\right)  \left(  z\otimes w\right)  =xz\otimes yw$. Hence,%
\begin{align*}
\left(  f\otimes\operatorname*{id}\nolimits_{B}\right)  \left(
\underbrace{ab}_{=xz\otimes yw}\right)   &  =\left(  f\otimes
\operatorname*{id}\nolimits_{B}\right)  \left(  xz\otimes yw\right)
=\underbrace{f\left(  xz\right)  }_{\substack{=xf\left(  z\right)  +f\left(
x\right)  z\\\text{(by (\ref{eq.def.deriv.axiom}), applied to}\\x\text{ and
}z\text{ instead of }a\text{ and }b\text{)}}}\otimes
\underbrace{\operatorname*{id}\nolimits_{B}\left(  yw\right)  }_{=yw}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the map }%
f\otimes\operatorname*{id}\nolimits_{B}\right) \\
&  =\left(  xf\left(  z\right)  +f\left(  x\right)  z\right)  \otimes
yw=xf\left(  z\right)  \otimes yw+f\left(  x\right)  z\otimes yw\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the tensor product is }%
\mathbf{k}\text{-bilinear}\right)  .
\end{align*}
Compared with%
\begin{align*}
&  \underbrace{a}_{=x\otimes y}\left(  f\otimes\operatorname*{id}%
\nolimits_{B}\right)  \left(  \underbrace{b}_{=z\otimes w}\right)  +\left(
f\otimes\operatorname*{id}\nolimits_{B}\right)  \left(  \underbrace{a}%
_{=x\otimes y}\right)  \underbrace{b}_{=z\otimes w}\\
&  =\left(  x\otimes y\right)  \cdot\underbrace{\left(  f\otimes
\operatorname*{id}\nolimits_{B}\right)  \left(  z\otimes w\right)
}_{\substack{=f\left(  z\right)  \otimes\operatorname*{id}\nolimits_{B}\left(
w\right)  \\\text{(by the definition of the map }f\otimes\operatorname*{id}%
\nolimits_{B}\text{)}}}+\underbrace{\left(  f\otimes\operatorname*{id}%
\nolimits_{B}\right)  \left(  x\otimes y\right)  }_{\substack{=f\left(
x\right)  \otimes\operatorname*{id}\nolimits_{B}\left(  y\right)  \\\text{(by
the definition of the map }f\otimes\operatorname*{id}\nolimits_{B}\text{)}%
}}\cdot\left(  z\otimes w\right) \\
&  =\left(  x\otimes y\right)  \cdot\left(  f\left(  z\right)  \otimes
\underbrace{\operatorname*{id}\nolimits_{B}\left(  w\right)  }_{=w}\right)
+\left(  f\left(  x\right)  \otimes\underbrace{\operatorname*{id}%
\nolimits_{B}\left(  y\right)  }_{=y}\right)  \cdot\left(  z\otimes w\right)
\\
&  =\underbrace{\left(  x\otimes y\right)  \cdot\left(  f\left(  z\right)
\otimes w\right)  }_{=xf\left(  z\right)  \otimes yw}+\underbrace{\left(
f\left(  x\right)  \otimes y\right)  \cdot\left(  z\otimes w\right)
}_{=f\left(  x\right)  z\otimes yw}=xf\left(  z\right)  \otimes yw+f\left(
x\right)  z\otimes yw,
\end{align*}
this shows that $\left(  f\otimes\operatorname*{id}\nolimits_{B}\right)
\left(  ab\right)  =a\left(  f\otimes\operatorname*{id}\nolimits_{B}\right)
\left(  b\right)  +\left(  f\otimes\operatorname*{id}\nolimits_{B}\right)
\left(  a\right)  b$. Thus, (\ref{pf.prop.der.tensid.a.goal}) is proven.

We thus have shown that $\left(  f\otimes\operatorname*{id}\nolimits_{B}%
\right)  \left(  ab\right)  =a\left(  f\otimes\operatorname*{id}%
\nolimits_{B}\right)  \left(  b\right)  +\left(  f\otimes\operatorname*{id}%
\nolimits_{B}\right)  \left(  a\right)  b$ for every $a\in A\otimes B$ and
$b\in A\otimes B$. Hence, Corollary \ref{cor.deriv.verlong.converse} (applied
to $A\otimes B$, $M\otimes B$ and $f\otimes\operatorname*{id}\nolimits_{B}$
instead of $A$, $M$ and $f$) shows that $f\otimes\operatorname*{id}%
\nolimits_{B}$ is a derivation. This proves Proposition \ref{prop.der.tensid}
\textbf{(a)}.

\textbf{(b)} The proof of Proposition \ref{prop.der.tensid} \textbf{(b)} is
analogous to the proof of Proposition \ref{prop.der.tensid} \textbf{(a)}
(indeed, Proposition \ref{prop.der.tensid} \textbf{(b)} differs from
Proposition \ref{prop.der.tensid} \textbf{(a)} only in the order of the
tensorands), and thus is left to the reader. (Alternatively, Proposition
\ref{prop.der.tensid} \textbf{(b)} can be deduced from Proposition
\ref{prop.der.tensid} \textbf{(a)} via the isomorphisms $A\otimes B\cong
B\otimes A$ and $M\otimes B\cong B\otimes M$.)
\end{proof}
\end{verlong}

As an example of how Propositions \ref{prop.der.compos1} and
\ref{prop.der.compos2} can be used, let us combine them with Proposition
\ref{prop.derivation.unique}:

\begin{corollary}
\label{cor.der.compos-unique}Let $A$ and $B$ be two $\mathbf{k}$-algebras. Let
$f:A\rightarrow B$ be a $\mathbf{k}$-algebra homomorphism. Let $d:A\rightarrow
A$ and $e:B\rightarrow B$ be two derivations. Let $S$ be a subset of $A$ which
generates $A$ as a $\mathbf{k}$-algebra. Assume that $\left(  f\circ d\right)
\mid_{S}=\left(  e\circ f\right)  \mid_{S}$. Then, $f\circ d=e\circ f$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.der.compos-unique}.]We make the $\left(
B,B\right)  $-bimodule $B$ into an $\left(  A,A\right)  $-bimodule via the
$\mathbf{k}$-algebra homomorphism $f:A\rightarrow B$. Proposition
\ref{prop.der.compos1} (applied to $B$ and $e$ instead of $M$ and $d$) shows
that $e\circ f:A\rightarrow B$ is a derivation.

\begin{vershort}
It is easy to see (using the definition of the $\left(  A,A\right)  $-bimodule
structure on $B$ and the fact that $f:A\rightarrow B$ is a $\mathbf{k}%
$-algebra homomorphism) that $f:A\rightarrow B$ is an $\left(  A,A\right)
$-bimodule homomorphism.
\end{vershort}

\begin{verlong}
On the other hand, $f:A\rightarrow B$ is an $\left(  A,A\right)  $-bimodule
homomorphism\footnote{\textit{Proof.} Recall that the $\left(  B,B\right)
$-module $B$ was made into an $\left(  A,A\right)  $-bimodule via the
$\mathbf{k}$-algebra homomorphism $f:A\rightarrow B$. Hence,%
\begin{equation}
\left(  xu=f\left(  x\right)  u\ \ \ \ \ \ \ \ \ \ \text{for every }x\in
A\text{ and }u\in B\right)  \label{pf.cor.der.compos-unique.fn1.1}%
\end{equation}
and%
\begin{equation}
\left(  ux=uf\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }x\in
A\text{ and }u\in B\right)  . \label{pf.cor.der.compos-unique.fn1.2}%
\end{equation}
\par
The map $f$ is a $\mathbf{k}$-algebra homomorphism, and thus $\mathbf{k}%
$-linear. Moreover, every $x\in A$ and $m\in A$ satisfy%
\begin{align*}
f\left(  xm\right)   &  =f\left(  x\right)  f\left(  m\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is a }\mathbf{k}%
\text{-algebra homomorphism}\right) \\
&  =xf\left(  m\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }xf\left(
m\right)  =f\left(  x\right)  f\left(  m\right)  \text{ (by
(\ref{pf.cor.der.compos-unique.fn1.1}), applied to }u=f\left(  m\right)
\text{)}\right)  .
\end{align*}
Hence, the map $f$ is left $A$-linear (since the map $f$ is $\mathbf{k}%
$-linear). Moreover, every $x\in A$ and $m\in A$ satisfy%
\begin{align*}
f\left(  mx\right)   &  =f\left(  m\right)  f\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is a }\mathbf{k}%
\text{-algebra homomorphism}\right) \\
&  =f\left(  m\right)  x\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\left(
m\right)  x=f\left(  m\right)  f\left(  x\right)  \text{ (by
(\ref{pf.cor.der.compos-unique.fn1.2}), applied to }u=f\left(  m\right)
\text{)}\right)  .
\end{align*}
Hence, the map $f$ is right $A$-linear (since the map $f$ is $\mathbf{k}%
$-linear). Now, the map $f$ is an $\left(  A,A\right)  $-bimodule homomorphism
(since $f$ is both left $A$-linear and right $A$-linear). Qed.}.
\end{verlong}

Hence, Proposition \ref{prop.der.compos2} (applied to $M=A$ and $N=B$) shows
that $f\circ d:A\rightarrow B$ is a derivation. Now, Proposition
\ref{prop.derivation.unique} (applied to $B$, $f\circ d$ and $e\circ f$
instead of $M$, $d$ and $e$) shows that $f\circ d=e\circ f$. This proves
Corollary \ref{cor.der.compos-unique}.
\end{proof}

\subsection{Composition powers of derivations}

If $A$ is a $\mathbf{k}$-algebra and $f:A\rightarrow A$ is a derivation, then
$f^{n}$ (for $n\in\mathbb{N}$) is usually not a derivation. However, this does
not mean that nothing can be said about $f^{n}\left(  ab\right)  $. The
following result generalizes the Leibniz law:

\begin{proposition}
\label{prop.der.fn}Let $A$ be a $\mathbf{k}$-algebra. Let $f:A\rightarrow A$
be a derivation. Let $n\in\mathbb{N}$. Let $a\in A$ and $b\in A$. Then,%
\[
f^{n}\left(  ab\right)  =\sum_{k=0}^{n}\dbinom{n}{k}f^{k}\left(  a\right)
f^{n-k}\left(  b\right)  .
\]

\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.der.fn}.]We shall prove Proposition
\ref{prop.der.fn} by induction over $n$:

\begin{vershort}
\textit{Induction base:} Proposition \ref{prop.der.fn} is easily verified in
the case when $n=0$ (because $f^{0}=\operatorname*{id}$). This completes the
induction base.
\end{vershort}

\begin{verlong}
\textit{Induction base:} We have
\[
\sum_{k=0}^{0}\dbinom{0}{k}f^{k}\left(  a\right)  f^{0-k}\left(  b\right)
=\underbrace{\dbinom{0}{0}}_{=1}\underbrace{f^{0}}_{=\operatorname*{id}%
}\left(  a\right)  \underbrace{f^{0-0}}_{=f^{0}=\operatorname*{id}}\left(
b\right)  =1\underbrace{\operatorname*{id}\left(  a\right)  }_{=a}%
\underbrace{\operatorname*{id}\left(  b\right)  }_{=b}=1ab=ab.
\]
Comparing this with $\underbrace{f^{0}}_{=\operatorname*{id}}\left(
ab\right)  =\operatorname*{id}\left(  ab\right)  =ab$, we obtain $f^{0}\left(
ab\right)  =\sum_{k=0}^{0}\dbinom{0}{k}f^{k}\left(  a\right)  f^{0-k}\left(
b\right)  $. In other words, Proposition \ref{prop.der.fn} holds in the case
when $n=0$. This completes the induction base.
\end{verlong}

\textit{Induction step:} Let $N$ be a positive integer. Assume that
Proposition \ref{prop.der.fn} holds in the case when $n=N-1$. We now must
prove that Proposition \ref{prop.der.fn} holds in the case when $n=N$.

Recall the recurrence relation for the binomial coefficients. It says that
\begin{equation}
\dbinom{a}{b}=\dbinom{a-1}{b}+\dbinom{a-1}{b-1} \label{pf.prop.der.fn.recbin}%
\end{equation}
for every $a\in\mathbb{Z}$ and every positive integer $b$. Hence, every
positive integer $k$ satisfies%
\begin{equation}
\dbinom{N}{k}=\dbinom{N-1}{k}+\dbinom{N-1}{k-1}
\label{pf.prop.der.fn.recbin.app}%
\end{equation}
(by (\ref{pf.prop.der.fn.recbin}), applied to $a=N$ and $b=k$).

We have assumed that Proposition \ref{prop.der.fn} holds in the case when
$n=N-1$. In other words, we have%
\[
f^{N-1}\left(  ab\right)  =\sum_{k=0}^{N-1}\dbinom{N-1}{k}f^{k}\left(
a\right)  f^{\left(  N-1\right)  -k}\left(  b\right)  .
\]


\begin{verlong}
Now, recall that $f:A\rightarrow A$ is a derivation. Hence, Corollary
\ref{cor.deriv.verlong} (applied to $M=A$) shows that $f$ is a $\mathbf{k}%
$-linear map and satisfies (\ref{eq.def.deriv.axiom}).
\end{verlong}

\begin{vershort}
Now,%
\begin{align}
&  \underbrace{f^{N}}_{=f\circ f^{N-1}}\left(  ab\right) \nonumber\\
&  =\left(  f\circ f^{N-1}\right)  \left(  ab\right)  =f\left(
\underbrace{f^{N-1}\left(  ab\right)  }_{=\sum_{k=0}^{N-1}\dbinom{N-1}{k}%
f^{k}\left(  a\right)  f^{\left(  N-1\right)  -k}\left(  b\right)  }\right)
\nonumber\\
&  =f\left(  \sum_{k=0}^{N-1}\dbinom{N-1}{k}f^{k}\left(  a\right)  f^{\left(
N-1\right)  -k}\left(  b\right)  \right) \nonumber\\
&  =\sum_{k=0}^{N-1}\dbinom{N-1}{k}\underbrace{f\left(  f^{k}\left(  a\right)
f^{\left(  N-1\right)  -k}\left(  b\right)  \right)  }_{\substack{=f^{k}%
\left(  a\right)  f\left(  f^{\left(  N-1\right)  -k}\left(  b\right)
\right)  +f\left(  f^{k}\left(  a\right)  \right)  f^{\left(  N-1\right)
-k}\left(  b\right)  \\\text{(since }f\text{ is a derivation)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is a }\mathbf{k}%
\text{-linear map}\right) \nonumber\\
&  =\sum_{k=0}^{N-1}\dbinom{N-1}{k}\left(  f^{k}\left(  a\right)  f\left(
f^{\left(  N-1\right)  -k}\left(  b\right)  \right)  +f\left(  f^{k}\left(
a\right)  \right)  f^{\left(  N-1\right)  -k}\left(  b\right)  \right)
\nonumber\\
&  =\sum_{k=0}^{N-1}\dbinom{N-1}{k}f^{k}\left(  a\right)  f\left(  f^{\left(
N-1\right)  -k}\left(  b\right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{k=0}^{N-1}\dbinom{N-1}{k}f\left(  f^{k}\left(
a\right)  \right)  f^{\left(  N-1\right)  -k}\left(  b\right)  .
\label{pf.prop.der.fn.short.1}%
\end{align}
But%
\begin{align*}
&  \sum_{k=0}^{N-1}\dbinom{N-1}{k}f^{k}\left(  a\right)  \underbrace{f\left(
f^{\left(  N-1\right)  -k}\left(  b\right)  \right)  }_{\substack{=\left(
f\circ f^{\left(  N-1\right)  -k}\right)  \left(  b\right)  \\=\left(  f\circ
f^{\left(  N-k\right)  -1}\right)  \left(  b\right)  \\=f^{N-k}\left(
b\right)  }}\\
&  =\sum_{k=0}^{N-1}\dbinom{N-1}{k}f^{k}\left(  a\right)  f^{N-k}\left(
b\right) \\
&  =\underbrace{\dbinom{N-1}{0}}_{=1}f^{0}\left(  a\right)  f^{N-0}\left(
b\right)  +\sum_{k=1}^{N-1}\dbinom{N-1}{k}f^{k}\left(  a\right)
f^{N-k}\left(  b\right) \\
&  =f^{0}\left(  a\right)  f^{N-0}\left(  b\right)  +\sum_{k=1}^{N-1}%
\dbinom{N-1}{k}f^{k}\left(  a\right)  f^{N-k}\left(  b\right)
\end{align*}
and
\begin{align*}
&  \sum_{k=0}^{N-1}\dbinom{N-1}{k}f\left(  f^{k}\left(  a\right)  \right)
f^{\left(  N-1\right)  -k}\left(  b\right) \\
&  =\sum_{k=1}^{N}\dbinom{N-1}{k-1}\underbrace{f\left(  f^{k-1}\left(
a\right)  \right)  }_{\substack{=\left(  f\circ f^{k-1}\right)  \left(
a\right)  \\=f^{k}\left(  a\right)  }}\underbrace{f^{\left(  N-1\right)
-\left(  k-1\right)  }}_{=f^{N-k}}\left(  b\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }k-1\text{ for
}k\text{ in the sum}\right) \\
&  =\sum_{k=1}^{N}\dbinom{N-1}{k-1}f^{k}\left(  a\right)  f^{N-k}\left(
b\right) \\
&  =\sum_{k=1}^{N-1}\dbinom{N-1}{k-1}f^{k}\left(  a\right)  f^{N-k}\left(
b\right)  +\underbrace{\dbinom{N-1}{N-1}}_{=1}f^{N}\left(  a\right)
f^{N-N}\left(  b\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}k=N\text{ from the sum}\right) \\
&  =\sum_{k=1}^{N-1}\dbinom{N-1}{k-1}f^{k}\left(  a\right)  f^{N-k}\left(
b\right)  +f^{N}\left(  a\right)  f^{N-N}\left(  b\right)  .
\end{align*}
Hence, (\ref{pf.prop.der.fn.short.1}) becomes%
\begin{align*}
&  f^{N}\left(  ab\right) \\
&  =\underbrace{\sum_{k=0}^{N-1}\dbinom{N-1}{k}f^{k}\left(  a\right)  f\left(
f^{\left(  N-1\right)  -k}\left(  b\right)  \right)  }_{=f^{0}\left(
a\right)  f^{N-0}\left(  b\right)  +\sum_{k=1}^{N-1}\dbinom{N-1}{k}%
f^{k}\left(  a\right)  f^{N-k}\left(  b\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum_{k=0}^{N-1}\dbinom{N-1}{k}f\left(
f^{k}\left(  a\right)  \right)  f^{\left(  N-1\right)  -k}\left(  b\right)
}_{=\sum_{k=1}^{N-1}\dbinom{N-1}{k-1}f^{k}\left(  a\right)  f^{N-k}\left(
b\right)  +f^{N}\left(  a\right)  f^{N-N}\left(  b\right)  }\\
&  =f^{0}\left(  a\right)  f^{N-0}\left(  b\right)  +\sum_{k=1}^{N-1}%
\dbinom{N-1}{k}f^{k}\left(  a\right)  f^{N-k}\left(  b\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{k=1}^{N-1}\dbinom{N-1}{k-1}f^{k}\left(
a\right)  f^{N-k}\left(  b\right)  +f^{N}\left(  a\right)  f^{N-N}\left(
b\right) \\
&  =f^{0}\left(  a\right)  f^{N-0}\left(  b\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum_{k=1}^{N-1}\dbinom{N-1}{k}%
f^{k}\left(  a\right)  f^{N-k}\left(  b\right)  +\sum_{k=1}^{N-1}\dbinom
{N-1}{k-1}f^{k}\left(  a\right)  f^{N-k}\left(  b\right)  }_{=\sum_{k=1}%
^{N-1}\left(  \dbinom{N-1}{k}+\dbinom{N-1}{k-1}\right)  f^{k}\left(  a\right)
f^{N-k}\left(  b\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ +f^{N}\left(  a\right)  f^{N-N}\left(  b\right) \\
&  =f^{0}\left(  a\right)  f^{N-0}\left(  b\right)  +\sum_{k=1}^{N-1}%
\underbrace{\left(  \dbinom{N-1}{k}+\dbinom{N-1}{k-1}\right)  }%
_{\substack{=\dbinom{N}{k}\\\text{(by (\ref{pf.prop.der.fn.recbin.app}))}%
}}f^{k}\left(  a\right)  f^{N-k}\left(  b\right) \\
&  \ \ \ \ \ \ \ \ \ \ +f^{N}\left(  a\right)  f^{N-N}\left(  b\right) \\
&  =f^{0}\left(  a\right)  f^{N-0}\left(  b\right)  +\sum_{k=1}^{N-1}%
\dbinom{N}{k}f^{k}\left(  a\right)  f^{N-k}\left(  b\right)  +f^{N}\left(
a\right)  f^{N-N}\left(  b\right)  .
\end{align*}
Comparing this with%
\begin{align*}
&  \sum_{k=0}^{N}\dbinom{N}{k}f^{k}\left(  a\right)  f^{N-k}\left(  b\right)
\\
&  =\underbrace{\dbinom{N}{0}}_{=1}f^{0}\left(  a\right)  f^{N-0}\left(
b\right)  +\sum_{k=1}^{N-1}\dbinom{N}{k}f^{k}\left(  a\right)  f^{N-k}\left(
b\right)  +\underbrace{\dbinom{N}{N}}_{=1}f^{N}\left(  a\right)
f^{N-N}\left(  b\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addends for
}k=0\text{ and for }k=N\text{ from the sum}\right) \\
&  =f^{0}\left(  a\right)  f^{N-0}\left(  b\right)  +\sum_{k=1}^{N-1}%
\dbinom{N}{k}f^{k}\left(  a\right)  f^{N-k}\left(  b\right)  +f^{N}\left(
a\right)  f^{N-N}\left(  b\right)  ,
\end{align*}
we obtain $f^{N}\left(  ab\right)  =\sum_{k=0}^{N}\dbinom{N}{k}f^{k}\left(
a\right)  f^{N-k}\left(  b\right)  $. In other words, Proposition
\ref{prop.der.fn} holds in the case when $n=N$. This completes the induction
step. Thus, the induction proof of Proposition \ref{prop.der.fn} is complete.
\end{vershort}

\begin{verlong}
Now, $f^{N}=f\circ f^{N-1}$. Hence,%
\begin{align}
&  \underbrace{f^{N}}_{=f\circ f^{N-1}}\left(  ab\right) \nonumber\\
&  =\left(  f\circ f^{N-1}\right)  \left(  ab\right)  =f\left(
\underbrace{f^{N-1}\left(  ab\right)  }_{=\sum_{k=0}^{N-1}\dbinom{N-1}{k}%
f^{k}\left(  a\right)  f^{\left(  N-1\right)  -k}\left(  b\right)  }\right)
\nonumber\\
&  =f\left(  \sum_{k=0}^{N-1}\dbinom{N-1}{k}f^{k}\left(  a\right)  f^{\left(
N-1\right)  -k}\left(  b\right)  \right) \nonumber\\
&  =\sum_{k=0}^{N-1}\dbinom{N-1}{k}\underbrace{f\left(  f^{k}\left(  a\right)
f^{\left(  N-1\right)  -k}\left(  b\right)  \right)  }_{\substack{=f^{k}%
\left(  a\right)  f\left(  f^{\left(  N-1\right)  -k}\left(  b\right)
\right)  +f\left(  f^{k}\left(  a\right)  \right)  f^{\left(  N-1\right)
-k}\left(  b\right)  \\\text{(by (\ref{eq.def.deriv.axiom}), applied to }%
f^{k}\left(  a\right)  \text{,}\\f^{\left(  N-1\right)  -k}\left(  b\right)
\text{ and }M\text{ instead of }a\text{, }b\text{ and }f\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is a }\mathbf{k}%
\text{-linear map}\right) \nonumber\\
&  =\sum_{k=0}^{N-1}\dbinom{N-1}{k}\left(  f^{k}\left(  a\right)  f\left(
f^{\left(  N-1\right)  -k}\left(  b\right)  \right)  +f\left(  f^{k}\left(
a\right)  \right)  f^{\left(  N-1\right)  -k}\left(  b\right)  \right)
\nonumber\\
&  =\sum_{k=0}^{N-1}\dbinom{N-1}{k}f^{k}\left(  a\right)  f\left(  f^{\left(
N-1\right)  -k}\left(  b\right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{k=0}^{N-1}\dbinom{N-1}{k}f\left(  f^{k}\left(
a\right)  \right)  f^{\left(  N-1\right)  -k}\left(  b\right)  .
\label{pf.prop.der.fn.1}%
\end{align}
But%
\begin{align*}
&  \sum_{k=0}^{N-1}\dbinom{N-1}{k}f^{k}\left(  a\right)  \underbrace{f\left(
f^{\left(  N-1\right)  -k}\left(  b\right)  \right)  }_{\substack{=\left(
f\circ f^{\left(  N-1\right)  -k}\right)  \left(  b\right)  \\=\left(  f\circ
f^{\left(  N-k\right)  -1}\right)  \left(  b\right)  \\\text{(since }\left(
N-1\right)  -k=\left(  N-k\right)  -1\text{)}}}\\
&  =\sum_{k=0}^{N-1}\dbinom{N-1}{k}f^{k}\left(  a\right)  \underbrace{\left(
f\circ f^{\left(  N-k\right)  -1}\right)  }_{=f^{N-k}}\left(  b\right) \\
&  =\sum_{k=0}^{N-1}\dbinom{N-1}{k}f^{k}\left(  a\right)  f^{N-k}\left(
b\right) \\
&  =\underbrace{\dbinom{N-1}{0}}_{=1}f^{0}\left(  a\right)  f^{N-0}\left(
b\right)  +\sum_{k=1}^{N-1}\dbinom{N-1}{k}f^{k}\left(  a\right)
f^{N-k}\left(  b\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}k=0\text{ from the sum}\right) \\
&  =f^{0}\left(  a\right)  f^{N-0}\left(  b\right)  +\sum_{k=1}^{N-1}%
\dbinom{N-1}{k}f^{k}\left(  a\right)  f^{N-k}\left(  b\right)
\end{align*}
and
\begin{align*}
&  \sum_{k=0}^{N-1}\dbinom{N-1}{k}\underbrace{f\left(  f^{k}\left(  a\right)
\right)  }_{=\left(  f\circ f^{k}\right)  \left(  a\right)  }f^{\left(
N-1\right)  -k}\left(  b\right) \\
&  =\sum_{k=0}^{N-1}\dbinom{N-1}{k}\left(  f\circ f^{k}\right)  \left(
a\right)  f^{\left(  N-1\right)  -k}\left(  b\right) \\
&  =\sum_{k=1}^{N}\dbinom{N-1}{k-1}\underbrace{\left(  f\circ f^{k-1}\right)
}_{=f^{k}}\left(  a\right)  \underbrace{f^{\left(  N-1\right)  -\left(
k-1\right)  }}_{\substack{=f^{N-k}\\\text{(since }\left(  N-1\right)  -\left(
k-1\right)  =N-k\text{)}}}\left(  b\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }k-1\text{ for
}k\text{ in the sum}\right) \\
&  =\sum_{k=1}^{N}\dbinom{N-1}{k-1}f^{k}\left(  a\right)  f^{N-k}\left(
b\right) \\
&  =\sum_{k=1}^{N-1}\dbinom{N-1}{k-1}f^{k}\left(  a\right)  f^{N-k}\left(
b\right)  +\underbrace{\dbinom{N-1}{N-1}}_{=1}f^{N}\left(  a\right)
f^{N-N}\left(  b\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}k=N\text{ from the sum}\right) \\
&  =\sum_{k=1}^{N-1}\dbinom{N-1}{k-1}f^{k}\left(  a\right)  f^{N-k}\left(
b\right)  +f^{N}\left(  a\right)  f^{N-N}\left(  b\right)  .
\end{align*}
Hence, (\ref{pf.prop.der.fn.1}) becomes%
\begin{align*}
&  f^{N}\left(  ab\right) \\
&  =\underbrace{\sum_{k=0}^{N-1}\dbinom{N-1}{k}f^{k}\left(  a\right)  f\left(
f^{\left(  N-1\right)  -k}\left(  b\right)  \right)  }_{=f^{0}\left(
a\right)  f^{N-0}\left(  b\right)  +\sum_{k=1}^{N-1}\dbinom{N-1}{k}%
f^{k}\left(  a\right)  f^{N-k}\left(  b\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum_{k=0}^{N-1}\dbinom{N-1}{k}f\left(
f^{k}\left(  a\right)  \right)  f^{\left(  N-1\right)  -k}\left(  b\right)
}_{=\sum_{k=1}^{N-1}\dbinom{N-1}{k-1}f^{k}\left(  a\right)  f^{N-k}\left(
b\right)  +f^{N}\left(  a\right)  f^{N-N}\left(  b\right)  }\\
&  =f^{0}\left(  a\right)  f^{N-0}\left(  b\right)  +\sum_{k=1}^{N-1}%
\dbinom{N-1}{k}f^{k}\left(  a\right)  f^{N-k}\left(  b\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{k=1}^{N-1}\dbinom{N-1}{k-1}f^{k}\left(
a\right)  f^{N-k}\left(  b\right)  +f^{N}\left(  a\right)  f^{N-N}\left(
b\right) \\
&  =f^{0}\left(  a\right)  f^{N-0}\left(  b\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum_{k=1}^{N-1}\dbinom{N-1}{k}%
f^{k}\left(  a\right)  f^{N-k}\left(  b\right)  +\sum_{k=1}^{N-1}\dbinom
{N-1}{k-1}f^{k}\left(  a\right)  f^{N-k}\left(  b\right)  }_{=\sum_{k=1}%
^{N-1}\left(  \dbinom{N-1}{k}+\dbinom{N-1}{k-1}\right)  f^{k}\left(  a\right)
f^{N-k}\left(  b\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ +f^{N}\left(  a\right)  f^{N-N}\left(  b\right) \\
&  =f^{0}\left(  a\right)  f^{N-0}\left(  b\right)  +\sum_{k=1}^{N-1}%
\underbrace{\left(  \dbinom{N-1}{k}+\dbinom{N-1}{k-1}\right)  }%
_{\substack{=\dbinom{N}{k}\\\text{(by (\ref{pf.prop.der.fn.recbin.app}))}%
}}f^{k}\left(  a\right)  f^{N-k}\left(  b\right) \\
&  \ \ \ \ \ \ \ \ \ \ +f^{N}\left(  a\right)  f^{N-N}\left(  b\right) \\
&  =f^{0}\left(  a\right)  f^{N-0}\left(  b\right)  +\sum_{k=1}^{N-1}%
\dbinom{N}{k}f^{k}\left(  a\right)  f^{N-k}\left(  b\right)  +f^{N}\left(
a\right)  f^{N-N}\left(  b\right)  .
\end{align*}
Comparing this with%
\begin{align*}
&  \sum_{k=0}^{N}\dbinom{N}{k}f^{k}\left(  a\right)  f^{N-k}\left(  b\right)
\\
&  =\dbinom{N}{0}f^{0}\left(  a\right)  f^{N-0}\left(  b\right)
+\underbrace{\sum_{k=1}^{N}\dbinom{N}{k}f^{k}\left(  a\right)  f^{N-k}\left(
b\right)  }_{\substack{=\sum_{k=1}^{N-1}\dbinom{N}{k}f^{k}\left(  a\right)
f^{N-k}\left(  b\right)  +\dbinom{N}{N}f^{N}\left(  a\right)  f^{N-N}\left(
b\right)  \\\text{(here, we have split off the addend for }k=N\text{ from the
sum}\\\text{(since }N\in\left\{  1,2,\ldots,N\right\}  \text{))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}k=0\text{ from the sum}\right) \\
&  =\underbrace{\dbinom{N}{0}}_{=1}f^{0}\left(  a\right)  f^{N-0}\left(
b\right)  +\sum_{k=1}^{N-1}\dbinom{N}{k}f^{k}\left(  a\right)  f^{N-k}\left(
b\right)  +\underbrace{\dbinom{N}{N}}_{=1}f^{N}\left(  a\right)
f^{N-N}\left(  b\right) \\
&  =f^{0}\left(  a\right)  f^{N-0}\left(  b\right)  +\sum_{k=1}^{N-1}%
\dbinom{N}{k}f^{k}\left(  a\right)  f^{N-k}\left(  b\right)  +f^{N}\left(
a\right)  f^{N-N}\left(  b\right)  ,
\end{align*}
we obtain $f^{N}\left(  ab\right)  =\sum_{k=0}^{N}\dbinom{N}{k}f^{k}\left(
a\right)  f^{N-k}\left(  b\right)  $. In other words, Proposition
\ref{prop.der.fn} holds in the case when $n=N$. This completes the induction
step. Thus, the induction proof of Proposition \ref{prop.der.fn} is complete.
\end{verlong}
\end{proof}

As a consequence of Proposition \ref{prop.der.fn}, we can easily see the following:

\begin{corollary}
\label{cor.der.fp}Let $A$ be a $\mathbf{k}$-algebra. Let $p$ be a prime number
such that $p1_{A}=0$. Let $f:A\rightarrow A$ be a derivation. Then,
$f^{p}:A\rightarrow A$ is a derivation.
\end{corollary}

\begin{vershort}
\begin{proof}
[Proof of Corollary \ref{cor.der.fp}.]A well-known property of binomial
coefficients\footnote{See, for example, \cite[Corollary 5.6]{Gri-fleck} for a
proof of this property (although what we call $q$ is denoted by $p$ in
\cite[Corollary 5.6]{Gri-fleck}).} states that if $q$ is a prime number, then
every $k\in\left\{  1,2,\ldots,q-1\right\}  $ satisfies $q\mid\dbinom{q}{k}$.
Applying this to $q=p$, we obtain the following: Every $k\in\left\{
1,2,\ldots,p-1\right\}  $ satisfies%
\begin{equation}
p\mid\dbinom{p}{k}. \label{pf.cor.der.fp.short.div}%
\end{equation}
Hence, every $k\in\left\{  1,2,\ldots,p-1\right\}  $ and every $a\in A$
satisfy%
\begin{equation}
\dbinom{p}{k}a=0 \label{pf.cor.der.fp.short.01}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.cor.der.fp.short.01}):} Let $k\in\left\{
1,2,\ldots,p-1\right\}  $ and $a\in A$. Then, (\ref{pf.cor.der.fp.short.div})
yields $p\mid\dbinom{p}{k}$. In other words, there exists an integer $z$ such
that $\dbinom{p}{k}=zp$. Consider this $z$. Now, $\underbrace{\dbinom{p}{k}%
}_{=zp}\underbrace{a}_{=1_{A}\cdot a}=z\underbrace{p1_{A}}_{=0}\cdot a=0$.
This proves (\ref{pf.cor.der.fp.short.01}).}.

Now, for every $a\in A$ and $b\in A$, we have%
\begin{align*}
&  f^{p}\left(  ab\right)  =\sum_{k=0}^{p}\dbinom{p}{k}f^{k}\left(  a\right)
f^{p-k}\left(  b\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition
\ref{prop.der.fn}, applied to }n=p\right) \\
&  =\underbrace{\dbinom{p}{0}}_{=1}\underbrace{f^{0}}_{=\operatorname*{id}%
}\left(  a\right)  \underbrace{f^{p-0}}_{=f^{p}}\left(  b\right)  +\sum
_{k=1}^{p-1}\underbrace{\dbinom{p}{k}f^{k}\left(  a\right)  }%
_{\substack{=0\\\text{(by (\ref{pf.cor.der.fp.short.01}), applied to}%
\\f^{k}\left(  a\right)  \text{ instead of }a\text{)}}}f^{p-k}\left(
b\right)  +\underbrace{\dbinom{p}{p}}_{=1}f^{p}\left(  a\right)
\underbrace{f^{p-p}}_{=f^{0}=\operatorname*{id}}\left(  b\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addends for
}k=0\text{ and for }k=p\text{ from the sum}\right) \\
&  =\underbrace{\operatorname*{id}\left(  a\right)  }_{=a}f^{p}\left(
b\right)  +\underbrace{\sum_{k=1}^{p-1}0f^{p-k}\left(  b\right)  }_{=0}%
+f^{p}\left(  a\right)  \underbrace{\operatorname*{id}\left(  b\right)  }%
_{=b}=af^{p}\left(  b\right)  +f^{p}\left(  a\right)  b.
\end{align*}
Hence, $f^{p}$ is a derivation. This proves Corollary \ref{cor.der.fp}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Corollary \ref{cor.der.fp}.]The map $f:A\rightarrow A$ is a
derivation. Hence, Corollary \ref{cor.deriv.verlong} (applied to $M=A$) shows
that $f$ is a $\mathbf{k}$-linear map and satisfies (\ref{eq.def.deriv.axiom}).

The map $f^{p}$ is $\mathbf{k}$-linear (since the map $f$ is $\mathbf{k}$-linear).

A well-known property of binomial coefficients\footnote{See, for example,
\cite[Corollary 5.6]{Gri-fleck} for a proof of this property (although what we
call $q$ is denoted by $p$ in \cite[Corollary 5.6]{Gri-fleck}).} states that
if $q$ is a prime number, then every $k\in\left\{  1,2,\ldots,q-1\right\}  $
satisfies $q\mid\dbinom{q}{k}$. Applying this to $q=p$, we obtain the
following: Every $k\in\left\{  1,2,\ldots,p-1\right\}  $ satisfies%
\begin{equation}
p\mid\dbinom{p}{k}. \label{pf.cor.der.fp.div}%
\end{equation}
Hence, every $k\in\left\{  1,2,\ldots,p-1\right\}  $ and every $a\in A$
satisfy%
\begin{equation}
\dbinom{p}{k}a=0 \label{pf.cor.der.fp.01}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.cor.der.fp.01}):} Let $k\in\left\{
1,2,\ldots,p-1\right\}  $ and $a\in A$. Then, (\ref{pf.cor.der.fp.div}) yields
$p\mid\dbinom{p}{k}$. In other words, there exists an integer $z$ such that
$\dbinom{p}{k}=zp$. Consider this $z$. Now, $\underbrace{\dbinom{p}{k}}%
_{=zp}\underbrace{a}_{=1_{A}\cdot a}=z\underbrace{p1_{A}}_{=0}\cdot a=0$. This
proves (\ref{pf.cor.der.fp.01}).}.

Now, for every $a\in A$ and $b\in A$, we have%
\begin{align*}
&  f^{p}\left(  ab\right)  =\sum_{k=0}^{p}\dbinom{p}{k}f^{k}\left(  a\right)
f^{p-k}\left(  b\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition
\ref{prop.der.fn}, applied to }n=p\right) \\
&  =\dbinom{p}{0}f^{0}\left(  a\right)  f^{p-0}\left(  b\right)
+\underbrace{\sum_{k=1}^{p}\dbinom{p}{k}f^{k}\left(  a\right)  f^{p-k}\left(
b\right)  }_{\substack{=\sum_{k=1}^{p-1}\dbinom{p}{k}f^{k}\left(  a\right)
f^{p-k}\left(  b\right)  +\dbinom{p}{p}f^{p}\left(  a\right)  f^{p-p}\left(
b\right)  \\\text{(here, we have split off the addend for }k=p\text{ from the
sum}\\\text{(since }p\in\left\{  1,2,\ldots,p\right\}  \text{))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}k=0\text{ from the sum}\right) \\
&  =\underbrace{\dbinom{p}{0}}_{=1}\underbrace{f^{0}}_{=\operatorname*{id}%
}\left(  a\right)  \underbrace{f^{p-0}}_{=f^{p}}\left(  b\right)  +\sum
_{k=1}^{p-1}\underbrace{\dbinom{p}{k}f^{k}\left(  a\right)  }%
_{\substack{=0\\\text{(by (\ref{pf.cor.der.fp.01}), applied to}\\f^{k}\left(
a\right)  \text{ instead of }a\text{)}}}f^{p-k}\left(  b\right)
+\underbrace{\dbinom{p}{p}}_{=1}f^{p}\left(  a\right)  \underbrace{f^{p-p}%
}_{=f^{0}=\operatorname*{id}}\left(  b\right) \\
&  =\operatorname*{id}\left(  a\right)  f^{p}\left(  b\right)
+\underbrace{\sum_{k=1}^{p-1}0f^{p-k}\left(  b\right)  }_{=0}+f^{p}\left(
a\right)  \operatorname*{id}\left(  b\right) \\
&  =\underbrace{\operatorname*{id}\left(  a\right)  }_{=a}f^{p}\left(
b\right)  +f^{p}\left(  a\right)  \underbrace{\operatorname*{id}\left(
b\right)  }_{=b}=af^{p}\left(  b\right)  +f^{p}\left(  a\right)  b.
\end{align*}
Hence, Corollary \ref{cor.deriv.verlong.converse} (applied to $A$ and $f^{p}$
instead of $M$ and $f$) shows that $f^{p}$ is a derivation. This proves
Corollary \ref{cor.der.fp}.
\end{proof}
\end{verlong}

\begin{example}
Let $A$ be a $\mathbf{k}$-algebra. Let $p$ be a prime number. Assume that
$p1_{A}=0$. Let $x\in A$. Then, Example \ref{exa.ad} (applied to $x$ instead
of $p$) defines a derivation $\operatorname*{ad}\nolimits_{x}:A\rightarrow A$
which sends every $a\in A$ to $xa-ax$. Corollary \ref{cor.der.fp} thus shows
that $\left(  \operatorname*{ad}\nolimits_{x}\right)  ^{p}$ is a derivation as
well. It can be shown that it is actually an inner derivation: $\left(
\operatorname*{ad}\nolimits_{x}\right)  ^{p}=\operatorname*{ad}%
\nolimits_{x^{p}}$. (Hint for the proof: Define two $\mathbf{k}$-linear maps
$L_{x}:A\rightarrow A,\ a\mapsto xa$ and $R_{x}:A\rightarrow A,\ a\mapsto ax$,
and prove that these two maps commute. Thus, the binomial formula can be used
to compute $\left(  L_{x}-R_{x}\right)  ^{p}$. Finally observe that
$\operatorname*{ad}\nolimits_{x}=L_{x}-R_{x}$.)
\end{example}

\subsection{A product formula for the Wronskian}

A consequence of Proposition \ref{prop.der.fn} is a certain property of the
Wronskian determinant, which we shall now define:\footnote{We are using the
notation $\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}$ for the
$n\times m$-matrix whose $\left(  i,j\right)  $-th entry is $a_{i,j}$.}

\begin{definition}
\label{def.wronski.gen}Let $A$ be a commutative $\mathbf{k}$-algebra. Let
$f:A\rightarrow A$ be a derivation. Let $n\in\mathbb{N}$. Let $a_{1}%
,a_{2},\ldots,a_{n}\in A$. Then, the $f$\textit{-Wronskian} of $a_{1}%
,a_{2},\ldots,a_{n}$ is defined to be the determinant%
\begin{align*}
&  \det\left(  \left(  f^{j-1}\left(  a_{i}\right)  \right)  _{1\leq i\leq
n,\ 1\leq j\leq n}\right) \\
&  =\det\left(
\begin{array}
[c]{cccc}%
f^{0}\left(  a_{1}\right)  & f^{1}\left(  a_{1}\right)  & \cdots &
f^{n-1}\left(  a_{1}\right) \\
f^{0}\left(  a_{2}\right)  & f^{1}\left(  a_{2}\right)  & \cdots &
f^{n-1}\left(  a_{2}\right) \\
\vdots & \vdots & \ddots & \vdots\\
f^{0}\left(  a_{n}\right)  & f^{1}\left(  a_{n}\right)  & \cdots &
f^{n-1}\left(  a_{n}\right)
\end{array}
\right)  \in A.
\end{align*}
This $f$-Wronskian is denoted by $W_{f}\left(  a_{1},a_{2},\ldots
,a_{n}\right)  $.
\end{definition}

\begin{theorem}
\label{thm.wronski.gen-prod}Let $A$ be a commutative $\mathbf{k}$-algebra. Let
$f:A\rightarrow A$ be a derivation. Let $n\in\mathbb{N}$. Let $a_{1}%
,a_{2},\ldots,a_{n}\in A$. Let $a\in A$. Then,%
\[
W_{f}\left(  aa_{1},aa_{2},\ldots,aa_{n}\right)  =a^{n}W_{f}\left(
a_{1},a_{2},\ldots,a_{n}\right)  .
\]

\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.wronski.gen-prod}.]The definition of the
$f$-Wronskian $W_{f}\left(  a_{1},a_{2},\ldots,a_{n}\right)  $ yields%
\begin{equation}
W_{f}\left(  a_{1},a_{2},\ldots,a_{n}\right)  =\det\left(  \left(
f^{j-1}\left(  a_{i}\right)  \right)  _{1\leq i\leq n,\ 1\leq j\leq n}\right)
. \label{pf.thm.wronski.gen-prod.Wf1}%
\end{equation}
The same argument (applied to $aa_{i}$ instead of $a_{i}$) yields%
\begin{equation}
W_{f}\left(  aa_{1},aa_{2},\ldots,aa_{n}\right)  =\det\left(  \left(
f^{j-1}\left(  aa_{i}\right)  \right)  _{1\leq i\leq n,\ 1\leq j\leq
n}\right)  . \label{pf.thm.wronski.gen-prod.Wf2}%
\end{equation}


Each $\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  \times\left\{
1,2,\ldots,n\right\}  $ satisfies $j-1\in\mathbb{N}$ (since $j\geq1$) and thus%
\begin{align}
f^{j-1}\left(  aa_{i}\right)   &  =\sum_{k=0}^{j-1}\dbinom{j-1}{k}f^{k}\left(
a\right)  f^{j-1-k}\left(  a_{i}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition \ref{prop.der.fn}, applied
to }n=j-1\text{ and }b=a_{i}\right) \nonumber\\
&  =\sum_{k=1}^{j}\dbinom{j-1}{k-1}f^{k-1}\left(  a\right)
\underbrace{f^{j-1-\left(  k-1\right)  }}_{\substack{=f^{j-k}\\\text{(since
}j-1-\left(  k-1\right)  =j-1\text{)}}}\left(  a_{i}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }k-1\text{ for
}k\text{ in the sum}\right) \nonumber\\
&  =\sum_{k=1}^{j}\dbinom{j-1}{k-1}f^{k-1}\left(  a\right)  f^{j-k}\left(
a_{i}\right)  . \label{pf.thm.wronski.gen-prod.faa=1}%
\end{align}


For each $\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  \times\left\{
1,2,\ldots,n\right\}  $, we define an element $c_{i,j}$ of $A$ by%
\begin{equation}
c_{i,j}=%
\begin{cases}
\dbinom{j-1}{i-1}f^{j-i}\left(  a\right)  , & \text{if }j\geq i;\\
0, & \text{if }j<i.
\end{cases}
\label{pf.thm.wronski.gen-prod.cij=}%
\end{equation}
Note that this is well-defined (because if $j\geq i$, then $j-i\in\mathbb{N}$
and therefore $f^{j-i}$ is a well-defined map).

\begin{vershort}
For each $i\in\left\{  1,2,\ldots,n\right\}  $, we have $i\geq i$ and thus%
\begin{align}
c_{i,i}  &  =\underbrace{\dbinom{i-1}{i-1}}_{\substack{=1}}\underbrace{f^{i-i}%
\left(  a\right)  }_{\substack{=f^{0}\left(  a\right)  \\\text{(since
}i-i=0\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}c_{i,i}\right) \nonumber\\
&  =\underbrace{f^{0}}_{=\operatorname*{id}\nolimits_{A}}\left(  a\right)  =a.
\label{pf.thm.wronski.gen-prod.short.cii=}%
\end{align}

\end{vershort}

\begin{verlong}
If $\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  \times\left\{
1,2,\ldots,n\right\}  $ satisfies $i\leq j$, then we have%
\begin{align}
c_{i,j}  &  =%
\begin{cases}
\dbinom{j-1}{i-1}f^{j-i}\left(  a\right)  , & \text{if }j\geq i;\\
0, & \text{if }j<i
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.wronski.gen-prod.cij=}%
)}\right) \nonumber\\
&  =\dbinom{j-1}{i-1}f^{j-i}\left(  a\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }j\geq i\text{ (because }i\leq j\text{)}\right)  .
\label{pf.thm.wronski.gen-prod.cij=1}%
\end{align}
Renaming $\left(  i,j\right)  $ as $\left(  k,j\right)  $ in this statement,
we obtain the following: If $\left(  k,j\right)  \in\left\{  1,2,\ldots
,n\right\}  \times\left\{  1,2,\ldots,n\right\}  $ satisfies $k\leq j$, then
we have%
\begin{equation}
c_{k,j}=\dbinom{j-1}{k-1}f^{j-k}\left(  a\right)  .
\label{pf.thm.wronski.gen-prod.ckj=1}%
\end{equation}


For each $i\in\left\{  1,2,\ldots,n\right\}  $, we have $i\geq i$ and thus%
\begin{align}
c_{i,i}  &  =\underbrace{\dbinom{i-1}{i-1}}_{\substack{=1\\\text{(since
}\dbinom{k}{k}=1\\\text{for each }k\in\mathbb{N}\text{)}}}\underbrace{f^{i-i}%
\left(  a\right)  }_{\substack{=f^{0}\left(  a\right)  \\\text{(since
}i-i=0\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.thm.wronski.gen-prod.cij=1}), applied to }j=i\right) \nonumber\\
&  =\underbrace{f^{0}}_{=\operatorname*{id}\nolimits_{A}}\left(  a\right)
=\operatorname*{id}\nolimits_{A}\left(  a\right)  =a.
\label{pf.thm.wronski.gen-prod.cii=}%
\end{align}


If $\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  \times\left\{
1,2,\ldots,n\right\}  $ satisfies $j<i$, then we have%
\begin{align}
c_{i,j}  &  =%
\begin{cases}
\dbinom{j-1}{i-1}f^{j-i}\left(  a\right)  , & \text{if }j\geq i;\\
0, & \text{if }j<i
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.wronski.gen-prod.cij=}%
)}\right) \nonumber\\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }j<i\right)  .
\label{pf.thm.wronski.gen-prod.cij=0}%
\end{align}
Hence, the $n\times n$-matrix $\left(  c_{i,j}\right)  _{1\leq i\leq n,\ 1\leq
j\leq n}$ is upper-triangular.

If $\left(  k,j\right)  \in\left\{  1,2,\ldots,n\right\}  \times\left\{
1,2,\ldots,n\right\}  $ satisfies $k>j$, then we have%
\begin{equation}
c_{k,j}=0 \label{pf.thm.wronski.gen-prod.ckj=0}%
\end{equation}
(by (\ref{pf.thm.wronski.gen-prod.cij=0}), applied to $\left(  k,j\right)  $
instead of $\left(  i,j\right)  $).
\end{verlong}

Define two $n\times n$-matrices $B\in A^{n\times n}$ and $C\in A^{n\times n}$
by%
\[
B=\left(  f^{j-1}\left(  a_{i}\right)  \right)  _{1\leq i\leq n,\ 1\leq j\leq
n}\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ C=\left(  c_{i,j}\right)
_{1\leq i\leq n,\ 1\leq j\leq n}.
\]
Then, the definition of the product of two matrices shows that%
\begin{equation}
BC=\left(  \sum_{k=1}^{n}f^{k-1}\left(  a_{i}\right)  \cdot c_{k,j}\right)
_{1\leq i\leq n,\ 1\leq j\leq n}. \label{pf.thm.wronski.gen-prod.BC=1}%
\end{equation}


\begin{vershort}
However, each $\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}
\times\left\{  1,2,\ldots,n\right\}  $ satisfies%
\begin{align*}
&  \sum_{k=1}^{n}f^{k-1}\left(  a_{i}\right)  \cdot c_{k,j}\\
&  =\sum_{k=1}^{j}f^{k-1}\left(  a_{i}\right)  \cdot\underbrace{c_{k,j}%
}_{\substack{=\dbinom{j-1}{k-1}f^{j-k}\left(  a\right)  \\\text{(by the
definition}\\\text{of }c_{k,j}\text{ (because }k\leq j\text{))}}}+\sum
_{k=j+1}^{n}f^{k-1}\left(  a_{i}\right)  \cdot\underbrace{c_{k,j}%
}_{\substack{=0\\\text{(by the definition}\\\text{of }c_{k,j}\text{ (because
}k>j\text{))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split the sum at
}k=j\text{, since }1\leq j\leq n\right) \\
&  =\sum_{k=1}^{j}f^{k-1}\left(  a_{i}\right)  \cdot\dbinom{j-1}{k-1}%
f^{j-k}\left(  a\right)  +\underbrace{\sum_{k=j+1}^{n}f^{k-1}\left(
a_{i}\right)  \cdot0}_{=0}\\
&  =\sum_{k=1}^{j}f^{k-1}\left(  a_{i}\right)  \cdot\dbinom{j-1}{k-1}%
f^{j-k}\left(  a\right)  =\sum_{k=1}^{j}\dbinom{j-1}{k-1}f^{k-1}\left(
a\right)  f^{j-k}\left(  a_{i}\right) \\
&  =f^{j-1}\left(  aa_{i}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.thm.wronski.gen-prod.faa=1})}\right)  .
\end{align*}

\end{vershort}

\begin{verlong}
However, each $\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}
\times\left\{  1,2,\ldots,n\right\}  $ satisfies%
\begin{align*}
&  \sum_{k=1}^{n}f^{k-1}\left(  a_{i}\right)  \cdot c_{k,j}\\
&  =\sum_{k=1}^{j}f^{k-1}\left(  a_{i}\right)  \cdot\underbrace{c_{k,j}%
}_{\substack{=\dbinom{j-1}{k-1}f^{j-k}\left(  a\right)  \\\text{(by
(\ref{pf.thm.wronski.gen-prod.ckj=1}) (because }k\leq j\text{))}}%
}+\sum_{k=j+1}^{n}f^{k-1}\left(  a_{i}\right)  \cdot\underbrace{c_{k,j}%
}_{\substack{=0\\\text{(by (\ref{pf.thm.wronski.gen-prod.ckj=0}) (because
}k\geq j+1>j\text{))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split the sum at
}k=j\text{, since }1\leq j\leq n\right) \\
&  =\sum_{k=1}^{j}f^{k-1}\left(  a_{i}\right)  \cdot\dbinom{j-1}{k-1}%
f^{j-k}\left(  a\right)  +\underbrace{\sum_{k=j+1}^{n}f^{k-1}\left(
a_{i}\right)  \cdot0}_{=0}\\
&  =\sum_{k=1}^{j}f^{k-1}\left(  a_{i}\right)  \cdot\dbinom{j-1}{k-1}%
f^{j-k}\left(  a\right)  =\sum_{k=1}^{j}\dbinom{j-1}{k-1}f^{k-1}\left(
a\right)  f^{j-k}\left(  a_{i}\right) \\
&  =f^{j-1}\left(  aa_{i}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.thm.wronski.gen-prod.faa=1})}\right)  .
\end{align*}

\end{verlong}

\noindent Hence,
\[
\left(  \sum_{k=1}^{n}f^{k-1}\left(  a_{i}\right)  \cdot c_{k,j}\right)
_{1\leq i\leq n,\ 1\leq j\leq n}=\left(  f^{j-1}\left(  aa_{i}\right)
\right)  _{1\leq i\leq n,\ 1\leq j\leq n}.
\]
Thus, (\ref{pf.thm.wronski.gen-prod.BC=1}) rewrites as%
\begin{equation}
BC=\left(  f^{j-1}\left(  aa_{i}\right)  \right)  _{1\leq i\leq n,\ 1\leq
j\leq n}. \label{pf.thm.wronski.gen-prod.BC=2}%
\end{equation}


\begin{vershort}
The definition of $c_{i,j}$ shows that $c_{i,j}=0$ whenever $j<i$. In other
words, the $n\times n$-matrix $\left(  c_{i,j}\right)  _{1\leq i\leq n,\ 1\leq
j\leq n}$ is upper-triangular. Since the determinant of an upper-triangular
matrix is the product of its diagonal entries, we thus conclude that%
\begin{equation}
\det\left(  \left(  c_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}\right)
=\prod_{i=1}^{n}\underbrace{c_{i,i}}_{\substack{=a\\\text{(by
(\ref{pf.thm.wronski.gen-prod.short.cii=}))}}}=\prod_{i=1}^{n}a=a^{n}.
\label{pf.thm.wronski.gen-prod.short.detC}%
\end{equation}

\end{vershort}

\begin{verlong}
Recall that the $n\times n$-matrix $\left(  c_{i,j}\right)  _{1\leq i\leq
n,\ 1\leq j\leq n}$ is upper-triangular. Thus, its determinant is%
\[
\det\left(  \left(  c_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}\right)
=\prod_{i=1}^{n}c_{i,i}%
\]
(since the determinant of an upper-triangular matrix is the product of its
diagonal entries). Hence,%
\begin{equation}
\det\left(  \left(  c_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}\right)
=\prod_{i=1}^{n}\underbrace{c_{i,i}}_{\substack{=a\\\text{(by
(\ref{pf.thm.wronski.gen-prod.cii=}))}}}=\prod_{i=1}^{n}a=a^{n}.
\label{pf.thm.wronski.gen-prod.detC}%
\end{equation}

\end{verlong}

\begin{vershort}
It is well-known that the determinant of the product of two square matrices
equals the product of their determinants. In other words, if $M$ and $N$ are
two $n\times n$-matrices (with entries in $A$), then $\det\left(  MN\right)
=\det M\cdot\det N$. Applying this to $M=B$ and $N=C$, we find%
\begin{align*}
\det\left(  BC\right)   &  =\det\underbrace{B}_{=\left(  f^{j-1}\left(
a_{i}\right)  \right)  _{1\leq i\leq n,\ 1\leq j\leq n}}\cdot\det
\underbrace{C}_{=\left(  c_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}}\\
&  =\underbrace{\det\left(  \left(  f^{j-1}\left(  a_{i}\right)  \right)
_{1\leq i\leq n,\ 1\leq j\leq n}\right)  }_{\substack{=W_{f}\left(
a_{1},a_{2},\ldots,a_{n}\right)  \\\text{(by
(\ref{pf.thm.wronski.gen-prod.Wf1}))}}}\cdot\underbrace{\det\left(  \left(
c_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}\right)  }_{\substack{=a^{n}%
\\\text{(by (\ref{pf.thm.wronski.gen-prod.short.detC}))}}}\\
&  =W_{f}\left(  a_{1},a_{2},\ldots,a_{n}\right)  \cdot a^{n}=a^{n}%
W_{f}\left(  a_{1},a_{2},\ldots,a_{n}\right)  .
\end{align*}

\end{vershort}

\begin{verlong}
It is well-known that the determinant of the product of two square matrices
equals the product of their determinants. In other words, if $M$ and $N$ are
two $n\times n$-matrices (with entries in $A$), then $\det\left(  MN\right)
=\det M\cdot\det N$. Applying this to $M=B$ and $N=C$, we find%
\begin{align*}
\det\left(  BC\right)   &  =\det\underbrace{B}_{=\left(  f^{j-1}\left(
a_{i}\right)  \right)  _{1\leq i\leq n,\ 1\leq j\leq n}}\cdot\det
\underbrace{C}_{=\left(  c_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}}\\
&  =\underbrace{\det\left(  \left(  f^{j-1}\left(  a_{i}\right)  \right)
_{1\leq i\leq n,\ 1\leq j\leq n}\right)  }_{\substack{=W_{f}\left(
a_{1},a_{2},\ldots,a_{n}\right)  \\\text{(by
(\ref{pf.thm.wronski.gen-prod.Wf1}))}}}\cdot\underbrace{\det\left(  \left(
c_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}\right)  }_{\substack{=a^{n}%
\\\text{(by (\ref{pf.thm.wronski.gen-prod.detC}))}}}\\
&  =W_{f}\left(  a_{1},a_{2},\ldots,a_{n}\right)  \cdot a^{n}=a^{n}%
W_{f}\left(  a_{1},a_{2},\ldots,a_{n}\right)  .
\end{align*}

\end{verlong}

Comparing this with%
\begin{align*}
\det\left(  BC\right)   &  =\det\left(  \left(  f^{j-1}\left(  aa_{i}\right)
\right)  _{1\leq i\leq n,\ 1\leq j\leq n}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{pf.thm.wronski.gen-prod.BC=2})}\right) \\
&  =W_{f}\left(  aa_{1},aa_{2},\ldots,aa_{n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.wronski.gen-prod.Wf2}%
)}\right)  ,
\end{align*}
we obtain $W_{f}\left(  aa_{1},aa_{2},\ldots,aa_{n}\right)  =a^{n}W_{f}\left(
a_{1},a_{2},\ldots,a_{n}\right)  $. This proves Theorem
\ref{thm.wronski.gen-prod}.
\end{proof}

\begin{example}
\label{exa.wronski.polys}Let $\mathbf{k}\left[  x\right]  $ and $\partial_{x}$
be as in Example \ref{exa.diff}. Recall that $\partial_{x}$ is a derivation.
Let $n\in\mathbb{N}$. Let $a_{1},a_{2},\ldots,a_{n}\in\mathbf{k}\left[
x\right]  $ be any $n$ polynomials in $\mathbf{k}\left[  x\right]  $. Then,
the $\partial_{x}$-Wronskian $W_{\partial_{x}}\left(  a_{1},a_{2},\ldots
,a_{n}\right)  $ is just called the \textit{Wronskian} of $a_{1},a_{2}%
,\ldots,a_{n}$, and is denoted by $W\left(  a_{1},a_{2},\ldots,a_{n}\right)
$. Thus, Theorem \ref{thm.wronski.gen-prod} (applied to $A=\mathbf{k}\left[
x\right]  $ and $f=\partial_{x}$) yields that%
\[
W\left(  aa_{1},aa_{2},\ldots,aa_{n}\right)  =a^{n}W\left(  a_{1},a_{2}%
,\ldots,a_{n}\right)  \ \ \ \ \ \ \ \ \ \ \text{for every polynomial }%
a\in\mathbf{k}\left[  x\right]  .
\]

\end{example}

\section{Derivations from the tensor and symmetric algebras}

\subsection{The tensor algebra}

We now turn to the topic of derivations from certain more specific algebras.
We begin with the tensor algebra of a $\mathbf{k}$-module. Let us recall its definition:

\begin{definition}
\label{def.T(V)}Let $V$ be a $\mathbf{k}$-module. For every $n\in\mathbb{N}$,
we let $V^{\otimes n}$ denote the $n$-th tensor power of $V$ (that is, the
$\mathbf{k}$-module $\underbrace{V\otimes V\otimes\cdots\otimes V}_{n\text{
times }V}$). We let $T\left(  V\right)  $ denote the tensor algebra of $V$.
This is the $\mathbf{k}$-algebra whose underlying $\mathbf{k}$-module is
$\bigoplus\limits_{n\geq0}V^{\otimes n}$, and whose multiplication is given by%
\begin{align*}
&  \left(  a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}\right)  \cdot\left(
b_{1}\otimes b_{2}\otimes\cdots\otimes b_{m}\right) \\
&  =a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}\otimes b_{1}\otimes
b_{2}\otimes\cdots\otimes b_{m}\\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{for every }n\in\mathbb{N}\text{, }%
m\in\mathbb{N}\text{, }a_{1},a_{2},\ldots,a_{n}\in V\text{ and }b_{1}%
,b_{2},\ldots,b_{m}\in V\right.  .
\end{align*}


We let $\iota_{T,V}$ be the canonical inclusion map of $V$ into $T\left(
V\right)  $. This map is the composition $V\overset{\cong}{\longrightarrow
}V^{\otimes1}\overset{\text{inclusion}}{\longrightarrow}\bigoplus
\limits_{n\geq0}V^{\otimes n}=T\left(  V\right)  $. (Here, the
\textquotedblleft$T$\textquotedblright\ in \textquotedblleft$\iota_{T,V}%
$\textquotedblright\ is not a variable, but stands for the letter
\textquotedblleft t\textquotedblright\ in \textquotedblleft\textbf{t}ensor
algebra\textquotedblright.)

We have $\iota_{T,V}\left(  V\right)  =V^{\otimes1}$. Some authors identify
$V$ with the $\mathbf{k}$-submodule $V^{\otimes1}$ of $T\left(  V\right)  $
via this map $\iota_{T,V}$ (so that for every $n\in\mathbb{N}$ and
$a_{1},a_{2},\ldots,a_{n}\in V$, we have $a_{1}a_{2}\cdots a_{n}=a_{1}\otimes
a_{2}\otimes\cdots\otimes a_{n}$ in the $\mathbf{k}$-algebra $T\left(
V\right)  $). This identification is harmless since $\iota_{T,V}$ is
injective; but we will not use this identification.
\end{definition}

The following fact is well-known as the universal property of the tensor algebra:

\begin{proposition}
\label{prop.T.univ}Let $V$ be a $\mathbf{k}$-module. Let $A$ be a $\mathbf{k}%
$-algebra. Let $f:V\rightarrow A$ be a $\mathbf{k}$-linear map. Then, there
exists a unique $\mathbf{k}$-algebra homomorphism $F:T\left(  V\right)
\rightarrow A$ such that $F\circ\iota_{T,V}=f$.
\end{proposition}

Our goal is now to prove an analogue of this property for derivations instead
of $\mathbf{k}$-algebra homomorphisms:

\begin{proposition}
\label{prop.T.deruniv}Let $V$ be a $\mathbf{k}$-module. Let $M$ be a $\left(
T\left(  V\right)  ,T\left(  V\right)  \right)  $-bimodule. Let
$f:V\rightarrow M$ be a $\mathbf{k}$-linear map. Then, there exists a unique
derivation $F:T\left(  V\right)  \rightarrow M$ such that $F\circ\iota
_{T,V}=f$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.T.deruniv}.]Define a map $g:V\rightarrow
\mathcal{R}_{T\left(  V\right)  }\left(  M\right)  $ by%
\begin{equation}
g=\mathbf{inl}_{T\left(  V\right)  ,M}\circ\iota_{T,V}+\mathbf{inr}_{T\left(
V\right)  ,M}\circ f. \label{pf.prop.T.deruniv.g}%
\end{equation}


\begin{vershort}
This map $g$ is $\mathbf{k}$-linear (since the maps $\mathbf{inl}_{T\left(
V\right)  ,M}$, $\iota_{T,V}$, $\mathbf{inr}_{T\left(  V\right)  ,M}$ and $f$
are $\mathbf{k}$-linear).
\end{vershort}

\begin{verlong}
The map $\mathbf{inl}_{T\left(  V\right)  ,M}\circ\iota_{T,V}$ is $\mathbf{k}%
$-linear (since it is the composition of the $\mathbf{k}$-linear maps
$\mathbf{inl}_{T\left(  V\right)  ,M}$ and $\iota_{T,V}$). The map
$\mathbf{inr}_{T\left(  V\right)  ,M}\circ f$ is $\mathbf{k}$-linear (since it
is the composition of the $\mathbf{k}$-linear maps $\mathbf{inr}_{T\left(
V\right)  ,M}$ and $f$). Now, the map $\mathbf{inl}_{T\left(  V\right)
,M}\circ\iota_{T,V}+\mathbf{inr}_{T\left(  V\right)  ,M}\circ f$ is
$\mathbf{k}$-linear (since it is the sum of the $\mathbf{k}$-linear maps
$\mathbf{inl}_{T\left(  V\right)  ,M}\circ\iota_{T,V}$ and $\mathbf{inr}%
_{T\left(  V\right)  ,M}\circ f$). Since $g=\mathbf{inl}_{T\left(  V\right)
,M}\circ\iota_{T,V}+\mathbf{inr}_{T\left(  V\right)  ,M}\circ f$, this
rewrites as follows: The map $g$ is $\mathbf{k}$-linear.
\end{verlong}

\begin{vershort}
From $g=\mathbf{inl}_{T\left(  V\right)  ,M}\circ\iota_{T,V}+\mathbf{inr}%
_{T\left(  V\right)  ,M}\circ f$, we obtain%
\begin{align*}
&  \mathbf{projl}_{T\left(  V\right)  ,M}\circ g\\
&  =\mathbf{projl}_{T\left(  V\right)  ,M}\circ\left(  \mathbf{inl}_{T\left(
V\right)  ,M}\circ\iota_{T,V}+\mathbf{inr}_{T\left(  V\right)  ,M}\circ
f\right) \\
&  =\mathbf{projl}_{T\left(  V\right)  ,M}\circ\left(  \mathbf{inl}_{T\left(
V\right)  ,M}\circ\iota_{T,V}\right)  +\mathbf{projl}_{T\left(  V\right)
,M}\circ\left(  \mathbf{inr}_{T\left(  V\right)  ,M}\circ f\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the maps }\mathbf{projl}_{T\left(
V\right)  ,M}\text{, }\mathbf{inl}_{T\left(  V\right)  ,M}\circ\iota
_{T,V}\text{ and }\mathbf{inr}_{T\left(  V\right)  ,M}\circ f\text{ are
}\mathbf{k}\text{-linear}\right) \\
&  =\underbrace{\mathbf{projl}_{T\left(  V\right)  ,M}\circ\mathbf{inl}%
_{T\left(  V\right)  ,M}}_{\substack{=\operatorname*{id}\nolimits_{T\left(
V\right)  }\\\text{(by (\ref{eq.prop.RAM.props.trivial.1}), applied to
}A=T\left(  V\right)  \text{)}}}\circ\iota_{T,V}+\underbrace{\mathbf{projl}%
_{T\left(  V\right)  ,M}\circ\mathbf{inr}_{T\left(  V\right)  ,M}%
}_{\substack{=0\\\text{(by (\ref{eq.prop.RAM.props.trivial.2}), applied to
}A=T\left(  V\right)  \text{)}}}\circ f\\
&  =\underbrace{\operatorname*{id}\nolimits_{T\left(  V\right)  }\circ
\iota_{T,V}}_{=\iota_{T,V}}+\underbrace{0\circ f}_{=0}=\iota_{T,V}.
\end{align*}

\end{vershort}

\begin{verlong}
Notice that
\begin{align*}
&  \mathbf{projl}_{T\left(  V\right)  ,M}\circ\underbrace{g}_{=\mathbf{inl}%
_{T\left(  V\right)  ,M}\circ\iota_{T,V}+\mathbf{inr}_{T\left(  V\right)
,M}\circ f}\\
&  =\mathbf{projl}_{T\left(  V\right)  ,M}\circ\left(  \mathbf{inl}_{T\left(
V\right)  ,M}\circ\iota_{T,V}+\mathbf{inr}_{T\left(  V\right)  ,M}\circ
f\right) \\
&  =\underbrace{\mathbf{projl}_{T\left(  V\right)  ,M}\circ\left(
\mathbf{inl}_{T\left(  V\right)  ,M}\circ\iota_{T,V}\right)  }%
_{=\mathbf{projl}_{T\left(  V\right)  ,M}\circ\mathbf{inl}_{T\left(  V\right)
,M}\circ\iota_{T,V}}+\underbrace{\mathbf{projl}_{T\left(  V\right)  ,M}%
\circ\left(  \mathbf{inr}_{T\left(  V\right)  ,M}\circ f\right)
}_{=\mathbf{projl}_{T\left(  V\right)  ,M}\circ\mathbf{inr}_{T\left(
V\right)  ,M}\circ f}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since composition of }\mathbf{k}\text{-linear maps is }\mathbf{k}%
\text{-bilinear, and since}\\
\text{the maps }\mathbf{projl}_{T\left(  V\right)  ,M}\text{, }\mathbf{inl}%
_{T\left(  V\right)  ,M}\circ\iota_{T,V}\text{ and }\mathbf{inr}_{T\left(
V\right)  ,M}\circ f\\
\text{are }\mathbf{k}\text{-linear}%
\end{array}
\right) \\
&  =\underbrace{\mathbf{projl}_{T\left(  V\right)  ,M}\circ\mathbf{inl}%
_{T\left(  V\right)  ,M}}_{\substack{=\operatorname*{id}\nolimits_{T\left(
V\right)  }\\\text{(by (\ref{eq.prop.RAM.props.trivial.1}), applied to
}A=T\left(  V\right)  \text{)}}}\circ\iota_{T,V}+\underbrace{\mathbf{projl}%
_{T\left(  V\right)  ,M}\circ\mathbf{inr}_{T\left(  V\right)  ,M}%
}_{\substack{=0\\\text{(by (\ref{eq.prop.RAM.props.trivial.2}), applied to
}A=T\left(  V\right)  \text{)}}}\circ f\\
&  =\operatorname*{id}\nolimits_{T\left(  V\right)  }\circ\iota_{T,V}%
+\underbrace{0\circ f}_{=0}=\operatorname*{id}\nolimits_{T\left(  V\right)
}\circ\iota_{T,V}=\iota_{T,V}.
\end{align*}

\end{verlong}

Recall that $\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  T\left(
V\right)  ,M\right)  $ was defined by%
\begin{align}
&  \operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  T\left(  V\right)
,M\right) \nonumber\\
&  =\left\{  \varphi\in\operatorname*{Hom}\left(  T\left(  V\right)
,\mathcal{R}_{T\left(  V\right)  }\left(  M\right)  \right)  \ \mid
\ \mathbf{projl}_{T\left(  V\right)  ,M}\circ\varphi=\operatorname*{id}%
\nolimits_{T\left(  V\right)  }\right\}  . \label{pf.prop.T.deruniv.HomR}%
\end{align}


\begin{vershort}
Proposition \ref{prop.T.univ} (applied to $g$ and $\mathcal{R}_{T\left(
V\right)  }\left(  M\right)  $ instead of $f$ and $A$) now shows that there
exists a unique $\mathbf{k}$-algebra homomorphism $F:T\left(  V\right)
\rightarrow\mathcal{R}_{T\left(  V\right)  }\left(  M\right)  $ such that
$F\circ\iota_{T,V}=g$. Let us denote this $F$ by $G$. Thus, $G$ is a
$\mathbf{k}$-algebra homomorphism $T\left(  V\right)  \rightarrow
\mathcal{R}_{T\left(  V\right)  }\left(  M\right)  $ and satisfies
$G\circ\iota_{T,V}=g$.
\end{vershort}

\begin{verlong}
Proposition \ref{prop.T.univ} (applied to $g$ and $\mathcal{R}_{T\left(
V\right)  }\left(  M\right)  $ instead of $f$ and $A$) now shows that there
exists a unique $\mathbf{k}$-algebra homomorphism $F:T\left(  V\right)
\rightarrow\mathcal{R}_{T\left(  V\right)  }\left(  M\right)  $ such that
$F\circ\iota_{T,V}=g$. Let us denote this $F$ by $G$. Thus, $G$ is a
$\mathbf{k}$-algebra homomorphism $F:T\left(  V\right)  \rightarrow
\mathcal{R}_{T\left(  V\right)  }\left(  M\right)  $ such that $F\circ
\iota_{T,V}=g$. In other words, $G$ is a $\mathbf{k}$-algebra homomorphism
$T\left(  V\right)  \rightarrow\mathcal{R}_{T\left(  V\right)  }\left(
M\right)  $ and satisfies $G\circ\iota_{T,V}=g$.
\end{verlong}

Proposition \ref{prop.RAM.props.alg} \textbf{(a)} (applied to $A=T\left(
V\right)  $) shows that the map $\mathbf{projl}_{T\left(  V\right)
,M}:\mathcal{R}_{T\left(  V\right)  }\left(  M\right)  \rightarrow T\left(
V\right)  $ is a $\mathbf{k}$-algebra homomorphism from $\mathcal{R}_{T\left(
V\right)  }\left(  M\right)  $ to $T\left(  V\right)  $. Hence, the map
$\mathbf{projl}_{T\left(  V\right)  ,M}\circ G$ is a $\mathbf{k}$-algebra
homomorphism (since it is the composition of the two $\mathbf{k}$-algebra
homomorphisms $\mathbf{projl}_{T\left(  V\right)  ,M}$ and $G$).

But Proposition \ref{prop.T.univ} (applied to $T\left(  V\right)  $ and
$\iota_{T,V}$ instead of $A$ and $f$) shows that there exists a unique
$\mathbf{k}$-algebra homomorphism $F:T\left(  V\right)  \rightarrow T\left(
V\right)  $ such that $F\circ\iota_{T,V}=\iota_{T,V}$. In particular, there
exists \textbf{at most one} such homomorphism. In other words, if $F_{1}$ and
$F_{2}$ are two $\mathbf{k}$-algebra homomorphisms $F:T\left(  V\right)
\rightarrow T\left(  V\right)  $ such that $F\circ\iota_{T,V}=\iota_{T,V}$,
then
\begin{equation}
F_{1}=F_{2}. \label{pf.prop.T.deruniv.F1=F2}%
\end{equation}


\begin{vershort}
Now, both $\operatorname*{id}\nolimits_{T\left(  V\right)  }$ and
$\mathbf{projl}_{T\left(  V\right)  ,M}\circ G$ are $\mathbf{k}$-algebra
homomorphisms $F:T\left(  V\right)  \rightarrow T\left(  V\right)  $ such that
$F\circ\iota_{T,V}=\iota_{T,V}$ (indeed, this is clear for $\operatorname*{id}%
\nolimits_{T\left(  V\right)  }$, whereas for $\mathbf{projl}_{T\left(
V\right)  ,M}\circ G$ it follows from $\mathbf{projl}_{T\left(  V\right)
,M}\circ\underbrace{G\circ\iota_{T,V}}_{=g}=\mathbf{projl}_{T\left(  V\right)
,M}\circ g=\iota_{T,V}$). Thus, we can apply (\ref{pf.prop.T.deruniv.F1=F2})
to $F_{1}=\mathbf{projl}_{T\left(  V\right)  ,M}\circ G$ and $F_{2}%
=\operatorname*{id}\nolimits_{T\left(  V\right)  }$. As a result, we obtain
$\mathbf{projl}_{T\left(  V\right)  ,M}\circ G=\operatorname*{id}%
\nolimits_{T\left(  V\right)  }$.
\end{vershort}

\begin{verlong}
But $\operatorname*{id}\nolimits_{T\left(  V\right)  }$ is a $\mathbf{k}%
$-algebra homomorphism $F:T\left(  V\right)  \rightarrow T\left(  V\right)  $
such that $F\circ\iota_{T,V}=\iota_{T,V}$ (because $\operatorname*{id}%
\nolimits_{T\left(  V\right)  }$ is a $\mathbf{k}$-algebra homomorphism
$T\left(  V\right)  \rightarrow T\left(  V\right)  $ and satisfies
$\operatorname*{id}\nolimits_{T\left(  V\right)  }\circ\iota_{T,V}=\iota
_{T,V}$). Also, $\mathbf{projl}_{T\left(  V\right)  ,M}\circ G$ is a
$\mathbf{k}$-algebra homomorphism $F:T\left(  V\right)  \rightarrow T\left(
V\right)  $ such that $F\circ\iota_{T,V}=\iota_{T,V}$ (because $\mathbf{projl}%
_{T\left(  V\right)  ,M}\circ G$ is a $\mathbf{k}$-algebra homomorphism
$T\left(  V\right)  \rightarrow T\left(  V\right)  $ and satisfies
$\mathbf{projl}_{T\left(  V\right)  ,M}\circ\underbrace{G\circ\iota_{T,V}%
}_{=g}=\mathbf{projl}_{T\left(  V\right)  ,M}\circ g=\iota_{T,V}$). Hence, we
can apply (\ref{pf.prop.T.deruniv.F1=F2}) to $F_{1}=\mathbf{projl}_{T\left(
V\right)  ,M}\circ G$ and $F_{2}=\operatorname*{id}\nolimits_{T\left(
V\right)  }$. As a result, we obtain $\mathbf{projl}_{T\left(  V\right)
,M}\circ G=\operatorname*{id}\nolimits_{T\left(  V\right)  }$.
\end{verlong}

\begin{vershort}
Now, $G$ is an element of $\operatorname*{Hom}\left(  T\left(  V\right)
,\mathcal{R}_{T\left(  V\right)  }\left(  M\right)  \right)  $ and satisfies
$\mathbf{projl}_{T\left(  V\right)  ,M}\circ G=\operatorname*{id}%
\nolimits_{T\left(  V\right)  }$. In other words,%
\begin{align*}
G  &  \in\left\{  \varphi\in\operatorname*{Hom}\left(  T\left(  V\right)
,\mathcal{R}_{T\left(  V\right)  }\left(  M\right)  \right)  \ \mid
\ \mathbf{projl}_{T\left(  V\right)  ,M}\circ\varphi=\operatorname*{id}%
\nolimits_{T\left(  V\right)  }\right\} \\
&  =\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  T\left(  V\right)
,M\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.T.deruniv.HomR}%
)}\right)  .
\end{align*}

\end{vershort}

\begin{verlong}
Now, $G$ is a $\mathbf{k}$-algebra homomorphism from $T\left(  V\right)  $ to
$\mathcal{R}_{T\left(  V\right)  }\left(  M\right)  $. Thus, in particular,
$G$ is a $\mathbf{k}$-linear map from $T\left(  V\right)  $ to $\mathcal{R}%
_{T\left(  V\right)  }\left(  M\right)  $. In other words, $G\in
\operatorname*{Hom}\left(  T\left(  V\right)  ,\mathcal{R}_{T\left(  V\right)
}\left(  M\right)  \right)  $. Hence, $G$ is an element $\varphi$ of
$\operatorname*{Hom}\left(  T\left(  V\right)  ,\mathcal{R}_{T\left(
V\right)  }\left(  M\right)  \right)  $ satisfying $\mathbf{projl}_{T\left(
V\right)  ,M}\circ\varphi=\operatorname*{id}\nolimits_{T\left(  V\right)  }$
(since $G$ is an element of $\operatorname*{Hom}\left(  T\left(  V\right)
,\mathcal{R}_{T\left(  V\right)  }\left(  M\right)  \right)  $ and satisfies
$\mathbf{projl}_{T\left(  V\right)  ,M}\circ G=\operatorname*{id}%
\nolimits_{T\left(  V\right)  }$). In other words,%
\begin{align*}
G  &  \in\left\{  \varphi\in\operatorname*{Hom}\left(  T\left(  V\right)
,\mathcal{R}_{T\left(  V\right)  }\left(  M\right)  \right)  \ \mid
\ \mathbf{projl}_{T\left(  V\right)  ,M}\circ\varphi=\operatorname*{id}%
\nolimits_{T\left(  V\right)  }\right\} \\
&  =\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  T\left(  V\right)
,M\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.T.deruniv.HomR}%
)}\right)  .
\end{align*}

\end{verlong}

\begin{vershort}
Consider the maps $\operatorname*{dth}\nolimits_{T\left(  V\right)
,M}:\operatorname*{Hom}\left(  T\left(  V\right)  ,M\right)  \rightarrow
\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  T\left(  V\right)
,M\right)  $ and $\operatorname*{htd}\nolimits_{T\left(  V\right)
,M}:\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  T\left(  V\right)
,M\right)  \rightarrow\operatorname*{Hom}\left(  T\left(  V\right)  ,M\right)
$ defined in Definition \ref{def.RAM.hombij}. Theorem \ref{thm.RAM.dth-htd}
\textbf{(b)} (applied to $A=T\left(  V\right)  $) shows that the maps
$\operatorname*{dth}\nolimits_{T\left(  V\right)  ,M}$ and
$\operatorname*{htd}\nolimits_{T\left(  V\right)  ,M}$ are mutually inverse.
In particular,%
\[
\operatorname*{dth}\nolimits_{T\left(  V\right)  ,M}\circ\operatorname*{htd}%
\nolimits_{T\left(  V\right)  ,M}=\operatorname*{id}%
\nolimits_{\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  T\left(
V\right)  ,M\right)  }.
\]

\end{vershort}

\begin{verlong}
Now, consider the maps $\operatorname*{dth}\nolimits_{T\left(  V\right)
,M}:\operatorname*{Hom}\left(  T\left(  V\right)  ,M\right)  \rightarrow
\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  T\left(  V\right)
,M\right)  $ and $\operatorname*{htd}\nolimits_{T\left(  V\right)
,M}:\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  T\left(  V\right)
,M\right)  \rightarrow\operatorname*{Hom}\left(  T\left(  V\right)  ,M\right)
$ defined in Definition \ref{def.RAM.hombij}. Theorem \ref{thm.RAM.dth-htd}
\textbf{(b)} (applied to $A=T\left(  V\right)  $) shows that the maps
$\operatorname*{dth}\nolimits_{T\left(  V\right)  ,M}$ and
$\operatorname*{htd}\nolimits_{T\left(  V\right)  ,M}$ are mutually inverse.
In other words, we have%
\begin{equation}
\operatorname*{dth}\nolimits_{T\left(  V\right)  ,M}\circ\operatorname*{htd}%
\nolimits_{T\left(  V\right)  ,M}=\operatorname*{id}%
\nolimits_{\operatorname*{Hom}\nolimits_{\mathcal{R}}\left(  T\left(
V\right)  ,M\right)  } \label{pf.prop.T.deruniv.those-inverses}%
\end{equation}
and%
\[
\operatorname*{htd}\nolimits_{T\left(  V\right)  ,M}\circ\operatorname*{dth}%
\nolimits_{T\left(  V\right)  ,M}=\operatorname*{id}%
\nolimits_{\operatorname*{Hom}\left(  T\left(  V\right)  ,M\right)  }.
\]

\end{verlong}

We shall now prove the following two statements:

\textit{Statement 1:} There exists \textbf{at least one} derivation
$F:T\left(  V\right)  \rightarrow M$ such that $F\circ\iota_{T,V}=f$.

\textit{Statement 2:} There exists \textbf{at most one} derivation $F:T\left(
V\right)  \rightarrow M$ such that $F\circ\iota_{T,V}=f$.

\textit{Proof of Statement 1:} Recall that $G\in\operatorname*{Hom}%
\nolimits_{\mathcal{R}}\left(  T\left(  V\right)  ,M\right)  $.

\begin{vershort}
Define an element $h\in\operatorname*{Hom}\left(  T\left(  V\right)
,M\right)  $ by $h=\operatorname*{htd}\nolimits_{T\left(  V\right)  ,M}\left(
G\right)  $. Thus,%
\[
h=\operatorname*{htd}\nolimits_{T\left(  V\right)  ,M}\left(  G\right)
=\mathbf{projr}_{T\left(  V\right)  ,M}\circ G\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\operatorname*{htd}\nolimits_{T\left(  V\right)
,M}\right)  ,
\]
so that%
\begin{align*}
&  \underbrace{h}_{=\mathbf{projr}_{T\left(  V\right)  ,M}\circ G}\circ
\iota_{T,V}\\
&  =\mathbf{projr}_{T\left(  V\right)  ,M}\circ\underbrace{G\circ\iota_{T,V}%
}_{=g=\mathbf{inl}_{T\left(  V\right)  ,M}\circ\iota_{T,V}+\mathbf{inr}%
_{T\left(  V\right)  ,M}\circ f}\\
&  =\mathbf{projr}_{T\left(  V\right)  ,M}\circ\left(  \mathbf{inl}_{T\left(
V\right)  ,M}\circ\iota_{T,V}+\mathbf{inr}_{T\left(  V\right)  ,M}\circ
f\right) \\
&  =\mathbf{projr}_{T\left(  V\right)  ,M}\circ\left(  \mathbf{inl}_{T\left(
V\right)  ,M}\circ\iota_{T,V}\right)  +\mathbf{projr}_{T\left(  V\right)
,M}\circ\left(  \mathbf{inr}_{T\left(  V\right)  ,M}\circ f\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the maps }\mathbf{projr}_{T\left(
V\right)  ,M}\text{, }\mathbf{inl}_{T\left(  V\right)  ,M}\circ\iota
_{T,V}\text{ and }\mathbf{inr}_{T\left(  V\right)  ,M}\circ f\text{ are
}\mathbf{k}\text{-linear}\right) \\
&  =\underbrace{\mathbf{projr}_{T\left(  V\right)  ,M}\circ\mathbf{inl}%
_{T\left(  V\right)  ,M}}_{\substack{=0\\\text{(by
(\ref{eq.prop.RAM.props.trivial.3}), applied to }A=T\left(  V\right)
\text{)}}}\circ\iota_{T,V}+\underbrace{\mathbf{projr}_{T\left(  V\right)
,M}\circ\mathbf{inr}_{T\left(  V\right)  ,M}}_{\substack{=\operatorname*{id}%
\nolimits_{M}\\\text{(by (\ref{eq.prop.RAM.props.trivial.4}), applied to
}A=T\left(  V\right)  \text{)}}}\circ f\\
&  =\underbrace{0\circ\iota_{T,V}}_{=0}+\underbrace{\operatorname*{id}%
\nolimits_{M}\circ f}_{=f}=f.
\end{align*}

\end{vershort}

\begin{verlong}
Define an element $h\in\operatorname*{Hom}\left(  T\left(  V\right)
,M\right)  $ by $h=\operatorname*{htd}\nolimits_{T\left(  V\right)  ,M}\left(
G\right)  $. Thus,%
\[
h=\operatorname*{htd}\nolimits_{T\left(  V\right)  ,M}\left(  G\right)
=\mathbf{projr}_{T\left(  V\right)  ,M}\circ G\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\operatorname*{htd}\nolimits_{T\left(  V\right)
,M}\right)  ,
\]
so that%
\begin{align*}
&  \underbrace{h}_{=\mathbf{projr}_{T\left(  V\right)  ,M}\circ G}\circ
\iota_{T,V}\\
&  =\mathbf{projr}_{T\left(  V\right)  ,M}\circ\underbrace{G\circ\iota_{T,V}%
}_{=g=\mathbf{inl}_{T\left(  V\right)  ,M}\circ\iota_{T,V}+\mathbf{inr}%
_{T\left(  V\right)  ,M}\circ f}\\
&  =\mathbf{projr}_{T\left(  V\right)  ,M}\circ\left(  \mathbf{inl}_{T\left(
V\right)  ,M}\circ\iota_{T,V}+\mathbf{inr}_{T\left(  V\right)  ,M}\circ
f\right) \\
&  =\underbrace{\mathbf{projr}_{T\left(  V\right)  ,M}\circ\left(
\mathbf{inl}_{T\left(  V\right)  ,M}\circ\iota_{T,V}\right)  }%
_{=\mathbf{projr}_{T\left(  V\right)  ,M}\circ\mathbf{inl}_{T\left(  V\right)
,M}\circ\iota_{T,V}}+\underbrace{\mathbf{projr}_{T\left(  V\right)  ,M}%
\circ\left(  \mathbf{inr}_{T\left(  V\right)  ,M}\circ f\right)
}_{=\mathbf{projl}_{T\left(  V\right)  ,M}\circ\mathbf{inr}_{T\left(
V\right)  ,M}\circ f}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since composition of }\mathbf{k}\text{-linear maps is }\mathbf{k}%
\text{-bilinear, and since}\\
\text{the maps }\mathbf{projr}_{T\left(  V\right)  ,M}\text{, }\mathbf{inl}%
_{T\left(  V\right)  ,M}\circ\iota_{T,V}\text{ and }\mathbf{inr}_{T\left(
V\right)  ,M}\circ f\text{ are }\mathbf{k}\text{-linear}%
\end{array}
\right) \\
&  =\underbrace{\mathbf{projr}_{T\left(  V\right)  ,M}\circ\mathbf{inl}%
_{T\left(  V\right)  ,M}}_{\substack{=0\\\text{(by
(\ref{eq.prop.RAM.props.trivial.3}), applied to }A=T\left(  V\right)
\text{)}}}\circ\iota_{T,V}+\underbrace{\mathbf{projr}_{T\left(  V\right)
,M}\circ\mathbf{inr}_{T\left(  V\right)  ,M}}_{\substack{=\operatorname*{id}%
\nolimits_{M}\\\text{(by (\ref{eq.prop.RAM.props.trivial.4}), applied to
}A=T\left(  V\right)  \text{)}}}\circ f\\
&  =\underbrace{0\circ\iota_{T,V}}_{=0}+\operatorname*{id}\nolimits_{M}\circ
f=\operatorname*{id}\nolimits_{M}\circ f=f.
\end{align*}

\end{verlong}

\begin{vershort}
Also,%
\begin{align*}
\operatorname*{dth}\nolimits_{T\left(  V\right)  ,M}\left(  \underbrace{h}%
_{=\operatorname*{htd}\nolimits_{T\left(  V\right)  ,M}\left(  G\right)
}\right)   &  =\operatorname*{dth}\nolimits_{T\left(  V\right)  ,M}\left(
\operatorname*{htd}\nolimits_{T\left(  V\right)  ,M}\left(  G\right)  \right)
\\
&  =\underbrace{\left(  \operatorname*{dth}\nolimits_{T\left(  V\right)
,M}\circ\operatorname*{htd}\nolimits_{T\left(  V\right)  ,M}\right)
}_{=\operatorname*{id}\nolimits_{\operatorname*{Hom}\nolimits_{\mathcal{R}%
}\left(  T\left(  V\right)  ,M\right)  }}\left(  G\right)  =G.
\end{align*}
Thus, $\operatorname*{dth}\nolimits_{T\left(  V\right)  ,M}\left(  h\right)  $
is a $\mathbf{k}$-algebra homomorphism (since $G$ is a $\mathbf{k}$-algebra homomorphism).
\end{vershort}

\begin{verlong}
Also,%
\begin{align*}
\operatorname*{dth}\nolimits_{T\left(  V\right)  ,M}\left(  \underbrace{h}%
_{=\operatorname*{htd}\nolimits_{T\left(  V\right)  ,M}\left(  G\right)
}\right)   &  =\operatorname*{dth}\nolimits_{T\left(  V\right)  ,M}\left(
\operatorname*{htd}\nolimits_{T\left(  V\right)  ,M}\left(  G\right)  \right)
\\
&  =\underbrace{\left(  \operatorname*{dth}\nolimits_{T\left(  V\right)
,M}\circ\operatorname*{htd}\nolimits_{T\left(  V\right)  ,M}\right)
}_{\substack{=\operatorname*{id}\nolimits_{\operatorname*{Hom}%
\nolimits_{\mathcal{R}}\left(  T\left(  V\right)  ,M\right)  }\\\text{(by
(\ref{pf.prop.T.deruniv.those-inverses}))}}}\left(  G\right) \\
&  =\operatorname*{id}\nolimits_{\operatorname*{Hom}\nolimits_{\mathcal{R}%
}\left(  T\left(  V\right)  ,M\right)  }\left(  G\right)  =G.
\end{align*}
Thus, $\operatorname*{dth}\nolimits_{T\left(  V\right)  ,M}\left(  h\right)  $
is a $\mathbf{k}$-algebra homomorphism (since $G$ is a $\mathbf{k}$-algebra homomorphism).
\end{verlong}

But Theorem \ref{thm.RAM.dth-htd} \textbf{(c)} (applied to $T\left(  V\right)
$ and $h$ instead of $A$ and $f$) shows that $h$ is a derivation if and only
if $\operatorname*{dth}\nolimits_{T\left(  V\right)  ,M}\left(  h\right)  $ is
a $\mathbf{k}$-algebra homomorphism from $T\left(  V\right)  $ to
$\mathcal{R}_{T\left(  V\right)  }\left(  M\right)  $. Thus, $h$ is a
derivation (since we know that $\operatorname*{dth}\nolimits_{T\left(
V\right)  ,M}\left(  h\right)  $ is a $\mathbf{k}$-algebra homomorphism from
$T\left(  V\right)  $ to $\mathcal{R}_{T\left(  V\right)  }\left(  M\right)  $).

So we know that $h$ is a derivation from $T\left(  V\right)  $ to $M$ and
satisfies $h\circ\iota_{T,V}=f$. Thus, there exists \textbf{at least one}
derivation $F:T\left(  V\right)  \rightarrow M$ such that $F\circ\iota
_{T,V}=f$ (namely, $F=h$). This proves Statement 1.

\textit{Proof of Statement 2:} Let $\varphi_{1}$ and $\varphi_{2}$ be two
derivations $F:T\left(  V\right)  \rightarrow M$ such that $F\circ\iota
_{T,V}=f$. We shall prove that $\varphi_{1}=\varphi_{2}$.

We know that $\varphi_{1}$ is a derivation $F:T\left(  V\right)  \rightarrow
M$ such that $F\circ\iota_{T,V}=f$. In other words, $\varphi_{1}$ is a
derivation $T\left(  V\right)  \rightarrow M$ and satisfies $\varphi_{1}%
\circ\iota_{T,V}=f$.

\begin{verlong}
Corollary \ref{cor.deriv.verlong} (applied to $T\left(  V\right)  $ and
$\varphi_{1}$ instead of $A$ and $f$) shows that $\varphi_{1}$ is a
$\mathbf{k}$-linear map and satisfies (\ref{eq.def.deriv.axiom}) (with $A$ and
$f$ replaced by $T\left(  V\right)  $ and $\varphi_{1}$). The map $\varphi
_{1}$ is a $\mathbf{k}$-linear map from $T\left(  V\right)  $ to $M$; in other
words, $\varphi_{1}\in\operatorname*{Hom}\left(  T\left(  V\right)  ,M\right)
$.
\end{verlong}

Let $\Phi_{1}=\operatorname*{dth}\nolimits_{T\left(  V\right)  ,M}\left(
\varphi_{1}\right)  $ and $\Phi_{2}=\operatorname*{dth}\nolimits_{T\left(
V\right)  ,M}\left(  \varphi_{2}\right)  $.

Theorem \ref{thm.RAM.dth-htd} \textbf{(c)} (applied to $T\left(  V\right)  $
and $\varphi_{1}$ instead of $A$ and $f$) shows that $\varphi_{1}$ is a
derivation if and only if $\operatorname*{dth}\nolimits_{T\left(  V\right)
,M}\left(  \varphi_{1}\right)  $ is a $\mathbf{k}$-algebra homomorphism from
$T\left(  V\right)  $ to $\mathcal{R}_{T\left(  V\right)  }\left(  M\right)
$. Thus, $\operatorname*{dth}\nolimits_{T\left(  V\right)  ,M}\left(
\varphi_{1}\right)  $ is a $\mathbf{k}$-algebra homomorphism from $T\left(
V\right)  $ to $\mathcal{R}_{T\left(  V\right)  }\left(  M\right)  $ (since we
know that $\varphi_{1}$ is a derivation). In other words, $\Phi_{1}$ is a
$\mathbf{k}$-algebra homomorphism from $T\left(  V\right)  $ to $\mathcal{R}%
_{T\left(  V\right)  }\left(  M\right)  $ (since $\Phi_{1}=\operatorname*{dth}%
\nolimits_{T\left(  V\right)  ,M}\left(  \varphi_{1}\right)  $). The same
argument (but applied to $\varphi_{2}$ and $\Phi_{2}$ instead of $\varphi_{1}$
and $\Phi_{1}$) shows that $\Phi_{2}$ is a $\mathbf{k}$-algebra homomorphism
from $T\left(  V\right)  $ to $\mathcal{R}_{T\left(  V\right)  }\left(
M\right)  $.

\begin{verlong}
The map $\mathbf{inr}_{T\left(  V\right)  ,M}\circ\varphi_{1}$ is $\mathbf{k}%
$-linear (since it is the composition of the $\mathbf{k}$-linear maps
$\mathbf{inr}_{T\left(  V\right)  ,M}$ and $\varphi_{1}$).
\end{verlong}

Now,%
\begin{align*}
\Phi_{1}  &  =\operatorname*{dth}\nolimits_{T\left(  V\right)  ,M}\left(
\varphi_{1}\right)  =\mathbf{inl}_{T\left(  V\right)  ,M}+\mathbf{inr}%
_{T\left(  V\right)  ,M}\circ\varphi_{1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname*{dth}%
\nolimits_{T\left(  V\right)  ,M}\right)  .
\end{align*}
Hence,%
\begin{align*}
&  \underbrace{\Phi_{1}}_{=\mathbf{inl}_{T\left(  V\right)  ,M}+\mathbf{inr}%
_{T\left(  V\right)  ,M}\circ\varphi_{1}}\circ\iota_{T,V}\\
&  =\left(  \mathbf{inl}_{T\left(  V\right)  ,M}+\mathbf{inr}_{T\left(
V\right)  ,M}\circ\varphi_{1}\right)  \circ\iota_{T,V}\\
&  =\mathbf{inl}_{T\left(  V\right)  ,M}\circ\iota_{T,V}+\underbrace{\left(
\mathbf{inr}_{T\left(  V\right)  ,M}\circ\varphi_{1}\right)  \circ\iota_{T,V}%
}_{=\mathbf{inr}_{T\left(  V\right)  ,M}\circ\varphi_{1}\circ\iota_{T,V}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since composition of }\mathbf{k}\text{-linear maps is }\mathbf{k}%
\text{-bilinear, and since}\\
\text{the maps }\mathbf{inl}_{T\left(  V\right)  ,M}\text{, }\mathbf{inr}%
_{T\left(  V\right)  ,M}\circ\varphi_{1}\text{ and }\iota_{T,V}\text{ are
}\mathbf{k}\text{-linear}%
\end{array}
\right) \\
&  =\mathbf{inl}_{T\left(  V\right)  ,M}\circ\iota_{T,V}+\mathbf{inr}%
_{T\left(  V\right)  ,M}\circ\underbrace{\varphi_{1}\circ\iota_{T,V}}_{=f}\\
&  =\mathbf{inl}_{T\left(  V\right)  ,M}\circ\iota_{T,V}+\mathbf{inr}%
_{T\left(  V\right)  ,M}\circ f=g\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.T.deruniv.g})}\right)  .
\end{align*}
The same argument (but applied to $\varphi_{2}$ and $\Phi_{2}$ instead of
$\varphi_{1}$ and $\Phi_{1}$) shows that $\Phi_{2}\circ\iota_{T,V}=g$.

Let us recall that there exists a unique $\mathbf{k}$-algebra homomorphism
$F:T\left(  V\right)  \rightarrow\mathcal{R}_{T\left(  V\right)  }\left(
M\right)  $ such that $F\circ\iota_{T,V}=g$. In particular, there exists
\textbf{at most one} such homomorphism. In other words, if $F_{1}$ and $F_{2}$
are two $\mathbf{k}$-algebra homomorphisms $F:T\left(  V\right)
\rightarrow\mathcal{R}_{T\left(  V\right)  }\left(  M\right)  $ such that
$F\circ\iota_{T,V}=g$, then
\begin{equation}
F_{1}=F_{2}. \label{pf.prop.T.deruniv.F1=F2second}%
\end{equation}


\begin{vershort}
Now, both $\Phi_{1}$ and $\Phi_{2}$ are $\mathbf{k}$-algebra homomorphisms
$F:T\left(  V\right)  \rightarrow\mathcal{R}_{T\left(  V\right)  }\left(
M\right)  $ such that $F\circ\iota_{T,V}=g$ (since $\Phi_{1}\circ\iota
_{T,V}=g$ and $\Phi_{2}\circ\iota_{T,V}=g$). Thus, we can apply
(\ref{pf.prop.T.deruniv.F1=F2second}) to $F_{1}=\Phi_{1}$ and $F_{2}=\Phi_{2}%
$. As a result, we obtain $\Phi_{1}=\Phi_{2}$.
\end{vershort}

\begin{verlong}
But $\Phi_{1}$ is a $\mathbf{k}$-algebra homomorphism $F:T\left(  V\right)
\rightarrow\mathcal{R}_{T\left(  V\right)  }\left(  M\right)  $ such that
$F\circ\iota_{T,V}=g$ (because $\Phi_{1}$ is a $\mathbf{k}$-algebra
homomorphism $T\left(  V\right)  \rightarrow\mathcal{R}_{T\left(  V\right)
}\left(  M\right)  $ and satisfies $\Phi_{1}\circ\iota_{T,V}=g$). But
$\Phi_{2}$ is a $\mathbf{k}$-algebra homomorphism $F:T\left(  V\right)
\rightarrow\mathcal{R}_{T\left(  V\right)  }\left(  M\right)  $ such that
$F\circ\iota_{T,V}=g$ (because $\Phi_{2}$ is a $\mathbf{k}$-algebra
homomorphism $T\left(  V\right)  \rightarrow\mathcal{R}_{T\left(  V\right)
}\left(  M\right)  $ and satisfies $\Phi_{2}\circ\iota_{T,V}=g$). Hence, we
can apply (\ref{pf.prop.T.deruniv.F1=F2second}) to $F_{1}=\Phi_{1}$ and
$F_{2}=\Phi_{2}$. As a result, we obtain $\Phi_{1}=\Phi_{2}$.
\end{verlong}

Recall that the maps $\operatorname*{dth}\nolimits_{T\left(  V\right)  ,M}$
and $\operatorname*{htd}\nolimits_{T\left(  V\right)  ,M}$ are mutually
inverse. Thus, the map $\operatorname*{dth}\nolimits_{T\left(  V\right)  ,M}$
is invertible, i.e., bijective, and therefore injective.

But $\operatorname*{dth}\nolimits_{T\left(  V\right)  ,M}\left(  \varphi
_{1}\right)  =\Phi_{1}=\Phi_{2}=\operatorname*{dth}\nolimits_{T\left(
V\right)  ,M}\left(  \varphi_{2}\right)  $. This shows that $\varphi
_{1}=\varphi_{2}$ (since the map $\operatorname*{dth}\nolimits_{T\left(
V\right)  ,M}$ is injective).

Let us now forget that we fixed $\varphi_{1}$ and $\varphi_{2}$. We thus have
shown that if $\varphi_{1}$ and $\varphi_{2}$ are two derivations $F:T\left(
V\right)  \rightarrow M$ such that $F\circ\iota_{T,V}=f$, then $\varphi
_{1}=\varphi_{2}$. In other words, there exists \textbf{at most one}
derivation $F:T\left(  V\right)  \rightarrow M$ such that $F\circ\iota
_{T,V}=f$. This proves Statement 2.

\begin{vershort}
Now, both Statement 1 and Statement 2 have been proven. Combining these two
statements, we obtain Proposition \ref{prop.T.deruniv}.
\end{vershort}

\begin{verlong}
Now, both Statement 1 and Statement 2 have been proven. Statement 1 shows that
there exists \textbf{at least one} derivation $F:T\left(  V\right)
\rightarrow M$ such that $F\circ\iota_{T,V}=f$, whereas Statement 2 proves
that there exists \textbf{at most one} such derivation. Combining these
results, we conclude that there exists a unique derivation $F:T\left(
V\right)  \rightarrow M$ such that $F\circ\iota_{T,V}=f$. This proves
Proposition \ref{prop.T.deruniv}.
\end{verlong}
\end{proof}

\begin{remark}
\label{rmk.prop.T.deruniv.otherproofs}Our proof of Proposition
\ref{prop.T.deruniv} above is not the only one possible; it is probably not
even the shortest one, and it is certainly not the most enlightening one. It
does, however, illustrate how a result can be proven using universal
properties and constructions such as $\mathcal{R}_{A}\left(  M\right)  $
without \textquotedblleft getting one's hands dirty\textquotedblright\ (e.g.,
computing with elements).

Let me sketch an alternative way to prove Proposition \ref{prop.T.deruniv}: We
subdivide the claim into a Statement 1 and a Statement 2 as in our above
proof. To prove Statement 1, we can construct a derivation $F:T\left(
V\right)  \rightarrow M$ such that $F\circ\iota_{T,V}=f$ explicitly, by
setting%
\begin{align*}
&  F\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right) \\
&  =\sum_{i=1}^{n}\left(  v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{i-1}\right)  \cdot f\left(  v_{i}\right)  \cdot\left(  v_{i+1}\otimes
v_{i+2}\otimes\cdots\otimes v_{n}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{for every }n\in\mathbb{N}\text{ and
}v_{1},v_{2},\ldots,v_{n}\in V\right.  .
\end{align*}
It is not hard to prove that this is well-defined (i.e., there exists a unique
$\mathbf{k}$-linear map $F:T\left(  V\right)  \rightarrow M$ satisfying this
equality), and indeed defines a derivation $F:T\left(  V\right)  \rightarrow
M$ satisfying $F\circ\iota_{T,V}=f$. To prove Statement 2, we could use
Proposition \ref{prop.derivation.unique} (applied to $A=T\left(  V\right)  $
and $S=\iota_{T,V}\left(  V\right)  $), since $\iota_{T,V}\left(  V\right)  $
generates the $\mathbf{k}$-algebra $T\left(  V\right)  $.

See \cite[Theorem 4.6.12]{etilie} for the details of this proof of Proposition
\ref{prop.T.deruniv}. (The actual statement of \cite[Theorem 4.6.12]{etilie}
is the particular case of Proposition \ref{prop.T.deruniv} for $\mathbf{k}%
=\mathbb{C}$; but the proof applies to the general case without changes.)
\end{remark}

\subsection{The symmetric algebra}

We now recall the definition of the symmetric algebra of a $\mathbf{k}$-module
$V$:

\begin{definition}
\label{def.Sym}Let $V$ be a $\mathbf{k}$-module. Let $J_{V}$ denote the
two-sided ideal of the $\mathbf{k}$-algebra $T\left(  V\right)  $ generated by
the tensors of the form $v\otimes w-w\otimes v$ with $v\in V$ and $w\in V$. We
define $\operatorname*{Sym}V$ to be the quotient algebra $T\left(  V\right)
/J_{V}$. We let $\pi_{\operatorname*{Sym},V}$ be the canonical projection from
$T\left(  V\right)  $ to its quotient $\mathbf{k}$-algebra
$\operatorname*{Sym}V$. The $\mathbf{k}$-algebra $\operatorname*{Sym}V$ is
known as the \textit{symmetric algebra} of $V$. It is well-known that the
$\mathbf{k}$-algebra $\operatorname*{Sym}V$ is commutative.

For every $n\in\mathbb{N}$, we write $\operatorname*{Sym}\nolimits^{n}V$ for
the $\mathbf{k}$-submodule $\pi_{\operatorname*{Sym},V}\left(  V^{\otimes
n}\right)  $ of $\operatorname*{Sym}V$. It is well-known that
$\operatorname*{Sym}V=\bigoplus\limits_{n\geq0}\operatorname*{Sym}%
\nolimits^{n}V$.

We let $\iota_{\operatorname*{Sym},V}$ denote the composition $\pi
_{\operatorname*{Sym},V}\circ\iota_{T,V}$. This composition $\iota
_{\operatorname*{Sym},V}$ is a $\mathbf{k}$-linear map $V\rightarrow
\operatorname*{Sym}V$. It is well-known that this map $\iota
_{\operatorname*{Sym},V}$ is injective and satisfies $\iota
_{\operatorname*{Sym},V}\left(  V\right)  =\operatorname*{Sym}\nolimits^{1}V$.
\end{definition}

The following fact is well-known as the universal property of the symmetric algebra:

\begin{proposition}
\label{prop.Sym.univ}Let $V$ be a $\mathbf{k}$-module. Let $A$ be a
commutative $\mathbf{k}$-algebra. Let $f:V\rightarrow A$ be a $\mathbf{k}%
$-linear map. Then, there exists a unique $\mathbf{k}$-algebra homomorphism
$F:\operatorname*{Sym}V\rightarrow A$ such that $F\circ\iota
_{\operatorname*{Sym},V}=f$.
\end{proposition}

This is clearly an analogue of Proposition \ref{prop.T.univ}. We can similarly
state an analogue of Proposition \ref{prop.T.deruniv}:

\begin{proposition}
\label{prop.Sym.deruniv}Let $V$ be a $\mathbf{k}$-module. Let $M$ be a
symmetric $\left(  \operatorname*{Sym}V,\operatorname*{Sym}V\right)
$-bimodule. Let $f:V\rightarrow M$ be a $\mathbf{k}$-linear map. Then, there
exists a unique derivation $F:\operatorname*{Sym}V\rightarrow M$ such that
$F\circ\iota_{\operatorname*{Sym},V}=f$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.Sym.deruniv}.]To obtain a proof of Proposition
\ref{prop.Sym.deruniv}, it suffices to make the following changes to our above
proof of Proposition \ref{prop.T.deruniv}:

\begin{itemize}
\item Add the following argument at the beginning of the proof:
\textquotedblleft The $\mathbf{k}$-algebra $\mathcal{R}_{\operatorname*{Sym}%
V}\left(  M\right)  $ is commutative (by Proposition \ref{prop.RAM.commut},
applied to $A=\operatorname*{Sym}V$).\textquotedblright.

\item Replace each appearance of \textquotedblleft$T\left(  V\right)
$\textquotedblright\ by \textquotedblleft$\operatorname*{Sym}V$%
\textquotedblright.

\item Replace each appearance of \textquotedblleft$\iota_{T,V}$%
\textquotedblright\ by \textquotedblleft$\iota_{\operatorname*{Sym},V}%
$\textquotedblright.

\item Replace each reference to Proposition \ref{prop.T.univ} by a reference
to Proposition \ref{prop.Sym.univ}. (Proposition \ref{prop.Sym.univ} can
indeed be applied since the two $\mathbf{k}$-algebras $\operatorname*{Sym}V$
and $\mathcal{R}_{\operatorname*{Sym}V}\left(  M\right)  $ are commutative.)
\end{itemize}
\end{proof}

\begin{remark}
It is possible to prove Proposition \ref{prop.Sym.deruniv} in more
down-to-earth manners, just as the same is possible for Proposition
\ref{prop.T.deruniv} (see Remark \ref{rmk.prop.T.deruniv.otherproofs} above).
\end{remark}

\begin{thebibliography}{99999999}                                                                                         %


\bibitem[EtiGri12]{etilie}Pavel Etingof, \textit{18.747: Infinite-dimensional
Lie algebras (Spring term 2012 at MIT)}, scribed by Darij Grinberg, version
0.44.\newline\url{https://sites.google.com/site/darijgrinberg/lie}

\bibitem[Becker14]{Becker}Hanno Becker, \textit{Extension of R-linear
derivation to localization}, math.stackexchange post \#1052192 (answer to
question \#1052153), \url{http://math.stackexchange.com/q/1052192}

\bibitem[CSA14]{CSA14}CSA and others, \textit{Contrasting definitions of
bimodules? An illusion?}, math.stackexchange post \#889130 and
answers,\newline\url{http://math.stackexchange.com/q/889130}

\bibitem[Grinbe16]{Gri-fleck}Darij Grinberg, \textit{Fleck's binomial
congruence using circulant matrices}, version 14 July 2019.\newline\url{http://www.cip.ifi.lmu.de/~grinberg/fleck.pdf}
\end{thebibliography}


\end{document}