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\begin{center}
\textbf{Rep\#1: Deformations of a bimodule algebra}

Darij Grinberg

[corrected 2026-06-03, proofread 2026-06-04]

\bigskip
\end{center}

The purpose of this short note is to generalize Problem 2.24 in \cite{1}. We
begin with a few definitions:

\begin{quote}
\textbf{Definition 1.} In the following, a \textit{ring} will always mean a
(not necessarily commutative) ring with unity. Ring homomorphisms are always
assumed to respect the unity. For every ring $R$, we denote the unity of $R$
by $1_{R}$. All $K$-algebras (for a commutative ring $K$) are understood to be
$K$-modules and simultaneously rings with a $K$-bilinear multiplication.

\textbf{Definition 2.} Let $A$ and $B$ be two rings, and let $\rho
:A\rightarrow B$ be a ring homomorphism. Then, for any $a\in A$ and any $b\in
B$, we shall denote the product $\rho\left(  a\right)  b$ by $ab$ and denote
the product $b\rho\left(  a\right)  $ by $ba$. Thus, $B$ becomes a left
$A$-module and a right $A$-module, and -- even better -- an $\left(
A,A\right)  $-bimodule (that is, the two actions of $A$ commute: $\left(
ab\right)  a^{\prime}=a\left(  ba^{\prime}\right)  $ for all $a\in A$, $b\in
B$ and $a^{\prime}\in A$).

\textbf{Definition 3.} Let $B$ be a ring. Then, we denote by $B\left[  \left[
t\right]  \right]  $ the ring of formal power series over $B$ in the
indeterminate $t$, where $t$ is supposed to commute with every element of $B$.
Formally, this means that we define $B\left[  \left[  t\right]  \right]  $ as
the ring of all sequences $\left(  b_{0},b_{1},b_{2},...\right)  \in
B^{\mathbb{N}}$ (where $\mathbb{N}$ means the set $\left\{  0,1,2,...\right\}
$), with addition defined by%
\[
\left(  b_{0},b_{1},b_{2},...\right)  +\left(  b_{0}^{\prime},b_{1}^{\prime
},b_{2}^{\prime},...\right)  =\left(  b_{0}+b_{0}^{\prime},b_{1}+b_{1}%
^{\prime},b_{2}+b_{2}^{\prime},...\right)
\]
and multiplication defined by%
\[
\left(  b_{0},b_{1},b_{2},...\right)  \cdot\left(  b_{0}^{\prime}%
,b_{1}^{\prime},b_{2}^{\prime},...\right)  =\left(  \sum_{\substack{\left(
i,j\right)  \in\mathbb{N}^{2};\\i+j=0}}b_{i}b_{j}^{\prime},\sum
_{\substack{\left(  i,j\right)  \in\mathbb{N}^{2};\\i+j=1}}b_{i}b_{j}^{\prime
},\sum_{\substack{\left(  i,j\right)  \in\mathbb{N}^{2};\\i+j=2}}b_{i}%
b_{j}^{\prime},...\right)  ,
\]
and we denote a sequence $\left(  b_{0},b_{1},b_{2},...\right)  \in
B^{\mathbb{N}}$ by $\sum\limits_{i=0}^{\infty}b_{i}t^{i}$. For every
$m\in\mathbb{N}$, the element $b_{m}\in B$ is called the $t^{m}$%
\textit{-coefficient of the power series }$\left(  b_{0},b_{1},b_{2}%
,...\right)  =\sum\limits_{i=0}^{\infty}b_{i}t^{i}$ (or the
\textit{coefficient of the power series }$\left(  b_{0},b_{1},b_{2}%
,...\right)  =\sum\limits_{i=0}^{\infty}b_{i}t^{i}$ \textit{before }$t^{m}$).
The element $b_{0}\in B$ is also called the \textit{constant term} of the
power series $\left(  b_{0},b_{1},b_{2},...\right)  =\sum\limits_{i=0}%
^{\infty}b_{i}t^{i}$. We embed the ring $B$ into $B\left[  \left[  t\right]
\right]  $ by identifying each $b\in B$ with the power series $\left(
b,0,0,0,\ldots\right)  \in B\left[  \left[  t\right]  \right]  $. (The latter
power series is known as a \textit{constant power series}, even though it is
not constant as a sequence!)

\textbf{Definition 4.} Let $B$ be a ring, and let $\left(  b_{m}%
,b_{m+1},...,b_{n}\right)  $ be a sequence of elements of $B$. Then, we denote
by $\overset{\longleftarrow}{\prod\limits_{i=m}^{n}}b_{i}$ the product
$b_{n}b_{n-1}...b_{m}$ (this product is supposed to mean $1$ if $m>n$).
\end{quote}

Now comes the generalization of Problem 2.24 (a) in \cite{1}\footnote{Problem
2.24 in \cite{1} is recovered from this generalization by setting
$B=\operatorname*{End}V$.}:

\begin{quote}
\textbf{Theorem 1.} Let $K$ be a commutative ring. Let $A$ and $B$ be two
$K$-algebras, and let $\rho:A\rightarrow B$ be a $K$-algebra homomorphism. As
in Definition 2, we thus conclude that $B$ becomes an $\left(  A,A\right)  $-bimodule.

Assume that%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{for every }K\text{-linear map }f:A\rightarrow B\text{ which satisfies}\\
\left(  f\left(  aa^{\prime}\right)  =af\left(  a^{\prime}\right)  +f\left(
a\right)  a^{\prime}\text{ for all }a\in A\text{ and }a^{\prime}\in A\right)
,\\
\text{there exists an element }s\in B\text{ such that}\\
\left(  f\left(  a\right)  =as-sa\text{ for all }a\in A\right)
\end{array}
\right)  \label{1ass}%
\end{equation}
(where we are using this $\left(  A,A\right)  $-bimodule structure on $B$ to
make sense of $af\left(  a^{\prime}\right)  $, $f\left(  a\right)  a^{\prime}%
$, $as$ and $sa$).

Let $B\left[  \left[  t\right]  \right]  $ be the ring of formal power series
over $B$ in the indeterminate $t$, where $t$ is supposed to commute with every
element of $B$.

Here and in the following, we let $1$ denote the unity $1_{B}$ of the ring $B$.

Let $\overline{\rho}:A\rightarrow B\left[  \left[  t\right]  \right]  $ be a
$K$-linear map such that any $a\in A$ and any $a^{\prime}\in A$ satisfy
\begin{equation}
\overline{\rho}\left(  aa^{\prime}\right)  =\overline{\rho}\left(  a\right)
\cdot\overline{\rho}\left(  a^{\prime}\right)  , \label{1rhoaa'}%
\end{equation}
and such that for every $a\in A$,
\begin{equation}
\text{the constant term of the power series }\overline{\rho}\left(  a\right)
\text{ equals }\rho\left(  a\right)  .\ \ \ \ \ \ \ \ \ \ \label{1rho0}%
\end{equation}


Then, there exists a power series $b\in B\left[  \left[  t\right]  \right]  $
with constant term $1$ such that for every $a\in A$, the power series
$b\overline{\rho}\left(  a\right)  b^{-1}\in B\left[  \left[  t\right]
\right]  $ equals the (constant) power series $\rho\left(  a\right)  $. (Note
that $b^{-1}$ is well-defined, since $b$ has constant term $1$ and thus has an
inverse\footnote{Here we are using the fact that any power series in $B\left[
\left[  t\right]  \right]  $ with constant term $1$ has an inverse. This is
well-known when $B$ is commutative, but the same proof applies in general (as
long as one remembers that an element of a monoid that has a left inverse and
a right inverse must always have an inverse).}.)
\end{quote}

\textit{Proof of Theorem 1.} First, we endow the ring $B\left[  \left[
t\right]  \right]  $ with the $\left(  t\right)  $-adic topology. This
topology is defined as the product topology on $B^{\mathbb{N}}$ (where each
copy of $B$ has the discrete topology), recalling that $B\left[  \left[
t\right]  \right]  =B^{\mathbb{N}}$ as a set. Equivalently, this topology is
defined in such a way that for every $f\in B\left[  \left[  t\right]  \right]
$, the family
\[
\left(  f+t^{0}B\left[  \left[  t\right]  \right]  ,\ f+t^{1}B\left[  \left[
t\right]  \right]  ,\ f+t^{2}B\left[  \left[  t\right]  \right]
,\ ...\right)
\]
is a basis of open neighbourhoods of $f$. This topology makes $B\left[
\left[  t\right]  \right]  $ a topological ring, since $t^{i}B\left[  \left[
t\right]  \right]  $ is a two-sided ideal of $B\left[  \left[  t\right]
\right]  $ for every $i\in\mathbb{N}$. Note that $t^{i}B\left[  \left[
t\right]  \right]  $ is the set of all power series whose first $i$
coefficients (starting with the constant term) are $0$.

The ring $B\left[  \left[  t\right]  \right]  $ can also be equipped with a
metric, in which any two distinct power series $f,g\in B\left[  \left[
t\right]  \right]  $ have distance $d\left(  f,g\right)  =2^{-i}$, where $i$
is the largest nonnegative integer satisfying $f\equiv g\operatorname{mod}%
t^{i}B\left[  \left[  t\right]  \right]  $ (and, of course, $d\left(
f,f\right)  =0$). The topology induced by this metric is exactly the $\left(
t\right)  $-adic topology. The metric space $B\left[  \left[  t\right]
\right]  $ (with this metric) is complete.

For any $k$ elements $u_{1}$, $u_{2}$, $...$, $u_{k}$ of $B$, and for every
$a\in A$, let us denote by $\overline{\rho}_{u_{1},u_{2},...,u_{k}}\left(
a\right)  $ the element%
\[
\left(  \overset{\longleftarrow}{\prod_{i=1}^{k}}\left(  1-u_{i}t^{i}\right)
\right)  \cdot\overline{\rho}\left(  a\right)  \cdot\left(
\overset{\longleftarrow}{\prod_{i=1}^{k}}\left(  1-u_{i}t^{i}\right)  \right)
^{-1}\in B\left[  \left[  t\right]  \right]
\]
(here, the inverse $\left(  \overset{\longleftarrow}{\prod\limits_{i=1}^{k}%
}\left(  1-u_{i}t^{i}\right)  \right)  ^{-1}$ exists because
$\overset{\longleftarrow}{\prod\limits_{i=1}^{k}}\left(  1-u_{i}t^{i}\right)
$ is a power series with constant term $1$). Clearly, $\overline{\rho}%
_{u_{1},u_{2},...,u_{k}}:A\rightarrow B\left[  \left[  t\right]  \right]  $ is
a $K$-linear map for any $k$ elements $u_{1}$, $u_{2}$, $...$, $u_{k}$ of $B$.

Now, we are going to recursively construct a sequence $\left(  u_{1}%
,u_{2},u_{3},...\right)  \in B^{\left\{  1,2,3,...\right\}  }$ of elements of
$B$ such that every $n\in\mathbb{N}$ satisfies%
\begin{equation}
\left(  \overline{\rho}_{u_{1},u_{2},...,u_{n}}\left(  a\right)  \equiv
\rho\left(  a\right)  \ \operatorname{mod}\ t^{n+1}B\left[  \left[  t\right]
\right]  \ \ \ \ \ \ \ \ \ \ \text{for every }\left.  a\in A\right.  \right)
. \label{1proof}%
\end{equation}


In fact, we first notice that the congruence (\ref{1proof}) is automatically
satisfied for $n=0$ (note that the product $\overset{\longleftarrow
}{\prod\limits_{i=1}^{n}}\left(  1-u_{i}t^{i}\right)  $ is an empty product
when $n=0$), because in the case $n=0$, we have $\overline{\rho}_{u_{1}%
,u_{2},...,u_{n}}\left(  a\right)  =\left(  \text{empty product}\right)
\cdot\overline{\rho}\left(  a\right)  \cdot\left(  \text{empty product}%
\right)  ^{-1}=\overline{\rho}\left(  a\right)  \equiv\rho\left(  a\right)
\ \operatorname{mod}\ tB\left[  \left[  t\right]  \right]  $ (since
(\ref{1rho0}) says that the constant term of the power series $\overline{\rho
}\left(  a\right)  $ equals $\rho\left(  a\right)  $). Now, we are going to
construct our sequence $\left(  u_{1},u_{2},u_{3},...\right)  \in B^{\left\{
1,2,3,...\right\}  }$ by recursion: Let $m$ be a positive integer. Assume that
we have constructed some elements $u_{1}$, $u_{2}$, $...$, $u_{m-1}$ of $B$
such that (\ref{1proof}) holds for $n=m-1$. Then, we are going to construct a
new element $u_{m}$ of $B$ such that (\ref{1proof}) holds for $n=m$.

In fact, applying (\ref{1proof}) to $n=m-1$ (we can do this since we have
\textit{assumed} that (\ref{1proof}) holds for $n=m-1$), we obtain%
\[
\overline{\rho}_{u_{1},u_{2},...,u_{m-1}}\left(  a\right)  \equiv\rho\left(
a\right)  \ \operatorname{mod}\ t^{m}B\left[  \left[  t\right]  \right]
\ \ \ \ \ \ \ \ \ \ \text{for every }\left.  a\in A\right.  .
\]
In other words, every $a\in A$ satisfies%
\[
\overline{\rho}_{u_{1},u_{2},...,u_{m-1}}\left(  a\right)  -\rho\left(
a\right)  \in t^{m}B\left[  \left[  t\right]  \right]  .
\]
Denoting the power series $\overline{\rho}_{u_{1},u_{2},...,u_{m-1}}\left(
a\right)  -\rho\left(  a\right)  $ by $q\left(  a\right)  $, we thus have
$q\left(  a\right)  \in t^{m}B\left[  \left[  t\right]  \right]  $ for every
$a\in A$. Hence, the first $m$ coefficients of the power series $q\left(
a\right)  $ are $0$ for every $a\in A$. In other words,
\begin{equation}
q_{0}\left(  a\right)  =q_{1}\left(  a\right)  =...=q_{m-1}\left(  a\right)
=0\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A, \label{1proof.pa0}%
\end{equation}
where $q_{i}\left(  a\right)  $ denotes the $t^{i}$-coefficient of the power
series $q\left(  a\right)  $ for every $i\in\mathbb{N}$. Thus, for every $a\in
A$, we have%
\begin{align*}
q\left(  a\right)   &  =\sum_{i=0}^{\infty}q_{i}\left(  a\right)  t^{i}%
=\sum_{i=0}^{m-1}\underbrace{q_{i}\left(  a\right)  }_{\substack{=0\\\text{(by
(\ref{1proof.pa0}))}}}t^{i}+q_{m}\left(  a\right)  t^{m}+\sum_{i=m+1}^{\infty
}\underbrace{q_{i}\left(  a\right)  t^{i}}_{\substack{\equiv
0\operatorname{mod}t^{m+1}B\left[  \left[  t\right]  \right]  \\\text{(since
}i\geq m+1\text{ yields}\\t^{i}\equiv0\operatorname{mod}t^{m+1}B\left[
\left[  t\right]  \right]  \text{)}}}\\
&  \equiv\sum_{i=0}^{m-1}0t^{i}+q_{m}\left(  a\right)  t^{m}+\sum
_{i=m+1}^{\infty}0=q_{m}\left(  a\right)  t^{m}\operatorname{mod}%
t^{m+1}B\left[  \left[  t\right]  \right]  .
\end{align*}
Hence,%
\begin{align}
\overline{\rho}_{u_{1},u_{2},...,u_{m-1}}\left(  a\right)   &
=\underbrace{\left(  \overline{\rho}_{u_{1},u_{2},...,u_{m-1}}\left(
a\right)  -\rho\left(  a\right)  \right)  }_{=q\left(  a\right)  \equiv
q_{m}\left(  a\right)  t^{m}\operatorname{mod}t^{m+1}B\left[  \left[
t\right]  \right]  }+\,\rho\left(  a\right) \nonumber\\
&  \equiv q_{m}\left(  a\right)  t^{m}+\rho\left(  a\right)
\operatorname{mod}t^{m+1}B\left[  \left[  t\right]  \right]  . \label{1proof4}%
\end{align}


However, the map $q:A\rightarrow B\left[  \left[  t\right]  \right]  $ (which
sends each $a\in A$ to $q\left(  a\right)  $) is $K$-linear (by its
definition, since the maps $\rho:A\rightarrow B$ and $\overline{\rho}%
_{u_{1},u_{2},...,u_{m-1}}:A\rightarrow B\left[  \left[  t\right]  \right]  $
are $K$-linear). Thus, the map $q_{m}:A\rightarrow B$ (which sends each $a\in
A$ to $q_{m}\left(  a\right)  $) is $K$-linear as well (since $q_{m}%
=\operatorname*{coeff}_{m}\circ\,q$, where $\operatorname*{coeff}_{m}:B\left[
\left[  t\right]  \right]  \rightarrow B$ is the map that takes every power
series to its $t^{m}$-coefficient, and thus $q_{m}$ is $K$-linear because both
$\operatorname*{coeff}_{m}$ and $q$ are $K$-linear).

Now, let $a\in A$ and $a^{\prime}\in A$. Then,%
\begin{align}
&  \overline{\rho}_{u_{1},u_{2},...,u_{m-1}}\left(  a\right)  \cdot
\overline{\rho}_{u_{1},u_{2},...,u_{m-1}}\left(  a^{\prime}\right) \nonumber\\
&  =\left(  \overset{\longleftarrow}{\prod_{i=1}^{m-1}}\left(  1-u_{i}%
t^{i}\right)  \cdot\overline{\rho}\left(  a\right)  \cdot\left(
\overset{\longleftarrow}{\prod_{i=1}^{m-1}}\left(  1-u_{i}t^{i}\right)
\right)  ^{-1}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \cdot\left(  \left(  \overset{\longleftarrow
}{\prod_{i=1}^{m-1}}\left(  1-u_{i}t^{i}\right)  \right)  \cdot\overline{\rho
}\left(  a^{\prime}\right)  \cdot\left(  \overset{\longleftarrow}{\prod
_{i=1}^{m-1}}\left(  1-u_{i}t^{i}\right)  \right)  ^{-1}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\overline{\rho}_{u_{1},u_{2},...,u_{m-1}}\left(  a\right)  \text{ and
}\overline{\rho}_{u_{1},u_{2},...,u_{m-1}}\left(  a^{\prime}\right)  \right)
\nonumber\\
&  =\left(  \overset{\longleftarrow}{\prod_{i=1}^{m-1}}\left(  1-u_{i}%
t^{i}\right)  \right)  \cdot\underbrace{\overline{\rho}\left(  a\right)
\cdot\overline{\rho}\left(  a^{\prime}\right)  }_{\substack{=\overline{\rho
}\left(  aa^{\prime}\right)  \\\text{(by (\ref{1rhoaa'}))}}}\cdot\left(
\overset{\longleftarrow}{\prod_{i=1}^{m-1}}\left(  1-u_{i}t^{i}\right)
\right)  ^{-1}\nonumber\\
&  =\left(  \overset{\longleftarrow}{\prod_{i=1}^{m-1}}\left(  1-u_{i}%
t^{i}\right)  \right)  \cdot\overline{\rho}\left(  aa^{\prime}\right)
\cdot\left(  \overset{\longleftarrow}{\prod_{i=1}^{m-1}}\left(  1-u_{i}%
t^{i}\right)  \right)  ^{-1}\nonumber\\
&  =\overline{\rho}_{u_{1},u_{2},...,u_{m-1}}\left(  aa^{\prime}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\overline{\rho}%
_{u_{1},u_{2},...,u_{m-1}}\left(  aa^{\prime}\right)  \right)  . \label{1rr}%
\end{align}
However,\footnote{We shall use the fact that $2m\geq m+1$ (which is true
because $m$ is a positive integer).}%
\begin{align*}
&  \underbrace{\overline{\rho}_{u_{1},u_{2},...,u_{m-1}}\left(  a\right)
}_{\substack{\equiv q_{m}\left(  a\right)  t^{m}+\rho\left(  a\right)
\operatorname{mod}t^{m+1}B\left[  \left[  t\right]  \right]  \\\left(
\text{by (\ref{1proof4})}\right)  }}\cdot\underbrace{\overline{\rho}%
_{u_{1},u_{2},...,u_{m-1}}\left(  a^{\prime}\right)  }_{\substack{\equiv
q_{m}\left(  a^{\prime}\right)  t^{m}+\rho\left(  a^{\prime}\right)
\operatorname{mod}t^{m+1}B\left[  \left[  t\right]  \right]  \\\left(
\text{by (\ref{1proof4}), applied to }a^{\prime}\text{ instead of }a\right)
}}\\
&  \equiv\left(  q_{m}\left(  a\right)  t^{m}+\rho\left(  a\right)  \right)
\cdot\left(  q_{m}\left(  a^{\prime}\right)  t^{m}+\rho\left(  a^{\prime
}\right)  \right) \\
&  =\underbrace{q_{m}\left(  a\right)  q_{m}\left(  a^{\prime}\right)  t^{2m}%
}_{\substack{\equiv0\operatorname{mod}t^{m+1}B\left[  \left[  t\right]
\right]  \text{ (since }2m\geq m+1\\\text{yields }t^{2m}\equiv
0\operatorname{mod}t^{m+1}B\left[  \left[  t\right]  \right]  \text{)}%
}}+\underbrace{q_{m}\left(  a\right)  \rho\left(  a^{\prime}\right)
t^{m}+\rho\left(  a\right)  q_{m}\left(  a^{\prime}\right)  t^{m}%
}_{\substack{=\left(  q_{m}\left(  a\right)  \rho\left(  a^{\prime}\right)
+\rho\left(  a\right)  q_{m}\left(  a^{\prime}\right)  \right)  t^{m}%
\\=\left(  q_{m}\left(  a\right)  a^{\prime}+aq_{m}\left(  a^{\prime}\right)
\right)  t^{m}\\\text{(since we are denoting }q_{m}\left(  a\right)
\rho\left(  a^{\prime}\right)  \text{ by }q_{m}\left(  a\right)  a^{\prime
}\\\text{and denoting }\rho\left(  a\right)  q_{m}\left(  a^{\prime}\right)
\text{ by }aq_{m}\left(  a^{\prime}\right)  \text{)}}}+\underbrace{\rho\left(
a\right)  \rho\left(  a^{\prime}\right)  }_{\substack{=\rho\left(  aa^{\prime
}\right)  \\\text{(since }\rho\text{ is a }K\text{-algebra}%
\\\text{homomorphism)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{since }t\text{
commutes with every element of }B\right) \\
&  \equiv0+\left(  q_{m}\left(  a\right)  a^{\prime}+aq_{m}\left(  a^{\prime
}\right)  \right)  t^{m}+\rho\left(  aa^{\prime}\right) \\
&  =\left(  q_{m}\left(  a\right)  a^{\prime}+aq_{m}\left(  a^{\prime}\right)
\right)  t^{m}+\rho\left(  aa^{\prime}\right)  \operatorname{mod}%
t^{m+1}B\left[  \left[  t\right]  \right]
\end{align*}
and%
\[
\overline{\rho}_{u_{1},u_{2},...,u_{m-1}}\left(  aa^{\prime}\right)  \equiv
q_{m}\left(  aa^{\prime}\right)  t^{m}+\rho\left(  aa^{\prime}\right)
\operatorname{mod}t^{m+1}B\left[  \left[  t\right]  \right]
\]
(by (\ref{1proof4}), applied to $aa^{\prime}$ instead of $a$). Thus,%
\begin{align*}
&  q_{m}\left(  aa^{\prime}\right)  t^{m}+\rho\left(  aa^{\prime}\right) \\
&  \equiv\overline{\rho}_{u_{1},u_{2},...,u_{m-1}}\left(  aa^{\prime}\right)
\\
&  =\overline{\rho}_{u_{1},u_{2},...,u_{m-1}}\left(  a\right)  \cdot
\overline{\rho}_{u_{1},u_{2},...,u_{m-1}}\left(  a^{\prime}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{1rr})}\right) \\
&  \equiv\left(  q_{m}\left(  a\right)  a^{\prime}+aq_{m}\left(  a^{\prime
}\right)  \right)  t^{m}+\rho\left(  aa^{\prime}\right)  \operatorname{mod}%
t^{m+1}B\left[  \left[  t\right]  \right]  .
\end{align*}
Subtracting $\rho\left(  aa^{\prime}\right)  $ from both sides of this
congruence, we obtain%
\[
q_{m}\left(  aa^{\prime}\right)  t^{m}\equiv\left(  q_{m}\left(  a\right)
a^{\prime}+aq_{m}\left(  a^{\prime}\right)  \right)  t^{m}\operatorname{mod}%
t^{m+1}B\left[  \left[  t\right]  \right]  .
\]
Hence, for each $i\in\left\{  0,1,...,m\right\}  $, the $t^{i}$-coefficient of
the power series $q_{m}\left(  aa^{\prime}\right)  t^{m}$ equals the $t^{i}%
$-coefficient of the power series $\left(  q_{m}\left(  a\right)  a^{\prime
}+aq_{m}\left(  a^{\prime}\right)  \right)  t^{m}$ (because if two power
series are congruent to each other modulo $t^{m+1}B\left[  \left[  t\right]
\right]  $, then their $t^{i}$-coefficients agree for each $i\in\left\{
0,1,...,m\right\}  $). Applying this to $i=m$, we see that the $t^{m}%
$-coefficient of the power series $q_{m}\left(  aa^{\prime}\right)  t^{m}$
equals the $t^{m}$-coefficient of the power series $\left(  q_{m}\left(
a\right)  a^{\prime}+aq_{m}\left(  a^{\prime}\right)  \right)  t^{m}$. In
other words, $q_{m}\left(  aa^{\prime}\right)  $ equals $q_{m}\left(
a\right)  a^{\prime}+aq_{m}\left(  a^{\prime}\right)  $ (since the $t^{m}%
$-coefficient of the power series $q_{m}\left(  aa^{\prime}\right)  t^{m}$ is
$q_{m}\left(  aa^{\prime}\right)  $, whereas the $t^{m}$-coefficient of the
power series $\left(  q_{m}\left(  a\right)  a^{\prime}+aq_{m}\left(
a^{\prime}\right)  \right)  t^{m}$ is $q_{m}\left(  a\right)  a^{\prime
}+aq_{m}\left(  a^{\prime}\right)  $). That is,%
\[
q_{m}\left(  aa^{\prime}\right)  =q_{m}\left(  a\right)  a^{\prime}%
+aq_{m}\left(  a^{\prime}\right)  =aq_{m}\left(  a^{\prime}\right)
+q_{m}\left(  a\right)  a^{\prime}.
\]


Forget that we fixed $a$ and $a^{\prime}$. Thus we have proved that
$q_{m}\left(  aa^{\prime}\right)  =aq_{m}\left(  a^{\prime}\right)
+q_{m}\left(  a\right)  a^{\prime}$ for all $a\in A$ and $a^{\prime}\in A$.
Since $q_{m}$ is a $K$-linear map, the assumption (\ref{1ass}) (applied to
$f=q_{m}$) thus yields that there exists an element $s\in B$ such that%
\[
\left(  q_{m}\left(  a\right)  =as-sa\text{ for all }a\in A\right)  .
\]
Consider this $s$.

Now, let $u_{m}$ be the element $-s\in B$. Then, we conclude that%
\begin{equation}
q_{m}\left(  a\right)  =u_{m}a-au_{m}\text{ for all }a\in A \label{1u_m}%
\end{equation}
(since $u_{m}=-s$ yields $s=-u_{m}$ and thus $q_{m}\left(  a\right)
=a\underbrace{s}_{=-u_{m}}-\underbrace{s}_{=-u_{m}}a=a\left(  -u_{m}\right)
-\left(  -u_{m}\right)  a=u_{m}a-au_{m}$). Now, we must prove that
(\ref{1proof}) holds for $n=m$. But every $a\in A$ satisfies\footnote{We shall
use the fact that $2m\geq m+1$ and $3m\geq m+1$ (which is true because $m$ is
a positive integer).}%
\begin{align*}
&  \overline{\rho}_{u_{1},u_{2},...,u_{m}}\left(  a\right) \\
&  =\left(  \underbrace{\overset{\longleftarrow}{\prod_{i=1}^{m}}\left(
1-u_{i}t^{i}\right)  }_{=\left(  1-u_{m}t^{m}\right)  \cdot
\overset{\longleftarrow}{\prod\limits_{i=1}^{m-1}}\left(  1-u_{i}t^{i}\right)
}\right)  \cdot\overline{\rho}\left(  a\right)  \cdot\left(
\underbrace{\overset{\longleftarrow}{\prod_{i=1}^{m}}\left(  1-u_{i}%
t^{i}\right)  }_{=\left(  1-u_{m}t^{m}\right)  \cdot\overset{\longleftarrow
}{\prod\limits_{i=1}^{m-1}}\left(  1-u_{i}t^{i}\right)  }\right)  ^{-1}\\
&  =\left(  \left(  1-u_{m}t^{m}\right)  \cdot\overset{\longleftarrow
}{\prod\limits_{i=1}^{m-1}}\left(  1-u_{i}t^{i}\right)  \right)
\cdot\overline{\rho}\left(  a\right)  \cdot\left(  \left(  1-u_{m}%
t^{m}\right)  \cdot\overset{\longleftarrow}{\prod\limits_{i=1}^{m-1}}\left(
1-u_{i}t^{i}\right)  \right)  ^{-1}\\
&  =\left(  1-u_{m}t^{m}\right)  \cdot\underbrace{\left(
\overset{\longleftarrow}{\prod_{i=1}^{m-1}}\left(  1-u_{i}t^{i}\right)
\right)  \cdot\overline{\rho}\left(  a\right)  \cdot\left(
\overset{\longleftarrow}{\prod_{i=1}^{m-1}}\left(  1-u_{i}t^{i}\right)
\right)  ^{-1}}_{\substack{=\overline{\rho}_{u_{1},u_{2},...,u_{m-1}}\left(
a\right)  \equiv q_{m}\left(  a\right)  t^{m}+\rho\left(  a\right)
\operatorname{mod}t^{m+1}B\left[  \left[  t\right]  \right]  \\\left(
\text{by (\ref{1proof4})}\right)  }}\cdot\underbrace{\left(  1-u_{m}%
t^{m}\right)  ^{-1}}_{\substack{\equiv1+u_{m}t^{m}\operatorname{mod}%
t^{m+1}B\left[  \left[  t\right]  \right]  \\\text{(since }\left(
1-u_{m}t^{m}\right)  \left(  1+u_{m}t^{m}\right)  \\=1-u_{m}^{2}t^{2m}%
\equiv1\operatorname{mod}t^{m+1}B\left[  \left[  t\right]  \right]
\\\text{(because }2m\geq m+1\text{ yields }t^{2m}\equiv0\operatorname{mod}%
t^{m+1}B\left[  \left[  t\right]  \right]  \text{))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(
\alpha\beta\right)  ^{-1}=\beta^{-1}\alpha^{-1}\text{ for any invertible
}\alpha,\beta\in B\left[  \left[  t\right]  \right]  \right) \\
&  \equiv\left(  1-u_{m}t^{m}\right)  \left(  q_{m}\left(  a\right)
t^{m}+\rho\left(  a\right)  \right)  \left(  1+u_{m}t^{m}\right) \\
&  =q_{m}\left(  a\right)  t^{m}+q_{m}\left(  a\right)  u_{m}t^{2m}%
+\rho\left(  a\right)  +\rho\left(  a\right)  u_{m}t^{m}\\
&  \ \ \ \ \ \ \ \ \ \ -u_{m}q_{m}\left(  a\right)  t^{2m}-u_{m}q_{m}\left(
a\right)  u_{m}t^{3m}-u_{m}\rho\left(  a\right)  t^{m}-u_{m}\rho\left(
a\right)  u_{m}t^{2m}\\
&  \equiv q_{m}\left(  a\right)  t^{m}+\rho\left(  a\right)  +\underbrace{\rho
\left(  a\right)  u_{m}}_{\substack{=au_{m}\\\text{(since we denote }%
\rho\left(  a\right)  u_{m}\text{ by }au_{m}\text{)}}}t^{m}-\underbrace{u_{m}%
\rho\left(  a\right)  }_{\substack{=u_{m}a\\\text{(since we denote }u_{m}%
\rho\left(  a\right)  \text{ by }u_{m}a\text{)}}}t^{m}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here we have removed all addends that contain a }t^{2m}\text{ or }%
t^{3m}\text{ factor,}\\
\text{because }2m\geq m+1\text{ yields }t^{2m}\equiv0\operatorname{mod}%
t^{m+1}B\left[  \left[  t\right]  \right]  \text{ and}\\
\text{because }3m\geq m+1\text{ yields }t^{3m}\equiv0\operatorname{mod}%
t^{m+1}B\left[  \left[  t\right]  \right]
\end{array}
\right) \\
&  =q_{m}\left(  a\right)  t^{m}+\rho\left(  a\right)  +au_{m}t^{m}%
-u_{m}at^{m}\\
&  =\left(  u_{m}a-au_{m}\right)  t^{m}+\rho\left(  a\right)  +au_{m}%
t^{m}-u_{m}at^{m}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{1u_m})}\right) \\
&  =\rho\left(  a\right)  \operatorname{mod}t^{m+1}B\left[  \left[  t\right]
\right]  .
\end{align*}
Hence, (\ref{1proof}) holds for $n=m$.

Thus we have shown that, if we have constructed some elements $u_{1}$, $u_{2}%
$, $...$, $u_{m-1}$ of $B$ such that (\ref{1proof}) holds for $n=m-1$, then we
can define a new element $u_{m}$ of $B$ in a way such that (\ref{1proof})
holds for $n=m$. This way, we can recursively construct elements $u_{1}$,
$u_{2}$, $u_{3}$, $...$ of $B$ which satisfy the congruence (\ref{1proof}) for
every $n\in\mathbb{N}$. Now, define a power series $b\in B\left[  \left[
t\right]  \right]  $ by $b=\lim\limits_{n\rightarrow\infty}%
\overset{\longleftarrow}{\prod\limits_{i=1}^{n}}\left(  1-u_{i}t^{i}\right)  $
(this power series $b$ is well-defined since the sequence $\left(
\overset{\longleftarrow}{\prod\limits_{i=1}^{n}}\left(  1-u_{i}t^{i}\right)
\right)  _{n\in\mathbb{N}}$ is a Cauchy sequence with respect to the
above-defined metric on the ring $B\left[  \left[  t\right]  \right]
\ \ \ \ $\footnote{This is because for every $k\in\mathbb{N}$, there exists
some $j\in\mathbb{N}$ such that
\[
\left(  \overset{\longleftarrow}{\prod\limits_{i=1}^{n}}\left(  1-u_{i}%
t^{i}\right)  \equiv\overset{\longleftarrow}{\prod\limits_{i=1}^{m}}\left(
1-u_{i}t^{i}\right)  \operatorname{mod}t^{k}B\left[  \left[  t\right]
\right]  \text{ for every }n\in\mathbb{N}\text{ and }m\in\mathbb{N}\text{
satisfying }n\geq j\text{ and }m\geq j\right)
\]
(namely, take $j=k$; then, any $n\geq j$ satisfies%
\begin{align*}
\overset{\longleftarrow}{\prod\limits_{i=1}^{n}}\left(  1-u_{i}t^{i}\right)
&  =\left(  \overset{\longleftarrow}{\prod\limits_{i=j+1}^{n}}%
\underbrace{\left(  1-u_{i}t^{i}\right)  }_{\substack{\equiv
1\operatorname{mod}t^{k}B\left[  \left[  t\right]  \right]  \text{,}%
\\\text{since }i\geq j+1\geq j=k\\\text{yields }t^{i}\equiv0\operatorname{mod}%
t^{k}B\left[  \left[  t\right]  \right]  }}\right)  \cdot\left(
\overset{\longleftarrow}{\prod\limits_{i=1}^{j}}\left(  1-u_{i}t^{i}\right)
\right)  \\
&  \equiv\left(  \overset{\longleftarrow}{\prod\limits_{i=j+1}^{n}}1\right)
\cdot\left(  \overset{\longleftarrow}{\prod\limits_{i=1}^{j}}\left(
1-u_{i}t^{i}\right)  \right)  =\overset{\longleftarrow}{\prod\limits_{i=1}%
^{j}}\left(  1-u_{i}t^{i}\right)  \operatorname{mod}t^{k}B\left[  \left[
t\right]  \right]  ,
\end{align*}
and similarly any $m\geq j$ satisfies%
\[
\overset{\longleftarrow}{\prod\limits_{i=1}^{m}}\left(  1-u_{i}t^{i}\right)
\equiv\overset{\longleftarrow}{\prod\limits_{i=1}^{j}}\left(  1-u_{i}%
t^{i}\right)  \operatorname{mod}t^{k}B\left[  \left[  t\right]  \right]  ,
\]
so that any $n\geq j$ and $m\geq j$ satisfy $\overset{\longleftarrow
}{\prod\limits_{i=1}^{n}}\left(  1-u_{i}t^{i}\right)  \equiv
\overset{\longleftarrow}{\prod\limits_{i=1}^{j}}\left(  1-u_{i}t^{i}\right)
\equiv\overset{\longleftarrow}{\prod\limits_{i=1}^{m}}\left(  1-u_{i}%
t^{i}\right)  \operatorname{mod}t^{k}B\left[  \left[  t\right]  \right]  $).}
and therefore converges\footnote{since $B\left[  \left[  t\right]  \right]  $
is complete for the above-defined metric}). Then, $b$ is a power series with
constant term $1$ (since $\overset{\longleftarrow}{\prod\limits_{i=1}^{n}%
}\left(  1-u_{i}t^{i}\right)  $ has constant term $1$ for each $n\in
\mathbb{N}$), and every $a\in A$ satisfies%
\begin{align*}
b\overline{\rho}\left(  a\right)  b^{-1} &  =\left(  \lim\limits_{n\rightarrow
\infty}\overset{\longleftarrow}{\prod\limits_{i=1}^{n}}\left(  1-u_{i}%
t^{i}\right)  \right)  \cdot\overline{\rho}\left(  a\right)  \cdot\left(
\lim\limits_{n\rightarrow\infty}\overset{\longleftarrow}{\prod\limits_{i=1}%
^{n}}\left(  1-u_{i}t^{i}\right)  \right)  ^{-1}\\
&  =\lim\limits_{n\rightarrow\infty}\underbrace{\left(
\overset{\longleftarrow}{\prod\limits_{i=1}^{n}}\left(  1-u_{i}t^{i}\right)
\cdot\overline{\rho}\left(  a\right)  \cdot\left(  \overset{\longleftarrow
}{\prod\limits_{i=1}^{n}}\left(  1-u_{i}t^{i}\right)  \right)  ^{-1}\right)
}_{\substack{=\overline{\rho}_{u_{1},u_{2},...,u_{n}}\left(  a\right)
\\\text{(by the definition of }\overline{\rho}_{u_{1},u_{2},...,u_{n}}\left(
a\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since taking reciprocals is a continuous map on}\\
\text{the set of all power series in }B\left[  \left[  t\right]  \right]
\text{ with constant term }1
\end{array}
\right)  \\
&  =\lim\limits_{n\rightarrow\infty}\overline{\rho}_{u_{1},u_{2},...,u_{n}%
}\left(  a\right)  =\rho\left(  a\right)
\end{align*}
(because of (\ref{1proof})\ \ \ \ \footnote{In fact, for every $i\in
\mathbb{N}$, there exists some $k\in\mathbb{N}$ such that
\[
\text{every }n\in\mathbb{N}\text{ satisfying }n\geq k\text{ satisfies
}\overline{\rho}_{u_{1},u_{2},...,u_{n}}\left(  a\right)  \equiv\rho\left(
a\right)  \operatorname{mod}t^{i}B\left[  \left[  t\right]  \right]
\]
(namely, set $k=i$; then, every $n\in\mathbb{N}$ satisfying $n\geq k$
satisfies $\overline{\rho}_{u_{1},u_{2},...,u_{n}}\left(  a\right)  \equiv
\rho\left(  a\right)  \operatorname{mod}t^{n+1}B\left[  \left[  t\right]
\right]  $ by (\ref{1proof}) and thus also satisfies $\overline{\rho}%
_{u_{1},u_{2},...,u_{n}}\left(  a\right)  \equiv\rho\left(  a\right)
\operatorname{mod}t^{i}B\left[  \left[  t\right]  \right]  $ because
$t^{n+1}B\left[  \left[  t\right]  \right]  \subseteq t^{i}B\left[  \left[
t\right]  \right]  $ (since $n+1\geq n\geq k=i$)). Thus, the sequence $\left(
\overline{\rho}_{u_{1},u_{2},...,u_{n}}\left(  a\right)  \right)
_{n\in\mathbb{N}}$ converges to $\rho\left(  a\right)  $. In other words,
$\lim\limits_{n\rightarrow\infty}\overline{\rho}_{u_{1},u_{2},...,u_{n}%
}\left(  a\right)  =\rho\left(  a\right)  $.}). This proves Theorem 1.

\begin{thebibliography}{9}                                                                                                %


\bibitem {1}Pavel Etingof, Oleg Golberg, Sebastian Hensel, Tiankai Liu, Alex
Schwendner, Elena Udovina and Dmitry Vaintrob, \textit{Introduction to
representation theory}, July 13, 2010.\newline%
\texttt{http://math.mit.edu/\symbol{126}etingof/replect.pdf}
\end{thebibliography}


\end{document}