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T_SkolemNoether.tex
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T_SkolemNoether.tex
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\section{Project: The Skolem--Noether theorem}
We now present an elementary proof of
the Skolem--Noether theorem. We refer to
\cite{MR3308118} for more information.
\begin{definition}
\index{Algebra!central}
Let $K$ be a field.
An algebra $A$ (over $K$) is \emph{central} if $Z(A)=K$.
\end{definition}
If $K$ is a field, then $M_n(K)$ is a central algebra.
\begin{proposition}
Let $A$ be a unitary algebra and $n\geq1$. Then
$A$ is central if and only if $M_n(A)$ is central.
\end{proposition}
\begin{proof}
If $M_n(A)$ is central and $z\in Z(A)$, then
$zI\in Z(M_n(A))=KI$. Thus
$z\in K$. Conversely, if $X\in Z(M_n(A))$, then, since
$XE_{kl}=E_{kl}X$ for all $k\ne l$, $X=aI$ for some $a\in A$.
Moreover,
$XaE_{11}=aE_{11}X$. Hence $a\in Z(A)=K1$.
\end{proof}
\begin{example}
$\mathbb{H}$ is a real central algebra.
\end{example}
\begin{example}
$\C$ is a complex central algebra but it is not a real central
algebra.
\end{example}
A celebrated theorem by Frobenius states that every
finite-dimensional
real central division algebra is isomorphic to $\R$ or $\mathbb{H}$ (see Theorem \ref{thm:Frobenius}).
\begin{proposition}
Every simple unitary ring is an algebra over its center.
\end{proposition}
\begin{proof}
Let $R$ be a simple unitary ring. It is enough to show that
$Z(R)$ is a field. If $z\in
Z(R)\setminus\{0\}$ then $zR$ is a non-zero ideal of $R$. Since $R$
is simple, $zR=R$. Thus $z$ is invertible.
\end{proof}
For an algebra $A$, let $L\colon A\to\End_K(A)$,
$a\mapsto L_a$, and $R\colon A\to\End_K(A)$, $a\mapsto R_a$, be given by
$L_a(x)=ax$ and $R_a(x)=xa$. Then both $L$ and $R$ are linear maps such that
\begin{align*}
L_{ab}=L_aL_b, && R_{ab}=R_bR_a, && L_aR_b=R_bL_a
\end{align*}
for all $a,b\in A$.
\begin{definition}
\index{Algebra!of multipliers}
Let $A$ be an algebra. The \emph{algebra of multipliers} of $A$
is
\[
M(A)=\left\{\sum_{j=1}^n L_{a_i}R_{b_i}:n\in\Z_{\geq0},\,a_1,\dots,a_n,b_1,\dots,b_n\in A\right\}.
\]
\end{definition}
It is an exercise to show that $M(A)$ is a subalgebra of $\End_K(A)$. Moreover,
if $A$ is unitary, then $M(A)$ is generated by the $L_a$ and the $R_b$ for
$a,b\in A$.
\begin{lemma}
\label{lem:SkolemNoether}
Let $A$ be an algebra and
$f\in M(A)$. Then there exists $n\geq0$ and
$a_1,\dots,a_n\in A$ and $b_1,\dots,b_n\in A$ such that
\[
f=\sum_{i=1}^n L_{a_i}R_{b_i}
\]
and $\{b_1,\dots,b_n\}$ is linearly independent.
\end{lemma}
\begin{proof}
Write $f=\sum_{i=1}^n L_{a_i}R_{b_i}$ with $n$ be minimal. If
$b_n=\sum_{j=1}^{n-1}\lambda_jb_j$, then
\[
f=\sum_{i=1}^{n-1}L_{a_i+\lambda_ia_n}R_{b_i},
\]
a contradiction.
\end{proof}
\begin{lemma}
\label{lem:SkolemNoether1}
Let $A$ be a central simple algebra.
If $\sum_{i=1}^n L_{a_i}R_{b_i}=0$ and $\{b_1,\dots,b_n\}$
(resp. $\{a_1,\dots,a_n\}$) is linearly independent,
then $a_i=0$ (resp. $b_i=0$) for all $i$.
\end{lemma}
\begin{proof}
The result holds for $n=1$. We want to prove that
if $a_1xb_1=0$ for all $x\in A$ and $b_1\ne0$, then
$a_1=0$. Assume that $a_1\ne 0$. The ideal of $A$ generated by
$a_1$ is non-zero, and hence it is equal to $A$. Thus there exist
$u_1,\dots,u_m,v_1,\dots,v_m\in A$ such that $1=\sum_{j=1}^m u_ja_1v_j$.
Write
\[
0=\sum_{j=1}^m L_{u_j}(L_{a_1}R_{b_1})L_{v_j}
=\sum_{j=1}^m L_{u_ja_1v_j}R_{b_1}=R_{b_1}.
\]
Hence $b_1=0$.
Assume that the lemma is not true and let $n>1$ be the smallest positive
integer where the lemma is false.
Assume that $a_n\ne 0$. Since $A$ is simple, the ideal generated by
$a_n$ is $A$. Then there exist
$u_1,\dots,u_m,v_1,\dots,v_m\in A$ such that $1=\sum_{j=1}^m u_ja_1v_j$ and
\[
0=\sum_{j=1}^m L_{u_j}\left(\sum_{i=1}^n L_{a_i}R_{b_i}\right)L_{v_j}=\sum_{i=1}^n\sum_{j=1}^m L_{u_ja_iv_j}R_{b_i}=\sum_{i=1}^n L_{c_i}R_{b_i},
\]
where $c_i=\sum_{j=1}^m u_ja_iv_j$ and $c_n=1$. Since
\[
0=L_x\left(\sum_{i=1}^n L_{c_i}R_{b_i}\right)-\left(\sum_{i=1}^n L_{c_i}R_{b_i}\right)L_x=\sum_{i=1}^{n-1}L_{xc_i-c_ix}R_{b_i}
\]
for all $x\in A$, it follows that $xc_i-c_ix=0$ for all
$x\in A$. Since $A$ is central, $c_i\in k$ for all
$i\in\{1,\dots,n-1\}$. Evaluate $0=\sum_{i=1}^n L_{c_i}R_{b_i}$ in $1_A$
we obtain that $0=c_1b_1+\cdots+c_nb_n$, a contradiction since
$\{b_1,\dots,b_n\}$ is linearly independent.
\end{proof}
\begin{lemma}
\label{lem:SkolemNoether2}
If $A$ is a finite-dimensional central simple algebra, then
\[
M(A)=\End_K(A).
\]
\end{lemma}
\begin{proof}
Let $\{a_1,\dots,a_n\}$ be a basis of $A$. We claim that
$\{L_{a_i}R_{a_j}:1\leq i,j\leq n\}$ is linearly independent. If
\[
\sum_{i,j=1}^n\lambda_{ij}L_{a_i}R_{a_j}=0,
\]
then
$\sum_{i=1}^nL_{a_i}R_{c_i}=0$, where
$c_i=\sum_{j=1}^n\lambda_{ij}R_{a_j}$. Since the $a_i$'s are linearly
independent, Lemma~\ref{lem:SkolemNoether1} implies that $c_i=0$ for all
$i\in\{1,\dots,n\}$, a contradiction since the $a_j$'s are linearly independent.
Hence $\dim_kM(A)\geq n^2=\dim\End_K(A)$.
\end{proof}
\begin{definition}
\index{Automorfism!inner}
Let $R$ be a unitary ring. An automorphism $f\in\Aut(R)$ is
\emph{inner} is there exists an invertible $r\in R$ such that
$f(x)=rxr^{-1}$ for all $x\in R$.
\end{definition}
For example, $\C\to\C$, $z\mapsto\overline{z}$, is not inner.
\begin{example}
Let $\lambda\in k\setminus\{0\}$ and $R=k[X]$. Then
\[
k[X]\to
k[X],\quad f(X)\mapsto f(X+\lambda),
\]
is not inner.
\end{example}
\begin{example}
Let $R$ be a ring. Then $R\times R\to R\times R$, $(x,y)\mapsto
(y,x)$, is not inner.
\end{example}
\begin{theorem}[Skolem--Noether]
\index{Skolem--Noether theorem}
\label{thm:SkolemNoether}
If $A$ is a finite-dimensional central simple algebra,
every automorphism of $A$ is inner.
\end{theorem}
\begin{proof}
Let $f\in\Aut(A)$. By Lemma~\ref{lem:SkolemNoether2},
$f=\sum_{i=1}^n L_{a_i}R_{b_i}$.
Without loss of generality, we may assume that $a_1\ne 0$ and
that $\{b_1,\dots,b_n\}$ is linearly independent.
Since $f$ is a homomorphism,
$L_{f(x)}f=fL_x$ for all $x\in A$. Then
\[
0=\sum_{i=1}^n L_{f(x)a_i-a_ix}R_{b_i}.
\]
By Lemma~\ref{lem:SkolemNoether1}, $f(x)a_1-a_1x=0$ for all
$x\in A$. We claim that $a_1$ is invertible.
Since $a_1\ne 0$ and $A$ is simple, the ideal of $A$ generated by $a_1$ is $A$.
Write $1=\sum_{i=1}^m u_ja_1v_j$. Thus $a_1$ is invertible,
as
\[
\left(\sum_{j=1}^m u_jf(v_j)\right)a_1=a_1\left(\sum_{j=1}^m f^{-1}(u_j)v_j\right)=1.\qedhere
\]
\end{proof}
%\begin{corollary}
% \label{cor:SkolemNoether1}
% Sea $A$ es un álgebra de dimensión finita unitaria. Si $a\in A$, entonces
% $a$ es inversible o es un divisor de cero.
%\end{corollary}
%
%\begin{proof}
% Como $A$ es de dimensión finita, $A$ es algebraica. Existe entonces un
% polinomio $f=\sum_{j=1}^n \lambda_jX^j\in k[X]$ (que podemos suponer de grado
% mínimo) tal que $f(a)=0$. Al escribir
% \[
% 0=f(a)=a(\lambda_na^{n-1}+\cdots+\lambda_2a+\lambda_1)+a_0
% \]
% vemos que existe un polinomio $g=\lambda_nX^{n-1}+\cdots+\lambda_1\in k[X]$
% tal que $g(a)\ne 0$ (por la minimalidad de $n$) y $ag(a)=-\lambda_0$. Si
% $a$ no es un divisor de cero, entonces $\lambda_0\ne 0$ y luego
% $a^{-1}=-\lambda_0^{-1}g(a)$.
%\end{proof}
%
%\begin{corollary}
% Sea $A$ un álgebra de dimensión finita unitaria y sean $a,b\in A$. Si
% $ab=1$, entonces $ba=1$.
%\end{corollary}
%
%\begin{proof}
% Es consecuencia inmediata del corolario~\ref{cor:SkolemNoether1}.
%\end{proof}
%
%\begin{corollary}
% Sea $D$ un álgebra de división de dimensión finita y sea $A$ una subálgebra
% de $D$. Entonces $A$ es un álgebra de división.
%\end{corollary}
%
%\begin{proof}
% Sea $a\in A\setminus\{0\}$. Como existe $d\in D$ tal que $ad=1$. Como $a$ es algebraico,
% existe $f\in k[X]$ de grado mínimo tal que $f(a)=0$. Luego $a$ es inversible con
% $a^{-1}=-\lambda_0^{-1}g(a)$ para algún $g\in k[X]$ tal que $g(a)\ne 0$. En
% particular, $a^{-1}\in A$ y además $A$ es unitaria.
%\end{proof}