\classheader{2012-11-19}

Now let's prove the theorem we stated last time,
\begin{theorem}[Jacobson density]
If $M$ is a semisimple $A$-module, then $A_M$ is pointwise dense in $A_M^{!!}$; in other words, for any $A_M^!$-linear map $f:M\to M$ and any collection $m_1,\ldots,m_n\in M$, there is an $a\in A$ such that $f(m_1)=am_1$ for all $i=1,\ldots,n$.
\end{theorem}
\begin{proof}
Let $M$ be a semisimple $A$-module, $f : M \to M$ an $A^!_M$-module map, and
$m_1,\dotsc,m_n \in M$ a collection of elements. We want to find
$a \in A$ such that $f(m_i) = am_i$ for all $i$. 

First, consider the case $n=1$: the submodule $Am \subset M$ has a direct summand $M' \subset M$
such that $M = Am \oplus M'$ since $M$ is semisimple. Let
$p : M \to Am$ denote the first projection, which is $A$-linear and 
hence $p \in A_M^!$. Therefore $f$ and $p$ commute, so $f(m) = f(p(m)) = p(f(m))$
implies $f(m) \in Am$, i.e., there exists $a \in A$ such that $f(m) = am$.

In the general case, we use a diagonal trick. Consider $M^n$, which is still
a semisimple $A$-module, and let $x = (m_1,\dotsc,m_n) \in M^n$. 
By the lemma from last lecture, the map $f \mapsto \diag(f,\dotsc,f) =:F$ is an isomorphism $A_M^{!!} \xrightarrow{\;\cong\;} A_{M^n}^{!!}$. The $n=1$ case implies there exists $a \in A$
such that $F(x) = ax$. Equivalently, $f(m_i) = am_i$ for all $i$. 
\end{proof}

\begin{corollary-N} Let $M$ be a semisimple $A$-module. Assume that $M$
is finitely generated over $A_M^!$. Then $A_M^{!!} = A_M$. 
\end{corollary-N}
\begin{proof}
Let $M = A_M^! m_1+ \dotsb + A_M^! m_n$. Let $f \in A_M^{!!}$. 
By Jacobson density, there exists $a \in A$ such that $f(m_i) = am_i$
for $i=1,\dotsc,n$. Given $m \in M$ we can write
$m = b_1 m_1 + \dotsb + b_n m_n$ for $b_i \in A_M^!$. 
Hence $f(m) = f(\sum b_i m_i) = \sum b_i f(m_i) = \sum b_i am_i =
\sum ab_i m_i = am$. 
\end{proof}

\begin{corollary-N} Let $A$ be an algebra over $k = \bar k$. 
Let $M$ be a simple finite dimensional $A$-module. Then $\act_M : A \to \End_k(M)$
is surjective. 
\end{corollary-N}
\begin{proof} Since $M$ is simple and $k$ is algebraically closed, 
$A_M^! = k$ by Schur's lemma. 
Corollary 1 implies $A_M^{!!}=A_M$ but $A_M^{!!} = k^! = \End_k(M)$,
so $A_M = \End_k(M)$. 
\end{proof}

\begin{theorem}[Burnside's theorem] Let $M$ be a finite-dimensional vector
space over an algebraically closed field $k$. Let $A \subset \End_k(M)$ be a subalgebra such that
there is no $A$-stable subspace in $M$ other than $0$ and $M$ itself. 
Then $A = \End_k(M)$. \end{theorem}
\begin{proof} Corollary 2 implies $\act_M:A \to \End_k(M)$ is
surjective, hence it is an isomorphism. 
\end{proof}

\subsection*{The Jacobson radical}
The following page was included in Homework 8:

\includepdf[pages={1}]{Jacobsonradical.pdf} 

\begin{lemma} $J(A/J(A))=0$. \end{lemma}
\begin{proof} Simple $A$-modules are also simple $A/J(A)$-modules, so 
we are done by definition (4L). 
\end{proof}

\begin{definition} Given a left/right/2-sided ideal $I \subset A$, 
\begin{enumerate} \item we say $I$ is nil if any $a \in I$ is nilpotent, and
\item we say $I$ is  nilpotent if there exists $n$ such that $I^n=0$, which is
equivalent to requiring $a_1 \dotsb a_n = 0$ for all $a_1,\dotsc,a_n \in I$. 
\end{enumerate}
\end{definition}

\begin{lemma}
$\text{}$

\begin{enumerate}
\item If $I,J$ are nilpotent, then $I+J$ is also nilpotent.
\item  Any nil ideal is contained in $J(A)$. 
\end{enumerate}
\end{lemma}
\begin{proof}[Proof of 1]
Suppose $I^m = 0$ and $J^n=0$. We claim that $(I+J)^{m+n}=0$. We can see this as follows: if we take $a_1,\dotsc,a_{m+n} \in I \cup J$, then there will be
either $\ge m$ elements $a_i \in I$ or $\ge n$ elements in $J$. Assume 
WLOG that $a_{i_1},\dotsc,a_{i_m} \in I$. Then 
$a_1\dotsb a_{m+n} = (\dotsb a_{i_1})(\dotsb a_{i_2})\dotsb (\dotsb a_{i_m})\dotsb
a_{m+n} = 0$ since $I$ is an ideal.
\end{proof}
\begin{proof}[Proof of 2]
If $I$ is a left nil ideal, then $a \in I$ implies that $xa \in I$ for all 
$x \in A$. Hence $xa$ is nilpotent, so $1+xa$ is invertible. Therefore by
(2L) $a \in J(A)$. 
\end{proof}