\classheader{2012-10-12}

Last time, we proved that for an algebraic element $a\in A$ in an algebra,
\[\spec(a)=\{\text{roots of the minimal polynomial }p_a\in k[t]\}.\]
Let $a^n=0$, so that $a$ is nilpotent. Then $p_a=t^n$, so that $\spec(a)=\{0\}$. This implies that $\lambda-a$ is invertible for any $\lambda\neq 0$. Let's see if we can find an explicit inverse.
\begin{align*}
(\lambda-a)^{-1}&=[\lambda(1-\lambda^{-1}a)]^{-1}\\
&=\lambda^{-1}(1-\lambda^{-1}a)^{-1}\\
&=\lambda^{-1}(1+(\lambda^{-1}a)+(\lambda^{-1}a)^2+\cdots)\\
&=\sum_{i=0}^\infty\lambda^{-(i+1)}a^i\\
&=\sum_{i=0}^{n-1}\lambda^{-(i+1)}a^i
\end{align*}
The intermediate steps aren't really allowed, but it gets us the right answer.

From now on we consider $k$-algebras over an algebraically closed field $k$.
\begin{proposition}
Let $A$ be a finite dimensional $k$-algebra.
\begin{enumerate}
\item $\spec(a)\neq \varnothing$ for any $a\in A$.
\item If $A$ is a division algebra, then $A\cong k$.
\end{enumerate}
\end{proposition}
\begin{proof}
For part 1, note that $\dim(A)<\infty$ implies that any $a\in A$ is algebraic, so that $\spec(a)=\{\text{roots of }p_a\}$, which is non-empty because $k$ is algebraically closed.

For part 2, note that for any $k$-algebra, we have an inclusion $k\hookrightarrow A$ by $\lambda\mapsto \lambda\cdot 1_A$. Suppose that $a\in A\setminus k$. Then for any $\lambda\in k$, we have that $\lambda-a\neq 0$, hence $\lambda-a$ is invertible for all $a\in k$, hence $\spec(a)=\varnothing$; but this contradicts part 1.
\end{proof}
\begin{proposition}[Schur lemma for algebras]
Let $M$ be a finite dimensional (over $k$) simple $A$-module. Then $\End_A(M)=k$, i.e. any endomorphism $f:M\to M$ is of the form $\lambda\cdot\id_M$ for some $\lambda\in k$.
\end{proposition}
\begin{proof}
We know that $\End_A(M)$ is a division algebra. The $A$-action on $M$ is the same as a map $A\to \End_k(M)$, which is a finite dimensional division algebra. Let $A'=\im(A)$, so $\dim(A')<\infty$ and $M$ is a simple $A'$-module, which implies $\End_{A'}(M)=k$, but $\End_A(M)=\End_{A'}(M)$.
\end{proof}
\begin{corollary}
$\text{}$

\begin{enumerate}
\item The center $Z(A)$ of $A$ acts by scalars in any finite-dimensional simple $A$-module.
\item If $A$ is commutative\footnote{A reader points out that $A$ also needs to be finite dimensional; otherwise a transcendental field extension of $k$ would provide a counterexample.}, then any simple $A$-module has dimension 1 over $k$.
\end{enumerate}
\end{corollary}
\begin{proof}
Let $M$ be a finite-dimensional simple $A$-module. Then consider the action map $\operatorname{act}:A\to\End_k(M)$. If $z\in Z(A)$, then $\operatorname{act}(z)\in\End_A(M)=k$ by the Schur lemma for algebras. Part 1 follows.

If $A=Z(A)$, then any element of $A$ acts on $M$ by scalars, so any vector subspace $N\subseteq M$ is $A$-stable. Thus, $\dim_k(M)=1$.
\end{proof}

Let $S_A$ be the isomorphism classes of simple $A$-modules.

Let $M$ be a finite direct sum of simple modules, so
\[M\cong\bigoplus_{L\in S_A}L^{n(L)}\]
For any $A$-module $N$, we have an evaluation map
$\operatorname{ev}:\Hom_A(N,M)\otimes_k N\to M$, defined by sending $f\otimes n$ to $f(n)$.

Now $M$ is a finite direct sum of simple modules, so we get an evaluation map
\[\bigoplus_{L\in S^A}\Hom_A(L,M)\otimes_k L\xrightarrow{\;\;\cong\;\;} M\]
by the Schur lemma.

\begin{theorem}[Wedderburn Theorem for Algebras]
$A$ is a finite-dimensional semi-simple $k$-algebra if and only if
\[A\cong \M_{r_1}(k)\oplus\cdots\oplus\M_{r_n}(k).\]
\end{theorem}
\begin{proof}
By Wedderburn's theorem, we have
\[A\cong \M_{r_1}(D_1)\oplus\cdots\oplus\M_{r_n}(D_n)\]
where the $D_i$ are division rings. The fact that $\dim_k(A)<\infty$ implies $\dim_k(D_i)<\infty$ which implies that $D_i=k$.
\end{proof}
\begin{corollary}
Let $A$ be a finite-dimensional semisimple algebra. Then
\begin{enumerate}
\item $S_A$ is a finite set and any $M\in S_A$ is finite dimensional over $k$. Moreover, $[A:L]=\dim(L)$ for all $L\in S_A$.
\item $\dim(A)=\sum_{L\in S_A}\dim(L)^2$
\end{enumerate}
\end{corollary}
\begin{proof}
Any simple $A$-module is cyclic, so that it is isomorphic to $A/J$ for some $J$. We have $\dim_k(A/J)\leq\dim_k(A)<\infty$, so any simple $A$-module is finite dimensional. 

Note that $A$ satisfies a universal property, $\Hom_A(A,L)\xrightarrow{\cong}L$ is an isomorphism of $k$-vector spaces. 

We can decompose $A=\bigoplus_{L\in S_A}L^{n_L}$. We get that, for any simple $L'$,
\[\Hom_A(L',A)=\Hom_A\bigg(L',\bigoplus_{L\in S_A} L^{n_L}\bigg)=\bigoplus_{L\in S_A}\Hom(L',L)^{n_L}=\bigoplus_{L\in S_A}\left(\genfrac{}{}{0pt}{}{k\text{ if }L=L'}{0\text{ if }L\neq L'}\right)^{n_L}= k^{n_{L'}}.\]
By the same argument we also have
\[\Hom_A(A,L')=\Hom_A\bigg(\bigoplus_{L\in S_A} L^{n_L},L'\bigg)=\bigoplus_{L\in S_A}\Hom(L,L')^{n_L}=\bigoplus_{L\in S_A}\left(\genfrac{}{}{0pt}{}{k\text{ if }L=L'}{0\text{ if }L\neq L'}\right)^{n_L}= k^{n_{L'}}.\]
Because $\Hom_A(A,L)\cong L$ as $k$-vector spaces, we therefore have that $\dim(L')=\dim(\Hom_A(A,L'))=n_{L'}$, and hence
\[\dim(A)=\sum n_L\dim(\Hom_A(L,A))=\sum n_L\dim(L)=\sum \dim(L)^2.\qedhere\]
\end{proof}