\classheader{2012-10-10}

A ring $A$ is said to be semisimple when any $A$-module is semisimple.

\begin{theorem}[Wedderburn Theorem]
For a ring $A$, the following conditions are equivalent:
\begin{enumerate}
\item $A$ is a finite direct sum
\[A=\M_{r_1}(D_1)\oplus\cdots\oplus \M_{r_n}(D_n)\]
where the $D_i$ are division rings.
\item $A$ is semisimple.
\item The rank 1 free $A$-module is semisimple.
\item Any left ideal in $A$ has the form $Ae$ where $e^2=e\in A$.
\item $A=Ae_1+\cdots+Ae_n$ where $e_i^2=e_i$ and $e_ie_j=0$ for $i\neq j$, and each $Ae_i$ is a simple $A$-module.
\end{enumerate}
\end{theorem}
\begin{proof}[(1 $\Rightarrow$ 2)] The fact that any module over $\M_r(D)$ is semisimple is something that is from your homework. 
Now note that if $A=A_1\oplus\cdots\oplus A_n$, then an $A$-module is equivalent to an $n$-tuple $\{M_i\}$ where $M_i$ is an $A_i$-module, so that because each $\M_{r_i}(D_i)$ is semisimple, so is $\M_{r_1}(D_1)\oplus\cdots\oplus \M_{r_n}(D_n)$.
\end{proof}
\begin{proof}[(2 $\Rightarrow$ 3)] Clear.
\end{proof}
\begin{proof}[(3 $\Rightarrow$ 4)]
$A$ is completely reducible as an $A$-module, so for any left ideal $I\subset A$, there exists a left ideal $I'\subset A$ such that $A=I\oplus I'$. Thus, there are $e\in I$ and $e'\in I'$ such that $1=e+e'$. Now note that for any $x\in I$,
\[\underbrace{x}_{\in I} = x\cdot 1=\underbrace{\;x\cdot e\;}_{\in I}+\underbrace{\;x\cdot e'\;}_{\in I'}\]
which implies that $x\cdot e'$ and $x=xe$. Thus, $I$ is of the form $Ae$, and $e=e^2$.
\end{proof}
\begin{proof}[(4 $\Rightarrow$ 5)]
Since any left ideal $I\subset A$ has the form $I=Ae$ where $e^2=e$, we have for any ideal $Ae$ that $A=Ae\oplus A(1-e)$. Thus, $A$ is completely reducible.

This implies that $A=\bigoplus L_i$, a possibly infinite direct sum of simple $A$-modules. But then $1=e_1+\cdots+e_n$, hence $x=x\cdot 1= xe_1+\cdots+xe_n$ where $xe_i\in L_i$. Thus $A=L_1\oplus\cdots\oplus L_n$.
\end{proof}
\begin{proof}[(5 $\Rightarrow$ 1)]
Reexpressing the statement of 5, we have that
\[A=\bigoplus_{j=1}^m L_i^{r_i},\qquad L_i\not\cong L_j\text{ for }i\neq j.\]
Therefore
\[A^{\text{op}}=\End_A(A)=\bigoplus_{j=1}^m\End(L_j^{r_j})=\bigoplus_{j=1}^m\M_{r_j}(\End_A(L_i))\]
and by Schur's lemma, $D_i=\End_A(L_i)$ is a division ring. But this gives a decomposition of $A^{\text{op}}$, not $A$. To fix this, note that
\[A=(A^{\text{op}})^{\text{op}}=\bigoplus_{j=1}^m \M_{r_j}(D_j)^{\text{op}}\]
that $\M_r(D)^{\text{op}}=\M_r(D^{\text{op}})$, and that $D^{\text{op}}$ is a division ring when $D$ is.
\end{proof}
\begin{corollary}
A commutative ring $A$ is semisimple if and only if it is a finite direct sum of fields.
\end{corollary}
\begin{proof}
There is no way to get a commutative ring if any of the $r_i>1$, nor if any of the $D_j$ are non-commutative division algebras.
\end{proof}

From this point forward, let $A$ be a $k$-algebra where $k$ is a field.
\begin{definition}
An element $a\in A$ is called algebraic if there is some monic $p\in k[t]$ such that $p(a)=0$.
\end{definition}
\begin{examples}
$\text{}$

\begin{itemize}
\item Any idempotent or nilpotent element is algebraic.
\item If $\dim_k(A)<\infty$ then any $a\in A$ is algebraic, because $1,a,a^2,\ldots$ cannot be linearly independent over $k$. Thus, there exist some $\lambda_i\in k$ such that $\sum \lambda_i a^{n_i}=0$.
\end{itemize}
\end{examples}

Let $a\in A$ be an algebraic element. Consider the $k$-algebra homomorphism $j:k[t]\to A$ defined by $j(f)=f(a)$. Because $k[t]$ is a PID, we have that $\ker(j)=(p_a)$ is a principal ideal. The monic polynomial $p_a$ is called the minimal polynomial for $a$. Note that we get an induced homomorphism $j:k[t]/(p_a)\hookrightarrow A$.

\begin{definition}
For any $a\in A$, we define
\[\spec(a)=\{\lambda\in k\mid \lambda-a\text{ is not invertible}\}.\]
\end{definition}
\begin{examples}
$\text{}$

\begin{itemize}
\item If $A=k\{X\}$, then $\spec(a)=\{\text{the values of }a\}$.
\item If $k$ is algebraically closed and $A=\M_n(k)$, then for an $a\in A$, $\spec(a)=\{\text{eigenvalues of }a\}$.
\end{itemize}
\end{examples}

\begin{lemma}
Let $a\in A$ be an algebraic element with minimal polynomial $p_a\in k[t]$. Then
\[\lambda-a\text{ is not invertible}\iff \lambda-a\text{ is a zero divisor}\iff p_a(\lambda)=0.\]
Thus, $\spec(a)=\{\text{roots of }p_a\}$.
\end{lemma}
\begin{proof}
First, we make a general remark: for any $\lambda\in k$, we have
\[p(t)-p(\lambda)=q(t)(\lambda-t),\]
and because $\deg(q)<\deg(p)$, we must have $q\notin (p)$, hence $q(a)\neq 0$. 

We have an injective homomorphism $j:k[t]/(p)\hookrightarrow A$ where we send $f$ to $f(a)$. If $p(\lambda)=0$, then
\[0=p(a)=q(a)(\lambda-a)\]
implies that $\lambda-a$ is a zero-divisor. Clearly, if $\lambda-a$ is a zero-divisor, it is not invertible.

Now we want to show that $\lambda-a$ is not invertible $\implies$ $p(\lambda)=0$. Assume for the sake of contradiction that $p(\lambda)\neq 0$; then
\[\underbrace{p(a)}_{=0} - \underbrace{p(\lambda)}_{\neq 0}=q(a)(\lambda-a)\]
demonstrates that $\lambda-a$ is invertible.
\end{proof}