\classheader{2012-10-08}

Today we'll talk about simple and semisimple modules.

\subsection*{Simple modules}

\begin{definition}
A module is simple if it is non-trivial and has no submodules except 0 and itself.
\end{definition}
\begin{example}
If our ring is a field $k$, then a $k$-module $M$ is simple if and only if $M$ is cyclic, which is the case if and only if $\dim_k(M)=1$.
\end{example}
Note that an $A$-module is simple if and only if $M=Am$ for any non-zero $m\in M$.
\begin{definition}
A (left) ideal $J\subsetneq A$ is called maximal if, for any ideal $I\supseteq J$, we have $A=I$.
\end{definition}
\begin{lemma}
$M$ is a simple $A$-module if and only if $M\cong A/J$ for some maximal (left) ideal $J$ of $A$.
\end{lemma}
\begin{proof}
There is a bijection between left submodules $I/J\subseteq A/J$ and ideals $J\subseteq I\subseteq A$:
\begin{center}
\begin{tikzpicture}[scale=1.5,>=angle 90]
\node (a) at (0,2) {$A$};
\node (b) at (1,2) {$A/J$};
\node (c) at (0,1) {$I$};
\node (d) at (1,1) {$I/J$};
\node (e) at (0,0) {$J$};
\node (f) at (1,0) {$0$};
\draw[->] (a) to (b);
\draw[->] (c) to (d);
\draw[->] (e) to (f);
\node[rotate=90] at (0,1.5) {$\subseteq$};
\node[rotate=90] at (0,0.5) {$\subseteq$};
\node[rotate=90] at (1,1.5) {$\subseteq$};
\node[rotate=90] at (1,0.5) {$\subseteq$};
\end{tikzpicture}
\end{center}
\vspace{-0.3in}

%\[\xymatrix{A\ar@{}[d]|{\cup} \ar[r] & A/J \ar@{}[d]|{\cup} \\ I\ar@{}[d]|{\cup} \ar@{-->}[r] & I/J \ar@{}[d]|{\cup}\\ J \ar[r] & 0}\]\qedhere
\end{proof}
\begin{proposition}
Any (left) ideal $I\subsetneq A$ is contained in a maximal ideal.
\end{proposition}
\begin{proof}
Zorn's lemma.
\end{proof}

\begin{corollary}
Any cyclic module has a simple quotient.
\end{corollary}

\begin{warning}
If $N$ is a submodule in $M$, it isn't necessarily true that there exists a maximal submodule $N'$ of $M$ with $N\subseteq N'$.
\end{warning}

\begin{exercise}
Show that $(\Q/\Z,+)$ has no maximal subgroups.
\end{exercise}

\begin{theorem}[Schur's lemma]
If $M$, $N$ are simple $A$-modules, then any morphism $f:M\to N$ is either 0 or is an isomorphism.
\end{theorem}
\begin{proof}
Assume that $f\neq 0$. Then $\im(f)\neq 0$ is a submodule in $N$, hence $\im(f)=N$, i.e. $f$ is surjective. Because $\ker(f)\neq M$, we must have $\ker(f)=0$, hence $f$ is injective.
\end{proof}
\begin{corollary}
For a simple module $M$, the ring $\End_A(M)$ is a division ring.
\end{corollary}
\begin{proof}
Every $f\neq 0$ in $\End_A(M)$ is an isomorphism.
\end{proof}
\begin{corollary}
If $A$ is commutative, then an ideal $I\subset A$ is maximal if and only if $A/I$ is a field.
\end{corollary}
\begin{proof}
If $I$ is maximal, then $A/I$ is a simple $A$-module, so that $\End_A(A/I)\cong A/I$ is a division ring, and hence a field.

If $A/I$ is a field, then $A/I$ is generated by every non-zero element, hence $A/I$ is simple, hence $I$ is maximal (by our lemma).
\end{proof}
\begin{corollary}
If $A$ is commutative, and $n\neq m$, then $A^n\not\cong A^m$ as $A$-modules.
\end{corollary}
\begin{proof}
Pick a maximal ideal $I\subset A$. If $M$ is an $A$-module, we can consider $M/IM$ as an $A/I$-module. In particular we get
\[(A/I)^n\cong A^n/I\cdot A^n\cong A^m/I\cdot A^m\cong (A/I)^m\]
but these are vector spaces over $A/I$, so we must have $m=n$.
\end{proof}
This is false for non-commutative rings in general. A good counterexample can be obtained by taking an infinite dimensional vector space $V$ over a field $k$, and letting $A=\End_k(V)$. Then $A^n\cong A$ for all $n\geq 1$.

\subsection*{Semisimple modules}

\begin{definition}[and Proposition]
For $M$ an $A$-module, the following are equivalent:
\begin{enumerate}
\item $M\cong \oplus \text{ simple modules}$
\item $M=\sum \text{simple submodules}$
\item $M$ is completely reducible, i.e. for any submodule $N\subset M$, there is an $N'\subset M$ such that $N\oplus N'=M$.
\end{enumerate}
\end{definition}
\begin{proof}
Zorn's lemma.
\end{proof}
If (1)-(3) hold then $M$ is called semisimple.
\begin{examples}
If $A=k$ is a field, then any module is semisimple.

If $A$ is a division ring, then any module is semisimple.

If $A=\M_n(D)$ where $D$ is a division ring, then any $A$-module is semisimple.
\end{examples}
\begin{corollary}
Any direct sum, quotient, or submodule of a semisimple module is semisimple.
\end{corollary}
\begin{proof}
The case of submodules follows from (3). The case of quotients follows from (2).  The case of direct sums follows from (1).
\end{proof}

\begin{notation}
If $M=\bigoplus_i M_i$, where the $M_i$'s are simple, then we set for a simple $L$
\[[M:L]=\#\{i\mid M_i\cong L\}.\]
\end{notation}
It's not necessarily clear this is well-defined; an alternative is to write $M$ as
\[M=\bigoplus_{L\in S_A}L^{[M:L]}\]
where $S_A$ is the set of isomorphism classes of simple $A$-modules.

For any $A$-modules $M$ and $N$, the ring $\End_A(N)$ acts (on the right) by composition on $\Hom_A(M,N)$. 

Let $L$ be a simple module. Then $D_L:=\End_A(L)$ is a division ring by Schur's lemma, so that $\Hom_A(M,L)$ is a $D_L$-module.

\begin{proposition}[Multiplicity formula]
For a semisimple module $M$,
\[[M:L]=\operatorname{rank}_{D_L}(\Hom(M,L)).\]
\end{proposition}
\begin{proof}
Assume $M=\bigoplus M_i$, where $M_i$ are simple. Then
\[\Hom_A(M,L)=\Hom_A\left(\bigoplus M_i,L\right)=\bigoplus \Hom_A(M_i,L)\cong\bigoplus_{\{i\mid M_i\cong L\}}\End_A(L)=D_L^{[M:L]}.\qedhere\]
\end{proof}

Let $M=L_1^{r_1}\oplus\cdots\oplus L_n^{r_n}$ where the $L_i$ are simple and $L_i\not\cong L_j$. Then
\begin{align*}
\End_A(M)&=\Hom(M,L_1^{r_1})\oplus\cdots\oplus \Hom(M,L_n^{r_n})\\
&=\Hom(M,L_1)^{r_1}\oplus\cdots\oplus\Hom(M,L_n)^{r_n}\\
&= \M_{r_1}(D_{L_1})\oplus \cdots\oplus \M_{r_n}(D_{L_n}).
\end{align*}