Semisimple module

A semisimple module is, informally, a module that is not far removed from simple modules. Specifically, it is a module $M$ with the following property: for every submodule $N \subset M$ , there exists a submodule $N' \subset M$ such that $N + N' = M$ and $N \cap N' = 0$ , where by 0 we mean the zero module.

Classification of semisimple modules

It happens that semisimple modules have a convenient classification (assuming the axiom of choice). To prove this classification, we first state some intermediate results.

Proposition. Let $M$ be a semisimple left $R$ -module. Then every submodule and quotient module of $M$ is also simple.

Proof. First, suppose that $N$ is a submodule of $M$ . Let $T$ be a submodule of $N$ , and let $N'$ be a submodule of $M$ such that $T \cap N' = 0$ and $T + N' = M$ . We note that if $\tau \in T$ and $\nu' \in N$ are elements such that $\tau + \nu' \in N$ , then $\nu' \in N$ . It follows that $N = M \cap N = (T + N') \cap N = T + (N' \cap N) .$ Since $T \cap (N' \cap N) = 0$ , it follows that $N$ is semisimple.

Now let us consider a quotient module $M/N$ of $M$ , with $\phi$ the canonical homomorphism $M \to M/N$ . Let $T$ be a submodule of $M/N$ . Then $\phi^{-1}(T)$ is a submodule of $M$ , so there exists a submodule $N' \subset M$ such that $\phi^{-1}(T) \cap N'=0$ and $\phi^{-1}(T) + N' = M$ . Then in $M/N$ , since $\phi$ is surjective, $\begin{align*} T + \phi(N') &= \phi \bigl( \phi^{-1}(T) + N' \bigr) = \phi(M) = M/N \\ T \cap \phi(N') &= \phi\bigl( \phi^{-1} (T) \bigr) \cap \phi(N') \subset \phi\bigl( \phi^{-1}(T) \cap N' \bigr) = \phi(0) = 0. \end{align*}$ Therefore $M/N$ is semisimple as well.

Lemma 1. Let $R$ be a ring, and let $M$ be a nonzero cyclic left $R$ -module. Then $M$ contains a maximal proper submodule.

Proof. Let $\alpha$ be a generator of $M$ . Let $\mathfrak{S}$ be the set of submodules that avoid $\alpha$ , ordered by inclusion. Then $\mathfrak{S}$ is nonempty, as $\{0 \} \in \mathfrak{S}$ . Also, if $( N_i )_{i \in I}$ is a nonempty chain in $\mathfrak{S}$ , then $\bigcup_{i \in I} N_i$ is an element of $\mathfrak{S}$ , as this is a submodule of $M$ that does not contain $\alpha$ . Then $\bigcup_{i\in I} N_i$ is an upper bound on the chain $(N_i)$ ; thus every chain has an upper bound. Then by Zorn's Lemma, $\mathfrak{S}$ has a maximal element. $\blacksquare$

Lemma 2. Every cyclic semisimple module has a simple submodule.

Proof. Let $M$ be a cyclic semisimple module, and let $\alpha$ be a generator for $M$ . Let $N$ be a maximal proper submodule of $M$ (as given in Lemma 1), and let $N'$ be a submodule such that $N+N' =M$ and $N \cap N' = 0$ . We claim that $N'$ is simple.

Indeed, suppose that $T$ is a nonzero submodule of $N'$ . Since the sum $N + N'$ is direct, it follows that the sum $N + T$ is direct. Since $N+T$ strictly contains $N$ , it follows that $N + T = M$ , so $N' \subset N + T$ ; it follows that $N' = T$ ; thus $N'$ is simple. $\blacksquare$

Theorem. Let $M$ be a left $R$ -module, for a ring $R$ . The following are equivalent:

$M$ is a semisimple $R$ -module;
$M$ is isomorphic to a direct sum of simple left $R$ -modules;
$M$ is isomorphic to an (internal) sum of $R$ -modules.

Proof. To prove that 2 implies 1, we suppose without loss of generality that $M$ is a direct sum $\bigoplus_{i \in I} M_i$ of simple left $R$ -modules $M_i$ . If $N$ is a submodule of $M$ , then for each $i \in I$ , $N \cap M_i$ is either 0 or $M_i$ ; if we take $J$ to be the family of $i$ such that $N \cap M_i = M_i$ , then we may take $N' = \bigoplus_{i \notin J} M_i$ .

To prove that 3 implies 2, we note that if $M_i$ is a simple submodule of any module $M$ , and $N$ is a submodule of $M$ , then $M_i \cap N$ is a submodule of $M_i$ , and hence equal to $M_i$ or 0. Now suppose that $M = \sum_{i \in I} M_i$ , where each of the $M_i$ is semisimple. Let us take a well-ordering on $I$ (such an ordering exists by the well-ordering theorem, a consequence of the axiom of choice), and let us define $J$ as the set of elements $j \in I$ such that $M_j \not\subset \sum_{i < j} M_i .$ It follows from transfinite induction that for each $j_0 \in J$ , the sum $\sum_{j\in J, j < j_0} M_j$ is direct, and that $\sum_{i \in I, i < j_0} M_i = \sum_{j\in J, j < j_0} M_j .$ Then $M = \sum_{j \in J} M_j$ , and the sum is direct.

To prove that 1 implies 3, let us take $N$ to be the sum of the cyclic submodules of $M$ , and let $N'$ be the module such that $N \cap N' = 0$ and $N+N' = M$ . Suppose that $N'$ has a nonzero element $\alpha$ ; then by our proposition, the cyclic submodule $\langle \alpha \rangle$ is semisimple, so by Lemma 2, it has a simple submodule that is a subset of $M$ , a contradiction. Therefore $N' = 0$ , so $M = N$ . $\blacksquare$

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Semisimple module

Classification of semisimple modules

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