Semisimple module

Revision as of 15:09, 17 August 2009 by Boy Soprano II (talk | contribs) (Created page with 'A '''semisimple module''' is, informally, a module that is not far removed from simple modules. Specifically, it is a module <math>M</math> with the following property: f…')
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 lemmas.

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. Let $M$ be a semisimple left $R$-module, and let $N$ be a submodule of $M$. Then $N$ is semisimple.

Proof. Let $T$ be a submodule of $N$, and let $T'$ be the submodule of $M$ such that $T \cap T' = 0$ and $T + T' = M$. Then $T \cap (T' \cap N) = 0$; furthermore, since $T \subset N$, if follows that $N = N \cap (T + T') = T + (T' \cap N)$. It follows that $N$ is semisimple. $\blacksquare$

Lemma 3. 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:

  1. $M$ is a semisimple $R$-module;
  2. $M$ is isomorphic to a direct sum of simple left $R$-modules;
  3. $M$ 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 Lemma 2, the cyclic submodule $\langle \alpha \rangle$ is semisimple, so by Lemma 3, it has a simple submodule that is a subset of $M$, a contradiction. Therefore $N' = 0$, so $M = N$. $\blacksquare$

See also