Semisimple module
A semisimple module is, informally, a module that is not far removed from simple modules. Specifically, it is a module with the following property: for every submodule , there exists a submodule such that and , 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 be a ring, and let be a nonzero cyclic left -module. Then contains a maximal proper submodule.
Proof. Let be a generator of . Let be the set of submodules that avoid , ordered by inclusion. Then is nonempty, as . Also, if is a nonempty chain in , then is an element of , as this is a submodule of that does not contain . Then is an upper bound on the chain ; thus every chain has an upper bound. Then by Zorn's Lemma, has a maximal element.
Lemma 2. Let be a semisimple left -module, and let
be a submodule of . Then is semisimple.
Proof. Let be a submodule of , and let be the submodule of such that and . Then ; furthermore, since , if follows that . It follows that is semisimple.
Lemma 3. Every cyclic semisimple module has a simple submodule.
Proof. Let be a cyclic semisimple module, and let be a generator for . Let be a maximal proper submodule of (as given in Lemma 1), and let be a submodule such that and . We claim that is simple.
Indeed, suppose that is a nonzero submodule of . Since the sum is direct, it follows that the sum is direct. Since strictly contains , it follows that , so ; it follows that ; thus is simple.
Theorem. Let be a left -module, for a ring . The following are equivalent:
- is a semisimple -module;
- is isomorphic to a direct sum of simple left -modules;
- is isomorphic to an (internal) sum of -modules.
Proof. To prove that 2 implies 1, we suppose without loss of generality that is a direct sum of simple left -modules . If is a submodule of , then for each , is either 0 or ; if we take to be the family of such that , then we may take .
To prove that 3 implies 2, we note that if is a simple submodule of any module , and is a submodule of , then is a submodule of , and hence equal to or 0. Now suppose that , where each of the is semisimple. Let us take a well-ordering on (such an ordering exists by the well-ordering theorem, a consequence of the axiom of choice), and let us define as the set of elements such that It follows from transfinite induction that for each , the sum is direct, and that Then , and the sum is direct.
To prove that 1 implies 3, let us take to be the sum of the cyclic submodules of , and let be the module such that and . Suppose that has a nonzero element ; then by Lemma 2, the cyclic submodule is semisimple, so by Lemma 3, it has a simple submodule that is a subset of , a contradiction. Therefore , so .