Difference between revisions of "Semisimple module"
(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…') |
(→Classification of semisimple modules) |
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It happens that semisimple modules have a convenient classification | It happens that semisimple modules have a convenient classification | ||
(assuming the [[axiom of choice]]). | (assuming the [[axiom of choice]]). | ||
− | To prove this classification, we first state some | + | To prove this classification, we first state some intermediate results. |
+ | |||
+ | '''Proposition.''' Let <math>M</math> be a semisimple left <math>R</math>-module. | ||
+ | Then every [[submodule]] and [[quotient]] module of <math>M</math> is also | ||
+ | simple. | ||
+ | |||
+ | ''Proof.'' First, suppose that <math>N</math> is a submodule of <math>M</math>. | ||
+ | Let <math>T</math> be a submodule of <math>N</math>, and let <math>N'</math> be a submodule of <math>M</math> | ||
+ | such that <math>T \cap N' = 0</math> and <math>T + N' = M</math>. We note that if | ||
+ | <math>\tau \in T</math> and <math>\nu' \in N</math> are elements such that <math>\tau + \nu' \in N</math>, | ||
+ | then <math>\nu' \in N</math>. It follows that | ||
+ | <cmath> N = M \cap N = (T + N') \cap N = T + (N' \cap N) . </cmath> | ||
+ | Since <math>T \cap (N' \cap N) = 0</math>, it follows that <math>N</math> is semisimple. | ||
+ | |||
+ | Now let us consider a quotient module <math>M/N</math> of <math>M</math>, with <math>\phi</math> the | ||
+ | canonical [[homomorphism]] <math>M \to M/N</math>. Let <math>T</math> be a submodule of | ||
+ | <math>M/N</math>. Then <math>\phi^{-1}(T)</math> is a submodule of <math>M</math>, so there exists | ||
+ | a submodule <math>N' \subset M</math> such that <math>\phi^{-1}(T) \cap N'=0</math> | ||
+ | and <math>\phi^{-1}(T) + N' = M</math>. Then in <math>M/N</math>, since <math>\phi</math> is surjective, | ||
+ | <cmath> \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*} </cmath> | ||
+ | Therefore <math>M/N</math> is semisimple as well. | ||
'''Lemma 1.''' Let <math>R</math> be a ring, and let <math>M</math> be a nonzero | '''Lemma 1.''' Let <math>R</math> be a ring, and let <math>M</math> be a nonzero | ||
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[[Zorn's Lemma]], <math>\mathfrak{S}</math> has a maximal element. <math>\blacksquare</math> | [[Zorn's Lemma]], <math>\mathfrak{S}</math> has a maximal element. <math>\blacksquare</math> | ||
− | + | '''Lemma 2.''' Every cyclic semisimple module has a simple submodule. | |
− | '''Lemma 2 | ||
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''Proof.'' Let <math>M</math> be a cyclic semisimple module, and let <math>\alpha</math> | ''Proof.'' Let <math>M</math> be a cyclic semisimple module, and let <math>\alpha</math> | ||
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of the cyclic submodules of <math>M</math>, and let <math>N'</math> be the module | of the cyclic submodules of <math>M</math>, and let <math>N'</math> be the module | ||
such that <math>N \cap N' = 0</math> and <math>N+N' = M</math>. Suppose that <math>N'</math> | such that <math>N \cap N' = 0</math> and <math>N+N' = M</math>. Suppose that <math>N'</math> | ||
− | has a nonzero element <math>\alpha</math>; then by | + | has a nonzero element <math>\alpha</math>; then by our proposition, the cyclic submodule |
− | <math>\langle \alpha \rangle</math> is semisimple, so by Lemma | + | <math>\langle \alpha \rangle</math> is semisimple, so by Lemma 2, it has a |
simple submodule that is a subset of <math>M</math>, a contradiction. | simple submodule that is a subset of <math>M</math>, a contradiction. | ||
Therefore <math>N' = 0</math>, so <math>M = N</math>. <math>\blacksquare</math> | Therefore <math>N' = 0</math>, so <math>M = N</math>. <math>\blacksquare</math> |
Revision as of 19:25, 17 August 2009
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 intermediate results.
Proposition. Let be a semisimple left -module. Then every submodule and quotient module of is also simple.
Proof. First, suppose that is a submodule of . Let be a submodule of , and let be a submodule of such that and . We note that if and are elements such that , then . It follows that Since , it follows that is semisimple.
Now let us consider a quotient module of , with the canonical homomorphism . Let be a submodule of . Then is a submodule of , so there exists a submodule such that and . Then in , since is surjective, Therefore is semisimple as well.
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. 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 our proposition, the cyclic submodule is semisimple, so by Lemma 2, it has a simple submodule that is a subset of , a contradiction. Therefore , so .