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Difference between revisions of "Semisimple module"

(→‎Classification of semisimple modules)
(→‎Classification of semisimple modules: bad proof)
 
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# <math>M</math> is isomorphic to an (internal) sum of <math>R</math>-modules.
 
# <math>M</math> is isomorphic to an (internal) sum of <math>R</math>-modules.
  
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''Proof.''  To prove that 2 implies 1, we suppose without loss of generality
+
<!--''Proof.''  To prove that 2 implies 1, we suppose without loss of generality
 
that <math>M</math> is a direct sum <math>\bigoplus_{i \in I} M_i</math> of simple left <math>R</math>-modules
 
that <math>M</math> is a direct sum <math>\bigoplus_{i \in I} M_i</math> of simple left <math>R</math>-modules
 
<math>M_i</math>.  If <math>N</math> is a submodule of <math>M</math>, then for each <math>i \in I</math>,
 
<math>M_i</math>.  If <math>N</math> is a submodule of <math>M</math>, then for each <math>i \in I</math>,
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<math>\langle \alpha \rangle</math> is semisimple, so by Lemma 2, it has a
 
<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.
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Therefore <math>N' = 0</math>, so <math>M = N</math>.  <math>\blacksquare</math>
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Therefore <math>N' = 0</math>, so <math>M = N</math>.  <math>\blacksquare</math> -->
  
 
== See also ==
 
== See also ==

Latest revision as of 15:02, 18 August 2009

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:

  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.


See also

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