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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 lemmas.
+
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
Line 26: Line 50:
 
[[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.''' Let <math>M</math> be a semisimple left <math>R</math>-module, and let
 
−
<math>N</math> be a submodule of <math>M</math>.  Then <math>N</math> is semisimple.
 
−
 
 
−
''Proof.''  Let <math>T</math> be a submodule of <math>N</math>, and let <math>T'</math> be the
 
−
submodule of <math>M</math> such that <math>T \cap T' = 0</math> and <math>T + T' = M</math>.
 
−
Then <math>T \cap (T' \cap N) = 0</math>; furthermore, since <math>T \subset N</math>,
 
−
if follows that <math>N = N \cap (T + T') = T + (T' \cap N)</math>.  It
 
−
follows that <math>N</math> is semisimple.  <math>\blacksquare</math>
 
−
 
 
−
'''Lemma 3.'''  Every cyclic semisimple module has a simple submodule.
 
  
 
''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>
Line 78: Line 92:
 
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 Lemma 2, the cyclic submodule
+
has a nonzero element <math>\alpha</math>; then by our proposition, the cyclic submodule
−
<math>\langle \alpha \rangle</math> is semisimple, so by Lemma 3, 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.
 
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 $M$ 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.

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 $M = N$. $\blacksquare$

See also

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