Semisimple Rings and Modules

After discussing simple modules, the next best thing is to look at semisimple modules, which are just direct sums of simple modules. Here’s a summary of the results we’ll prove:

  • A module is semisimple iff it is a sum of simple submodules.
  • Quotients, sums, direct sums and submodules of semisimple modules are also semisimple.

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Semisimple Modules

Again, we fix a base ring R, possibly non-commutative. All modules are left modules.

Definition. An R-module M is said to be semisimple if it is a sum of simple submodules.

The key theorem we wish to prove is the following.

Theorem. Let M be a semisimple R-module and N \subseteq M a submodule. Then we can find simple submodules M_i \subseteq M (indexed by i\in I) such that

M = N \oplus \left( \oplus_{i\in I} M_i \right).

The “direct sum” ⊕ means that every element m of M is uniquely writable as a sum n + \sum_i m_i, where n\in N, m_i\in M_i and only finitely many terms are non-zero.

Proof

This will be by Zorn’s lemma. Consider collections ∑ of simple submodules S of M such that:

M_\Sigma := N \oplus \left( \oplus_{S\in \Sigma} S\right)

is a direct sum. Note that at least one ∑ exists, i.e. ∑ = ∅ is valid (in which case we get M_\emptyset = N). [ For those who worry about set-theoretic validity, note that the collection of all such ∑ forms a bona fide set. ]

To apply Zorn’s, we need to prove that every chain of ∑’s has an upper bound.

Suppose \{\Sigma_\alpha\}_\alpha is a chain: i.e. for any \Sigma_\alpha, \Sigma_\beta, either \Sigma_\alpha\subseteq \Sigma_\beta or \Sigma_\beta \subseteq \Sigma_\alpha. Let \Sigma = \cup_\alpha \Sigma_\alpha; let us show that M_\Sigma = N \oplus \left(\oplus_{S\in\Sigma} S\right) is a direct sum.

  • If not, then n + \sum_{S\in \Sigma} m_S = 0 for some n\in N, m_S \in S. But this is a finite sum, so the equality already holds in some \Sigma_\alpha (since the \Sigma_\alpha‘s form a chain), which is a contradiction.

Thus, the chain \{\Sigma_\alpha\}_\alpha has an upper bound. Zorn’s lemma tells us there is a maximal ∑. If M_\Sigma \ne M, pick m \in M-M_\Sigma. Since M is a sum of simple submodules, write

m = m_1 + m_2 + \ldots + m_r, \qquad m_k \in M_k, where each Mk is simple.

Since m\not\in M_\Sigma we have M_k \not\subseteq M_\Sigma for some k. But this means M_k \cap M_J is a proper submodule of Mk, and must be zero (since Mk is simple). Hence M_k \oplus M_\Sigma is a direct sum, so we could have added the simple module Mk to the collection ∑, contradicting its maximality. Thus, M_\Sigma = M and we’re done. ♦

Now we’re ready to prove all the necessary properties of semisimple modules.

Corollary 1. Every semisimple module M is a direct sum of simple modules.

Proof. Pick N = 0 in the theorem. ♦

Corollary 2. If each N_i\subseteq M is a semisimple submodule of a module M, then so is N :=\sum N_i.

Proof. Each N_i is a sum of simple modules; by definition so is N. ♦

Corollary 3. If N is a submodule of a semisimple M, then there is a submodule P of M such that M = N\oplus P.

Proof. Apply the theorem and let P := \oplus_{i\in I} M_i. ♦

Corollary 4. A submodule and quotient of a semisimple module M is semisimple.

Proof. Submodule follows from the definition of semisimplicity; quotient follows from the theorem, since M/N \cong \oplus_{i\in I} M_i is a direct sum of simple modules. ♦

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Semisimple Rings

Definition. The ring R is semisimple if it is a semisimple module over itself.

The main result we want to show is:

Theorem. Any module over a semisimple ring R is semisimple.

Proof

Let M be a module. If m is a non-zero element of M, take the homomorphism fR → M, which takes r → rm. Then Rm is a submodule of M isomorphic to R/ker(f), which is a semisimple R-module since R is. Thus Rm is semisimple. Since M is a sum of semisimple submodules, M is also semisimple. ♦

Let us look at some ways to create semisimple rings.

Proposition. (i) If I is a (two-sided) ideal of semisimple ring R, then R/I is a semisimple ring.

(ii) If R and S are semisimple rings, so is R × S.

Proof

(i) Any left ideal of R/I corresponds to a left ideal of R containing I, which is a sum of simple submodules J. The image of J in R/I, i.e. (J + I)/IJ / (J ∩ I), is thus either 0 or J. Either way, R/I is a sum of simple submodules.

(ii) Any left ideal M of R × S is of the form I × J, for left ideal I of R and J of S. [ To see why, multiply elements of M by (1, 0) and (0, 1). ] Since I and J are both sums of simple submodules, so is I × J. ♦

Finally, decomposing R gives us a complete list of simple R-modules.

Proposition. Let R be a semisimple ring; write R = \oplus_i N_i as a direct sum of simple left ideals. Then any simple module M is isomorphic to some N_i. In particular, there are only finitely many simple R-modules up to isomorphism.

[ Note: the N_i which occur may repeat; in the extreme case, we can even have R \cong N^k for a single simple module N; this just means N is the only simple R-module up to isomorphism. ]

Proof

We know that any simple module M is isomorphic to quotient R/I for a maximal left ideal I of R. For each i, consider

f_i : N_i \to \oplus_i N_i \to (\oplus_i N_i)/I = M.

Since N_i, M are both simple, f_i = 0 or an isomorphism. If all f_i=0, then so is \sum_i f_i : R = \oplus_i N_i \to M which is absurd. Hence some f_i is an isomorphism, which proves the first statement.

The second statement follows from the following lemma. ♦

Lemma. Writing the base ring as a direct sum of submodules R = \oplus_i N_i, only finitely many of the modules are non-zero. 

Proof

Indeed, write 1 as a finite sum x_1 + x_2 + \ldots + x_k where x_i \in N_i. For an N_i not in this list, any y\in N_i gives:

y = y\cdot 1 = y x_1 + y x_2 + \ldots + y x_k \in N_1 + N_2 + \ldots + N_k.

So y = 0 since N_i does not lie in the list of N_1, \ldots, N_k. ♦

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Examples

1. Every field K is semisimple, since K itself is a simple K-module.

2. The ring Z is not semisimple. [ Why? ]

3. Every finite abelian group M is a product of cyclic groups. Thus M is semisimple as a Z-module if and only if it is a product of prime cyclic groups. Indeed, each prime cyclic group is clearly simple, hence so is their product. Conversely, if M contains a subgroup N isomorphic to Z/pr for r>1, then M has a unique simple subgroup pr-1Z/pr so it is not semisimple.

4. Is the ring RR[x]/(x2) semisimple? What about SR[x]/(x2 – 1) [ Answer: no to the first question, because its only simple left ideal is Rx, so 1 does not lie in the sum of all simple left ideals. Yes to the second, since it is isomorphic to R × R.  ]

5. Let R be the ring of upper triangular real 2 × 2 matrices \begin{pmatrix} a & b \\ 0 & d\end{pmatrix}. Is this semisimple? [ Answer: no, let R act on the space of column vectors R2. Call this R-module M. The space N := R(1, 0)t is a simple submodule of M, but we cannot find a submodule P for which MN ⊕ P. Thus we get a non-semisimple R-module. ]

6. Let R be the ring of real 2 × 2 matrices. Prove that R is semisimple. [ Hint: write R as a direct sum of two simple modules, which are spaces of column vectors. ]

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