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.

## Semisimple Modules

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

Definition.AnR-moduleMis said to besemisimpleif 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 a submodule. Then we can find simple submodules (indexed by ) such that

The “direct sum” ⊕ means that every element *m* of *M* is uniquely writable as a sum where 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:

is a direct sum. Note that at least one ∑ exists, i.e. ∑ = ∅ is valid (in which case we get ). [ *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 is a chain: i.e. for any either or Let let us show that is a direct sum.

- If not, then for some But this is a finite sum, so the equality already holds in some (since the ‘s form a chain), which is a contradiction.

Thus, the chain has an upper bound. Zorn’s lemma tells us there is a maximal ∑. If , pick Since *M* is a sum of simple submodules, write

, where each *M _{k}* is simple.

Since we have for some *k*. But this means is a proper submodule of *M _{k}*, and must be zero (since

*M*is simple). Hence is a direct sum, so we could have added the simple module

_{k}*M*to the collection ∑, contradicting its maximality. Thus, and we’re done. ♦

_{k}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 is a semisimple submodule of a module M, then so is

**Proof**. Each 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 .

**Proof**. Apply the theorem and let . ♦

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 is a direct sum of simple modules. ♦

## Semisimple Rings

Definition.The ringRissemisimpleif 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 *f* : *R* → *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*)/*I* ≅ *J* / (*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 as a direct sum of simple left ideals. Then any simple module M is isomorphic to some . In particular, there are only finitely many simple R-modules up to isomorphism.

[ Note: the which occur may repeat; in the extreme case, we can even have 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

Since are both simple, or an isomorphism. If all , then so is which is absurd. Hence some 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 only finitely many of the modules are non-zero.

**Proof**

Indeed, write 1 as a finite sum where For an not in this list, any gives:

So *y* = 0 since does not lie in the list of . ♦

## 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**/*p*^{r} for *r*>1, then *M* has a unique simple subgroup *p*^{r-1}**Z**/*p*^{r} so it is not semisimple.

4. Is the ring *R* = **R**[*x*]/(*x*^{2}) semisimple? What about *S* = **R**[*x*]/(*x*^{2} – 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 . Is this semisimple? [ Answer: no, let *R* act on the space of column vectors **R**^{2}. 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 *M* = *N* ⊕ *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. ]