In this post I want to say something on semisimple rings and semisimple modules.

###### Basic settings

Suppose that is a ring with unit and is a module over . We say that is a simple module if it does not contain other submodule other than and . Then what is a semisimple module over ? We say that is semisimple if any submodule of is a direct summand of . Note that this is a very strong condition on the module. Recall that this kind of phenomena occurs most naturally to vector spaces over a field. So, in this way, we can say that semisimple modules are the modules most similar to vector spaces. Besides, it is easy to see that a simple module is automatically a semisimple module.

Let’s then define what a semisimple ring is. We say that is a semisimple ring if the -module (the ring acts by left multiplication on the abelian group ) is a semisimple module over . So here we should really be careful about semisimple rings because we are apt to thinking that is a semisimple ring if any subring of is a direct summand of , which is completely false according to the definition. In fact, is a semisimple ring if any ideal is a direct summand of the module .

Now it is not difficult to see that: is a semisimple ring if and only if any module over is semisimple. The ‘if’ part is clear. For the ‘only if’ part, note that is a quotient of some , which is again semisimple. Now consider the exact sequence

Note that is a submodule of , thus is a direct summand of , so is also a direct summand of , thus is also semisimple.

Here we have used a crucial property of semisimple modules over : if is semisimple, then its submodules and quotient modules are again semisimple.

So in this way, a semisimple ring looks very like a field in that all the modules over it are semisimple.

###### Structure theory of semisimple rings

What is fascinating about the theory of semisimple rings is that there is a nice decomposition of a semisimple ring into some well known algebraic objects. Let’s briefly have a look at this part of the theory.

Suppose that is a semisimple ring and is a simple -module. Now choose any element , we have a morphism of -modules

It is easy to see that is a submodule of , and since is simple, thus either or . If , then , this shows that for . Thus is a quotient of , and since the latter is a semisimple -module, thus . Note that this is an isomorphism of -modules, thus we can view as an ideal of . In this way, we find that all simple modules come from ideals of , so now we shall restrict our attention to simple ideals of (which is defined in an obvious way to be an ideal of , which, viewed as an -module, is a simple module).

Note that if we do not assume that is commutative, and we consider for the present the left -modules, then the resulting fact is to consider all the left simple ideals of .

We write for the set of equivalence classes of left simple ideals of . For any , we set to be the (direct) sum of the left ideals of isomorphic to as -modules. We next show that is a two-sided ideal of . For any two representatives , using Schur’s lemma, we see that if . Thus we have that in this case. So, we see that these are two-sided ideals. Now that is semisimple, we can write . Using the (finite) decomposition of , we see that is in fact a finite sum and . The by-product of this process is the elements for each (which is now a finite set). These elements have very nice properties(suppose now are isomorphism classes, not representatives of these classes):

Note that these properties give us another description of the above decomposition . This is an isomorphism of -modules. Yet we have something more: each is itself a ring. Let’s examine this nice result. Note that we have , thus for any , . Since this is a direct sum, thus according to the uniqueness of the decomposition, we get that for all . Now for any , we have that , which, combined with some other simple verifications, shows that is indeed a ring. So, in this sense,

is also an isomorphism of rings(all of these use essentially the fact that is semisimple).

###### Simple rings and their structures

So to study the properties of , it suffices to study these . These are in fact simple rings in the following sense: a ring is said to be simple if is semisimple and it contains exactly one isomorphism class of left simple ideals. We will write a simple ring as , just to emphasize that fact that the isomorphism class of left simple ideals is .

Next we will start from to construct a division ring. This is one of the most interesting part of the theory. We fix a left simple ideal and we set . We next show that is this division ring. Indeed, for any , since is simple, thus according to Schur’s lemma, is an isomorphism. Clearly has a ring structure, so we are done: is indeed a division ring.

Now let’s give an -module structure: for any , we define . It is straightforward that this defines an -module structure on . Note that is already a division ring, so as an -module, behave very much like a vector space over a field: we have the concept of dimensions. We will come to this point later.

Note until present we only consider one single left simple ideal of . Next we consider the whole ring , which is a direct sum of the left simple ideals, thus is also an -module. Surprisingly, it has a much nicer description: is isomorphic to the endomorphism ring .

Let’s first give a map from one to the other. For any , we set simply the left multiplication, which indeed maps to since is a left ideal. We should verify that . For any , we should show that . Indeed, , now is a -linear map, thus , so we are done: is indeed an element in . Next we have to show that this map

is an isomorphism. Note that is a two-sided ideal of , thus as the above argument, .where both summands have ring structures and contain different isomorphism classes of left simple ideals, thus we must have or . If it is the second case, , which is impossible since is non-zero, thus , showing that is injective. It is also easy to show that is surjective, thus we get an isomorphism

Note that if is actually a field(not simply a division ring), then is a vector space over . Let’s suppose that , then this result shows that the simple ring is the matrix ring . This is really a nice description of simple rings.

So, let’s return to the semisimple case: suppose again is a semisimple ring, then , each summand is a matrix algebra over some division ring .

Now we want to concentrate ourselves to some special (semi)simple rings: rings that are algebras over some field.

###### One special case

Let’s fix a field . Suppose is a finite dimensional -algebra and a semisimple ring, then each is a finite dimensional -algebra and a simple ring. Now each is a division ring and at the same time a -algebra, which is finite dimensional since .

Now let’s make a simple yet illuminating calculation. Suppose that . Then we have that

Yet , thus

Combining these two results, we get that

So we have that , the matrix ring with coefficients in . And in general, .

If we suppose that is algebraically closed, we should have that for all .

###### One simple application

As an application of this result, we consider the representation theory of a finite group with . We set , the group algebra of dimension . We can show that is semisimple(we will talk about this proof in another post). We supple also that is algebraically closed, then according to the above result, we have

In other words, we have . Besides, each is an irreducible representation of , so in the end we get a fundamental result in the representation theory of finite groups

where runs through all the irreducible representations of .