Notes on semisimple rings and modules

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

Basic settings

Suppose that R is a ring with unit and M is a module over R. We say that M is a simple module if it does not contain other submodule other than 0 and M. Then what is a semisimple module over R? We say that M is semisimple if any submodule M' of M is a direct summand of M. 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 R is a semisimple ring if the R-module R(the ring R acts by left multiplication on the abelian group R) is a semisimple module over R. So here we should really be careful about semisimple rings because we are apt to thinking that R is a semisimple ring if any subring of R is a direct summand of R, which is completely false according to the definition. In fact, R is a semisimple ring if any ideal R is a direct summand of the module R.

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

0\rightarrow Ker(p)\rightarrow R^n\xrightarrow{p}M\rightarrow 0

Note that Ker(p) is a submodule of R^n, thus is a direct summand of R^n, so M is also a direct summand of R^n, thus is also semisimple.

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

So in this way, a semisimple ring R 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 R is a semisimple ring and M is a simple R-module. Now choose any element m\in M, we have a morphism of R-modules

f_m:R\rightarrow M,r\mapsto rm

It is easy to see that Im(f_m) is a submodule of M, and since M is simple, thus either Im(f_m)=M or Im(f_m)=0. If m\neq 0, then 1_Rm=m\in Im(f_m), this shows that Im(f_m)=M for m\neq 0. Thus M is a quotient of R, and since the latter is a semisimple R-module, thus R\simeq M\bigoplus Ker(f_m). Note that this is an isomorphism of R-modules, thus we can view M as an ideal of R. In this way, we find that all simple modules come from ideals of R, so now we shall restrict our attention to simple ideals of R(which is defined in an obvious way to be an ideal of R, which, viewed as an R-module, is a simple module).

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

We write \mathfrak{I} for the set of equivalence classes of left simple ideals of R. For any I\in \mathfrak{I}, we set M_I to be the (direct) sum of the left ideals of R isomorphic to I as R-modules. We next show that M_I is a two-sided ideal of R. For any two representatives I,J\in\mathfrak{I}, using Schur’s lemma, we see that IJ=0 if I\not\simeq J. Thus we have that M_IM_J=0 in this case. So, we see that these M_I are two-sided ideals. Now that R is semisimple, we can write R=\sum_{I\in\mathfrak{I}}M_I. Using the (finite) decomposition of 1_R=\sum_I e_I, we see that R=\sum_IM_I is in fact a finite sum and R=\bigoplus_I M_I. The by-product of this process is the elements e_I for each I\in\mathfrak{I}(which is now a finite set). These elements have very nice properties(suppose now I,J\in\mathfrak{I} are isomorphism classes, not representatives of these classes):

e_Ie_J=0,I\neq J; e_Ie_I=e_I;\sum_I e_I=1_R

Note that these properties give us another description of the above decomposition R=\bigoplus_I Re_I. This is an isomorphism of R-modules. Yet we have something more: each Re_I is itself a ring. Let’s examine this nice result. Note that we have 1_R=\sum_I e_I, thus for any r\in R, r=1_Rr=\sum_Ie_Ir=e1_R=\sum_Ire_I. Since this is a direct sum, thus according to the uniqueness of the decomposition, we get that e_Ir=re_I for all I\in\mathfrak{I}. Now for any re_I,r'e_I\in Re_I, we have that (re_I)(r'e_I)=r(e_Ir')e_I=r(r'e_I)e_I=rr'e_I, which, combined with some other simple verifications, shows that Re_I is indeed a ring. So, in this sense,

R\simeq \bigoplus_I Re_I

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

Simple rings and their structures

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

Next we will start from I to construct a division ring. This is one of the most interesting part of the theory. We fix a left simple ideal I\neq0 and we set E=End_{M_I}(I). We next show that E is this division ring. Indeed, for any 0\neq f\in E, since I is simple, thus according to Schur’s lemma, f is an isomorphism. Clearly E has a ring structure, so we are done: E is indeed a division ring.

Now let’s give I an E-module structure: for any m\in I,f\in E=End_{M_I}(I), we define f\cdot m=f(m). It is straightforward that this defines an E-module structure on I. Note that E is already a division ring, so as an E-module, I 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 M_I. Next we consider the whole ring M_I, which is a direct sum of the left simple ideals, thus is also an E-module. Surprisingly, it has a much nicer description: M_I is isomorphic to the endomorphism ring E'=End_E(I).

Let’s first give a map from one to the other. For any m\in M_I, we set f_m:I\rightarrow I, r\mapsto mr simply the left multiplication, which indeed maps I to I since I is a left ideal. We should verify that f_m\in End_E(I). For any h\in E=End_{M_I}(I), we should show that h\cdot(f_m(r))=f_m(h\cdot r). Indeed, h\cdot(f_m(r))=h(mr), now h is a M_I-linear map, thus h(mr)=m(h(r))=f_m(h(r))=f_m(h\cdot r), so we are done: f_m is indeed an element in End_E(I). Next we have to show that this map

\phi: M_I\rightarrow End_E(I),m\mapsto f_m

is an isomorphism. Note that Ker(\phi) is a two-sided ideal of M_I, thus as the above argument, M_I=Ker(\phi)\bigoplus M_I/Ker(\phi).where both summands have ring structures and contain different isomorphism classes of left simple ideals, thus we must have Ker(\phi)=0 or Ker(\phi)=M_I. If it is the second case, I=1\cdot I=0, which is impossible since I is non-zero, thus Ker(\phi)=0, showing that \phi is injective. It is also easy to show that \phi is surjective, thus we get an isomorphism

M_I\simeq End_E(I)

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

So, let’s return to the semisimple case: suppose again R is a semisimple ring, then R\simeq \bigoplus_I M_I, each summand is a matrix algebra over some division ring E_I=End_{M_I}(I).

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 k. Suppose R is a finite dimensional k-algebra and a semisimple ring, then each M_I is a finite dimensional k-algebra and a simple ring. Now each E_I=End_{M_I}(I) is a division ring and at the same time a k-algebra, which is finite dimensional since End_{M_I}(I)\subset End_k(I).

Now let’s make a simple yet illuminating calculation. Suppose that M_I\simeq I^{\bigoplus d_I}, n_I=dim_{E_I}(I). Then we have that


Yet M_I\simeq End_{E_I}(I), thus


Combining these two results, we get that

n_Id_I=n_I^2\implies n_I=d_I

So we have that M_I\simeq M_{n_I}(E_I), the matrix ring with coefficients in E_I. And in general, R\simeq\bigoplus_I M_{n_I}(E_I).

If we suppose that k is algebraically closed, we should have that k=E_I for all I.

One simple application

As an application of this result, we consider the representation theory of a finite group G with char(k)\not| |G|. We set R=k[G], the group algebra of dimension dim_k(R)=|G|. We can show that R is semisimple(we will talk about this proof in another post). We supple also that k is algebraically closed, then according to the above result, we have

k[G]\simeq\bigoplus M_{n_I}(k)

In other words, we have |G|=\sum_In_I^2. Besides, each I is an irreducible representation of G, so in the end we get a fundamental result in the representation theory of finite groups


where V runs through all the irreducible representations of G.


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