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

$dim_k(M_I)=d_Idim_k(I)=d_In_Idim_k(E_I)$

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

$dim_k(M_I)=n_I^2dim_k(E_I)$

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

$|G|=\sum_Vdim_k(V)^2$

where $V$ runs through all the irreducible representations of $G$.