This entry is a direct continuation of this post and is part of a larger series on representation theory that starts with that post. In this blog post, we continue our study of semisimple rings and bring it to a conclusion by proving a classification. Then we apply this insight into the structure of semisimple algebras to the the group algebra to get more information on irreducible representations.

## Endomorphism Rings of Semisimple Modules

The goal behind the following lemmas and corollaries is to compute the endomorphism ring of a semisimple module that is a finite direct sum of simple modules, the motivation being that to finite-dimensional representations in the semisimple case (cf. 3.20) and to a semisimple rings as a module over itself (cf. 3.24).

The idea behind the proof is simply to “multiply out” the hom set that gives us endomorphisms by pulling out (finite) direct sums from both arguments and then use Schur’s lemma to understand the hom sets between simple modules.

**Definition 4.1** For a ring and a natural number , denotes the matrix ring with entries in with the usual formulas for addition and multiplication.

The following lemma is a generalization of a well-known fact from linear algebra:

**Lemma 4.2** Let be a right module over a ring , then

**Proof** Lets index the direct summands in (so for all , but we’re keeping track of which component it is)

Note that as abelian groups, the isomorphism is clear, since we have

Now for each summand at least as an abelian group, which we can use to idenitfy elements in with elements in by sending the component living in with the entry at the position . The verification that this respects multiplication amounts to the same calculation that shows that composition of linear maps corresponds to matrix multiplication.

This gives a description of the endomorphism rings of finite direct sums of simple modules, where each simple submodule is in the same isomorphism class. To properly “multiply out” our endomorphism rings, we just need another application of Schur’s lemma:

**Lemma 4.3** Let and be modules over a ring and suppose that , then we have as rings

If is a -algebra for some field , then this an isomorphism of -algebras.

**Proof** We always have an injection given by . Here means that we apply on the first, and on the second summand. Since addition and composition of such endomorphisms can be carried out component-wise, this is a ring homomorphism and if everything is a -vector space, it is also -linear. The image of that map is the set of endomorphisms of that map and to itself. Given our conditions, if we have any endomorphism of , restricting to and composing with the projection , gives a homomorphism which is zero by assumption. This shows that the endomorphism maps to itself, and by the same argument , thus the map is surjective.

**Corollary 4.4 **Let be a right module over a ring that is a finite sum of simple submodules such that the are pairwise non-isomorphic. Let be the endomorphism rings, then

**Proof** By Schur’s lemma, there are no nonzero homomorphisms between semisimple modules that don’t have a simple submodule in common, so that we can apply 4.3 times to get that . Now apply 4.2 to each factor.

Using this, we get a partial converse to Schur’s lemma:

**Corollary 4.5** Let be a module that is a finite sum of simple submodules, then is simple if and only if the endomorphism ring of is a division ring.

**Proof** One direction is just Schur’s lemma. The other direction follows from 4.4 by noting that the only way that is a division ring is if .

## Enter the Matrix Ring

Corollary 4.4 already gives us a description of endomorphism rings of modules which are finite sums of simple submodules. In the description, matrix rings over division algebras appear. This means to deepen our understanding of those endomorphism rings, we should study matrix rings.

We begin by studying their ideals:

**Lemma 4.6** Let be a ring and , then the map from two-sided ideals of to two-sided ideals of , sending to is a bijection. (Here is the subset of all elements in where each entry is contained in )

**Proof** It’s clear that is a two-sided ideal in when is a two-sided ideal in . It’s also clear that the map is injective (we can recover from just by looking at the set of elements of that appear in the entries of the matrices in .

For surjectivity, let be the matrix that is zero everwhere except at , where it is . Then if is a twosided ideal and , we compute that is the matrix that agrees with at the place and is zero everywhere else. A calculation shows that the set of elements in that appear as an entry in the place is a two-sided ideal in , but then , because we can permute the entries of a matrix that has only one non-zero entry by multiplying from the left and right by permutation matrices (and then by taking sums, we get that every element in is contained in .

On the other hand , because we can first multiply a matix from the left and right by permutating matrices and then multiply by from the left and right to get that we could have also defined as the set of elements which appear as any matrix entry for elements in . Thus .

This lemma implies in particular that if is a division ring, then has no proper non-zero two-sided ideals. We can use this to together with another property that holds in general for finite-dimensional modules:

**Lemma 4.7 **Let be a module over a ring and suppose that contains a division ring such that is a finite-dimensional -vector space. (We don’t require that is a -algebra, i.e. that is commutative and in the center of ), then contains a simple submodule.

**Proof** Start with any non-zero submodule of . e.g. itself. If is not simple, choose a proper non-zero submodule of , then if that submodule is not simple choose a proper non-zero submodule etc.

Sinc ethe dimension over has to decrease at every step, this process has to terminate, which gives us a simple submodule.

**Lemma 4.8 **Let be a ring that has no proper nonzero two-sided ideals that has a minimal right ideal , then is semisimple and every simple module over is isomorphic to

**Proof** Let be such a ring and let be a minimal right ideal, then we can make a two-sided ideal out of by taking the product .

Since doesn’t contain proer nonzero two-sided ideals and , we get that . For each fixed , is a homomorphic image of the simple module under the image of the right -linear map , so the image is either isomorphic to (and hence simple) or zero by Schur’s lemma.

The implication (2.) implies (3.) (compare the proof or the remark after the proof) in 3.14 gives us that there is a subset such that . Thus is semisimple. 3.25 implies that every simple module is isomorphic to a direct summand in the sum , but every summand of that is isomorphic to which implies that every simple module is isomorphic to .

A natural question is how to recover from . The following lemma provides this, if we also have the module :

**Lemma** **4.9** Let be a ring and consider as the column space, which is a right -module, then we get that . If we think of as a submodule of , then is given by all scalar multiples of the identity.

Let be the matrix that has zeroes everywhere except a at , let be the vector in that has zeroes everyhwere except a one in the -th component. For any let be the permutation matrix associated to such that applying that matrix corresponds to permuting the entries with .

Then for any , we get that . Now multiplying with corresponds to making all entries zero except the first entry which is left untouched. Thus is a multiple of , so we can find a unique such that .

For every let be the transposition that switches and , then we have .

Since the form a basis, we have seen that is just scalar multiplication (from the left) with , or as a matrix , where denotes the identity matrix.

This proves .

Putting the last few lemmas together, we obtain the main result on modules over matrix rings over a division algebra:

**Lemma 4.10 **Let be a division algebra and , then is semismple and the unique simple right module over is and . (Here we’re regarding as row vectors so that the right -action makes sense.)

**Proof** 4.6 implies that has no nonzero proper two-sided ideals and 4.7 implies that it has a simple submodule, so that 4.8 applies and is semisimple with only one isomorphism class of simple modules.

The statement that is a simple -module is just linear algebra:

Given any non-zero vector in , for all , there’s a linear transformations (i.e. a matrix) that sends to . Thus is a simple module over as every non-zero submodule is the whole module. The part about the endomorphism ring follows from 4.9.

Now we come to the main theorem about semisimple rings that classifies them and also relates their ring-theoretic properties to their modules.

**Theorem 4.11 **(Artin-Wedderburn) A ring is a semisimple iff there exist natural numbers and division rings such that .

Here the is the number of simple modules up to isomorphism, is the endomorphism ring of and . In particular, is uniquely determined and the and are unique up to isomorphism and permutation of the factors.

**Proof** Note that for every ring , considered as a right module over itself, we have an isomorphism of rings by applying 4.9 with . Let by a decomposition of into simple right submodules such that the are pairwise non-isomorphic. Let .

By 4.4, we have an isomorphism . Here is uniquely determined as number of simple modules over up to isomorphism. The together with the are uniquely determined as the endomorphism rings of the simple modules and the dimension of the simple modules over their endomorphism ring (cf. lemma 4.10 for modules over matrix rings over a division algebra and 2.5 for modules over a product).

For the reverse direction, note that matrix rings over division rings are semisimple by 4.10 and 3.15 implies that finite products of semisimple rings are semisimple.

After having finally proved the main theorem for semisimple rings, we can give lots of applications:

**Corollary 4.12 **Semisimplicity for rings is left-right symmetric, i.e. if is symmetric, so it is opposite ring .

**Proof** Using 4.11, it’s enough to show that since the opposite ring of a division ring will be a division rings as well.

An isomorphism is given by transposition .

**Corollary 4.13 **Let be a semisimple ring, then the following are equivalent:

- is commutative
- For all simple -modules , the endomorphism ring is a field and .

**Proof** We apply 4.11: is commutative iff all are one and all are commutative, i.e. fields.

**Corollary 4.14** Let be an abelian group, then every irreducible real representation of is at most 2-dimensional.

**Proof** Let be an abelian finite group. Since is the algebraic closure of and has dimension 2 over , we get that all finite field extensions of have dimension at most 2. By 4.13 all simple -modules are one-dimensional over their endomorphism rings, which are finite field extensions of by 3.27 and 4.13, so the endomorphism rings are at most two-dimensional, thus all simple -modules have dimension at most 2 over .

**Corollary 4.15 **Let be a finite group and be an algebraically closed field such that the characteristic of doesn’t divide the order of , then the following are equivalent:

- is abelian
- All irreducible representations of are one-dimensional
- The number of irreducible representations is equal to the order of

**Proof** The equivalence of (1) and (2) follows from 4.13 and 3.27 and 3.29, since 3.27 and 3.29 together imply that the endomorphism ring of any simple -module is , so the statement follows from 4.13.

The equivalence of (2) and (3) follows from 3.30

**Reminder** For a group , the abelianization is the largest commutative quotient of . Explicitly, it is given by the quotient of the subgroup generated by all commutators. It has the universal property that every morphism from to an abelian group factors uniquely through the map .

**Corollary 4.16 **Let be a finite group and be an algebraically closed field such that the characteristic of doesn’t divide the order of , then the number of one-dimensional representations of up to isomorphism is equal to the order of the abelianization

**Proof** One-dimensional representations are homomorphisms .

Since is commutative, these correspond to homomorphisms , i.e. one-dimensional representations of . One-dimensional representations are automatically irreducible and 4.15 tells us that since is abelian, there are exactly up to isomorphism.

## The Center of the Group Algebra

The last corollary gives a partially answers the question for the number of irreducible representations of a group, by characterizing the number of one-dimensional representations in terms of the group theory of . In this section, we give a group-theoretic characterization of the number of all irreducible representations. (Given suitable assumptions on the field)

**Definition** **4.17 **If is a ring, then the center is defined as . It is a commutative subring of .

**Lemma 4.18** If is a ring and , then consits of those diagonal matrices where all diagonal entries are the same value that lies in . So we have an isomorphism of rings .

**Proof** We rergard as row vectors which is naturally a right -module. Then for , the map is -linear, because for , we have .

By 4.9 is a diagonal matrix where all diagonal entries are the same. Clearly the diagonal entry must lie in , as commutes with other such diagonal matrices.

**Lemma 4.19** If and are rings, then .

**Proof** This is an easy computation. Multiplication in is defined component-wise.

**Corollary 4.20** If is a semisimple ring with the Artin-Wedderburn decomposition , then .

**Corollary 4.21** If is a semisimple algebra over a field , then where is the number of simple -modules up to isomorphism. We have equality if is algebraically closed.

**Proof** By 4.11, the number of simple -modules up to isomorphism is if . By 4.20, we have . Each factor has least dimension , which shows the inequality.

If is algebraically closed, then for all by 3.29 which shows that equality holds.

**Corollary 4.22 **If is semisimple real algebra and is the number of simple -modules up to isomorphism, then

**Proof** Argue as in the proof of 2.21 and use the fact that where is a finite-dimensional division algebra over has to be isomorphic to or , since it is a finite field extension of , so the dimension is at most two.

These corollaries relate the number of simple modules of a semisimple algebra and their center. Thus the next thing to do is to study the center of group algebras.

**Reminder** If is a group, then for , the conjugacy class of consists of all elements in that are conjugate to . In other words, if we consider the action , then this is the orbit of under this action. The different conjugacy classes are the minimal subsets of that are closed under conjugation and form a partition of .

**Lemma 4.23** Let be a field and let be a group (we don’t need it to be finite). Then has a -basis given by the set of elements of the form where varies over all conjugacy classes of with finitely many elements.

**Proof** Since is a -algebra with a -basis given by , we can check if an elements is in the center just by checking if it commutes with every element of . Let be an element of (so all but finitely many are zero. Then is in the center of if and only if for all , we have .

Writing out , this equation becomes . Here we used that is a bijection.

Comparing coefficients, this is equivalent to for all .

If this holds for all , then we get that the (finite) set of elements such that is closed under conjugation and if two elements inside are conjugate, their coefficients are equal. Take a system of representatives for where acts on by conjugation. Then by the above considerations, we get that , where is the conjugacy class of . This shows that the set of elements of the form where is a finite conjugacy class is a generating system for . It is linearly independent because distinct conjugacy classes are distinct.

**Theorem 4.24 **If is a finite group and is a field such that the characteristic of doesn’t divide the order of and let be the number of irreduible representations, up to isomorphism. Let be the number of conjugacy classes of . Then with equality if is algebraically closed. If , then we have .

**Proof** By 4.23 is the dimension of the center of . Now apply 4.21 and 4.22.

Thus we have obtained a relation between the number of irreducible representations and the element structure of by heavily employing the structure theory for semisimple rings. This is a nice illustration of the power of ring-theoretic tools in representation theory.