# Semisimplicity and Representations, Part 1

This post is the third one in a series on representation theory. The previous posts are this one and that one (in this order.) The nature of this post is mostly ring-theoretic, but we will give applications to representation theory throughout the development of the general theory.

## Semisimple Modules

Under suitable assumption on $G$ and $K$, Maschke’s theorem (1.22) tells us that any submodule of a $K[G]$-module is a direct summand, i.e. we can find a complement.
One can try to apply this repeatedly to decompose a $K[G]$-module into smaller submodules. If the dimension is finite, then at some point we have to end up with a direct sum of modules that don’t have a non-zero proper submodule. This is because if one direct summand had a non-zero proper submodule, we could just decompose it further by Maschke’s theorem. The assumption of finite dimension implies that this process has to terminate, as the dimension of the summands decreases every time we decompose something.
This motivates the following definition to give a name to the modules we obtained as summands in the end:

Definition 3.1 A non-zero module over a ring is called simple if it doesn’t have a proper non-zero submodule.

Example 3.2 If $K$ is a field, or more generally a division ring, then a vector space over $K$ is simple iff it is one-dimensional.

Example 3.3 If we consider modules over $\mathbb{Z}$, i.e. abelian groups, then simple modules are just simple abelian groups. It’s known that simple abelian groups are the groups that are cyclic of prime order: $\Bbb{Z}/p\Bbb{Z}$.

Example 3.4 If $K$ is a field, and we consider $K[X]$-modules, i.e. $K$-vector spaces equipped with a choice of endomorphism $A$ (cf. the first section of the last entry), then a module is simple iff it doesn’t have a non-zero proper $A$-invariant subspace. One can show that this equivalent to being isomorphic to $K[X]/(f)$ for some irreducible $f \in K[X]$. In particular, if $K$ is algebraically closed, then simple $K[X]$-modules are precisely the one-dimensional ones. (Where the endomorphism necessarily acts by scalar multiplication.) This means that over an algebraically closed field, an endomorphism of a finite-dimensional vector space is diagonalizable, if and only if the associated $K[X]$-module is a direct sum of simple modules. (And in general, the associated $K[X]$-module is a direct sum of simple modules if and only if the the endomorphism is diagonalizable over an algebraic closure.) We will encounter the condition of being a direct sum of simple modules later in this post.

Generalizing the last two examples, we have the following result:

Lemma 3.5  If $R$ is any ring and $\mathfrak{m}$ is a maximal left ideal, then $R/\mathfrak{m}$ is a simple module. Conversely, every simple module is of that form

Proof If $\mathfrak{m}$ is a maximal left ideal, then submodules of $R/\mathfrak{m}$ correspond to submodules of $R$ containing $\mathfrak{m}$, so $R/\mathfrak{m}$ is simple by definition of $\mathfrak{m}$ being maximal.
Conversely, if $M$ is a simple module and $m \in M$ is nonzero, then $Rm$ is a non-zero submodule of $M$, so $M=Rm$. This means the map $R \to M, r \mapsto rm$ is surjective so we get an isomorphism $R/I \cong M$ for some proper ideal $I$. If $I$ is not maximal, then there’s a proper submodule containing $I$, which corresponds to a proper non-zero submodule of $M$, which is impossible.

Example/Definition 3.6 If $K$ is a field and $G$ is a group, then similar to example 3.4, simple $K[G]$-modules are representations with no non-zero proper $G$-invariant subspace. These are called irreducible representations.

Example 3.7 The representations of a cyclic group of order $n$, corresponding to irreducible factors of $X^n-1$ that we have constructed in 2.6. are irreducible. The reason is that they’re irreducible $K[X]$-modules, where the action of $X$ corresponds to the action of a generator of the group. (cf. 3.4 and the proof of 2.6)

We have seen in 3.5 that simple modules are generated by one element, let’s give this property a label (generalizing the notion of cyclic groups):

Definition 3.8 Modules that are generated by a single element are called cyclic modules.

By our considerations in the beginning of the section, we see that when Maschke’s theorem applies and we have a finite-dimensional representation, it decomposes as a direct sum of irreducible subrepresentations. The purpose of the following lemmas is to generalize this. (Because we work without any finiteness conditions, we will need some form of the axiom of choice. If one is only interested in modules that satisfy some finiteness condition (e.g. finite-dimensional modules for an algebra over a field), then the dependence on choice can be eliminated and the arguments are much easier.)

Definition 3.9 A module $M$ over a ring is called semisimple if every submodule $N \leq M$ has a complement, i.e. there exists a submodule $N' \leq M$ such that $M=N\oplus N'$

Lemma 3.10 Submodules and quotients of a semisimple modules are semisimple.

Proof If $M$ is semisimple and $M/N$ is a quotient, then for submodule $\overline{K} \leq M/N$, we can take the preimage under the projection $M \to M/N$ to get a submodule $K \leq M$ that projects onto $\overline{K}$. Then the image under the projection of a complement of $K$ will be a complement for $\overline{K}$. If $N \leq M$ is a submodule, then we can find a complement $N'$, but then $M/N' \cong N$ so that $N$ is quotient of $M$, so the previous case applies.

We will need the following result for an important property of semisimple modules. Most readers will probably be familiar with this, at least in the commutative case:

Lemma 3.11 Let $R$ be a ring. Then every proper left ideal is contained in a maximal left ideal.

Proof Let $I$ be a proper left ideal and let $\mathcal{P}$ be the set of all proper left ideals containing $I$. If we have an ascending chain $(I_{i})_{i \in \Omega}$, where $I_i \in \mathcal{P}$, then $\displaystyle \cup_{i \in \Omega} I_i$ is an upper bound. This is a proper ideal, because if it wasn’t, some $I_i$ would contain $1$, which is impossible. So Zorn’s lemma applies and we get a maximal element in $\mathcal{P}$

Corollary 3.12 Every non-zero cyclic module contains a maximal submodule

Proof Any non-zero cyclic module is of the form $R/I$ where $I$ is a proper left ideal. Now apply 3.11 to $I$. We get a maximal left ideal $\mathfrak{m}$ contaning $I$. Then $\mathfrak{m}/I$ is a maximal submodule of $R/I$.

Lemma 3.13 Any non-zero semisimple module contains a simple submodule.

Proof Let $M$ be a semisimple module over a ring $R$. As by 3.10, submodules of semisimple modules are semisimple, it suffices to treat the case where $M$ is cyclic. In that case, $M$ contains a maximal submodule $N \leq M$ by 3.12.
As $M$ is semisimple, we can find a submodule $S \leq M$ such that $M = N \oplus S$.
If $S$ is not simple, then there is a non-zero proper submodule $S' \subsetneq S$, but then $N \subsetneq N \oplus S' \subsetneq N \oplus S = M$, which contradicts the maximality of $N$.

We now come to the main result on semisimple modules, the proof is a little technical.
The most important part of the statement for us is the implication (1)=>(2) (cf. 3.16), but we give the full result for completeness.

Proposition 3.14 For a module $M$, the following statements are equivalent:

1. $M$ is semisimple
2. $M$ is a sum of simple submodules
3. $M$ is a direct sum of simple submodules

Proof
(1.) implies (2.): Let $M$ be semisimple and let $\mathrm{soc}(M)$ be the sum of all simple submodules, then as $M$ is semisimple, we get that $M=\mathrm{soc}(M) \oplus N$ for some $N \leq M$.
If $N$ is non-zero, we get that $N$ contains a simple submodule by 3.10 and 3.13, but this contradicts the definition of $\mathrm{soc}(M)$ and the fact that $\mathrm{soc}(M) \cap N = 0$.

(2.) implies (3.): Suppose $M = \sum_{i \in I} M_i$ where all $M_i$ are simple.
Consider the set of subsets $J \leq I$ such that $\sum_{i \in J}M_i = \bigoplus_{i \in J}M_i$. This is partially ordered by inclusion and the usual Zorn’s lemma argument works (just take unions of chains as upper bounds) so that we get a maximal element $J_{\omega}$. Suppose that $\bigoplus_{i \in J_\omega} M_i = \sum_{i \in J_\omega} M_i \subsetneq \sum_{i \in I} M_i = M$, then for some $i_0 \in I$, we get that $M_{i_0} \not \subset \bigoplus_{i \in J_\omega} M_i$, which implies that $M_{i_0} \cap \oplus_{i \in J_\omega} M_i = 0$, since $M_{i_0}$ is simple and that intersection is a proper submodule. But then we get
$M_{i_0} + \oplus_{i \in J_\omega} M_i = M_{i_0} \oplus \oplus_{i \in J_\omega} M_i$ which contradicts the maximality of $J_\omega$.

(3.) implies (1.): Let $M = \bigoplus_{i\in I} M_i$ with all $M_i$ simple. Let $N \leq M$ be a submodule. We may assume that $N$ is a proper submodule.
Consider the set of subsets $J \subset I$ such that $N \cap \bigoplus_{i \in J} M_i = 0$. This is non-empty, as $N$ is a proper submodule, it doesn’t contain some $M_i$, but then $N \cap M_i = 0$, as $M_i$ is simple.
Now we apply (surprisingly!) Zorn’s lemma to this set, partially ordered by inclusion by taking unions as upper bounds for chains. Let $J_\omega$ be a maximal element.
Then consider $N+\bigoplus_{i \in J_\omega} M_i = N \oplus \bigoplus_{i \in J_\omega} M_i$. If this is a proper submodule of $M$, then it must have zero intersection with some $M_{i_o}$ for $i_0 \in I$. It follows that $N \cap M_{i_0} = 0$, $i_0 \not \in J_\omega$ and $M_{i_0} \cap M_i = 0$ for all $i \in J_\omega$, thus $M_{i_0} + \bigoplus_{i \in J_\omega}M_i= M_{i_0} \oplus \bigoplus_{i \in J_\omega} M_i$, so that by maximality of $J_\omega$, we get $N \cap (M_{i_0} \oplus \bigoplus_{i \in J_\omega} M_i) \neq 0$, so we can choose $n$ non-zero in that intersection. Write $n=m+m'$ for some $m \in M_{i_0}$ and $m' \in \bigoplus_{i \in J_\omega} M_i$
Then $m=n-m'$ is contained in $M_{i_0} \cap (N \oplus \bigoplus_{i \in J_\omega} M_i)$ which is zero by the choice of $i_0$.
Thus $n=m'$ is a non-zero element of $N \cap \bigoplus_{i \in J_\omega} M_i = 0$ which is impossible, thus $M=N \oplus \bigoplus_{i \in J_\omega} M_i$.

Note that the proof for the implication from (2) to (3) actually shows that if a module is a sum of simple submodules, one can find a subset of the index set such that the sum is direct and still gives the whole module.

Corollary 3.15 Direct sums of semisimple modules are semisimple.

Proof Use the equivalence between (1) and (3) in 3.14

Corollary 3.16 If $G$ is a finite group and $K$ is a field such that the characteristic of $K$ doesn’t divide the order of $G$, then any representation of $G$ over $K$ is a direct sum of irreducible subrepresentations.

Proof Follows from 1.22 and 3.14

We have already seen an instance of this phenomenon in our study of cyclic groups in the semisimple case. (cf. 2.6 and 3.7)

Let’s first record an observation, so that the statements we’re about to prove make sense:

Lemma 3.17 Let $R$ be any ring and let $M$ and $N$ be modules over $R$, then $\mathrm{End}_R(M)$ and $\mathrm{End}_R(N)$ are rings and if $R$ is a $K$-algebra, they are also $K$-algebras.
$\mathrm{Hom}_R(M,N)$ is a left module over $\mathrm{End}_R(M)$ and a right module over $\mathrm{End}_R(N)$ and these actions are compatible, i.e. $\mathrm{Hom}_R(M,N)$ is a $(\mathrm{End}_R(M),\mathrm{End}_R(N))$-bimodule.

Proof The statement might look complicated, but all we’re doing here is just composing maps: $\mathrm{End}_R(M)$ is a ring (or $K$-algebra) under composition of maps and the module structures on $\mathrm{Hom}_R(M,N)$ are given by composing with endomorphisms from the left or the right. All properties we need follow from properties of composing linear maps

Lemma 3.18 (Schur) Let $M$ and $N$ be modules over a ring $R$ and let $f:M \to N$ be a linear map, then:

1. If $M$ is simple, then $f$ is either zero or injective.
2. If $N$ is simple, then $f$ is either zero or surjective.
3. If both $M$ and $N$ are simple, then $f$ is either zero or an isomorphism.
4. If $M$ is simple, then $\mathrm{End}_R(M)$ is a division ring. (cf. 3.17)

Proof
(1): As $M$ is simple, $\mathrm{ker}(f)$ is either $M$ or $0$.
(2): As $N$ is simple, $\mathrm{im}(f)$ is either $N$ or $0$.
(3): Follows from (1) and (2).
(4): Follows from (3).

Despite the easy proof, Schur’s lemma is quite useful and will be a constant companion while dealing with simple modules. We give a first application.

Lemma 3.19 Let $M$ be a semisimple module that is a finite direct sum of simple submodules, write $M \cong \bigoplus_{i=1}^n M_i^{e_i}$ where the $M_i$ are pairwise non-isomorphic. Then for every $i$, set $D_i=\mathrm{End}_R(M_i)$. Then we have an equality $e_i = \mathrm{dim}_{D_i}(\mathrm{Hom}_R(M_i,M))= \mathrm{dim}_{D_i}(\mathrm{Hom}_R(M,M_i))$. In particular, the exponent $e_i$ is independent of the decomposition, so the decomposition is unique up isomorphism and permutation of the factors.

Proof  $\mathrm{Hom}_R (M_i,M)$ $=\mathrm{Hom}_R(M_i,\bigoplus_{j=1}^n M_j^{e_j})$ $\cong \bigoplus_{j=1}^n \mathrm{Hom}_R(M_i, M_j)^{e_j}$
Note that this isomorphism is $D_i$-linear, because the action of $D_i$ is given by composition in the first argument.
Schur’s lemma implies $\mathrm{Hom}_R(M_i,M_j) = 0$ unless $i=j$, so we get $\bigoplus_{j=1}^n \mathrm{Hom}_R(M_i, M_j)^{e_j} \cong \mathrm{Hom}_R(M_i, M_i)^{e_i}= D_i^{e_i}$. The case with switched arguments works the in the same way.

Corollary 3.20 Let $G$ be a finite group and let $K$ be a field such that the characteristic of $K$ does not divide the order of $G$, then every finite-dimensional representation can be written as a direct sum of irreducible subrepresentations which are uniquely determined up to isomorphism, including their multiplicity.

Proof Existence follows from 3.16 and uniqueness from 3.19

The last corollary justifies why one pays a lot of attention to irreducible representations, especially when Maschke’s theorem applies.

## Semisimple Rings and Algebras

So far, we have just studied (semi)simple modules. A general philosophy in ring theory is to study relations between the internal structure of a ring and the structure of its modules. Whenever there’s a notion for modules, one possible definition for a ring-theoretic property is obtained by just considering a ring as a left, right or two-sided module over itself. (For technical reasons, we will work with right ideals and modules in this section. It will allow us to skip some passage from a ring to its opposite ring in a future post. One can dualize all statements by using that left $R$-modules are right $R^{op}$-modules, where $R^{op}$ is the opposite ring which has reversed order of multiplication. Note that the group algebra $K[G]$ is isomorphic to its own opposite ring, via the map given on the basis $G$ by $g \mapsto g^{-1}$.)

If we apply this to the properties we’ve been studying, we get that a ring that is simple as a right module over itself is just a division ring. The way to see this is that every non-zero element must generate the whole ring as a right deal (and by group theory it’s enough to have all right inverses.). We already have a name for that, so that’s nothing new. This doesn’t happen with the following definition:

Definition 3.21 A ring $R$ is called semisimple if it is semisimple as a right module over itself.

We’re deliberately not being careful with the chirality here: Theoretically, one should define left and right semisimple, but as we shall see that they are equivalent.

We can apply the theory we have developed for semisimple modules to show how this property is reflected in the structure of the modules over a ring:

Lemma 3.22 A ring $R$ is semisimple if and only if all right modules over $R$ are semisimple.

Proof One direction is obvious. For the other one, note that if $R$ is semisimple, 3.15 implies that all direct sums of $R$, i.e. all free modules are semisimple. By 3.10, this also shows that all quotients of free modules are semisimple. But every module is a quotient of a free module.

Remarkably, this tells us that it would have been sufficient to prove Maschke’s theorem for just for one single representation, the one corresponding to the $K[G]$-module $K[G]$ to get decompositions into irreducible representations. (Even infinite-dimensional ones.)

Lemma 3.23 If a ring is a direct sum of non-zero right ideals, then the sum is finite.

Proof Suppose $R=\bigoplus_{i \in I} J_i$, then we have $1=(a_i)_{i \in I}$ where all but finitely many $a_i$ are zero. Let $I' \subset I$ be the subset of $I$ consisting of the indices $i$ such that $a_i \neq 0$. Then for any $r \in R$, we have $r= 1 \cdot r = \sum_{i \in I'} a_ir$ because the sum is direct, this expression is the unique way to write $r$ as a sum from elements in $J_i$ where $i$ ranges over $I$. since we assumed that all $J_i$ are non-zero, this implies $I=I'$, so that $I$ is finite.

Corollary 3.24 A semisimple ring is a finite direct sum of simple right $R$-modules (also called minimal right ideals in this case.)

Proof Apply 3.14 and then 3.23.

Corollary 3.25 Let $R$ be a semisimple ring, then every simple right $R$-modules $M_i$ occurs as a direct summand of $R$ (as a right $R$-module over itself) and the multiplicity is equal to the dimension of $M_i$ over its endomorphism ring (which is a division ring). In particular, that dimension is finite.

Proof Note that 3.24 implies that 3.19 is applicable to $R$ (by which we always mean as a right module over itself in this proof).
Let $e_i$ be the multiplicity with which $M_i$ occurs in the decomposition of $R$ as a direct sum of simple submodules. By 3.19 $e_i$ is independent of the decomposition, but it might be zero. But 3.19 tells us that $e_i=\mathrm{dim}_{D_i}(\mathrm{Hom}_R(R,M_i))=\mathrm{dim}_{D_i}(M_i)$ which also tells us two things:
1) The RHS is finite
2) The LHS is non-zero, as $M_i \neq 0$.

We want to apply this to the case where $R$ is an algebra over a field $K$, but for this it would be nice to know that the $D_i$ are finite-dimensional over $K$. We need some easy results on finiteness conditions.

Lemma 3.26 Let $K$ be a field and let $M$ and $N$ be modules over a finite-dimensional algebra $A$, then if $M$ and $N$ are finitely-generated over $A$, they are finite-dimensional over $K$ and so is $\mathrm{Hom}_A(M,N)$.

Proof $M$ being finitely-generated means that we can find a $A$-linear surjection $A^n \to M$. As $A$ is a $K$-algebra, this surjection is also $K$-linear. $A^n$ is finite-dimensional over $K$, because $A$ is, this implies that $M$ is finite-dimensional over $K$. If $M$ is finitely generated, let $S$ be a finite-generating system, then the map $\mathrm{Hom}_R(M,N) \to N^S, f \mapsto (f(s))_{s \in S}$ is $K$-linear. It is also injective, because any map from $M$ is determined by where it sends the generating system $S$. $N^S$ is a finite-dimensional vector space by the previous part, thus $\mathrm{Hom}_R(M,N)$ is finite-dimensional.

Corollary 3.27 Let $A$ is a finite-dimensional algebra over a field $K$, then all simple modules and their endomorphism rings are finite-dimensional over $K$.

Lemma 3.28 Let $A$ be a semisimple algebra over a field and let $M_1, \dots, M_n$ be a list of all simple modules, up to isomorphism. Let $D_i=\mathrm{End}_A(M_i)$ be their endomorphism rings. Then $\mathrm{dim}_K(A)=\sum_{i=1}^n \mathrm{dim}_K(M_i)^2/\mathrm{dim}_K(D_i)$ (where all dimensions are finite.)

Proof 3.25 implies that $A \cong \bigoplus_{i=1}^n M_i^{e_i}$ where $e_i=\mathrm{dim}_{D_i}(M_i)$, this implies that $\mathrm{dim}_K(A)= \sum_{i=1}^n \mathrm{dim}_K(M_i)\mathrm{dim}_{D_i}(M_i)$. (3.27 tells us that we don’t have to worry about infinite dimension.)
So the only thing left to show is that $\mathrm{dim}_K(D_i) \mathrm{dim}_{D_i}(M_i)=\mathrm{dim}_K(M_i)$. But this is clear: We have $M_i=D_i^{e_i}$, so we just compare the $K$-dimension of both sides.

The following lemma tells us that we can leave out the factors if $\mathrm{dim}_K(D_i)$ if $K$ is algebraically closed.

Lemma 3.29 If $K$ is an algebraically closed field, then every finite-dimensional division algebra over $K$ is one-dimensional, i.e. $K$ itself.

Proof Let $D$ be finite-dimensional division algebra over $K$.
Let $d \in D$, then consider the $K$-subalgebra $K[d]$ generated by $d$. Every element in $K[d]$ is a polynomial in $d$, so $K[d]$ is a quotient of the polynomial ring $K[x]$ via the map $ev_x: x \mapsto d$. But then $\mathrm{ker}(ev_x)$ is a non-zero prime ideal, as the image is finite-dimensional over $K$ and doesn’t contain zero divisors. This implies $\mathrm{ker}(ev_x)=(x-\lambda)$ for some $\lambda \in K$, so $d=\lambda \in K$.

We note the following corollary to 3.28 and 3.29

Corollary 3.30  Let $A$ be a semisimple algebra over a field and let $M_1, \dots, M_n$ be a list of all simple modules, up to isomorphism. Then $\mathrm{dim}_K(A)\leq \sum_{i=1}^n \mathrm{dim}_K(M_i)^2$ and we have equality if $K$ is algebraically closed.

If we put in some knowledge about finite-dimensional divison algebras over $\Bbb R$ (namely, the fact that the only ones are $\Bbb R, \Bbb C, \Bbb H$, so the dimension is at most 4), we also get the following:

Corollary 3.31 Let $A$ be a semisimple algebra over $\Bbb R$ and let $M_1, \dots, M_n$ be a list of all simple modules, up to isomorphism. Then $\frac{1}{4} \sum_{i=1}^n \mathrm{dim}_\Bbb{R}(M_i)^2 \leq \mathrm{dim}_\Bbb{R}(A)\leq \sum_{i=1}^n \mathrm{dim}_\Bbb{R}(M_i)^2$

These corollaries translate into statements about representations when we apply them to the group algebra $K[G]$.

Let’s close this post by recapitulating what we have shown about representations in the case where $G$ is finite and the characteristic of $K$ doesn’t divide the order of $G$:

• There are finitely many irreducible representations up to isomorphism(which are all finite-dimensional)
• Every irreducible representation occurs as a direct summand of the so called “regular representation”, which is the representation corresponding to $K[G]$ as a module over itself.
• Every representation is a direct sum of copies of irreducible subrepresentations, even infinite-dimensional ones.
• We know that for finite-dimensional representations the decomposition into a direct sum of irreducibles is unique up to isomorphism of the factors, including multiplicities. (I didn’t want to deal with cardinals for the infinite-dimensional case)
• We have a nice formula that relates the dimension of irreducibles, the dimension of their endomorphism rings and the order of $G$ (which is obviously the dimension of $K[G]$). If we don’t want to talk about the endomorphism rings, we still have an inequality, which is an equality in the algebraically closed case.

In the next post, we will continue our study of semisimple rings and give applications, e.g. by describing the number of irreducible representations in terms of the group $G$.