In this blog post, my goal is to familiarize you with the idea that simple $R$ modules are entirely determined by the submodule structure of $R$ (i.e., the lattice of left ideals). Furthermore, you will see in which cases we can even identify all simple modules as submodules of $R$ – an important corollary of this is that the left regular representation of a finite group $G$ as a $ℂ$-vector space (i.e., a $ℂ[G]$-module) already contains all the irreducible representations.
In the lifetime of a student interested in algebra, modules start out being something quite obscure, and knowledge about them usually comes in these steps:
I believe I'm somewhat starting to get to point four, which is why I'm not sure of which points the last one actually consists.
The key thing I want to talk about is the question
Given a ring $R$, how many simple $R$-Modules are there?,
Which might seem difficult at first glance – the term „simple“ reminds one of the classification of finite simple groups, which is everything but trivial. However, then you encounter the following:
THM Let $D$ be a division ring. Then there is only one isotype of simple $M_r(D)$-Modules, namely $D^r$.
Here, “isotype“ means “classes of modules up to isomorphism” Of course, $D^r$ is taken to be equipped with the action of left matrix multiplication.
Of course, simple modules are extremely restricted in terms of what homomorphisms can go into or out of them. For instance, we have the following (which directly implies Schur's lemma):
LEM Let $S$ be a simple $R$-Module. Then every $R$-Module-Homomorphism (from now on just “morphism”) into $S$ is either trivial or surjective, and every morphism from $S$ is either trivial
The proof is trivial once you recall that kernel and image are necessarily submodules.
Let $S$ be a simple $R$-Module, and $0\neq v\in S$. It is then clear that the submodule $Rv$ generated by $v$ must be $S$ in its entirety – so in particular, every morphism into $S$ is given by its value on $v$ completely. In particular, for every such $v$, we have a unique morphism $ρ_v\colon R\to S$ – where we take $R$ as na $R$-module as usual by left multiplication – satisfying $ρ_v(1)=v$. Its surjectivity implies that $S\simeq R/\operatorname{Ker}ρ_v$. Since (proper) submodules containing the kernel correspond to (proper) submodules of the image, and there are none of the latter, we see that $\operatorname{Ker}ρ_v$ must be in fact maximal.
Now, let me just remind you that submodules of ($R$ with left multiplication) correspond to left ideals of $R$. The observations above lead us to the conclusion:
OBS. Isotypes of simple $R$-Modules are in one-to-one correspondence to maximal left ideals $I$ of $R$ by means of taking the quotient $R/I$.
Proof We did the direction simple module → maximal left ideal above. Conversely, if we have a maximal left ideal (resp. maximal submodule $M\leq R$), quotienting by that gives us a simple module, again by the correspondence of submodules containing the kernel (of the quotient homomorphism) vs. submodules in the image.
For this section, recall that the complement to a submodule $S≤M$ is a submodule $T≤M$ such that $S\cap T = 0$ and $S+T=M$.
Once you think about it, if we have the ability to complement the ideal $I$, this (now minimal) ideal must actually be isomorphic to our simple module by the following argument:
LEM Let $S$ be a simple $R$-Module and $φ\colon M\to S$ a nontrivial – hence surjective – morphism with kernel $N$. If $N$ admits a complement $N^\prime$, the latter must be isomorphic to $S$.
Proof Denote the injection of $N^\prime$ into $M$ by $ι$. Composing $φ\circ ι\colon N^\prime\to M\to S$, we see that it must be nontrivial, and hence surjective. However, since $N^\prime$ complements a maximal submodule, it must be minimal, and hence, as an $R$-Module on its own, simple. Therefore, $φ\circ ι$ must also be injective, and hence is an isomorphism.
A module in which every submodule admits a complement is called semisimple – and if our $R$-module happens to be $R$, we call that artinian semisimple. However, do note that you may encounter a different – albeit equivalent – definition, such as “M decomposes into a direct sum of simple modules” and so on.
Consider the matrix ring $R:=M_r(D)$ over a division ring $D$ (also “skew field” – that is, a field that is not necessarily commutative; think Hamilton's quaternions. We can show with some nontrivial effort that the maximal submodules correspond to sets of matrices collectively annihilating some onedimensional “subspace” (read: rank one submodule) of $D^r$. However, that identification is not necessary due to the following observation:
LEM If $I$ is a maximal submodule of $R:=M_r(D)$, one of the minimal submodules $Re_{11},\ldots,Re_{rr}$ must be a complement to $I$.
Proof Let $φ$ denote the projection $R\to R/I$. Since it is nontrivial, $φ(1)≠0$. Because $1=\sum_i e_{ii}$, at least one $e_{jj}$ must be mapped to something nonzero, i.e. not be contained in $I$. It remains to show that the submodule generated by this element, namely $Re_{jj}$, is a complement to $I$. For that, note that $Re_{jj}\cap I\leq Re_{jj}$. However, we can see that the latter is simple (see the following lemma), which implies that either $Re_{jj}=0$ or $Re_{jj}\leq I$ – but the latter is impossible, because $e_{jj}\not\in I$. Finally, because every submodule which trivially intersects a maximal submodule must be a complement, we are done.
To see that (without loss of generality) $Re_{11}$ is simple, note that the generated submodule consists of all the matrices with entries only in the first column– which is easily identified with the $R$-Module $D^r$ equipped with matrix multiplication from the left – and since for every pairs of vectors $v,w\in D^r$, there is a matrix $A$ such that $Av=w$, we see that any one vector generates all of $D^r$, and hence no proper submodule of $D^r$ can exist.
To whom the last sentence is not clear, let me make this a precise lattice-theoretic argument. Remember that the set of submodules of a module $M$ (in general, most subobjects of universal algebra-esque structures) always forms a complete bounded lattice with meet being the intersection, join being the submodule addition (i.e., “smallest submodule containing both”), $0$ being our minimum and $M$ our maximum (“$1$”). In this lattice, we will call an element “maximal” (resp. “minimal”) if it is a maximal element $≠1$ (resp. a minimal element $≠0$). Also recall that submodules satisfy the modular identity: If $N^\prime≤N$, we have $N^\prime+(L\cap N)=(N^\prime+L)\cap N$ for all $L$. Furthermore, remember that the complement of $x\in L$ is an element $y\in L$ such that $x\wedge y = 0$ and $x\vee y = 1$.
LEM Let $(L,\wedge,\vee,0,1)$ be a bounded modular lattice, $x,y\in L$, and $x$ maximal. TFAE:
Proof $(3)\implies (4)$ is clear. $(4)\implies (1)$: Let $y≠0$ with $x\wedge y = 0$. Because we're in a bounded lattice, maximality of $x$ is equivalent to stating that joins $x\vee -$ must be in $\{x,1\}$. Hence, $x≤x\vee y\in {x, 1}$. The first option means that $y\leq x$, which would imply that $x\wedge y = y \neq 0$, which is impossible. Hence $x\vee y = 1$ remains and we have a complement. $(1)\implies (2)$: Let $x\wedge y = 0$ and $x\vee y = 1$. For any $z\leq y$, we can apply the modular law with “middleman” $x$ to see that \begin{align} z &= z \vee 0 = z\vee(x\wedge y) \stackrel{!}{=} (z\vee x)\wedge y\ &\in \{x,1\}\wedge y = \{0,y\} \end{align} where the restriction to precisely two options is caused by the maximality of $x$. Therefore, we have shown that any element below $y$ must be either $0$ or $y$, proving minimality of $y$. $(2)\implies (3)$: Let $y≠0$ be minimal and $x\wedge y = 0$. Since any other element $z≤y$ which is nonzero must equal $y$, there are no smaller elements in $\{z\in L\mid z≠0, z\wedge x = 0\}$.
The “trivial” example of a ring is $ℤ$, which basically asks us to use $ℤ$-modules as the first example of mirroring a new concept. Since $ℤ$-Modules are seen to be abelian groups – and, just as importantly, $ℤ$-Module homomorphisms are seen to be precisely the homomorphisms of abelian groups, we conclude that submodules must coincide with subgroups. Furthermore, it is easy to see by simple group theoretical arguments that the simple modules must precisely be the groups $C_p$ where $p$ is a prime.
If we instead follow our argumentation above, it makes sense to identify the ideals of $ℤ$: The maximal ideals are easily seen to be the subgroups $pℤ$, which precisely matches our observation, since $ℤ/pℤ\simeq C_p$. However, given $H≤ℤ$, we can always find a strictly smaller subgroup – for instance $2H$. This absence of minimal subgroup implies that there cannot be any complements, and hints at the fact that we cannot find the simple abelian groups inside of $ℤ$. This makes complete sense: Of course we cannot find a copy of $C_p$ in $ℤ$ – every element of the latter has order zero!
The conclusion from this is: $ℤ$ is not semisimple.
In this section, let $G$ be a finite group and $K$ a field of characteristic zero (prototypically $K=ℂ$).
In the theory of group representations, one is concerned about ways in which we can represent a group $G$ as, say, a group of operators over a $K$-Vector spacej There is one very simple way to do this: Taking $G$ as a $K$-Basis, we can assign every $g\in G$ the operator permuting the basis elements by left multiplication. This is called the left regular representation.
There is a nice way to formulate this in the language of module theory:
DEF Let $G$ be a group and $R$ be a ring. The Group ring $R[G]$ is defined to consist of formal sums $\sum_{g\in G}r_g g$, where addition is defined component-wise, and multiplication defined such that $(1g)(1h) = 1(gh)$ for all $g, h\in G$.
It is straightforward (and a good exercise) to prove that this indeed defines a unique ring structure on $R[G]$.
In general, we can see that any set $X$ equipped with a $G$-action can be made into a $K[G]$-Module by taking $X$ as a $K$-Basis and by $K$-linearly extending the action.
In particular, we see that the left regular representation is just our group ring $K[G]$ viewed as a $K[G]$-Module.
If $G$ is finite, we can find a $K$-bilinear form on $K[G]$ which is invariant under the action of $G$ – this will enable us to find complements to submodules of $K[G]$ by means of the classical orthogonal complement!
DEF/LEM Define $β\colon K[G]\times K[G]\to K$ to be the standard scalar product on the basis $\{1g\mid g\in G\}$. Then $β$ is invariant under the action of $K[G]$. Furthermore, the orthogonal complement induced by $β$ maps $K[G]$-Submodules to $K[G]$-Submodules.
Proof That $β$ is $K[G]$-invariant is easily seen because multiplication by any $h\in G$ maps the orthonormal basis $\{1g\mid g\in G\}$ onto itself, hence leaves $β$ invariant, and extending this argument to $K[G]$ follows by observing $K$-bilinearity.
Now let $S≤K[G]$, and let $\sum_g k_g g$ be in the orthogonal complement. Then, for every $s\in S$ and $h\in G$, we have $$ β(h \sum_g k_g g, s) = β(\sum_g (hk_g) g, s) = \sum_g k_g β(hg, s) = \sum_g k_g β(g, h^{-1}s)\ = β(\sum_g k_g g, h^{-1}s) = 0, $$ because $h^{-1}s$ is still in $S$. Hence, the orthogonal complement of $S$ is invariant under the action of $G$, and since orthogonal complements with respect to a $K$-bilinear form are always $K$-vector spaces, it must be invariant under the entire $K[G]$-action.
The last missing piece is the following:
OBS $β$ is nondegenerate, and hence the orthogonal complement gives a subspace which is a complement in the usual lattice-theoretic sense.
Ergo, setting $K=ℂ$, we have that $ℂ[G]$ is semisimple, which again implies that any simple $ℂ[G]$-Module (also called „irreducible complex representation“) must be a submodule of $ℂ[G]$.