## Simple Modules

We briefly talked about modules over a (possibly non-commutative) ring R. An important aspect of modules is that unlike vector spaces, modules are usually not free, i.e. they don’t have a basis. For example, take the Z-module given by Z/2Z.

[ Recall: a Z-module is just the same as an abelian group, and a Z-module homomorphism corresponds precisely to a homomorphism of abelian groups. ]

Another annoying aspect is that for a submodule $N\subseteq M$, we often cannot find a $P\subseteq M$ such that $M = N\oplus P$ (i.e. $N\cap P = 0$ and N + P = M, which just means that every element of M is uniquely writable as x+y, with x in N and y in P). A good example is given by N = 2ZM = Z; for any other submodule P of Z, we either get P=0 or a non-zero module which must necessarily intersect N.

Another problem is that a submodule of a free module is not necessarily free. This is less intuitive, since any subgroup of a free abelian group is actually free (even the non-finitely generated case!), so a beginning student may have problems grasping it.

Exercise : let R be the ring Z × Z, where multiplication and addition are both component-wise (e.g. (2, 1) × (-5, 3) = (-10, 3)). Prove that the R-module R has a submodule which is not free. [ Answer: take the submodule Z × {0}. ]

In this article, we will look at a simple case where modules are rather well-behaved. This is probably as good as it gets, next to vector spaces over division rings.

[ Convention: all modules are left modules, i.e. for $r\in R$ and $m\in M$ we have the multiplication $rm\in M$. This ensures that multiplication-by-r followed by multiplication-by-s is simply multiplication-by-sr, since s(rm) = (sr)m. In the case of right modules, this would have been multiplication-by-rs. ] ## Simple Modules

First we have:

Definition. An R-module M is said to be simple if it is non-zero, and has no submodules except 0 and itself.

Simple modules are akin to prime numbers, and just like we don’t accept 1 as a prime, nor do we accept 0 as a simple module (it messes up the factorization terms). In the case where R is a field (or even a division ring), a simple module is the same as a vector space of dimension 1. However, in the general case not all simple modules are isomorphic:

Exercse: let RZ. Find all simple Z-modules. [ Answer: they’re modules of the form Z/for prime p. ]

Even in the case where RZ it’s clear not every module has a simple submodule. E.g. take MZ. So the following theory only works in rather limited cases, specifically when the ring is not too “complicated”.

[ Aside: what constitutes a “complicated” ring? Roughly speaking, these are rings which are “hard to understand”. We’ll give some examples here: fields are the easiest types of rings, and division rings aren’t too bad, although division ring extensions are much harder to describe than field extensions. Commutative rings can be classified by a value of “dimension”, which is roughly the number of parameters required to describe the ring, but these will come later. ]

A trivial but important observation:

Let N be a simple submodule of M. For any submodule N’ of M, either N ∩ N’ = 0 or N’ contains N.

If N’ also simple, then N ∩ N’ = 0 or N = N’.

Since N ∩ N’ is a submodule of N, a simple module, it is either 0 or N itself; this proves the first statement. The second statement follows easily from the first. ♦

Another easy observation:

Any simple module is isomorphic to R/I, for a maximal left ideal I of R.

[ Note: a left ideal I of R is simply a left submodule. In other words, a left ideal is a subset I of R such that (i) (I, +) is an abelian subgroup, and (ii) for any $r\in R, x \in I$, we have $rx \in I.$ ]

[ A maximal submodule of M is a submodule N of M such that N ≠ M and, whenever $N \subseteq P \subseteq M$, we must have PN or PM. From the correspondence between (a) submodules of M/N and (b) submodules of M containing N, we conclude that N is maximal iff M/N is simple. ]

Proof

If I is a maximal left ideal (and thus maximal submodule) of R, then as noted above, R/I is a simple module. Conversely, let M be a simple R-module; pick a non-zero element m of M. Take the R-module homomorphism $f:R \to M, r \mapsto rm.$ Its image is a non-zero submodule of M so must be the whole of M. Thus f is surjective and $M \cong R/\text{ker} f.$ ♦

Finally, the following result is extremely important!

Schur’s Lemma. If $f:M \to N$ is a homomorphism of simple modules, then either f=0 or f is an isomorphism. In particular, the ring $\text{End}_R(M)$ is a division ring for a simple module M.

Proof

The kernel of f is a submodule of M, so it is either 0 or whole of M. Likewise, the image of f must be 0 or the whole of N. By considering various cases, we see that either f = 0 (in which case ker(f) = M and im(f) = 0) or f is an isomorphism (in which case ker(f) = 0 and im(f) = N). The second statement follows from the first, since the product in End(M) is just composition of endomorphisms. ♦ ## Some Examples

Let’s consider some concrete cases.

• Take RZ and MZ/5Z, which is a simple R-module. To find the endomorphism ring End(M), consider an fM → M. This is wholly determined by f(1), so we obtain a map End(M) → M which takes f to f(1). You can check that this is surjective, so End(M) = Z/5, a field.
• Take R = R × R, where R is the field of real numbers, and MR × {0}, an R-module which is clearly simple. Again an endomorphism fM → M is uniquely determined by f((1,0)) so we have End(M) = R, a field.
• Let RR × R again, and MR × {0}, N = {0} × R be simple R-modules. Are they isomorphic as R-modules? What about $M' = \{(x, x) : x\in\mathbf{R}\}$ and $N' = \{(x, 2x): x\in\mathbf{R}\}$?  [ Answer: no for the first question, we have (0, 1)·M = 0 but (0, 1)·N ≠ 0. Yes to the second question. ]
• Let R be the ring of upper-triangular 2 × 2 matrices with real entries, i.e. $\begin{pmatrix} a & b \\ 0 & d\end{pmatrix}$, $a,b,d\in\mathbf{R}.$ Let M be the space R2 and R acts on M via multiplying matrix with vector. Find a simple submodule of M. [ Answer: take the R-vector space spanned by (1, 0)’. ]
• So far all End(M) we’ve seen are commutative. For a cheap way to get a non-commutative case, let R be a division ring and MR. Any R-linear fM → M is of the form f(m) = mr for some r. [ Question: why mr and not rm? ] Thus composing f(m) = mr followed by g(m) = ms gives us gf(m) = mrs and we have End(M) = Rop (the opposite ring!). Not every division ring is isomorphic to its opposite ring, but it’s hard to construct an example.
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### 2 Responses to Simple Modules

1. Jack says:

I don’t understand the second last example–why is M a simple R-module? Isn’t $\mathbb{R}\times 0$ is non-trivial submodule?

2. limsup says:

Sorry that was a bad mistake. Corrected.