The n-Category Café

It’s clear how the simple modules might be seen as “atomic”. They’re the nonzero modules that have no nontrivial submodules.

But what claim do the projective indecomposables have to be the “atoms” of the module world? Indecomposability, the nonexistence of a nontrivial direct summand, is a weaker condition than simplicity. And what does being projective have to do with it?

The answer comes from the Krull-Schmidt theorem. This says that over a finite enough ring $A$, every finitely generated module is isomorphic to a finite direct sum of indecomposable modules, uniquely up to reordering and isomorphism.

In particular, we can decompose the $A$-module $A$ as a sum $P_1 \oplus \cdots \oplus P_n$ of indecomposables. Now the $A$-module $A$ is projective (being free), and each $P_i$ is a direct summand of $A$, from which it follows that each $P_i$ is projective indecomposable. We’ve therefore decomposed $A$, uniquely up to isomorphism, as a direct sum of projective indecomposables.

But that’s not all. The Krull-Schmidt theorem also implies that every projective indecomposable $A$-module appears on this list $P_1, \ldots, P_n$. That’s not immediately obvious, but you can find a proof in my note, for instance. And in this sense, the projective indecomposables are exactly the “pieces” or “atoms” of $A$.

Here and below, I’m assuming that $A$ is a finite-dimensional algebra over a field. And in case any experts are reading this, I’m using “atomic” in an entirely informal way (hence the quotation marks). Inevitably, someone has given a precise meaning to “atomic module”, but that’s not how I’m using it here.

One of the first things we learn in linear algebra is the rank-nullity formula. This says that for an endomorphism $\theta$ of a finite-dimensional vector space $V$, the dimensions of the image and kernel are complementary:

$$dim\, im \theta + dim\, ker \theta = dim V.$$

Fitting’s lemma says that when you raise $\theta$ to a high enough power, the image and kernel themselves are complementary:

$$im \theta^n \oplus ker \theta^n = V \qquad (n \gg 0).$$

I’ve written about this before, calling $im \theta^n$ the eventual image, $im^\infty \theta$, and calling $ker\theta^n$ the eventual kernel, $ker^\infty \theta$, for $n \gg 0$. (They don’t change once $n$ gets high enough.) But what I hadn’t realized is that Fitting’s lemma is incredibly useful in the representation theory of finite-dimensional algebras.

For instance, Fitting’s lemma can be used to show that every projective indecomposable module is finitely generated — and indeed, cyclic (that is, generated as a module by a single element). Simple modules are cyclic too, since the submodule generated by any nonzero element must be the module itself. So, both projective indecomposable and simple modules are “small”, in the sense of being generated by a single element. In other words:

Atoms are small.

Whatever “atom” means, they should certainly be small!

But also, “atoms” shouldn’t have much internal structure. For instance, an atom shouldn’t have enough complexity that it admits lots of interesting endomorphisms. There are always going to be some, namely, multiplication by any scalar, and this means that the endomorphism ring $End(M)$ of a nonzero module $M$ always contains a copy of the ground field $K$. But it’s a fact that when $M$ is atomic in either of the two senses I’m talking about, $End(M)$ isn’t too much bigger than $K$.

Let me explain that first for simple modules, since that’s, well, simpler.

A basic fact about simple modules is:

Every endomorphism of a simple module is invertible or zero.

Why? Because the kernel of such an endomorphism is a submodule, so it’s either zero or the whole module. So the endomorphism is either zero or injective. But it’s a linear endomorphism of a finite-dimensional vector space, so “injective” and “surjective” and “invertible” all mean the same thing.

Assume from now on that $K$ is algebraically closed. Let $S$ be a simple module and $\theta$ an endomorphism of $S$. Then $\theta$ has an eigenvalue, $\lambda$, say. But then $(\theta - \lambda\cdot id)$ is not invertible, and must therefore be zero.

What we’ve just shown is that the only endomorphisms of a simple module are the rescalings $\lambda\cdot id$ (which are always there for any module). So $End(S) = K$:

A simple module has as few endomorphisms as could be.

Now let’s do it for projective indecomposables. Fitting’s lemma can be used to show:

Every endomorphism of an indecomposable finitely generated module is invertible or nilpotent.

That’s easy to see: writing $M$ for the module and $\theta$ for the endomorphism, we can find $n \geq 1$ such that $im \theta^n \oplus ker \theta^n = M$. Since $M$ is indecomposable, $im \theta^n$ is either $0$, in which case $\theta$ is nilpotent, or $M$, in which case $\theta$ is surjective and therefore invertible. Done!

I said earlier that (by Fitting’s lemma) every projective indecomposable is finitely generated. So, every endomorphism of a projective indecomposable is invertible or nilpotent.

Let’s try to classify all the endomorphisms of a projective indecomposable module $P$. We’re hoping there aren’t many.

Exactly the same argument as for simple modules — the one with the eigenvalues — shows that every endomorphism of a projective indecomposable module is of the form $\lambda\cdot id + \varepsilon$, where $\lambda$ is a scalar and $\varepsilon$ is a nilpotent endomorphism. So if you’re willing to regard nilpotents as negligible (and why else would I have used an $\varepsilon$?):

A projective indecomposable module has nearly as few endomorphisms as could be.

(If you want to be more precise about it, $End(P)$ is a local ring with residue field $K$. All that’s left to prove here is that $End(P)$ is local, or equivalently that for every endomorphism $\theta$, either $\theta$ or $id - \theta$ is invertible. We can prove this by contradiction. If neither is invertible, both are nilpotent — and that’s impossible, since the sum of two commuting nilpotents is again nilpotent.)

So all in all, what this means is that for “atoms” in either of our two senses, there are barely more endomorphisms than the rescalings. More poetically:

Atoms have very little internal structure.

My note covers a few more things than I’ve mentioned here, but I’ll mention just one more. There is, as I’ve said, a canonical bijection between isomorphism classes of indecomposable modules and isomorphism classes of simple modules. But how big are these two sets of isomorphism classes?

The answer is that they’re finite. In other words, there are only finitely many “atoms”, in either sense.

Why? Well, I mentioned earlier that as a consequence of the Krull-Schmidt theorem, the $A$-module $A$ is a finite direct sum $P_1 \oplus \cdots \oplus P_n$ of projective indecomposables, and that every projective indecomposable appears somewhere on this list (up to iso, of course). So, there are only finitely many projective indecomposables. It follows that there are only finitely many simple modules too.

An alternative argument comes in from the opposite direction. The Jordan-Hölder theorem tells us that the $A$-module $A$ has a well-defined set-with-multiplicity $S_1, \ldots, S_r$ of composition factors, which are simple modules, and that every simple module appears somewhere on this list. So, there are only finitely many simple modules. It follows that there are only finitely many projective indecomposables too.