### The Atoms of the Module World

#### Posted by Tom Leinster

In many branches of mathematics, there is a clear notion of “atomic” or “indivisible” object. Examples are prime numbers, connected spaces, transitive group actions, and ergodic dynamical systems.

But in the world of modules, things aren’t so clear. There are at least two competing notions of “atomic” object: simple modules and, less obviously, projective indecomposable modules. Neither condition implies the other, even when the ring we’re over is a nice one, such as a finite-dimensional algebra over a field.

So it’s a wonderful fact that when we’re over a nice ring, there is a canonical bijection between $\{$isomorphism classes of simple modules$\}$ and $\{$isomorphism classes of projective indecomposable modules$\}$.

Even though neither condition implies the other, modules that are “atoms” in one sense correspond one-to-one with modules that are “atoms” in the other. And the correspondence is defined in a really easy way: a simple module $S$ corresponds to a projective indecomposable module $P$ exactly when $S$ is a quotient of $P$.

This fact is so wonderful that I had to write a short expository note
on it *(update — now arXived)*. I’ll explain the best bits here — including how it all
depends on one of my favourite things in linear algebra, the eventual
image.

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.

## Re: The Atoms of the Module World

Hi. One might also consider the indecomposable injectives as possible ‘atoms’, and these generalise somewhat better than projectives to other classes of algebras.

Also, I think there is a mistake in Lemma 4.4 of your note. Every finitely generated (hence finite dimensional) module has a simple quotient, but it seems you are not assuming this to start with. In this case, one cannot just apply Zorn’s Lemma to deduce the existence of a maximal submodule, otherwise this would work for all modules, where the result is definitely false.

Added by Tom Leinster on 2014-10-13: I’ve now edited the note to fix the mistake, in the way suggested by Andrew below. What was Lemma 4.4 is now Lemma 5.4.