## October 5, 2014

### 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.

Posted at October 5, 2014 12:20 PM UTC

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### 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.

Posted by: Andrew Hubery on October 5, 2014 6:36 PM | Permalink | Reply to this

### Re: The Atoms of the Module World

Hi Andrew, thanks for the feedback. I’d be interested to hear from you (or anyone else) more on what the story is with indecomposable injectives.

Re Lemma 4.4, yes, of course you’re right. I was too cavalier with Zorn’s lemma. For others reading: I thought a simple application of Zorn’s lemma implied that every nonzero module has a maximal submodule (just as every nontrivial ring has a maximal ideal). The problem is that the union of a chain of proper submodules need not be proper. (For rings, the union of a chain of proper ideals is proper, since an ideal is proper iff it doesn’t contain $1$.) And there are counterexamples: e.g. the $\mathbb{Z}$-module $\mathbb{Q}$ has no maximal submodule.

I appreciate you pointing it out! I’ll have to think about a fix.

Posted by: Tom Leinster on October 5, 2014 10:01 PM | Permalink | Reply to this

### Re: The Atoms of the Module World

For a finite dimensional algebra $A$ over a field $k$, or more generally an Artin algebra, the indecomposable injectives are also in bijection with the simples, the relation being given by taking the socle of the injective, or the injective envelope of a simple.

There is then a nice bijection taking the indecomposable projective with simple top $S$ to the indecomposable injective with simple socle $S$. This extends to a functor, called the Nakajama functor, given by $\nu = D\mathrm{Hom}_A(-,A) \cong -\otimes_A D(A),$ where $D=\mathrm{Hom}_k(-,k)$ is the vector space duality. The Nakayama functor yields an equivalence between the subcategories of projective and injective modules.

In general, there may be very few projectives, but lots of injectives.

The example I was thinking of was a principal ideal domain, for example the integers, or the polynomial ring $k[X]$ if we want to work over a field. Here there is a unique indecomposable projective, the ring itself. On the other hand the simple modules are parametrised by the primes. Each simple has an injective envelope (necessarily indecomposable), and then there is the ‘generic module’, which is the field of fractions of the ring. This yields all indecomposable injectives.

Posted by: Andrew Hubery on October 6, 2014 9:39 AM | Permalink | Reply to this

### Re: The Atoms of the Module World

Nice! Thanks very much.

I’m still struggling with fixing that mistake. Corollary 4.6 of these notes of Serganova says that every nonzero module over an Artinian ring has a simple quotient.

This appears as a consequence of a corollary of a corollary of a corollary. Moreover, I only need the very special case of an indecomposable projective module over a finite-dimensional algebra over a field. I’m trying to extract a more direct proof. Do you know one?

Posted by: Tom Leinster on October 7, 2014 12:06 AM | Permalink | Reply to this

### Re: The Atoms of the Module World

Another possible approach: according to someone on the internet, every nonzero projective module has a maximal submodule. This is over an arbitrary noncommutative ring with identity. But I haven’t been able to find a proof or reference.

Posted by: Tom Leinster on October 7, 2014 1:05 AM | Permalink | Reply to this

### Re: The Atoms of the Module World

I found a reference:

Hyman Bass, Finitistic Dimension and a homological generalization of semi-primary rings, Trans. Amer. Math. Soc. 95 (1960), 466-488

Proposition 2.7 shows the following (and the proof is not too difficult). Suppose we have a free module $F$ over a ring $A$. Let $J$ be the Jacobson radical of $A$. Then no nonzero summand $P$ of $F$ is contained in $J F$.

It follows that for any nonzero projective $P$ we have $P\neq J P$. Since $J P$ can also be expressed as the intersection of all maximal submodules of $P$, we conclude that $P$ has a maximal submodule.

Posted by: Andrew Hubery on October 7, 2014 9:23 AM | Permalink | Reply to this

### Re: The Atoms of the Module World

Thanks very much; that’s really helpful.

I’m now wondering whether it’s possible to simplify Bass’s proof by taking advantage of either the fact that we’re over a particularly nice ring (a finite-dimensional algebra over a field) or the fact that the module is indecomposable.

Posted by: Tom Leinster on October 7, 2014 3:03 PM | Permalink | Reply to this

### Re: The Atoms of the Module World

This is the best I can do, but it still requires using the Jacobson radical.

Let $A$ be any ring, and $M$ and $A$-module. Define $\mathrm{rad}(A)$ to be the intersection of all maximal submodules of $M$.

(1) $\mathrm{rad}(A)$ is a two-sided ideal. For, it can be expressed as the intersection of all annihilators of all simple modules.

(2) $\mathrm{rad}(\bigoplus_i M_i)=\bigoplus_i\mathrm{rad}(M_i)$ for all modules $M_i$.

(3) $\mathrm{rad}(A)M\subseteq\mathrm{rad}(M)$, with equality for all projective modules (clear for free modules, now use (2)).

Now specialise to $A$ a finite dimensional algebra over a field (or artinian ring).

(4) $\mathrm{rad}(A)$ is a nilpotent ideal. For, if $\mathrm{rad}(A)^n\neq0$, then by dimension it has a maximal submodule, and so $\mathrm{rad}(A)^{n+1}\subset\mathrm{rad}(A)^n$ is strict.

(5) Every non-zero projective module has a maximal submodule. Use (3) and (4).

Posted by: Andrew Hubery on October 9, 2014 8:28 AM | Permalink | Reply to this

### Re: The Atoms of the Module World

Thanks! Amateur that I am, it takes me a while to do the simplest things in algebra, so I’m still working out all the proofs of your (1)-(5). Just one difficulty left to resolve now, but I won’t ask for help just yet!

Something else I should be able to figure out but haven’t done yet: where do you need that $rad(A)$ is a two-sided ideal?

Posted by: Tom Leinster on October 10, 2014 7:03 PM | Permalink | Reply to this

### Re: The Atoms of the Module World

Ha! You probably don’t need it.

I guess I was thinking of taking a quotient at some stage and using that $A/\mathrm{rad}(A)$ is semisimple, but in the end this wasn’t necessary. (BTW using this one can easily get that when $A$ is artinian, (3) is an equality for all modules $M$.)

Posted by: Andrew Hubery on October 10, 2014 8:11 PM | Permalink | Reply to this

### Re: The Atoms of the Module World

Here’s a proof, in full, that every nonzero projective module over a finite-dimensional algebra has a maximal ideal. This is almost precisely your argument, but I’m including all the details — mostly so that I don’t have to work them out again when I come back to this.

Let $A$ be a finite-dimensional algebra over a field. (We won’t need the full strength of that assumption until later.) “Module” will mean left module, and “ideal” will mean left ideal.

(I)(a) For each ideal $I$ of $A$, there is a subfunctor $I\cdot - : {}_A Mod \to {}_A Mod$ of the identity, where $I \cdot M = I M$ is the submodule generated by $\{ i m : i \in I, m \in M\}$.

For given a homomorphism $f: M \to N$, we have $f(i m) = i f(m) \in I N$ for each $i \in I$ and $m \in M$. Hence $f(I M) \subseteq I N$.

(I)(b) There is also a subfunctor $rad : {}_A Mod \to {}_A Mod$ of the identity, where $rad(M)$ is the intersection of all maximal submodules of $M$.

For given a homomorphism $f: M \to N$, let $K$ be a maximal submodule of $N$. Then $N/K$ is simple, so the composite $M \stackrel{f}{\to} N \to N/K$ is either zero or surjective, so $ker(M \stackrel{f}{\to} N \to N/K)$ is either $M$ or maximal, and in either case contains $rad(M)$. So $f(rad(M)) \subseteq K$. Hence $f(rad(M)) \subseteq rad(N)$.

(II)(a) $rad(A)M \subseteq rad(M)$ for all modules $M$.

For let $m \in M$. We have a homomorphism $-\cdot m: A \to M$, so by I(b), $rad(A)m \subseteq \rad(M)$.

(II)(b) $rad(A)P = rad(P)$ for all projective modules $P$.

First let $F$ be the free module on a set $S$. Write $(e_s)_{s \in S}$ for the canonical basis of $F$. Take $x = \sum_s x_s e_s \in rad(F)$, where all but finitely many $x_s$ are zero. For each $s \in S$, we have the $s$-projection map $F \to A$, so by I(b), $x_s \in rad(A)$. Hence $x \in rad(A)F$, giving $rad(F) \subseteq rad(A)F$.

Now let $P$ be any projective module. We can choose a free module $F$ and an epimorphism $\pi: F \to P$, which by projectivity has a section $\iota$. So $rad(P) = \pi\iota rad(P) \subseteq \pi rad(F) = \pi (rad(A) F) \subseteq rad(A) P,$ using (respectively) $\pi\iota = 1$, I(b), the previous paragraph, and I(a).

(III) The ideal $rad(A)$ of $A$ is nilpotent.

Since $A$ is finite-dimensional, we can choose $n \geq 0$ that minimizes the dimension of the $A$-module $rad(A)^n$. Suppose that dimension is nonzero. By finite-dimensionality again, $rad(A)^n$ has a maximal submodule, so $rad(rad(A)^n) \subset rad(A)^n$ (where $\subset$ means strict inclusion). But then $rad(A)^{n + 1} = rad(A) \cdot rad(A)^n \subseteq rad(rad(A)^n) \subset rad(A)^n$ by (II)(a), a contradiction.

(IV) Every nonzero projective $A$-module has a maximal ideal.

Let $P$ be a projective module with no maximal ideal. Then $P = rad(P) = rad(A)P$ by (II)(b), so $P = rad(A)^n P$ for all $n \geq 0$, so $P = 0$ by (III).

Posted by: Tom Leinster on October 11, 2014 5:10 PM | Permalink | Reply to this

### Re: The Atoms of the Module World

I’ve just updated my note, removing the mistake that Andrew identified and adding the argument that he suggested (in the form laid out in my last comment).

Posted by: Tom Leinster on October 13, 2014 2:01 PM | Permalink | Reply to this

### Re: The Atoms of the Module World

… but I seem to remember there’s a fun kind of matrix $A_{i j}$ you can form where you take two simple modules, say the $i$th one and the $j$th one, and count how many times the $j$th one shows up as a composition factor in the projective indecomposable corresponding to the $i$th one. Or something like that. Does this ring a bell?

Actually I think I’m distorting some idea that’s a bit fancier… but anyway, is there anything interesting to say about the matrix I described? Might it sometimes be symmetric, even when that’s not utterly obvious? (When our algebra is semisimple I guess it’s obvious!)

Posted by: John Baez on October 6, 2014 2:10 AM | Permalink | Reply to this

### Re: The Atoms of the Module World

Right — that’s the so-called Cartan matrix. As you said, its entries count how many times each simple module appears as a composition factor in each projective indecomposable.

I haven’t thought about the question of symmetry, but I do have something else interesting to say. It’s related to the magnitude/Euler characteristic of the category of projective indecomposables, which turns out to be a recognizable homological invariant of our algebra. That’s why I’m thinking about this stuff. But I’ll save it for another day.

Posted by: Tom Leinster on October 6, 2014 2:50 AM | Permalink | Reply to this

### Re: The Atoms of the Module World

Cartan matrix! Okay, those usually aren’t symmetric.

It’s gradually coming back to me. A really nice example of an algebra to illustrate this whole theory is the category algebra of a quiver, especially an ADE quiver. Once upon a time Jim Dolan and I studied this stuff…

Posted by: John Baez on October 9, 2014 1:45 AM | Permalink | Reply to this

### Re: The Atoms of the Module World

In algebraic geometry atoms are points. In the noetherian case, Gabriel’s reconstruction theorem shows that they correspond to the indecomposable injectives. In general, there is a theorem due to Rosenberg, completed by Gabber and made available by Brandenburg that gives a procedure to reconstruct a quasi-seprated scheme (i.e. with quasi-compact diagonal) from its abelian category of quasi-coherent sheaves. There, a kind of “atom” is used, namely that of spectral objects. The procedure continues up to giving a structure sheaf on the “spectrum” of the abelian category. Full details in http://arxiv.org/abs/1310.5978.

I am curious what kind of ringed spaces arise when an abelian category not arising from a quasi-seprated schemes is plugged into the construction.

Posted by: Leo Alonso on October 7, 2014 12:39 PM | Permalink | Reply to this

### Re: The Atoms of the Module World

I’m only just coming to this thread, and was immediately struck by “ergodic dynamical system” being referred to as an ‘atom’.

The only thing I half-know about ergodic systems (involving a measure space $X$ and a measure-preserving trasformation $T: X \to X$ satisfying a suitable condition) are the various ergodic theorems (see sections that start here and following). What you say about ‘atomistic’ is new to me and I’m not immediately seeing a connection. Could you enlighten me, please?

Posted by: Todd Trimble on October 13, 2014 3:17 AM | Permalink | Reply to this

### Re: The Atoms of the Module World

It’s entirely possible that I’m overextending myself, and I know very little about ergodic theory. All I meant was the following.

An ergodic system is a probability space $X$ together with a measure-preserving transformation $T: X \to X$, with the following “irreducibility” property: it has no completely invariant subspaces except for the trivial ones. That is, if $E$ is a measurable subset of $X$ with $T^{-1}E = E$ then $E$ has measure $0$ or $1$.

I don’t know to what extent this is in analogy with primes, connected spaces, etc. For instance, I don’t know whether arbitrary measure-preserving endomorphisms of probability/measure spaces can be “built out of” ergodic ones, whatever that might mean.

Posted by: Tom Leinster on October 13, 2014 11:16 AM | Permalink | Reply to this

### Re: The Atoms of the Module World

Assuming this structural point of view and taking in consideration pre-categorical and composition ontological theories , which describe sampling perspectives , oriented to form and shape in space-associated universalization ( as columns ) , may be ascertained in the article by Ziv Ran in Proc Lond Math Soc - 2006 , III , Ser 92 , n 3 , p 545 , considering Lie atom associated to Mx , the Hitchin connection and local geommetries in moduli spaces .

Posted by: Sabino Guillermo Echebarria Mendieta on October 13, 2014 11:08 AM | Permalink | Reply to this

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