I’ve been trying to re-understand some of the proofs in commutative and homological algebra. I never really had a good feeling for spectral sequences, but they seemed to crop up in purely theoretical proofs quite frequently. (Of course, they crop up in computations quite frequently, too.) After learning about derived categories it became possible to re-interpret many of these proofs. That’s what I’d like to do in this post.

Here is a toy example of a result, which does not use spectral sequences in its usual proof, but which can be interpreted in terms of the derived category.

Proposition 1Let be a local noetherian ring with residue field . Then a finitely generated -module such that is free.

Let’s try to understand the usual proof in terms of the derived category. Throughout, this will mean the *bounded-below* derived category of -modules: in other words, this is the category of bounded-below complexes of projectives and homotopy classes of maps. Any module can be identified with an object of by choosing a projective resolution.

So, suppose satisfies . Another way of saying this is that the *derived* tensor product

has no homology in negative degrees (it is in degree zero). Choose a free -module with a map which induces an isomorphism . Then we have that

by hypothesis. In particular, if is the cofiber (in ) of , then .

We’d like to conclude from this that is actually zero, or that : this will imply the desired freeness. Here, we have:

Lemma 2 (Derived Nakayama)Let have finitely generated homology. Suppose . Then .

*Proof:* In fact, suppose , and let be maximal such that . In this case, we find that by the usual Nakayama lemma; the highest homology group of a derived tensor product is the ordinary tensor product. This contradicts the assumption that .

It’s a standard fact that more is true: one only really needs to conclude that is free. We can still run most of the same proof through. We still get a map

inducing an isomorphism , and the map

now induces an isomorphism in degrees . It follows that the cofiber of has the property that is concentrated in degrees . The next (refined) version of the derived Nakayama lemma implies that itself is concentrated in these degrees, which means that is an isomorphism.

Lemma 3 (Derived Nakayama, II)Let have finitely generated homology. Suppose is concentrated in degrees . Then is concentrated in degrees .

“Concentrated in degrees ” means that the homology is nonzero only in these degrees. The proof is the same as before. Anyway, we find that we can recover the classical result that vanishing of is enough for freeness of a finitely generated module over a noetherian local ring.

**Regular ideals**

So far, this was a little silly: we’ve more or less taken usual proof of this result, and made no changes other than to add one or two slightly fancier words. Let’s try something a little different.

Let be an ideal in the noetherian ring . Then we have an isomorphism

Cup-product gives a map

When is generated by a regular sequence, this map is an isomorphism; in fact, when is generated by a regular sequence, we have an isomorphism of graded rings

which can be seen by using the Koszul resolution of and tensoring that with .

Proposition 4 (Quillen)The converse is also true: if is projective over and is a surjection, then is locally generated by a regular sequence.

In other words, for each containing , the ideal is generated by a regular sequence.

*Proof:* To see this, we may as well assume that is local and is contained in the maximal ideal. Let be a set of generators for over . We can form the Koszul complex , a complex of finitely generated free -modules together with a map

which is an isomorphism in degree zero. (We have graded most of the Koszul complex in negative degrees.)

Anyway, this induces a map

In degree zero, this is an isomorphism. In degree 1, this is the map , and hence an isomorphism. In degree two, the map is a surjection on homology by hypothesis.

In particular, if we form the cofiber of , we find that is concentrated in degrees : in particular, by an analog of the “derived Nakayama lemma” above, we find that itself is concentrated in degrees . Thus, is an isomorphism in degrees .

It now follows that the Koszul complex has no homology in degree : by a general lemma, it follows that is a regular sequence.

**Flatness criteria**

Here’s another example. Let be a local homomorphism of local noetherian rings, and let be a finitely generated -module. There are a whole bunch of criteria for when is flat over ; see for instance the CRing project’s chapter on them. I’d like to re-interpret some of them using the derived category.

Theorem 5 (Local criterion)is -flat if and only if .

*Proof:* To say that is -flat is to say that

is concentrated in degree zero, for any finitely generated -module . In fact, we just need to show that has no homology in degree for any finitely generated -module . What we are given is that has that property. Using filtrations, we find that

whenever has *finite length:* that is, whenever has a finite filtration whose subquotients are .

The idea is now to bootstrap to all finitely generated -modules. The key observation is that

where the completion (of the homology) is with respect to the -adic topology. In fact, to see this, one may replace by a complex of finitely generated projective -modules and drop the ‘s. In this case, it is just the ordinary tensor product, and the result becomes the exactness of completions.

So, anyway, if is any finitely generated -module, we find that

because each is of *finite length*. Since completion is faithfully flat, it follows that has no homology in degree .

As before, this example isn’t really anything new: all that happened was that the original proof was tweaked a little, so that some aspects were emphasized and others (such as the implicit use of Artin-Rees) de-emphasized.

May 17, 2012 at 11:47 pm

Hey you gave a great talk today. But the monad nonsense is a bit too general for me (and other people), do you mind writing something explain it a bit (or you may have done that already) in this blog?

Is it safe to think that as the simplicial resolution given by a pair of adjoint functors? (That’s all you needed for the bar resolution.)

May 22, 2012 at 6:42 pm

Hi, I could definitely say something about the bar resolution and simplicial methods shortly. (I found it pretty confusing myself.) The simplicial resolution just comes from a monad: whenever you have a pair of adjoint functors, you get a monad.