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 1 Let ${(A, \mathfrak{m})}$ be a local noetherian ring with residue field ${k}$. Then a finitely generated ${A}$-module ${M}$ such that ${\mathrm{Tor}_i(M, k) = 0, i > 0}$ 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 ${D^-(A)}$ of ${A}$-modules: in other words, this is the category of bounded-below complexes of projectives and homotopy classes of maps. Any module ${M}$ can be identified with an object of ${D^-(A)}$ by choosing a projective resolution.

So, suppose ${M}$ satisfies ${\mathrm{Tor}_i(M, k) = 0, i > 0}$. Another way of saying this is that the derived tensor product

$\displaystyle M \stackrel{\mathbb{L}}{\otimes} k$

has no homology in negative degrees (it is ${M \otimes k}$ in degree zero). Choose a free ${A}$-module ${P}$ with a map ${P \rightarrow M}$ which induces an isomorphism ${P \otimes k \simeq M \otimes k}$. Then we have that

$\displaystyle P \stackrel{\mathbb{L}}{\otimes} k \simeq M \stackrel{\mathbb{L}}{\otimes} k$

by hypothesis. In particular, if ${C}$ is the cofiber (in ${D^-(A)}$) of ${P \rightarrow M}$, then ${C \stackrel{\mathbb{L}}{\otimes} k = 0}$.

We’d like to conclude from this that ${C}$ is actually zero, or that ${P \simeq M}$: this will imply the desired freeness. Here, we have:

Lemma 2 (Derived Nakayama) Let ${C \in D^-(A)}$ have finitely generated homology. Suppose ${C \stackrel{\mathbb{L}}{\otimes} k = 0}$. Then ${C = 0}$.

Proof: In fact, suppose ${C \neq 0}$, and let ${i}$ be maximal such that ${H_i(C) \neq 0}$. In this case, we find that ${H_i (C \stackrel{\mathbb{L}}{\otimes} k) = H_i(C) \otimes_A k \neq 0}$ by the usual Nakayama lemma; the highest homology group of a derived tensor product is the ordinary tensor product. This contradicts the assumption that ${C \stackrel{\mathbb{L}}{\otimes} k \simeq 0}$. $\Box$

It’s a standard fact that more is true: one only really needs ${\mathrm{Tor}_1(M, k) = 0}$ to conclude that ${M}$ is free. We can still run most of the same proof through. We still get a map

$\displaystyle P \rightarrow M$

inducing an isomorphism ${P \otimes k \rightarrow M \otimes k}$, and the map

$\displaystyle P \stackrel{\mathbb{L}}{\otimes} k \rightarrow M \stackrel{\mathbb{L}}{\otimes} k$

now induces an isomorphism in degrees ${-1, 0}$. It follows that the cofiber ${C}$ of ${P \rightarrow M}$ has the property that ${C \stackrel{\mathbb{L}}{\otimes} k}$ is concentrated in degrees ${\leq -2}$. The next (refined) version of the derived Nakayama lemma implies that ${C}$ itself is concentrated in these degrees, which means that ${P \rightarrow M}$ is an isomorphism.

Lemma 3 (Derived Nakayama, II) Let ${C \in D^-(A)}$ have finitely generated homology. Suppose ${C \stackrel{\mathbb{L}}{\otimes} k}$ is concentrated in degrees ${\leq m}$. Then ${C}$ is concentrated in degrees ${\leq m}$.

“Concentrated in degrees ${\leq m}$” 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 ${\mathrm{Tor}_1}$ 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 ${I \subset A}$ be an ideal in the noetherian ring ${A}$. Then we have an isomorphism

$\displaystyle \mathrm{Tor}_1^A(A/I, A/I) \simeq I/I^2.$

Cup-product gives a map

$\displaystyle \bigwedge^2 (I/I^2) \rightarrow \mathrm{Tor}_2^A(A/I, A/I).$

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

$\displaystyle \bigwedge^\bullet (I/I^2) \simeq \mathrm{Tor}_\bullet^A(A/I, A/I),$

which can be seen by using the Koszul resolution of ${A/I}$ and tensoring that with ${A/I}$.

Proposition 4 (Quillen) The converse is also true: if ${I/I^2}$ is projective over ${A/I}$ and ${\bigwedge^2 (I/I^2) \rightarrow \mathrm{Tor}_2^A(A/I, A/I)}$ is a surjection, then ${I}$ is locally generated by a regular sequence.

In other words, for each ${\mathfrak{p} \in \mathrm{Spec} A}$ containing ${I}$, the ideal ${I_{\mathfrak{p}} \subset A_{\mathfrak{p}}}$ is generated by a regular sequence.

Proof: To see this, we may as well assume that ${A}$ is local and ${I}$ is contained in the maximal ideal. Let ${x_1, \dots, x_n}$ be a set of generators for ${I/I^2}$ over ${A/I}$. We can form the Koszul complex ${K(x) = K(x_1, \dots, x_n)}$, a complex of finitely generated free ${A}$-modules together with a map

$\displaystyle K(x) \rightarrow A/I$

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

Anyway, this induces a map

$\displaystyle K(x) \stackrel{\mathbb{L}}{\otimes} A/I \rightarrow A/I \stackrel{\mathbb{L}}{\otimes} A/I.$

In degree zero, this is an isomorphism. In degree 1, this is the map ${I/I^2 \rightarrow \mathrm{Tor}_1(A/I, A/I)}$, and hence an isomorphism. In degree two, the map is a surjection on homology by hypothesis.

In particular, if we form the cofiber ${C}$ of ${K(x) \rightarrow A/I}$, we find that ${C \stackrel{\mathbb{L}}{\otimes} A/I}$ is concentrated in degrees ${\leq -2}$: in particular, by an analog of the “derived Nakayama lemma” above, we find that ${C}$ itself is concentrated in degrees ${\leq -2}$. Thus, ${K(x) \rightarrow A/I}$ is an isomorphism in degrees ${-1, 0}$.

It now follows that the Koszul complex ${K(x)}$ has no homology in degree ${-1}$: by a general lemma, it follows that ${x_1, \dots, x_n}$ is a regular sequence. $\Box$

Flatness criteria

Here’s another example. Let ${(R, \mathfrak{m}) \rightarrow (S, \mathfrak{n})}$ be a local homomorphism of local noetherian rings, and let ${M}$ be a finitely generated ${S}$-module. There are a whole bunch of criteria for when ${M}$ is flat over ${R}$; 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) ${M}$ is ${R}$-flat if and only if ${\mathrm{Tor}_1^R(M, R/\mathfrak{m}) =0 }$.

Proof: To say that ${M}$ is ${R}$-flat is to say that

$\displaystyle M \stackrel{\mathbb{L}}{\otimes}_R N$

is concentrated in degree zero, for any finitely generated ${R}$-module ${N}$. In fact, we just need to show that ${M \stackrel{\mathbb{L}}{\otimes}_R N}$ has no homology in degree ${-1}$ for any finitely generated ${R}$-module ${N}$. What we are given is that ${M \stackrel{\mathbb{L}}{\otimes} R/\mathfrak{m}}$ has that property. Using filtrations, we find that

$\displaystyle H_{-1}(M \stackrel{\mathbb{L}}{\otimes}_R N) = 0$

whenever ${N}$ has finite length: that is, whenever ${N}$ has a finite filtration whose subquotients are ${R/\mathfrak{m}}$.

The idea is now to bootstrap to all finitely generated ${R}$-modules. The key observation is that

$\displaystyle \varprojlim H_i (M \stackrel{\mathbb{L}}{\otimes}_R N/\mathfrak{m}^t N) = \widehat{H_i(M \stackrel{\mathbb{L}}{\otimes}_R N)}$

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

So, anyway, if ${N}$ is any finitely generated ${R}$-module, we find that

$\displaystyle \varprojlim H_{-1} (M \stackrel{\mathbb{L}}{\otimes}_R N/\mathfrak{m}^t N) = \widehat{H_{-1}(M \stackrel{\mathbb{L}}{\otimes}_R N)} = 0,$

because each ${N/\mathfrak{m}^t N}$ is of finite length. Since completion is faithfully flat, it follows that ${M \stackrel{\mathbb{L}}{\otimes}_R N}$ has no homology in degree ${-1}$.

$\Box$

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.