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 ${C = 0}$. (more…)

Ultimately, we are headed towards a characterization of formal smoothness for reasonable morphisms (e.g. the types one encounters in classical algebraic geometry): we want to show that they are precisely the flat morphisms whose fibers are smooth varieties. This will be a much more usable criterion in practice (formal smoothness is given by a somewhat abstract lifting property, but checking that a concrete variety is smooth is much easier).  This is the intuition between smoothness: one should think of a flat map is a “continuously varying” family of fibers, and one wishes the fibers to be regular. This corresponds to the fact from differential topology that a submersion has submanifolds as its fibers.

It is actually far from obvious that a formally smooth (and finitely presented) morphism is even flat. Ultimately, the idea of the proof is going to be write the ring as a quotient of a localization of a polynomial ring. The advantage is that this auxiliary ring will be clearly flat, and it will also have fibers that are regular local rings.  In a regular local ring, we have a large supply of regular sequences, and the point is that we will be able to lift the regularity of these sequences from the fiber to the full ring.

Thus we shall use the following piece of local algebra.

Theorem Let ${(A, \mathfrak{m}) \rightarrow (B, \mathfrak{n})}$ be a local homomorphism of local noetherian rings. Let ${M}$ be a finitely generated ${B}$-module, which is flat over ${A}$.

Let ${f \in B}$. Then the following are equivalent:

1. ${M/fM}$ is flat over ${A}$ and ${f: M \rightarrow M}$ is injective.
2. ${f: M \otimes k \rightarrow M \otimes k}$ is injective where ${k = A/\mathfrak{m}}$.

This is a useful criterion of checking when an element is ${M}$-regular by checking on the fiber. That is, what really matters is that we can deduce the first statement from the second. (more…)

I’ve not been a very good MaBloWriMo participant this time around. Nonetheless, coursework does tend to sap the time and energy I have for blogging. I have been independently looking as of late at the formal function theorem in algebraic geometry, which can be phrased loosely by saying that the higher direct images under a proper morphism of schemes commute with formal completions. This is proved in Hartshorne for projective morphisms by first verifying it for the standard line bundles and then using a (subtle) exactness argument, but EGA III.4 presents an argument for general proper morphisms. The result is quite powerful, with applications for instance to Zariski’s main theorem (or at least a weak version thereof), and I would like to say a few words about it at some point, at least after I have a fuller understanding of it than I do now. So I confess to having been distracted by algebraic geometry.

For today, I shall continue with the story on the Koszul complex, and barely begin the connection between Koszul homology and regular sequences. Last time, we were trying to prove:

Proposition 24 Let ${\lambda: L \rightarrow R, \lambda': L' \rightarrow R}$ be linear functionals. Then the Koszul complex ${K_*(\lambda \oplus \lambda')}$ is the tensor product ${K_*(\lambda) \otimes K_*(\lambda')}$ as differential graded algebras.

So in other words, not only is the algebra structure preserved by taking the tensor product, but when you think of them as chain complexes, ${K_*(\lambda \oplus \lambda') \simeq K_*(\lambda) \oplus K_*(\lambda')}$. This is a condition on the differentials. Here ${\lambda \oplus \lambda'}$ is the functional ${L \oplus L' \stackrel{\lambda \oplus \lambda'}{\rightarrow} R \oplus R \rightarrow R}$ where the last map is addition.

So for instance this implies that ${K_*(\mathbf{f}) \otimes K_*(\mathbf{f}') \simeq K_*(\mathbf{f}, \mathbf{f}')}$ for two tuples ${\mathbf{f} = (f_1, \dots, f_i), \mathbf{f}' = (f'_1, \dots, f'_j)}$. This implies that in the case we care about most, catenation of lists of elements corresponds to the tensor product.

Before starting the proof, let us talk about differential graded algebras. This is not really necessary, but the Koszul complex is a special case of a differential graded algebra.

Definition 25 A differential graded algebra is a graded unital associative algebra ${A}$ together with a derivation ${d: A \rightarrow A}$ of degree one (i.e. increasing the degree by one). This derivation is required to satisfy a graded version of the usual Leibnitz rule: ${d(ab) = (da)b + (-1)^{\mathrm{deg} a} a (db) }$. Moreover, ${A}$ is required to be a complex: ${d^2=0}$. So the derivation is a differential.

So the basic example to keep in mind here is the case of the Koszul complex. This is an algebra (it’s the exterior algebra). The derivation ${d}$ was immediately checked to be a differential. There is apparently a category-theoretic interpretation of DGAs, but I have not studied this.

Proof: As already stated, the graded algebra structures on ${K_*(\lambda), K_*(\lambda')}$ are the same. This is, I suppose, a piece of linear algebra, about exterior products, and I won’t prove it here. The point is that the differentials coincide. The differential on ${K_*(\lambda \oplus \lambda')}$ is given by extending the homomorphism ${L \oplus L' \stackrel{\lambda \oplus \lambda'}{\rightarrow} R}$ to a derivation. This extension is unique. (more…)

Thanks to all who responded to the bleg yesterday. I’m still haven’t completely decided on the topic owing to lack of time (I actually wrote this post last weekend), but the suggestions are interesting. My current plan is, following Omar’s comment, to look at Proofs from the Book tomorrow and pick something combinatorial or discrete-ish, like the marriage problem or Arrow’s theorem. I think this will be in the appropriate spirit and will make for a good one-hour talk.

4. ${\mathrm{Ext}}$ and depth

One of the first really nontrivial facts we need to prove is that the lengths of maximal ${M}$-sequences are all the same. This is a highly useful fact, and we shall constantly use it in arguments (we already have, actually). More precisely, let ${I \subset R}$ be an ideal, and ${M}$ a finitely generated module. Assume ${R}$ is noetherian.

Theorem 12 Suppose ${M}$ is a f.g. ${R}$-module and ${IM \neq M}$. All maximal ${M}$-sequences in ${I}$ have the same length. This length is the smallest value of ${r}$ such that ${\mathrm{Ext}^r(R/I, M) \neq 0}$.

I don’t really have time to define the ${\mathrm{Ext}}$ functors in any detail here beyond the fact that they are the derived functors of ${\hom}$. So for instance, ${\mathrm{Ext}(P, M)=0}$ if ${P}$ is projective, and ${\mathrm{Ext}(N, Q) = 0}$ if ${Q}$ is injective. These ${\mathrm{Ext}}$ functors can be defined in any abelian category, and measure the “extensions” in a certain technical sense (irrelevant for the present discussion).

So the goal is to prove this theorem. In the first case, let us suppose ${r = 0}$, that is there is a nontrivial ${R/I \rightarrow M}$. The image of this must be annihilated by ${I}$. Thus no element in ${I}$ can act as a zerodivisor on ${M}$. So when ${r = 0}$, there are no ${M}$-sequences (except the “empty” one of length zero).

Conversely, if all ${M}$-sequences are of length zero, then no element of ${I}$ can act as a nonzerodivisor on ${M}$. It follows that each ${x \in I}$ is contained in an associated prime of ${M}$, and hence by the prime avoidance lemma, that ${I}$ itself is contained in an associated prime ${\mathfrak{p}}$ of ${M}$. This prime avoidance argument will crop up quite frequently.

(I don’t know why these <br >’s appear below. If anyone with better HTML knowledge than I could explain what I’m doing wrong, I’d appreciate it!)

Today, we will continue with our goal of understanding some aspects of commutative algebra, by defining depth.

0.3. Depth

Constructing regular sequences sequences is a useful task. We often want to ask how long we can make them subject to some constraint. For instance,

Definition 8
Suppose ${I}$ is an ideal such that ${IM \neq M}$. Then we define the ${I}$-depth of ${M}$ to be the maximum length of a maximal ${M}$-sequence contained in ${I}$. When ${R}$ is a local ring and ${I}$ the maximal ideal, then that number is simply called the depth of ${M}$.

The depth of a proper ideal ${I \subset R}$ is its depth on ${R}$.

The definition is slightly awkward, but it turns out that all maximal ${M}$-sequences in ${I}$ have the same length. So we can use any of them to compute the depth.

The first thing we can prove using the above machinery is that depth is really a “geometric” invariant, in that it depends only on the radical of ${I}$. (more…)

Regular sequences don’t necessarily behave well with respect to permutation or localization under additional hypotheses. However, in all cases they behave well with respect to taking powers. The upshot of this is that the invariant called depth that we will soon introduce is invariant under passing to the radical. We shall deduce this from the following easy fact.

Lemma 4 Suppose we have an exact sequence of ${R}$-modules

$\displaystyle 0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0.$

Suppose the sequence ${x_1, \dots, x_n \in R}$ is ${M'}$-regular and ${M''}$-regular. Then it is ${M}$-regular.

The converse is not true, of course.

Proof: Morally, this is the snake lemma. For instance, the fact that multiplication by ${x_1}$ is injective on ${M', M''}$ implies by the snake diagram that ${M \stackrel{x_1}{\rightarrow} M}$ is injective. However, we don’t a priori know that a simple inductive argument on ${n}$ will work to prove this. The reason is that it needs to be seen that quotienting each term by ${(x_1, \dots, x_{n-1})}$ will preserve exactness. However, a general fact will tell us that this is indeed the case. See below. Anyway, this general fact now lets us induct on ${n}$. If we assume that ${x_1, \dots, x_{n-1}}$ is ${M}$-regular, we need only prove that ${x_{n}: M/(x_1, \dots, x_{n-1})M \rightarrow M/(x_1, \dots, x_{n-1})}$ is injective. (It is not surjective or the sequence would not be ${M''}$-regular.) But we have the exact sequence by the next lemma,

$\displaystyle 0 \rightarrow M'/(x_1 \dots x_{n-1})M' \rightarrow M/(x_1 \dots x_{n-1})M \rightarrow M''/(x_1 \dots x_{n-1})M'' \rightarrow 0$

and the injectivity of ${x_n}$ on the two ends implies it at the middle by the snake lemma. (more…)

I’m just going to start MaBloWriMo here. Anyway, as Qiaochu observes in the comments, National Blog Writing Month (as opposed to National Novel Writing Month) was in October. Before I start, I think it might be appropriate to post my half-complete notes on commutative algebra from a course I’m taking. Since I’m not going to be self-contained, they might be a helpful reference for readers.

For the next few posts, we shall always assume that all rings are commutative and noetherian. Commutativity is always needed. Noetherianness won’t be needed for now, but it will soon become indispensable. So let ${R}$ be a ring, and ${M}$ an ${R}$-module. We want to talk about the definition of a regular sequence on ${M}$. This is going to be a sequence of elements of ${R}$ that act “independently” on ${M}$ in some sense. We are going to make this precise below when we interpret a consequence of this condition via associated graded modules. This definition is basically the groundwork for everything that follows, as you need it for the definitions of notions such as Cohen-Macaulayness and regularity. Let us start by stating the definition.

Definition 1 A sequence ${x_1, \dots, x_n \in M}$ is ${M}$-regular (or is an ${M}$-sequence if for each ${k \leq n}$, ${x_k}$ is a nonzerodivisor on the ${R}$-module ${M/(x_1, \dots, x_{k-1}) M}$ and also ${(x_1, \dots, x_n) M \neq M}$.

So ${x_1}$ is a nonzerodivisor on ${M}$, by the first part. That is, the homothety ${M \stackrel{x_1}{\rightarrow} M}$ is injective. The last condition is also going to turn out to be necessary for us. (more…)