# Descent of locally free modules

Locally free modules do not satisfy descent for fpqc coverings. I have an example involving a countable “product” of affine curves, which I will upload to the stacks project soon.

But what about fppf descent? Suppose A —> B is a faithfully flat ring map of finite presentation. Let M be an A-module such that M ⊗_A B is free. Is M a locally free A-module? (By this I mean locally free on the spectrum of A.) It turns out that if A is Noetherian, then the answer is yes. This follows from the results of Bass in his paper on “big” projective modules. But in general I don’t know the answer. If you do know the answer, or have a reference, please email me.

# Nonzero kernel

Another fun algebra lemma: If R is a ring and φ : M —> N is a map of finite free R-modules with rank(M) > rank(N), then the kernel of φ is not zero.

# Finite fibres

Suppose that f : X —> Y is a morphism of projective varieties and y is a point of Y such that there are only finitely many points x_1, …, x_r in X mapping to y. Then there exists an affine open neighborhood V of y in Y such that f^{-1}(V) —> V is finite.

How do you prove this? Here is a fun argument. First you prove that f is a projective morphism, and hence we can generalize the statement to arbitrary projective morphism. This is good because then we can localize on Y and reach the situation where Y is affine. In this case X is quasi-projective and we can find an affine open U of X containing x_1, …, x_r, see Lemma Tag 01ZY. Then f(X \ U) is closed and does not contain y. Hence we can find a principal open V of Y such that f^{-1}(V) \subset U. In particular f^{-1}(V) = U ∩ f^{-1}(V) is a principal open of U, whence affine. Now f^{-1}(V) —> V is a projective morphism of affines. There is a cute argument proving that a universally closed morphism of affines is an integral morphism, see Lemma Tag 01WM. Finally, an integral morphism of finite type is finite.

Of course, the same thing is true for proper morphisms… see Lemma Tag 02UP.

# Dimension of varieties

This semester I am continuing my course on algebraic geometry. I wanted to list here the steps I used to get a useful dimension theory for varieties so that the next time I teach I can look it up:

1. Prove going up for finite ring maps (done last semester).
2. For a finite surjective morphism of schemes X —> Y you prove that dim(X) = dim(Y) using going up and the fact that the fibres are discrete.
3. Prove the Krull Hauptidealsatz: In a Noetherian ring a prime minimal over a principal ideal has height at most 1. For a proof see [E, page 232].
4. Generalize to longer sequences: In a Noetherian ring a prime minimal over (f_1, …, f_r) has height at most r. For a proof see [E, page 233].
5. If A is a Noetherian local ring and x ∈ m_A then dim(A/xA) ∈ {dim(A), dim(A) – 1} and is equal to dim(A) – 1 if and only if x is not contained in any of the minimal prime ideals of A. In particular if x is a nonzero divisor then dim(A/xA) = dim(A) – 1.
6. Prove that if A is a Noetherian local ring, then dim(A) is equal to the minimal number of elements generating an ideal of definition.
7. If Z is irreducible closed in a Noetherian scheme X show that codim(Z, X) is the dimension of O_{X, ξ} where ξ is the generic point of Z.
8. A closed subvariety Z of an affine variety X has codimension 1 if and only if it is an irreducible component of V(f) for some nonzero f ∈ Γ(X, O_X).
9. Prove Noether normalization.
10. If Z is a closed subvariety of X of codimension 1 show that trdeg_k k(Z) = trdeg_k k(X) – 1. This you do using Tate’s argument which you can find in Mumford’s red book: Namely you first do a Zariski shrinking to get to the situation where Z = V(f). Then you choose a finite dominant map Π : X —> A^d_k by Noether normalization. Then you let g = Nm(f) and you show that V(g) = Π(V(f)). Hence k(Z) is a finite extension of k(V(g)) and it is easy to show that k(V(g)) has transcendence degree d – 1.

At this point you know that if you have ANY maximal chain of irreducible subvarieties {x} = X_0 ⊂ X_1 ⊂ X_2 ⊂ … ⊂ X_d = X, then the transcendence degree drops by exactly 1 in each step. Therefore we see that not only is the dimension equal to the transcendence degree of the function field, but also each maximal chain has the same length. This implies that dim(Z) + codim(Z, X) = dim(X) for any irreducible closed subvariety Z and in particular it implies that dim(O_{X, x}) = dim(X) for each closed point x ∈ X.

Let me know if I neglected to mention a “biggish” step in the outline above.

What is missing in this account of the theory is the link between dimension of a Noetherian local ring A and the degree of the Hilbert polynomial of the graded algebra Gr_{m_A}(A). Which is just so cool! Oh well, you can’t do everything…

[E] Eisenbud, Commutative Algebra with a View Toward Algebraic Geometry.

# Conditions on diagonal not needed

In a recent contribution of Jonathan Wang to the stacks project we find the following criterion of algebraicity of stacks (see Lemma Tag 05UL):

If X is a stack in groupoids over (Sch/S)_{fppf} such that there exists an algebraic space U and a morphism u : U —> X which is representable by algebraic spaces, surjective, and smooth, then X is an algebraic stack.

In other words, you do not need to check that the diagonal is representable by algebraic spaces. The analogue of this statement for algebraic spaces is Lemma Tag 046K (for etale maps) and Theorem Tag 04S6 (for smooth maps).

The quoted result is closely related to the statement that the stack associated to a smooth groupoid in algebraic spaces is an algebraic stack (Theorem Tag 04TK). Namely, given u : U —> X as above you can construct a groupoid by taking R = U x_X U and show that X is equivalent to [U/R] as a stack. But somehow the statements have different flavors. Finally, the result as quoted above is often how one comes about it in moduli theory: Namely, given a moduli stack M we often already have a scheme U and a representable smooth surjective morphism u : U —> M. Please try this out on your favorite moduli problem!

# Universal flattening

In this post I talked a bit about flattening of morphisms. Meanwhile I have written some more about this in the stacks project which led to a change in definitions. Namely, I have formally introduced the following terminology:

1. Given a morphism of schemes X —> S we say there exists a universal flattening of X if there exists a monomorphism of schemes S’ —> S such that the base change X_{S’} of X is flat over S’ and such that for any morphism of schemes T —> S we have that X_T is flat over T if and only if T —> S factors through S’.
2. Given a morphism of schemes X —> S we say there exists a flattening stratification of X if there exists a universal flattening S’ —> S and moreover S’ is isomorphic as an S-scheme to the disjoint union of locally closed subschemes of S.

Of course the definition of “having a flattening stratification” this is a bit nonsensical, since we really want to know how to “enumerate” the locally closed subschemes so obtained. Please let me know if you think this terminology isn’t suitable.

Perhaps the simplest case where a universal flattening doesn’t exist is the immersion of A^1 – {0} into A^2. Currently the strongest existence result in the stacks project is (see Lemma Tag 05UH):

If f : X —> S is of finite presentation and X is S-pure then a universal flattening S’ —> S of X exists.

Note that the assumptions hold f is proper and of finite presentation. It is much easier to prove that a flattening stratification exists if f is projective and of finite presentation and I strongly urge the reader to always use the result on projective morphisms, and only use the result quoted above if absolutely necessary.

PS: I recently received a preprint by Andrew Kresch where, besides other results, he gives examples of cases where the universal flattening exists (he call this the “flatification”) but where there does not exist a flattening stratification.

# A challenge

Here is a challenge to an commutative algebraist out there. Give a direct algebraic proof of the following statement (see Lemma Tag 05U9):

Let A —> B be a local ring homomorphism which is essentially of finite type. Let N be a finite type B-module. Let M be a flat A-module. Let u : N —> N be an A-module map such that N/m_AN —> M/m_AM is injective. Then u is A-universally injective, N is a B-module of finite presentation, and N is flat as an A-module.

To my mind it is at least conceivable that there is a direct proof of this (not using the currently used technology). It wouldn’t directly imply all the wonderful things proved by Raynaud and Gruson but it would go a long way towards verifying some of them. In particular, it would give an independent proof of the following result (see Theorem Tag 05UA):

Let f : X —> S be a finite type morphism of schemes. Let x ∈ X with s = f(x) ∈ S. Suppose that X is flat over S at all points x’ ∈ Ass(X_s) which specialize to x. Then X is flat over S at x.

This result is used in an essential way in the main result on universal flattening which I will explain in the next blog post.

# Purity

Let f : X —> S be a morphism of finite type. The relative assassin Ass(X/S) of X/S is the set of points x of X which are embedded points of their fibres. So if f has reduced fibres or if f has fibres which are S_1, then these are just the generic points of the fibres, but in general there may be more. If T —> S is a morphism of schemes then it isn’t quite true that Ass(X_T/T) is the inverse image of Ass(X/S), but it is almost true, see Remark Tag 05KL.

Definition: We say X is S-pure if for any x ∈ Ass(X/S) the image of the closure {x} is closed in S, and if the same thing remains true after any etale base change.

Clearly if f is proper then X is pure over S. If f is quasi-finite and separated then X is S-pure if and only if X is finite over S (see Lemma Tag 05K4). It turns out that if S is Noetherian, then purity is preserved under arbitrary base change (see Lemma Tag 05J8), but in general this is not true (see Lemma Tag 05JK). If f is flat with geometrically irreducible (nonempty) fibres, then X is S-pure (see Lemma Tag 05K5).

A key algebraic result is the following statement: Let A —> B be a flat ring map of finite presentation. Then B is projective as an A-module if and only if Spec(B) is pure over Spec(A), see Proposition Tag 05MD. The current proof involves several bootstraps and starts with proving the result in case A —> B is a smooth ring map with geometrically irreducible fibres.

I challenge any commutative algebraist to prove this statement without using the language of schemes. You will find another challenge in the next post.

# Update

This morning I finished incorporating the material from sections 1 through 4 of the paper by Raynaud and Gruson into the stacks project. Most of it is  in the chapter entitled More on Flatness. There is a lot of very interesting stuff contained in this chapter and I will discuss some of those results in the following blog posts. Note that I previously blogged about this paper here, here, here, here, here, here, and here.

It turns out that it was kind of a mistake to do this, as the payoff wasn’t as great as I had hoped for. Moreover, I don’t think you are going to find the chapter easy to read. So the benefit of having done this is mainly that I now understand this material very well, but I’m not sure if it is going to help any one else. Maybe the lesson is that I should stick to the strategy I have used in the past: only prove those statements that are actually needed to build foundations for algebraic stacks. This will sometimes require us to go back and generalize previous results but (1) we can do this as the stacks project is a “live” book, and (2) it is probably a good idea to rewrite earlier parts in order to improve them anyway.

The long(ish) term plan for what I want work on for the stacks project now is the following: I will first add a discussion of Hilbert schemes/spaces/stacks parameterizing finite closed subscheme/space/stacks. I will prove just enough so I can prove this theorem of Artin: A stack which has a flat and finitely presented cover by a scheme is an algebraic stack. A preview for the argument is a write-up of Bhargav Bhatt you can find here.

Curiously, Artin’s result for algebraic spaces is already in the stacks project: It is Theorem TAG 04S6. It was proved by a completely different method, namely using a Keel-Mori type argument whose punch line is explained on the blog here.

# Update

In the last two and and a half weeks I’ve updated the material on derived categories and derived functors. You can now find this material in a new chapter entitled Derived Categories.

The original exposition defined the bounded below derived category as the homotopy category of bounded below complexes of injectives. This is actually a very good way to think about derived categories if you are mainly interested in computing cohomology of sheaves on spaces and/or sites. On the other hand, it does not tell you which problem derived functors really solve. Let’s discuss this a bit more in the setting of sheaves of modules on a ringed space (X, O_X). I will assume you know how to define cohomology of sheaves by injective resolutions, left derived functors by projective resolutions, you have heard that D(A) is complexes up to quasi-isomorphism, but you don’t yet know exactly why one makes this choice.

Let F : Mod(O_X) —> A be a right exact functor from the abelian category of O_X-modules into an abelian category A. The category Mod(O_X) usually does not have enough projectives. Hence it wouldn’t work to define the bounded above derived category in terms of bounded above complexes of projectives. You could still make this definition but there wouldn’t be a functor from the category of modules into it and hence it wouldn’t suffice to compute left derived functors of F. In fact, what should be the “left derived functors” of F in this setting? Grothendieck, Verdier, and Deligne’s solution is the following: Let M be an O_X-module. Consider the category of all resolutions

… —> K^{-1} —> K^0 —> M —> 0

where K^i is an arbitrary O_X-module. For any such resolution we can consider the complex

F(K^*) = ( … —> F(K^{-1}) —> F(K^0) —> 0 )

in the abelian category A. We say that LF is defined at M if and only if the system of all F(K^*) is essentially constant up to quasi-isomorphism, i.e., essentially constant in the bounded above derived category D^-(A). If one can choose K^* so that F(K^*) is actually equal to this essentially constant value, then one says that K^* computes LF(M). These definitions are motivated by the case where there do exist enough projectives: in that case one shows that given a projective resolutions P^* there always exists a map P^* —> K^*, hence the system is essentially constant with value F(P^*). We say an object M is left acyclic for F if M computes LF. Note that this makes sense without knowing that LF is everywhere defined! It turns out that LF is defined for any M which has a resolution K^* where all K^n are left acyclic for F and that in this case F(K^*) is the value of RF(M) in D^-(A). For example, why is one allowed to use bounded above flat resolutions to compute tors? The reason is that flat modules are left acyclic for tensoring with a sheaf (this is not a triviality — it is something you have to prove; hint: use Lemma Tag 05T9).

I started rewriting the material on derived categories because I gave 2 lectures about derived categories and derived functors in my graduate student seminar, and I wanted to understand the details. Let me know if you find any typos, errors, or lack of clarity. Also, there is still quite a bit missing, for example a discussion of derived categories of dg-modules would be cool.