It turns out that the result mentioned in this post is especially useful in theoretical considerations. For example consider the following statement

Given an 1-morphism \cX —> \cY of stacks in groupoids which is representable by algebraic spaces such that \cY is an algebraic stack, then \cX is an algebraic stack.

Proof: pick a scheme Y and a surjective smooth morphism Y —> \cY. By assumption the 2-fibre product Y x_{\cY} \cX is representable by an algebraic space X. The projection map X —> \cX is surjective and smooth as a base change and we win by the result mentioned above.

Of course the result can be proved in other ways as well, but it is quite pleasing how short the argument above is. This kind of thing is especially helpful because we intend to prove many results of this kind!

[Edit March 08: Here are some links to the result mentioned above and its improvement suggested by David Rydh in the comment below.]

Let f : X —> Y be a morphism over a base B. Let P be a property of morphisms. We often want to know that

1. there is a maximal open W of B such that the restriction f_W : X_W —> Y_W of f has property P, and
2. formation of W commutes with arbitrary base change B’ —> B.

Of course this usually isn’t the case without further assumptions on X,Y,f, and B. One of the reasons this type of result is useful, is that you can check whether a point b of B is in W by looking at the base change of the morphism f to a morphism f_b : X_b —> Y_b of schemes (or algebraic spaces or algebraic stacks) over the point b.

A well known and useful case is the following result

If X is proper, flat, of finite presentation over B, Y is proper over B, and P = “being an isomorphism”, then 1 and 2 hold. 

I recently added this to the stacks project for relative maps of algebraic spaces, see Lemma Tag 05XD. When you analyze the proof you find two more basic results that lead to the above. The first is that

If X is proper over B, Y is separated over B, and P = “being a closed immersion”, then 1 and 2 hold. 

see Lemma Tag 05XA. This first result is in some sense elementary (although its proof in the current exposition is not). The second is that

If X is proper, flat, of finite presentation over B, Y is locally of finite type over B, and P = “being flat”, then 1 and 2 hold. 

see Lemma Tag 05XB. The current proof of this second result uses the “critère de platitude par fibres” which is nontrivial. Does anybody know how to prove the result on the locus where f is an isomorphism without appealing to this criterion?

Let S be a scheme. There are many ways to turn the category of schemes over S into a site, but some of the things you can do lead to the same topos, i.e., the category of sheaves are identical. In this case we say these sites define the same topology. Here are some examples of comparisons of topologies:

1. The smooth topology and the etale topology are the same. See Lemma Tag 055V.
2. The fppf topology is the same as the one you get by considering fppf coverings {T_i —> T} such that each T_i —> T is locally quasi-finite. See Lemma Tag 0572.
3. The topology generated by Zariski coverings and {f : T —> S} with f surjective finite locally free is finer than the etale topoloy, see Lemma Tag 02LH and Remark Tag 02LI.
4. The fppf topology is the same as the one generated by Zariski coverings and finite surjective locally free morphisms. See Lemma Tag 05WN.

Somehow I hadn’t realized 4 earlier. What made me think of it today was this comment by David Rydh.

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.

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.

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.

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.

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!

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.

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.