# Fpqc coverings

On Mathoverflow Anton Geraschenko asks the following question:

Suppose F : Sch^{opp}→Set is a sheaf in the fpqc topology, has quasi-compact representable diagonal, and has an fpqc cover by a scheme. Must F be an algebraic space? That is, must F have an étale cover by a scheme?

The question isn’t well posed as the question does not specify _exactly_ what is meant by an “fpqc cover by a scheme”. Does it mean a morphism which is a surjection of sheaves in the fpqc topology? In that case you get counter examples by looking at ind schemes (for example the functor of morphisms A^1 —> A^1). Does it mean a flat morphism which is a surjection of sheaves in the fpqc topology? In that case, I think there is a counter example by taking an ind-scheme where the transition morphisms are flat closed immersions (I can explain but it isn’t interesting). Does it mean a flat, surjective, quasi-compact morphism? In this case it is more difficult to give a counter example, but I think I have one. Before I get into it, note that algebraic spaces in general do not have such coverings, so that the resulting category of “fpqc-spaces” does not contain the category of algebraic spaces.

Here is my idea for a counter example. (I’m having a kind of deja vu here, so it is perhaps somebody else’s idea? Please let me know if so.) Consider the functor F = (P^1)^∞, i.e., for a scheme T the value F(T) is the set of f = (f_1, f_2, f_3, …) where each f_i : T —> P^1 is a morphism. A product of sheaves is a sheaf, so F is a sheaf. The diagonal is representable: if f : T —> F and g : S —> F, then T ×_F S is the scheme theoretic intersection of the closed subschemes T ×_{f_i, P^1, g_i} S inside the scheme T × S. Consider U = (SL_2)^∞ with its canonical morphism U —> F. Note that U is an affine scheme. OK, and now you can show that the morphism U —> F is flat, surjective, and even open. Without giving all the details, if f : T —> F is a morphism, then you show that Z = T &times_F U is the infinite fibre product of the schemes Z_i = T ×_{f_i, P^1} SL_2 over T. Each of the morphisms Z_i —> T is surjective, smooth, and affine which implies the assertions. In particular, if F where an algebraic space it would be a quasi-compact and separated (by our description of fibre products over F) algebraic space. Hence cohomology of quasi-coherent sheaves would vanish above a certain cutoff (see Proposition Tag 072B and remarks preceding it). But clearly by taking O(-2,…,-2,0,…) on F = (P^1)^∞ we get a quasi-coherent sheaf whose cohomology is nonzero in an arbirary positive degree.

## 9 thoughts on “Fpqc coverings”

1. Brian Conrad on said:

Hey Johan, I had the same idea, with a different argument for why it cannot have an etale scheme cover (which I emailed to Anton and he may post on MO), and later I noticed another good property: instead of working with SL_2, work with the disjoint union of the two standard affine lines that cover P^1 to get a scheme cover of this beast which is even formally etale. Neat-o. A way I like to explain what “goes wrong” with the P^1 in contrast with A^1 variant is that it is a manifestation of the utter failure of Proj to be a good functor for injective graded ring maps in general.

• Johan on said:

Since you had the same idea is it possible we discussed (P^1)^∞ together at some point? Or that it is in some paper?

2. Brian Conrad on said:

Hmm, I don’t think we ever discussed it. While walking home I was just trying to make a non-stacky version of the stacky example [Spec(k)/Gal(k_s/k)] I’d made with Galois groups for the earlier MO question on this theme, and so the idea of G/H for “big” G and H naturally came to mind. Since semisimple groups don’t come in particularly large towers, due to the simply connected covers, taking big products seemed the only reasonable alternative, so product of SL_2’s as G and product of their upper Borels as H (to avoid affine quotient) seemed quite natural. But I didn’t want to “fpqc sheafify”, so directly considering the product of P^1’s is the thing to do.

I will guess that you probably were led to it in essentially the same way, since you also focus on the SL_2 viewpoint to make an fpqc scheme cover. Is my guess correct?

• Johan on said:

Nope, it was the realization that a (countable) product of quasi-compact and quasi-separated algebraic spaces is one of these “fpqc-spaces” that led me to it. I still wonder if maybe the geometric Langlands guys have thought this kind of example through since they often like to think about “big” things.

3. I’m having deja vu as well! Brian Conrad suggested the same counterexample; it took me a while to convince myself it should work, and I haven’t yet written it up for MO. He also pointed out that instead of taking U to be an infinite product of SL_2’s, you can use an infinite product of (A^1 ⊔ A^1)’s, in which case the U–>F is actually formally etale. However, it sounds like there’s a trick (that I haven’t understood) to see that U–>F is flat if you use SL_2’s. If you use (A^1 ⊔ A^1)’s, you have to roll up your sleeves and show that an infinite product of flat affine morphisms is flat (this basically reduces to the case of a finite product since any element of an infinite tensor product can use only a finite number of the factors).

The vanishing of cohomology trick is different, and definitely awesomer than what I had in mind. Essentially, I was hoping to get something like the following to work to show F is not an algebraic space. I think this is not what Brian had in mind, so I take credit for all errors. Suppose S–>F is an etale cover. As you said, F is quasi-compact, so S can be assumed affine. Then pulling back to U, using the fact that F has affine diagonal, keeping track of descent data, and using EGA IV section 8 magic for affine etale morphisms, show that S–>F must actually be obtained from an etale cover V of (P^1)^n (n finite) by base change along the projection F–>(P^1)^n. But then S is the product of V and (P^1)^∞. In particular, S contains a copy of P^1 as a closed subscheme, contradicting affineness.

• Whoops, I see I failed to refresh the page to see what happened before posting my comment.

• Brian Conrad on said:

Anton, the “trick” I had in mind (and perhaps Johan too?) to see the faithful flatness at a glance in the SL_2 setting is that an injection between Hopf algebras over a field is always faithfully flat (proved in Waterhouse’s book on affine group schemes, for example). That’s no longer available when using the covering built from an infinite product of disjoint union of two affine lines, hence the need to “roll up one’s sleeves” in that situation to get flatness in another way.

• Johan on said:

If Z_i —> T, i=1,2,3,… are flat and affine morphisms of schemes, then Z = Z_1 ×_T Z_2 ×_T … is representable by a scheme which is flat and affine over T. Proof: write Z as the limit of the finite fibre products and use that colimits of flat ring maps are flat.

• Brian Conrad on said:

Johan, initially we have a product map \prod Z_i –> \prod T_i with Z_i flat over T_i all affine over an affine base S (such as Spec(k)), so I suppose your point is that if we define T = \prod T_i and

Z’_i = Z_i x \prod_{j \ne i} T_j

then the T-scheme \prod Z_i with products over S is identified with \prod Z’_i where now the products over over T. Then your above simple limit argument applies.

That is much better than what I had in mind (hence the “roll up one’s sleeves” aspect), which was to express infinite tensor products in the source and target as limits of finite tensor products to reduce to a result on preservation of flatness when one is forming limits in both the source *and* target. That seemed to require an auxiliary flatness hypothesis on the transition maps, which is satisfied when working over a field. Your way is much better since it works even when the base isn’t a field.