Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

References and background on universal homeomorphisms

Definition [EGA I (2d ed.) 3.8.1]. A morphism $f:V\to U$ is a universal homeomorphism if for any morphism $U'\to U$, the pullback $V\times_UU'\to U'$ is a homeomorphism.

Theorem [EGA IV 18.12.11]. A morphism is a universal homeomorphism if and only if it is surjective, integral, and radicial.

Theorem ["Topological invariance of the étale topos," SGA I Exp IX, 4.10 and SGA IV Exp. VIII, 1.1] If $f:V\to U$ is a universal homeomorphism, then the induced morphism $f:V_{\textrm{ét}}\to U_{\textrm{ét}}$ of the small étale topoi is an equivalence.

General examples. Any nilimmersion, any purely inseparable field extension (or any base change thereby), the geometric Frobenius of an $\mathbf{F}_p$-scheme [SGA V Exp. XIV=XV, § 1, No. 2, Pr. 2(a)].

Theorem. Suppose $X$ a reduced scheme with finitely many irreducible components. Denote by $X'$ its normalization. Then the natural morphism $X'\to X$ is a universal homeomorphism if and only if $X$ is geometrically unibranch.

Specific example. Suppose $k$ an algebraically closed field of characteristic $2$. Consider the subring $k[x^2,xy,y]\subset k[x,y]$. The induced morphism

$$\mathrm{Spec}k[x,y]\to\mathrm{Spec}k[x^2,xy,y]$$

is a universal homeomorphism.

Question

Do pushouts along universal homeomorphisms exist in the category of schemes?

In more detail. Suppose $f:V\to U$ a universal homeomorphism, and suppose $p:V\to W$ a morphism. Everything here is a scheme; I can assume $W$ quasicompact and quasiseparated, but I have no control over the map $V\to W$. Now of course I can construct the pushout $P$ of $V\to U$ along $V\to W$ as a locally ringed space with no trouble (just take the underlying space of $W$ along with the fiber product $O_W \times_{p_{\star}O_V}p_{\star}O_U$), but I can't show that $P$ is a scheme. Is it?

Thoughts

Of course the key point here is that $f$ is a universal homeomorphism, not just some run-of-the-mill morphism. So one can try to treat the cases where $f$ is schematically dominant or a nilimmersion separately.

Update

If $f$ is a nilimmersion, then I now see how to prove this completely. I still have no idea how to proceed the schematically dominant case.

[EDIT: I removed the additional question.]

share|improve this question
add comment

1 Answer

The answer to your question is yes in positive characteristic. Check Kollár's paper "Quotient Spaces Modulo Algebraic Groups" (Ann. of Math. 1997), Lem 8.4. The reason is that universal homeomorphisms behave as nil-immersions in positive characteristic (due to Frobenius). I think the answer is no in characteristic zero.

A general comment is that without assuming that one of the two morphisms in a push-out is a monomorphism the "correct" push-out should be a stack. This partly explains why most results on push-outs are when one of the maps is a monomorphism (e.g. push-outs of a finite map and a closed immersion, push-outs of an etale morphism and an open immersion).

share|improve this answer
    
Hi, David. Thanks for the answer! I know that between noetherian schemes, any finite universal homeomorphism factors a sufficiently high-powered Frobenius. In other words, every universal homeomorphism is a section up to Frobenius. I wrote about this in my Voevodsky space paper, before I learned that Kollar had the same theorem. The only proof I know really uses noetherian induction. Do you have something better? –  Clark Barwick Feb 10 '10 at 18:52
    
Also, for Kollar here, a pushout is any commutative square (satisfying some finiteness conditions); he uses the term "universal pushout" for the colimit. He only asserts that his diagram is a pushout in the weaker sense, and I don't see that pushouts (in the strong sense) along Frobenii exist in general. Is it true? –  Clark Barwick Feb 10 '10 at 18:55
    
A basic fact is that if the presumable push-out has the correct underlying set, topology and functions, then it is the categorical push-out. This is in the category of schemes where it is obvious (as everything are locally ringed spaces). More generally, this should hold in the category of algebraic space under suitable assumptions (e.g. to deal with non-locally separated spaces). In Kollár's case, the "pushout" is universally homeomorphic to one of the other involved schemes, so we get the categorical pushout (poss. non-noetherian) by adding some functions (but the space is the same). –  David Rydh Feb 11 '10 at 4:02
    
I'm sorry; I'm afraid I'm not following you here. Again, I know how to form the categorical pushout in the category LRS of locally ringed spaces; that's easy. And I know that if that LRS-pushout is a scheme, it will be the (categorical) pushout in schemes. But the question is exactly whether one can show that that LRS-pushout is a scheme. Put differently, how does one know that when you add functions to get to the LRS-pushout, what you're left with is still a scheme? Unfortunately, I still don't see how Kollár's construction addresses this question. –  Clark Barwick Feb 11 '10 at 16:00
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.