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

I have seen the following claim without proof in more than one paper, but it is sufficiently general that I suspect it is stated too strongly to be true:

Let $G$ be an affine group scheme (say, over a field of characteristic zero), let $X$ be a scheme (smooth over the same field), and let $P \to X$ be a $G_X$-torsor. If $Y$ is a scheme with a $G$-action, then the associated bundle $P \times^G Y$ is a scheme.

I can use fpqc descent to show that this is true when $Y$ is affine (or some other effective fpqc descent class), and I can use Zariski descent when $P$ is Zariski-locally trivial. In full generality, I suspect one can assemble known counterexamples of descent to falsify this claim, but I have been unable to do so.

Question: Is there a counterexample known? Failing that, is there a proof of the claim?

I'm somewhat more interested in the case where $G$ is connected, but here is a candidate that I don't know how to prove: Take $X$ to be a smooth curve with a nontrivial étale double cover $P$ (with $G$ constant of order 2), and set $Y$ to be Hironaka's 3-fold with involution (whose quotient sheaf is not a scheme).

share|improve this question
    
I don't know about counterexamples. But there are positive results in SGA 3 and in follow-up work by Raynaud. –  Jason Starr Aug 5 '11 at 12:45
    
Thanks. I looked in SGA3, but didn't find anything this strong. I'll have a look at Raynaud. –  S. Carnahan Aug 5 '11 at 15:47

1 Answer 1

up vote 7 down vote accepted

This is false; there is a counterexample (with $G$ a finite group of order $2$) in my notes on descent theory http://homepage.sns.it/vistoli/descent.pdf, subsection 4.4.2.

share|improve this answer
    
Thank you. Somehow, I had missed that part before. –  S. Carnahan Aug 8 '11 at 2:35

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.