Timeline for Weil restriction of abelian schemes along finite étale (resp. finite locally free) morphisms
Current License: CC BY-SA 3.0
9 events
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| Apr 15, 2013 at 18:21 | comment | added | Simon Pepin Lehalleur | @Jason Starr: which paper of Murre are you referring to ? Is it the Bourbaki seminar on unramified functors ? @nosr: because of Raynaud's theorem, most of the algebraic spaces (e.g. semi-abelian algebraic spaces of constant rank) I need are schemes. On the other hand, I can work with algebraic spaces, since I am ultimately interested in the complex of sheaves and the associated objects in triangulated categories of mixed motives. It is more a matter of not being familiar with the technology. | |
| Apr 14, 2013 at 4:52 | comment | added | user28172 | Do you seek a proof that avoids algebraic spaces entirely? That seems unlikely to be possible, and if one is going to be using algebraic spaces then the Faltings-Chai proof is an extremely natural one (building off of the normal case in a clever way). For your purposes with Deligne 1-motives, is it a problem to work throughout with algebraic spaces (which are well-suited to questions related to quotients in algebraic geometry)? | |
| Apr 11, 2013 at 0:53 | comment | added | Jason Starr | @Simon: Cf. Murre. | |
| Apr 10, 2013 at 20:12 | history | edited | Simon Pepin Lehalleur | CC BY-SA 3.0 |
added 158 characters in body; added 2 characters in body; edited title
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| Apr 10, 2013 at 20:05 | comment | added | Simon Pepin Lehalleur | @Emerton: this shows that the restriction is an algebraic space, but how do you prove that it is a scheme ? Indeed, I see that 1) is overkill, even in the case where $S'/S$ is only finite locally free. I will edit the question accordingly. | |
| Apr 10, 2013 at 19:21 | answer | added | Angelo | timeline score: 3 | |
| Apr 10, 2013 at 18:07 | comment | added | Emerton | Dear Simon, Can you check this by etale descent? Namely, if you let $S''$ over $S$ be a Galois etale cover lying over $S'$, and pull-back the restriction of scalars to $S''$, then it should just a product of the appropriate number of copies of the original $A$. So after this etale pull-back we do get an abelian scheme. Hopefully descent now applies to conclude that the restriction of scalars itself was an ab. scheme. Regards, Matthew | |
| Apr 10, 2013 at 18:03 | history | edited | Michael Stoll |
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| Apr 10, 2013 at 16:49 | history | asked | Simon Pepin Lehalleur | CC BY-SA 3.0 |