Timeline for Is tensoring with a module representable iff it is locally free of finite rank?
Current License: CC BY-SA 2.5
17 events
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jun 15, 2011 at 12:27 | comment | added | naf | I think Corollary 2 of the article "Nitsure, Nitin: Representability of Hom implies flatness. Proc. Indian Acad. Sci. Math. Sci. 114 (2004), no. 1, 7–14" gives a proof in the case $A$ is noetherian. (He considers the functor on all schemes as in EGA but I think the proof should also work for the functor restricted to affine schemes.) | |
| Dec 6, 2009 at 20:20 | answer | added | user2035 | timeline score: 1 | |
| Dec 6, 2009 at 19:24 | comment | added | user2035 | If the canonical map Hom(N,A)⊗B → Hom(N,B) is an isomorphism for all A-algebras B, this is true in particular for A-algebras of the type A⊕T for some A-module T, so it is true for Hom(N,A)⊗T → Hom(N,T). Exactness of Hom(N,T) for variable T is equivalent to N projective. | |
| Dec 6, 2009 at 18:20 | comment | added | Jonathan Wise | By the way, an interesting way to think of more general quasi-coherent sheaves as geometric objects is with $\mathbf{A}^1$-linear Picard stacks (and higer Picard stacks). | |
| Dec 6, 2009 at 11:22 | comment | added | Andrew Critch | @a-fortiori, I'm not sure I understand your argument about projectivity... it seems like the arrows go the wrong way to make use of the $A\oplus B$ algebras you suggest, but maybe I'm missing something... | |
| Dec 6, 2009 at 11:16 | history | edited | Andrew Critch | CC BY-SA 2.5 |
clarification "by a scheme"
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| Dec 6, 2009 at 10:14 | answer | added | Jonathan Wise | timeline score: 1 | |
| Dec 6, 2009 at 9:37 | history | edited | Andrew Critch | CC BY-SA 2.5 |
title fix
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| Dec 6, 2009 at 9:30 | history | edited | Andrew Critch | CC BY-SA 2.5 |
reformulated
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| Dec 6, 2009 at 9:15 | history | edited | Andrew Critch | CC BY-SA 2.5 |
corrections
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| Dec 6, 2009 at 8:52 | answer | added | Kevin Buzzard | timeline score: 0 | |
| Dec 6, 2009 at 8:27 | answer | added | Kevin Buzzard | timeline score: 0 | |
| Dec 6, 2009 at 8:21 | comment | added | user2035 | It is not sufficient for M to have a pre-dual: V(N) represents Hom_B(N⊗B,B)=Hom(N,B), not Hom(N,A)⊗B. In fact, if Hom(N,A)⊗B=Hom(N,B), this functor is exact in the variable A-module B (consider algebras of the form B'=A⊕B with B²=0), so N is projective. | |
| Dec 6, 2009 at 7:18 | history | edited | Andrew Critch | CC BY-SA 2.5 |
elaboration
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| Dec 6, 2009 at 7:10 | history | asked | Andrew Critch | CC BY-SA 2.5 |