Timeline for When will the pushforward of a structure sheaf still be a structure sheaf?
Current License: CC BY-SA 3.0
15 events
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| Apr 6, 2017 at 7:10 | comment | added | David Holmes | @R.vanDobbendeBruyn Hi Remy, you're quite right, thanks for pointing this out. In case it is helpful for someone, a reference for the statement `For a flat projective [or proper] morphism, it suffices to have geometrically connected and geometrically reduced fibres' is [FGA explained, ex. 9.3.11] - solution towards the end of the book. I will try to think more about the non-flat case at some point... | |
| Apr 5, 2017 at 5:36 | comment | added | R. van Dobben de Bruyn | @DavidHolmes: actually there are much more pathological counterexamples. For example, consider the relative (wrt $\bar{\mathbb F}_p$) Frobenius morphism $F \colon \mathbb P^1_{\bar{\mathbb F}_p} \to \mathbb P^1_{\bar{\mathbb F}_p}$. It is projective with geometrically connected fibres, and both sides are smooth geometrically integral varieties. Yet $F_*\mathcal O_X \neq \mathcal O_Y$, since it is a degree $p$ finite flat morphism. For a flat projective morphism, it suffices to have geometrically connected and geometrically reduced fibres, but I'm not sure what to do without flatness. | |
| Oct 17, 2016 at 12:43 | comment | added | David Holmes | Ps. I know this is an old question, but it seems quite a few people (including students) are still reading it, hence the comments. | |
| Oct 17, 2016 at 12:37 | comment | added | David Holmes | After reading the answers below, I want to add a small warning that some of them are false unless one also assumes that $X$ is integral, or at least reduced. For example, if $Y = Spec k[t]$ and $X = Spec k[t,x]/(x^2,xt)$ then $f:X \to Y$ is finite hence projective, and has connected fibres, and $Y$ is normal, but it does not hold that $f_*\mathcal O_X = \mathcal O_S$. To fix this, one checks easily that $\mathcal O_X(X)$ is a domain if $X$ is integral, then modifies the statements below by adding the assumption that $X$ be integral in many places. | |
| May 1, 2011 at 17:00 | vote | accept | YOURS | ||
| May 1, 2011 at 17:00 | vote | accept | YOURS | ||
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| May 1, 2011 at 16:59 | vote | accept | YOURS | ||
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| Apr 30, 2011 at 3:10 | history | edited | Charles Staats | CC BY-SA 3.0 |
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| Apr 29, 2011 at 8:13 | comment | added | Laurent Moret-Bailly | In the first claim, the assumption on Picard groups is not needed: if $f_*\mathcal{O}_X$ is invertible, then the natural map $\mathcal{O}_Y\to f_*\mathcal{O}_X$ is an isomorphism. Working locally, this just means that if $g:A\to B$ is a ring homomorphism such that $B$ is a free $A$-module of rank $1$, then $f$ is an isomorphism. If $b\in B$, multiplication by $b$ is $A$-linear, hence is multiplication by some $f(a)$. In particular $b1_B=f(a)1_B$, hence $b=f(a)$. | |
| Apr 29, 2011 at 3:19 | answer | added | roy smith | timeline score: 25 | |
| Apr 29, 2011 at 2:49 | answer | added | Sándor Kovács | timeline score: 4 | |
| Apr 28, 2011 at 21:14 | answer | added | Karl Schwede | timeline score: 17 | |
| Apr 28, 2011 at 16:34 | answer | added | J.C. Ottem | timeline score: 13 | |
| Apr 28, 2011 at 15:49 | history | edited | YOURS | CC BY-SA 3.0 |
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| Apr 28, 2011 at 15:42 | history | asked | YOURS | CC BY-SA 3.0 |