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Let $X$ be a scheme and $\mathcal{F}$ be a sheaf on $X$ which is torsion $\mathcal{O}_X-$module (i.e., every local section is annihilated by an element of the ring $\mathcal{O}_X(U)$) or nilpotent (i.e., a power of any local section is zero). When can we say that $H^i(\mathcal{F})$ or $H^i(Hom_X(\mathcal{F},\mathcal{O}_X))$ vanishes for $i>0$. For example if $X$ is a curve then does the first cohomology groups given above vanish?

Let us take for example $X$ is a non-reduced curve in $\mathbb{P}^3$ ($X_{red}$ not smooth). Can we say that $H^i(Hom_X(I_X/I_X^2,I_{X_{red}}/I_X))$ equal to zero for $i$ equal to $0$ or $1$? (Here $I_X, I_{X_{red}}$ denotes respectively the ideal sheaves of $X, X_{red}$ in $\mathbb{P}^3$.)

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  • $\begingroup$ If $X$ is a curve and $\mathcal F$ is torsion and coherent, then its support is 0-dimensional, so the higher cohomology automatically vanishes. If $\mathcal F$ is torsion and quasicoherent, it is a filtered colimit of torsion coherent sheaves so its higher cohomology still vanishes. $\endgroup$
    – Eric Wofsey
    Commented Dec 3, 2012 at 16:26
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    $\begingroup$ @Naga: you should think about this question a little more. The notion of torsion does not include the notion of nilpotent for the simple reason that $\mathcal F$ may not have a multiplication. Perhaps you want to look at $\mathcal O_X$-algebras? Torsion and nilpotent are still two different notions, but perhaps you can figure out what it is that you are asking. Furthermore, in your particular example you are asking if the $0^{\mathrm{th}}$ or $1^{\mathrm{st}}$ cohomology of a sheaf on a curve vanishes. It is unlikely that anything like that could happen. If it is torsion... (cont'd) $\endgroup$
    – Sándor Kovács
    Commented Dec 3, 2012 at 18:14
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    $\begingroup$ @Eric Wofsey & Sandor Kovacs: even if $F$ is torsion it might have support on the whole curve. Consider the ${\cal O}_X$-module $(\epsilon)$ on the curve $C[\epsilon]=C\times_k{\rm Spec}\, k[\epsilon]/\epsilon^2$. This has support $=C$. $\endgroup$
    – Damian Rössler
    Commented Dec 3, 2012 at 18:33
  • $\begingroup$ @Damian: yes, of course, you're right, I actually wanted to say that, but then got lost in my own neverending comment. $\endgroup$
    – Sándor Kovács
    Commented Dec 3, 2012 at 19:28
  • $\begingroup$ @Naga: I also forgot to add this: What is $I_C$ in your specific example/question? Is it just $I_X$? $\endgroup$
    – Sándor Kovács
    Commented Dec 3, 2012 at 19:29

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As Sandor points out, you must think about it and ask a more precise question. I felt that you are interested in space curves, so let me give a quick example. Let $C\subset\mathbb{P}^3$ be any smooth curve of genus at least 1 and let $I$ denote its ideal sheaf. Then take any surjective map $I/I^2\to L$ where $L\in\mathrm{Pic}\ C$ (one can always do this if $\deg L>>0$ since $I/I^2$ is a rank two bundle on $C$) and take the push out to get a scheme $X$ supported on $C$ with the following exact sequence, $0\to L\to \mathcal{O}_X\to\mathcal{O}_C\to 0$. Then $L$ is torsion in your sense as an $\mathcal{O}_X$ module and $Hom_X(L,\mathcal{O}_X)=\mathcal{O}_C$. In particular, its $H^1$ is not zero.

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  • $\begingroup$ You mean $L \in \operatorname{Pic}(C)$? $\endgroup$
    – red_trumpet
    Commented Sep 2, 2021 at 6:30
  • $\begingroup$ @red_trumpet yes, will correct it. $\endgroup$
    – Mohan
    Commented Sep 2, 2021 at 15:04

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