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

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 $\mathbb{P}^3$.)

share|improve this question
    
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. –  Eric Wofsey Dec 3 '12 at 16:26
1  
@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) –  Sándor Kovács Dec 3 '12 at 18:14
1  
@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 $=C$. –  Damian Rössler Dec 3 '12 at 18:33
    
@Damian: yes, of course, you're right, I actually wanted to say that, but then got lost in my own neverending comment. –  Sándor Kovács Dec 3 '12 at 19:28
    
@Naga: I also forgot to add this: What is $I_C$ in your specific example/question? Is it just $I_X$? –  Sándor Kovács Dec 3 '12 at 19:29
show 2 more comments

1 Answer

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 $L\in\mathrm{Pic}\ X$ (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.

share|improve this answer
add comment

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.