Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $Y$ be a closed subscheme of the scheme $X$ and let $i : Y \rightarrow X$ be the inclusion morphism. Then the sheaf of ideals of $Y$ is defined to be the kernel of the morphism of sheafs $i^{\#}: \mathcal{O}_X \rightarrow i_* \mathcal{O}_Y$. Denote the sheaf of ideals by $J$. Then $J$ is an $\mathcal{O}_X$-module, since for any open set $U$ in $X$, $J(U)$ is an ideal of $\mathcal{O}_X(U)$. Is it also true that $J$ is $\mathcal{O}_Y$-module? How about $J/J^2$?

Remark: This question is motivated by the observation in Hartshorne (Remark 8.9.1,p. 175). In particular (using the notation pertaining to the reference) if $f:X \rightarrow Y$ is a morphism of schemes and $\Delta: X \rightarrow X \times_Y X$ is the diagonal morphism, then $\Delta(X)$ is locally closed, i.e. closed in an open subset $W$ of $X \rightarrow X \times_Y X$. Let $J$ be the sheaf of ideals of $\Delta(X)$ inside $W$. Then $J$ is an $\mathcal{O}_W$-module. Why is $J/J^2$ an $\mathcal{O}_{\Delta(X)}$-module?

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

The sheaf $J$ is a sheaf on $X$, whereas $\mathscr{O}_Y$ is a sheaf on $Y$, so no, $J$ is not generally an $\mathscr{O}_Y$-module (unless, e.g., $X=Y$). The way to get an $\mathscr{O}_Y$-module from $J$ is to pull it back: $i^*J$ is a sheaf of modules on $\mathscr{O}_Y$. In general, given any $\mathscr{O}_X$-module, the naturally associated $\mathscr{O}_Y$-module is the pullback via $i$.

Regarding Hartshorne's remarks, he is abusing notation (a little bit, in my opinion). Since $\Delta:X\rightarrow X\times_YX$ is an immersion, there is an open subscheme $U$ of the target of $\Delta$ such that $\delta$ factors as a closed immersion $X\rightarrow U$ followed by the open immersion $U\hookrightarrow X\times_YX$. In fact there is a largest open subscheme for which this factorization is possible, namely the complement in $X\times_YX$ of $\overline{\Delta(X)}\setminus\Delta(X)$. Anyway, choosing an open subscheme $U$ such that a factorization of the type above exists, we get a quasi-coherent ideal sheaf $J$ of $\mathscr{O}_U$, and $J/J^2$ is a quasi-coherent $\mathscr{O}_U$-module. It is clear that as an $\mathscr{O}_U$-module, $J/J^2$ is killed by $J$. Because $j:X\rightarrow U$ is a closed immersion (identifying $X$ with $\Delta(X)$), it is a general fact that pushforward along $j$ is fully faithful on quasi-coherent sheaves with essential image the quasi-coherent $\mathscr{O}_U$-modules killed by $J$ (the ideal sheaf). So when Hartshorne says $J/J^2$ can be regarded as a sheaf of modules on $\mathscr{O}_X$ (he actually says $\mathscr{O}_{\Delta(X)}$ but I'm sticking with $X$), he really means there is a unique $\mathscr{O}_X$-module $F$ which, when pushed forward to $U$, is isomorphic to $J/J^2$. In fact $F=j^*(J/J^2)$. So Hartshorne is basically leaving out the $j^*$.

Also, the open subscheme $U$ used to get the sheaf on $X$ (which is $\Omega_{X/Y}^1$) doesn't matter in the end. For an arbitrary immersion (not just the diagonal), the sheaf $j^*(J/J^2)$ obtained in the manner above is called the conormal sheaf of the immersion. For a proof of the assertion about pushforward of quasi-coherent sheaves along an immersion, see the section called ``Closed immersions and quasi-coherent sheaves" in the Stacks Project Chapter on morphisms of schemes.

A down to earth manifestation of this situation is the following fact. If $A$ is a ring and $J$ is an ideal of $A$, then an $A/J$-module is the ``same thing" as an $A$-module killed by $J$, and if $M$ is an $A$-module killed by $J$, then the natural map $M\otimes_A(A/J)\rightarrow M$ is an isomorphism. More precisely, the restriction functor from $A/J$-modules to $A$-modules (which corresponds to pushforward along $\mathrm{Spec}(A/J)\rightarrow\mathrm{Spec}(A)$) is fully faithful with essential image the $A$-modules killed by $J$, and its quasi-inverse (on the essential image) is given by base change, $M\rightsquigarrow M\otimes_A(A/J)$ (which corresponds to pullback along $\mathrm{Spec}(A/J)\rightarrow\mathrm{Spec}(A)$). In fact, the assertion I alluded to above about quasi-coherent sheaves pretty much amounts to this statement about modules.

share|improve this answer
    
Thank you very much, that's a very rich answer and it might take a while to understand. I might be back with some questions :) –  Manos Nov 6 '12 at 16:02
    
Dear @Manos, You're welcome! I'm happy to be able to help. –  Keenan Kidwell Nov 6 '12 at 17:52
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.