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?


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.

  • $\begingroup$ 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 :) $\endgroup$
    – Manos
    Nov 6 '12 at 16:02
  • $\begingroup$ Dear @Manos, You're welcome! I'm happy to be able to help. $\endgroup$ Nov 6 '12 at 17:52

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