7
$\begingroup$

Let $Y$ be a closed subscheme of a scheme $X$ and let $i:Y \rightarrow X$ be the inclusion morphism. Then the ideal sheaf of $Y$ is defined to be the kernel of the morphism of sheafs $i^{\#}: \mathcal{O}_X \rightarrow i_* \mathcal{O}_Y$ (Hartshorne p.115). My question is: why is the ideal sheaf defined for a closed subscheme, and not more generally, i.e. for any morphism of schemes $f:Y \rightarrow X$ as the kernel of the morphism of sheafs $f^{\#}: \mathcal{O}_X \rightarrow f_* \mathcal{O}_Y$?

  • 0
    In general, $i_*\mathcal O_Y$ is an $\mathcal O_X$-module, and need not be a sheaf of ideals. It is of course still possible to consider the kernel.2012-11-05
  • 0
    @Andrew: I am sorry, i may not be understanding your point. I can see that $i_*\mathcal{O}_Y$ is an $\mathcal{O}_X$-module. Why would it be a sheaf of ideals? How is this related to my question?2012-11-05
  • 0
    Dear Manos, sorry for the confusion. That should have said "quotient by a sheaf of ideals."2012-11-05
  • 0
    I see now, thanks. The point is that when $Y$ is a closed subscheme, $f^{\#}(U)$ is surjective, for a general $f$ this is not true, right?2012-11-05
  • 0
    That's right. We like to talk about ideal sheaves, which correspond naturally to closed subschemes. We can talk about kernels of morphisms of arbitrary $\mathcal O_X$-modules, but their geometric meaning is not quite so obvious (although these are ideals, and so do also define closed subschemes).2012-11-05
  • 0
    Ok that's clarifying. If you could make it into an answer i will happily vote and accept it. Thanks.2012-11-05
  • 0
    By the way, since you mentioned it, what is the geometric meaning of the quotient of a sheaf by a sheaf of ideals?2012-11-05
  • 0
    When you start with an ideal sheaf $I$, the closed subscheme $Y$ you cook up from $I$ can be thought of as the vanishing locus of $I$. One can then naturally consider the quotient sheaf $O_X/I$ as the natural set of functions on $Y$. (But of course, since $O_X$ and $I$ are sheaves on $X$, the quotient is really the PUSHFORWARD of the sheaf of functions on $Y$) If you still feel confused, try your favorite example of closed subscheme in the affine space.2012-11-05

1 Answers 1

4

A closed subscheme of $X$ corresponds to a closed subspace $|Y|$ together with a quasicoherent sheaf of ideals $\mathcal I_Y$ of $\mathcal O_X.$ In this situation, the sheaf of ideals is naturally identified as the kernel of the sheaf morphism $\mathcal O_X\to f_*\mathcal O_Y\cong\mathcal O_X/\mathcal I_Y.$

More generally, any morphism of schemes $f:Y\to X$ comes with a sheaf morphism $f^\sharp:\mathcal O_X\to f_*\mathcal O_Y,$ but this need not be surjective, and the geometric meaning of the kernel may not be obvious (we are now considering $\operatorname{Ann}(f_*\mathcal O_Y)$). For a very simple example, consider the structure morphism $f:\mathbb A^1_k\to\operatorname{Spec}(k)$ induced by $f^\sharp:k\to k[x]$ over a field $k.$ The sheaf morphism $f^\sharp$ has trivial kernel, so the corresponding closed subscheme is $\operatorname{Spec}(k)$ which corresponds to the support of $k[x]$ as an $\mathcal O_{\operatorname{Spec}(k)}$-module.

Have a look at Hartshorne Exercise II.5.6 for more details.