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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 $f^{\#}: \mathcal{O}_X \rightarrow f_* \mathcal{O}_Y$?

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    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

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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.