# Understanding the proof of Hartshorne Prop. II.7.13.b

Suppose $$X$$ be a Noetherian scheme and $$\mathscr{I}$$ be a coherent sheaf of ideal on $$X$$. Let $$Y$$ be an closed subscheme of $$X$$ corresponding to $$\mathscr{I}$$.

Hartshorne said that it holds $$\mathscr{I}|U \cong \mathscr{O}_U$$ for open complement $$U=X-Y$$ in the proof of proposition.

I tried it to prove for the affine case, $$X = \operatorname{Spec} A$$ and $$\mathscr{I} \cong \tilde{I}$$ for some ideal $$I$$ of $$A$$.

However, it seems nonsense since RHS is about (a sheaf of) ring and LHS is about ideal.

I guess that $$I$$ is of rank 1, but it can't be guaranteed. We just know it is finitely generated.

So, how could I understand this statement?

EDIT : Let investiate the Affine case deeply.

Suppose $$X = \operatorname{Spec} A$$, $$\mathscr{I} \cong \tilde{I}$$, $$Y \cong V(I) \cong \operatorname{Spec} A/I$$. Since $$\mathscr{I}|U$$ be an $$\mathscr{O}|U$$-module, it's enough to see that $$\mathscr{I}|U$$ and $$\mathscr{O}_U$$ has same stalk at any point in $$U$$ with $$\mathscr{I}$$.

For $$\mathfrak{p} \notin V(I)$$, we can get $$\mathscr{O}_{U,\mathfrak{p}} \cong A_\mathfrak{p}$$ and $$(\mathscr{I}|U)_{,\mathfrak{p}} \cong I_{\mathfrak{p}}$$. Then, is $$I_{\mathfrak{p}} \cong A_{\mathfrak{p}}$$ as $$\mathcal{O}_U$$-module? How could we guarantee this?

• The two sheaves are to be both viewed as sheaves of $\mathscr{O}_U$-modules (the RHS being, of course, the free $\mathscr{O}_U$-module). – Wojowu Apr 10 '20 at 16:39
• @Wojowu Then, is $\mathscr{I}$ a free module of rank $1$? I mean, is it an invertible sheaf? – ChoMedit Apr 10 '20 at 16:57
• $\mathscr I$ itself need not be. However, when restricted to $U$, it is. – Wojowu Apr 10 '20 at 17:04
• @Wojowu Okay, then it makes sense. I think it's enough to see the stalks of each sheaf. Am I right? – ChoMedit Apr 10 '20 at 17:15
• @Wojowu I tried to prove that they are locally isomorphic as $\mathscr{O}_U$-module, but it seems still not clear.. – ChoMedit Apr 11 '20 at 5:08

Intuitively, you can see ot this way: for any closed subset $$Y\subset X$$, there is a maximal sheaf of quasi-coherent sheaves defining $$Y$$: this is the reduced structure on $$Y$$. This sheaf $$\mathcal{I}$$ is precisely the sheaf of functions vanishing on $$Y$$. If $$U = X\setminus Y$$, this condition is empty on $$U$$, and we get $$\mathcal{I}_{|U} = {O_X}_{|U}$$.
More generally, if $$\mathcal{I}$$ is any quasi-coherent sheaf defining $$Y$$, we just need to show that $$\forall x \in U$$, $$1\in \mathcal{I}_x$$. But saying that $$x$$ is not in $$Y$$ is the same as saying that there is a function $$f$$ of $$\mathcal{I}$$ defined on a neighbourhood of $$x$$, which does not vanish at $$x$$. This function is invertible in the stalk $$O_{X,x}$$, so $$\mathcal{I}_x = O_{X,x}$$.