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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?

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

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  • $\begingroup$ It seems right. I'll check it. $\endgroup$ – ChoMedit Apr 11 '20 at 10:39
  • $\begingroup$ Okay. I think this is right. It's just a simple argument if we know the quotient of stalk is the stalk of quotient. $\endgroup$ – ChoMedit Apr 11 '20 at 11:06

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