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?
Best Answer
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}$.