Let $X$ be a scheme. Let $\mathcal{I}$ be a quasi-coherent ideal of $\mathcal{O}_X$. Let $Y = Supp(\mathcal{O}_X/\mathcal{I})$. Let $f\colon Y \rightarrow X$ be the canonical injection. Then how do we prove that $(Y, f^{-1}(\mathcal{O}_X/\mathcal{I}))$ is a scheme and $f$ is a closed immersion?
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Have you looked at [Hartshorne, Ch. II, Prop. 5.9]? – Zhen Lin Nov 03 '12 at 19:32
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@ZhenLin I've just looked at it. I think the proof has a gap we need to fill. Suppose $X = Spec(A)$ and $\mathcal{I} = \tilde I$. How do we prove that $f^{-1}(\mathcal{O}_X/\mathcal{I}) = \tilde {(A/I)}$? – Makoto Kato Nov 03 '12 at 20:02
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Let Spec $A$ be affine open in $X$. The restriction of $\mathcal I$ to Spec $A$ is the sheafification of some ideal $I$ of $A$ (by quasi-coherence). The sheaf $\mathcal O_X/\mathcal I$ restricts to the sheaf associated to $A/I$, and so its support, intersected with Spec $A$, is precisely the image of Spec $A/I$ in Spec $A$. Furthermore, the restriction of the sheaf attached to $A/I$ to Spec $A/I$ is precisely the structure sheaf of Spec $A/I$.
(This is a special case of a more general fact: if $M$ is an $A$-module which is annihilated by the ideal $I$ of $A$, then we can regard $M$ as an $A/I$-module too, and so we can get associated sheaves on both Spec $A$ and on Spec $A/I$. These are canonically identified via $f^{-1}$ and $f_*$.)
Since Spec $A/I$, with its structure sheaf, is an open subset of $(Y,f^{-1}(\mathcal O_X/\mathcal I))$, and $Y$ is covered by such open sets, we see that $(Y,(\mathcal O_X/\mathcal I))$ admits an open cover by affine schemes, and so is a scheme.
Matt E
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It seems there is a typo in your last sentence. You want to conclude that $(Y,(\mathcal{O}_X/\mathcal{I}))$ is a scheme (but you write "affine"). – Fredrik Meyer Nov 04 '12 at 11:00
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Why is that the "support intersected with Spec A is precisely the image of Spec $A/I$"? I came upon this question reading Hartshorne 5.9 where he similarly claims this. The image of Spec $A/I$ in Spec $A$ is $V(I)$ so we want to show that $V(I) = {p \mid (A/I)_p \ne 0}$. Did I translate that correctly into commutative algebra? – Future Dec 22 '15 at 06:26
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The following is a bit long for a comment.
Matt E wrote: "Furthermore, the restriction of the sheaf attached to $A/I$ to Spec $A/I$ is precisely the structure sheaf of Spec $A/I$."
Let me explain this. We can assume $X =$ Spec $A$. The canonical morphism $\mathcal{O}_X/\mathcal{I} \rightarrow f_*f^{-1}(\mathcal{O}_X/\mathcal{I})$ is an isomorphism by this result. Hence $\Gamma(D(f), \mathcal{O}_X/\mathcal{I})$ is canonically isomorphic to $\Gamma(D(f) \cap Y, f^{-1}(\mathcal{O}_X/\mathcal{I}))$ for $f \in A$. $\Gamma(D(f), \mathcal{O}_X/\mathcal{I}) = (A/I)_f = (A/I)_{\bar f}$, where $\bar f$ is the image of$f$ in $A/I$. $D(f) \cap Y = D(\bar f)$. Hence $f^{-1}(\mathcal{O}_X/\mathcal{I})$ is the structure sheaf of Spec $A/I$.
Makoto Kato
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It is thankful that this answer is still helpful in 20s.. But I'm stll stuck in one thing, in your arguement, why the underlying space $Y$ is the same as Spec$\frac{A}{I}$. (I understood now the structure sheaf is the same by your comment, but still wondering why the underlying top space is the same). – jingyey Nov 09 '20 at 12:56