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Let $X$ be an open subscheme of an affine scheme $\operatorname{Spec} A$, let $f : X \to \operatorname{Spec} A$ be the inclusion, and let $\mathscr{I}$ be a quasicoherent ideal sheaf on $X$. Since $\mathscr{O}_X = f^{-1} \mathscr{O}_{\operatorname{Spec} A}$, there is a canonical ring homomorphism $\mathscr{O}_{\operatorname{Spec} A} \to f_* \mathscr{O}_X$. Let $\mathscr{J} \subseteq \mathscr{O}_{\operatorname{Spec} A}$ be the preimage of $f_* \mathscr{I} \subseteq f_* \mathscr{O}_X$. Clearly, $\mathscr{J}$ is an ideal sheaf, and moreover $\mathscr{I} = f^{-1} \mathscr{J}$ (as subsheaves of $\mathscr{O}_X$).

Question. Is $\mathscr{J}$ quasicoherent?

This is certainly true when $X$ is $\operatorname{Spec} A[S^{-1}]$ for some finitely-generated multiplicative set $S$: the question is then a simple matter of commutative algebra. On the other hand, at the level of topological spaces, every closed subset of $X$ (affine or not) can be obtained as the preimage of some closed subset of $\operatorname{Spec} A$, but unfortunately the map from ideal sheaves to closed subsets is not necessarily a bijection.

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Dear Zhen, I'm not sure that you get notified about the edit, so just in case: I added a counter-example to my answer. Regards, –  Matt E Jun 19 '13 at 13:22

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up vote 7 down vote accepted

If $f$ is quasi-compact and quasi-separated, then $f_*$ preseves quasi-coherence, and so $f_*\mathscr I$ is quasi-coherent, as is $f_* \mathscr O_X$. Thus $\mathscr J$ is a fibre product of quasi-coherent sheaves (the fibre product of $f_*\mathscr I$ and $\mathscr O_{\mathrm{Spec} A}$ over $f_*\mathscr O_X$), and so is quasi-coherent.

More geometrically, the ideal $\mathscr I$ cuts out a closed subscheme $Z$ of $X$. The composite $Z \hookrightarrow X \buildrel f \over \to \mathrm{Spec} A$ is quasi-compact and separated (provided that the second map is), and so has a well-defined scheme-theoretic image. The ideal $\mathscr J$ is then the quasi-coherent ideal sheaf cutting out this scheme-theoretic image.

Off the top of my head, I'm not sure about the case when $f$ is not quasi-compact and quasi-separated.

(Actually, I guess the map is separated, being an open immersion, so the issue, if there is one, is quasi-compactness.)

Additional remarks: If $\mathscr I$ is radical (i.e. the closed subscheme of $X$ it cuts out is reduced), then $\mathscr J$ is just the ideal sheaf cutting out the closure of its zero locus (with the induced reduced structure), and so is quasi-coherent.

Also, if we let $J$ denote the global sections of $\mathscr J$, this is an ideal in $A$, and if $\mathscr J$ is quasi-coherent, then it is the ideal sheaf attached to $J$. In any case, we may base-change the whole situation from $A$ to $A/J$, and so assume that $\mathscr I$ is nilpotent, and that the only global sections of $\mathscr J$ are zero.

So the question (about whether there exists a counterexample) becomes: can we find an affine scheme Spec $A$, and a non-zero quasi-coherent nilpotent ideal sheaf $\mathscr I$ on an open subset of $X$, so that the only element of $A$ restricting to a section of $\mathscr I$ on $X$ is $0$?

(Added later:) Okay, here is a counterexample.

We need Spec $A$ with an open subset $X$ so that $X$ is not quasi-compact (so we will set things up so $X$ is the union of countable number of connected components $X_n$) and a quasi-coherent nilpotent ideal sheaf $\mathscr I$ on $X$ (which will be given by a quasi-coherent nilpotent ideal sheaf $\mathscr I_n$ on each $X_n$) so that there are no non-zero elements of $A$ giving a section of $\mathscr I$ (so if we let $I_n$ denote the ideal of elements in $A$ that restrict to elements of $\mathscr I_n$ on $X_n$, we need $\bigcap_n I_n = 0$).

Take $A = \mathbb C[x_1,x_2,\ldots]/(x_i x_j^{i+1}, i \neq j).$

Let $O$ be the point $x_i = 0$ ("the origin") and let $X = $ Spec $A \setminus \{ O \}.$ Then $X$ is the union of the various "thickened lines" $X_n$, where $X_n$ is the locus where $x_n \neq 0,$ and hence $x_i^{n+1} = 0$ for $i \neq n$. Let $\mathscr I_n$ be the nilpotent ideal sheaf on $X_n$ generated by $x_i^n$ ($i \neq n$). Then $I_n$, the preimage of $\mathscr I_n$ in $A$, is equal to $(x_i^n, i \neq n)$, and the intersection of all these $I_n$ is indeed equal to $0$.

(I'm pretty sure this is a standard example.)

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