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Let $X$ be a scheme, $p : Z\to X$ a closed immersion, $\mathcal{F}$ a locally free sheaf of modules on $Z$ of finite rank.

Assume both $\mathcal{O}_Z$ and $\mathcal{O}_X$ are coherent sheaves of $\mathcal{O}_Z$, resp. $\mathcal{O}_X$-modules.

Is $p_*\mathcal{F}$ finitely presented, as an $\mathcal{O}_X$-module?

Of course, it is not locally free. It is not so clear that it should be finitely presented. Already for $\mathcal{F} = \mathcal{O}_Z$, is this true?

Remark.

(1) I do not want to assume $X$ is Noetherian.

(2) Given the assumption on $\mathcal{O}_Z$ and $\mathcal{O}_X$, the question is asking whether $p_*\mathcal{F}$ is coherent.

Improve this question
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  • $\begingroup$ No, that is not true. Let $X$ be $\text{Spec}\ k[x_1,x_2,x_3,\dots]$ and let $Z$ be the closed subscheme associated to the ideal $\langle x_1, x_2, x_3, \dots \rangle.$ $\endgroup$
    – Jason Starr
    Commented May 9, 2018 at 13:32
  • $\begingroup$ I missed the hypothesis that $X$ is a coherent $\mathcal{O}_X$-module. $\endgroup$
    – Jason Starr
    Commented May 9, 2018 at 13:33
  • $\begingroup$ Yes but a similar counterexample should work $\endgroup$
    – user120812
    Commented May 9, 2018 at 13:33
  • $\begingroup$ For instance, if $X$ is a Noetherian $\mathbf{F}_p$ scheme that is smooth, and $Z\subset X$ a smooth closed subscheme, then $Z^{perf}$ and $X^{perf}$ should give a counterexample choosing $\mathcal{F} = \mathcal{O}_{Z^{perf}}$. Similar to your example, in a way. I think I saw this on an answer, some time ago. $\endgroup$
    – user120812
    Commented May 9, 2018 at 13:35

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