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Let $f : X \to Y$ be a finitely presented proper morphism. Let $\mathcal{F}$ be a quasi-coherent sheaf on $X$. Do the functors $R^i f_* \mathcal{F}$ preserve any of the following properties:

(1) finite type

(2) finitely presented

(3) coherent?

Even if $\mathcal{F}$ is coherent I don't know if we can say anything at all about finiteness properties of $R^i f_* \mathcal{F}$?

Of course if $Y$ is locally Noetherian then these properties are equivalent and it is well-known that $R^i f_*$ preserves coherence.

However, I don't see how to reduce to the Noetherian case because of failures for cohomology to commute with base change.

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According to Kiehl, R. in

Ein ``Descente''-Lemma und Grothendiecks Projektionssatz für nichtnoethersche Schemata (Math. Ann. 198 (1972), 287–316.)

If $f \colon X \to Y$ is a proper finitely presented morphism of schemes and $\mathcal{K}$ is pseudo-coherent complex on $X$ then $\mathsf{R}f_*\mathcal{K}$ is also pseudo-coherent.

If, further $Y$ is noetherian, then "pseudo-coherent" just means that $\mathcal{K}$ is bounded above with coherent cohomology. In the non-noetherian case is a slightly stronger condition meaning that locally it has a resolution by finitely generated locally free modules.

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    $\begingroup$ I think that in Kiehl's theorem you may want the morphism to be pseudocoherent as in stacks.math.columbia.edu/tag/067X. For example, if the morphism is finite, then you would want the pushforward of the structure sheaf to be pseudocoherent, so if the morphism is finitely presented but not pseudocoherent this could break, no? (If the morphism is flat, then finitely presented implies pseudocoherent, so we would be good in that case). $\endgroup$
    – afh
    Commented Feb 7 at 13:06
  • $\begingroup$ Could be worth pointing out: it is unclear to me how to control the cohomology of a pseudocoherent complex. One knows that the top cohomology is finitely presented, but I think that the rest of the cohomologies need not be even of finite type. For example, take a complex of the form $R^{\oplus k} \to R^{\oplus n}$ for some ring $R$, where $R^{\oplus k} \to R^{\oplus k} \to M \to 0$ is a presentation of an $R$-module that is finitely presented but not (-2)-pseudocoherent. Then I think that the kernel of $R^{\oplus k} \to R^{\oplus n}$ is not of finite type, so the left-most cohomology is nasty. $\endgroup$
    – afh
    Commented Feb 7 at 17:45

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