Removing Noetherian condition from cohomology and base change

This question is related to a question I asked a few days ago. Since there seems to be no (at least for me) satisfying reference for cohomology and base change as stated by Vakil in his script in exercise 28.2.M (or below), I would like to record a proof in this post. But I am no expert and not sure about some steps. Let me start with the statement.

Let π:XY be a proper finitely presented morphism, F coherent over X and flat over Y and the base change map ϕpy be surjective at yY. Then the following hold.

  1. There is an open neighborhood U of y such that for any ϕ:ZU, the base change ϕpZ is an isomorphism.
  2. Furthermore, ϕp1y is surjective if and only if RpπF is locally free in some neighborhood of y.

The idea of the proof (as indicated by Vakil) is to reduce to the case of a locally Noetherian base using techniques from Grothendieck. As the statement is local we can assume that Y=SpecR is affine. Now we get the following Cartesian diagram
X @>\alpha>> X_0\\
@V \pi V V @VV \pi_0 V\\
Y @>>\alpha_0> Y_0

where Y_0 is Noetherian and affine (in fact a \mathbb Z-subalgebra R_0 of R) and \pi_0 is proper. Moreover there is a coherent sheaf \mathscr F_0 on X_0 flat over Y_0 such that \mathscr F is the pullback of \mathscr F_0 by \alpha. Now let y \in Y be a point (we identify y with \operatorname{Spec} k(y)) and let y_0 denote the image under \alpha_0. We get the two Cartesian squares
X_y @>>> X_{0,y_0} @>>> X_0\\
@VVV @VVV @VV\pi_0V\\
y @>>> y_0 @>>> Y_0

Since the map at the bottom left is flat, the base change map of the left square is an isomorphism. Hence the base change map of the right square is an isomorphism if and only the base change map of the outer diagram is an isomorphism. This should imply (here I am not sure) that the morphism \phi_y in the statement of the theorem is surjective / an isomorphism if and only if \phi_{y_0} is surjective / an isomorphism (this would certainly be the case if it is true that \phi_y = \phi_{y_0} \otimes k(y)).

Hence the assumption of the theorem holds at y if and only if it holds at y_0 and we assume now that this is the case. By cohomology and base change for a Noetherian base there is an open subset U of Y_0 such that the base change map with any morphism mapping to U is an isomorphism. Then the same statement is certainly true for \alpha_0^{-1}(U) and after possibly shrinking Y and Y_0 we may assume that the property holds for those open sets.

It is left to show that the second assertion of the theorem holds. Similarly to before we get that \phi_y^{p-1} is surjective if and only if \phi_{y_0}^{p-1} is surjective. Since the base change map associated with \alpha_0 is an isomorphism we see that R^p\pi_*\mathscr F is locally free if R^p\pi_{0*}\mathscr F_0 is locally free (since it is a pullback of the last sheaf). But I have no idea how I would get the reverse implication.

I would be really grateful if someone could fill in the missing steps (or point out where I did something completely wrong). I am especially insecure about what I am doing since Conrad in Conrad, Brian, Grothendieck duality and base change, Lecture Notes in Mathematics. 1750. Berlin: Springer. x, 296 p. (2000). ZBL0992.14001. seems to use more sophisticated arguments to arrive at weaker statement in section 5.1


Since the question is about non-noetherian base change, I have just released a preprint with some collaborators that has a result in this direction (apologies for the self-promotion). The paper is titled Relative perfect complexes.

In concrete, one of the results (Corollary 5.6.) says:

Let f : X \to Y be a quasi-proper morphism of schemes and assume further that Y is reduced. The function h^p ( f , \mathcal{E}) is locally constant for some p if, and only if, R^pf_*\mathcal{E} is locally free and the canonical map
(R^pf_*\mathcal{E})_y \otimes_{\mathcal{O}_{Y,y}} \kappa(y) \longrightarrow H^p(X_y,\mathcal{E}(y)) is an isomorphism for every y \in Y.

Notice that there is no noetherian hypothesis on the schemes, but beware the derived fiber \mathcal{E}(y) has a special definition that allows to bypass flatness conditions on \mathcal{E}. I don’t know if you keep interested in this topic; I hope this is of some help.

Source : Link , Question Author : Fabian Ruoff , Answer Author : Leo Alonso

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