Removing Noetherian condition from cohomology and base change

Question

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 $\pi \colon X \to Y$ be a proper finitely presented morphism, $\mathscr F$ coherent over $X$ and flat over $Y$ and the base change map $\phi^p_y$ be surjective at $y \in Y$. Then the following hold.

1. There is an open neighborhood $U$ of $y$ such that for any $\phi \colon Z \to U$, the base change $\phi^p_Z$ is an isomorphism.
2. Furthermore, $\phi^{p-1}_y$ is surjective if and only if $R^p\pi_*\mathscr F$ 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 $Y = \operatorname{Spec} R$ is affine. Now we get the following Cartesian diagram $\require{AMScd}$ \begin{CD} X @>\alpha>> X_0\\ @V \pi V V @VV \pi_0 V\\ Y @>>\alpha_0> Y_0 \end{CD} 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 $\require{AMScd}$ \begin{CD} X_y @>>> X_{0,y_0} @>>> X_0\\ @VVV @VVV @VV\pi_0V\\ y @>>> y_0 @>>> Y_0 \end{CD} 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

Answer 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.

Answer 2

I'll fill in the gaps in your proof. The following is the proof I found while going through Vakil's book. The first issue is using the given surjectivity of $\phi_y$ to show the surjectivity of $\phi_{y_0}$. The main claim is this:

Claim 1: Assume $\phi_y$ is surjective. We can increase $N$ (by a lot), so that $\phi_{y_0}$ will also be surjective.

Proof

Let $\{U_i\}$ be a Cech cover of $X_0$, which pulls back to a Cech cover $\{V_i\}$ of $X$.

Take a basis $v_i$ for the finite dimensional vector space $H^p(X_y, \mathscr{F}\otimes \kappa (y))$. By surjectivity we can find for each $i$ an element $u_i\in H^p(X, \mathscr{F})\otimes_R \kappa(y)$ which maps to $v_i$. By clearing denominators, we can assume $u_i\in H^p(X, \mathscr{F})\otimes_R(R/\mathfrak{q})$, where $\mathfrak{q}$ is the ideal corresponding to $y$ (the images of the $u_i$ will still form a basis). We can further replace the $u_i$ by their representatives in $H^p(X,\mathscr{F})$.

Each $u_i$ has a representative in the module $C^p(\{V_j\}, \mathscr{F})$ of the Cech complex on $X$. Recall $$C^p(\{V_j\},\mathscr{F})=\prod_{|I|=p+1}\mathscr{F}(V_I)=R\otimes_{R_0}C^p(\{U_j\}, \mathscr{F}_0)$$ Thus we can write the representative of $u_i$ as $\sum_{k}r_{ik}\otimes u_{ik}$, where $r_{ik}\in R$ and $u_{ik}\in C^p(\{U_j\}, \mathscr{F}_0)$. Notice that $u_i$ is by assumption in the cohomology, so the representative is in the kernel of $C^p(X, \mathscr{F})\to C^{p+1}(X, \mathscr{F})$. This fact corresponds to finitely many relations among finitely many elements $r\in R$. We can add all of these elements as indeterminates to $R_0$ (including the $r_{ik}$) and quotient out by the ideal of relations among them. This corresponds to a base change so that the map $\pi_0$ remains proper and $\mathscr{F}_0$ remains flat over $Y_0$. After this, by construction we have elements $u_i^0\in H^p(X_0, \mathscr{F}_0)$ that pull back to the $u_i$. Indeed, the elements corresponding to $\sum r_i\otimes u_{ik}$ work: these will be in the kernel of $C^p(X_0, \mathscr{F}_0)\to C^{p+1}(X_0,\mathscr{F}_0)$ because we added all the relations needed.

Suppose the $u_i^0$ pull-back to elements $v_i^0\in H^p(X_{0, y_0}, \mathscr{F}_0\otimes \kappa(y_0))$. These in turn must pull back to the $v_i$, so they are linearly independent. But $y\to y_0$ is a flat morphism, so $$h^p(X_{0, y_0}, \mathscr{F}_0\otimes \kappa(y_0))=h^p(X_y, \mathscr{F}\otimes\kappa(y))$$ This implies the $v^0_i$ form a basis of $H^p(X_{0,y_0},\mathscr{F}_0\otimes \kappa(y_0))$, so the base change morphism is surjective. $\blacksquare$

Your proof of part $(i)$ of the theorem using this claim is now correct. I'll now move to part $(ii)$ of the theorem.

Using the first part, we have an open neighborhood $U_0$ of $y_0$ so that for every $Z\to U_0$ the map $\phi^p_{Z}$ is an isomorphism. Replace $Y_0$ by $U_0$ (and $Y$ with $\alpha_0^{-1}(U_0)$). We want to show that $\phi^{p-1}_y$ (the base change map of $y\to Y$) is surjective iff $R^p\pi_*\mathscr{F}$ is locally free near $y$.

Claim 2: If $\phi_{y_0}^p$ is surjective, this remains true after base-changing $R_0$ by some intermediate $R_0\to B\to R$.

Proof: We have a base change diagram \begin{CD} X@>>> X_{B}@>>> X_0\\ @V{\pi}VV @V{\pi_B}VV @V{\pi_0}VV\\ \text{Spec}\ R@>>> \text{Spec}\ B @>{\alpha}>> \text{Spec}\ R_0 \end{CD} Let $y_B$ be the image of $y$ in $\text{Spec}\ B$. Then the base change diagram \begin{CD} X_{B,y_B}@>>> X_{0,y_0}@>>> X_0\\ @VVV @VVV @V{\pi_0}VV\\ y_B @>>> y_0 @>{\alpha}>> \text{Spec}\ R_0 \end{CD} shows that $\phi^p$ of the big rectangle is an isomorphism ($y_B\to y_0$ is flat). Now, the diagram \begin{CD} X_{B,y_B}@>>> X_{B}@>>> X_0\\ @VVV @V{\pi_B}VV @V{\pi_0}VV\\ y_B@>>> \text{Spec}\ B @>{\alpha}>> \text{Spec}\ R_0 \end{CD} shows that $\phi^p$ of the left square is an isomorphism (because it's true for the right square and for the big rectangle). Thus we may replace $Y_0$ with $\text{Spec}\ B$: $\phi^p_{y_0}$ will still be an isomorphism. $\blacksquare$

In one direction, assume $\phi^{p-1}_y$ is surjective. As at the beginning of the proof of the previous part, add indeterminates so that $\phi^{p-1}_{y_0}$ will also be surjective (this doesn't hurt the surjectivity of $\phi^p_{y_0}$, by Claim 2). By the Noetherian case, $R^p(\pi_0)_*\mathscr{F}_0$ is locally free around $y_0$.

By the construction of $U_0$, the map $\phi^p_{Y}$ is an isomorphism.
This implies $$R^p\pi_* \mathscr{F}\cong \alpha_0^*R^p(\pi_0)_*\mathscr{F}_0$$ As $R^p(\pi_0)_*\mathscr{F}_0$ is locally free around $y_0$, we get that $R^p\pi_* \mathscr{F}$ is locally free around $y$, which finishes the proof.

Now for the converse. Assume that $R^p\pi_* \mathscr{F}$ is locally free near $y$. We could have at the beginning chosen $Y$ to be any affine neighborhood of $y$, so we can actually assume $R^p\pi_* \mathscr{F}$ is free (on all of $Y$).
Write $$R^p(\pi_0)_*\mathscr{F}_0=\widetilde{M}$$ for some finite $R_0$-module $M$. Thus $R^p\pi_* \mathscr{F}=\widetilde{R\otimes_{R_0}M}$. Choose generators $e_i$ of the free module $R\otimes_{R_0}M$. We can write $$e_i=\sum_{k}r_{ik}\otimes m_{ik}$$ where $r_{ik}\in R, m_{ik}\in M$. Let $f_j$ be any finite set of generators of $M$: by assumptions that the $e_i$ generate $M\otimes R$, we can write $$f_j\otimes 1=\sum s_{ij}e_i$$ where $s_{ij}\in R$. These relations are true because they are linear combinations with coefficients in $R$ of the defining relations of the tensor product of modules. We adjoin all the elements of $R$ appearing in these linear combinations to $R_0$ as indeterminates. We also add indeterminates corresponding to all the $r_{ik}, s_{ij}$, (and add all the relations between them). Call the resulting ring $B$. Let $\mathscr{F}_B$ be the pullback of $\mathscr{F}$ to $X_B$. By construction of $U_0$, we have $$R^p(\pi_B)_* \mathscr{F}_B\cong \widetilde{M\otimes_{R_0} B}$$ But this is a free module: The $e_i$ are in it, they generate it (because they generate each $f_j\otimes 1$), and they are linearly independent (any linear dependence among them will also be a linear dependence over $R$). We may now replace $R_0$ by $B$, according to Claim 2.

In this situation, the module $R^p(\pi_0)_*\mathscr{F}_0$ is a free module. Thus by part $(ii)$ of the theorem, $\phi^{p-1}_{y_0}$ is an isomorphism. The diagram \begin{CD} X_y@>>> X_{0,y_0}@>>>X_0\\ @VVV @VVV @VVV\\ y@>>> y_0@>>>Y_0 \end{CD} implies that $\phi^{p-1}$ of the big rectangle is an isomorphism. Now the diagram \begin{CD} X_y@>>> X@>>>X_0\\ @VVV @VVV @VVV\\ y@>>> Y@>>>Y_0 \end{CD} implies that $\phi^{p-1}$ of the left square is an surjective (because $\phi^{p-1}$ of the whole rectangle is an isomorphism). This finishes the proof. $\blacksquare$

This was much longer than it had to be ;). The main idea is simple: If some property holds for y, then it is "because" of finitely many relations between elements $R$, and we can translate that to a property of $y_0$ by adding all the needed elements and relations between them.