Counterexample to flat base change for $\mathcal{O}_X$-modules

Question

Consider a cartesian diagram

$$\require{AMScd} \begin{CD} X' @>{f'}>> X\\ @V{p'} VV @VV{p} V\\ S' @>{f}>> S \end{CD}$$

of schemes (or even locally ringed spaces). If $\mathcal{F}$ is any $\mathcal{O}_X$-module, then there is a base change morphism $$ f^* R^i p_* \mathcal F \to R^i(p')_* (f')^* \mathcal F. $$ The flat base change theorem says that if all objects are schemes, $f$ is flat, and $\mathcal F$ is quasicoherent, then base change is an isomorphism. Is there a nice counterexample if $\mathcal F$ is not quasicoherent?

Answer 1

I think that the following might be a counterexample. Take $S = \mathrm{Spec}(\mathbb{R})$ and $X = \mathbb{A}^1_{\mathbb{R}}$. I will take $S' - \mathrm{Spec}(\mathbb{C})$, so $X' = \mathbb{A}^1_{\mathbb{C}}$. Consider the point $x: \mathrm{Spec}(\mathbb{R}[t]/(t^2+1)) \to \mathbb{A}^1_{\mathbb{R}}$. Let $\mathcal{F}$ be the sheaf of $\mathcal{O}_{\mathbb{A}^1_{\mathbb{R}}}$-modules defined as follows for opens $U \subset \mathbb{A}^1_{\mathbb{R}}$:

$$ \mathcal{F}(U) = \begin{cases} 0 \; \text{ if $x \notin U$} \\ \mathbb{R}[t]_{(t^2+1)} \; \text{ if $x \in U$} \end{cases} $$

(Note that this is actually a sheaf of $\mathcal{O}_{\mathbb{A}^1_{\mathbb{R}}}$-algebras in the evident way).

The pushforward $p_*(\mathcal{F})$ can be viewed as the $\mathbb{R}$-algebra $\mathbb{R}[t]_{(t^2+1)}$. The pullback $f^*p_*(\mathcal{F})$ is the base change $\mathbb{C}$-algebra $A:= \mathbb{C} \otimes_{\mathbb{R}} \mathbb{R}[t]_{(t^2+1)}$. Note that $A$ is a localization of $\mathbb{C}[t]$, and so it is a domain.

Let $x_1, x_2 \in \mathbb{A}^1_{\mathbb{C}}$ be the two preimages of $x$ (corresponding to $\pm \sqrt{-1}$). I think that the pullback $(f')^*(\mathcal{F})$ is the sheafification of the following presheaf $\mathcal{G}$ of $\mathcal{O}_{\mathbb{A}_{\mathbb{C}}^1}$-algebras:

$$ \mathcal{G}(V) = \begin{cases} A \; \text{ if $x_1 \in V$ and $x_2 \in V$} \\ A_{(t-t(x_1))} \; \text{ if $x_1 \in V$ and $x_2 \notin V$}\\ A_{(t-t(x_2))} \; \text{ if $x_2 \in V$ and $x_1 \notin V$} \\ 0 \; \text{ otherwise} \end{cases} $$

I believe that the sheafification is the following:

$$ (f')^*(\mathcal{F})(V) = \begin{cases} A_{(t-t(x_1))} \times A_{(t-t(x_2))} \; \text{ if $x_1 \in V$ and $x_2 \in V$} \\ A_{(t-t(x_1))} \; \text{ if $x_1 \in V$ and $x_2 \notin V$}\\ A_{(t-t(x_2))} \; \text{ if $x_2 \in V$ and $x_1 \notin V$}\\ 0 \; \text{ otherwise} \end{cases} $$

Therefore, the pushforward $(p')_*((f')^*(\mathcal{F}))$ is the product $A_{(t-t(x_1))} \times A_{(t-t(x_2))}$, which is not a domain. I believe that the base change map is given by the diagonal $A \to A_{(t-t(x_1))} \times A_{(t-t(x_2))}$, so it is not an isomorphism.

(Edited: corrected the computation of the pullback).