If a morphism from a commutative absolutely flat ring has integral fibers, does it induce an embedding of spectra?

Question

All rings are commutative and unital.

Let $A$ be an absolutely flat ring and $A \rightarrow B$ a ring monomorphism with integral fibers (i.e. for each $\mathfrak{p} \in \operatorname{Spec}(A), B \otimes_A \kappa(\mathfrak{p})$ is a domain).

Then there is a continuous bijection $\operatorname{minSpec}(B) \rightarrow \operatorname{Spec}(A)$ (where $\operatorname{minSpec}$ denotes the space of minimal primes with the subspace topology).

Question: Is it necessarily the case that $\operatorname{Spec}(A)$ is homeomorphic to $\operatorname{minSpec}(B)$ (the space of minimal primes of $B$)?

Because $\operatorname{Spec}(A)$ is compact Hausdorff, this is equivalent to asking if $\operatorname{minSpec}(B)$ is necessarily compact.

I suspect the answer is no, but am having difficulty producing a counterexample.

In the following two cases the answer is clearly yes: (1) If $A$ is a finite product of fields (2) If $A \rightarrow B$ exhibits $B$ as a projective $A$-module.

Answer 1

Here's a counterexample. Let $A=k^{\mathbb{N}}$ where $k$ is a field, and $B=A[X]/I$ with $I$ generated by the $e_nX$ for $n\in\mathbb{N}$, where $e_n\in A$ has $1\in k$ at the $n$th coordinate and $0$ elsewhere. The annihilator in $A$ of $X\in B$ is $k^{(\mathbb{N})}$, the subset of $A$ consisting of the elements that have a non-zero value at only finitely many coordinates.

For $\mathfrak{p}\in\operatorname{spec}(A)$ we have $A\to A_{\mathfrak{p}}$ surjective with kernel $\mathfrak{p}$, since $A$ is absolutely flat (von Neumann regular), and so $\kappa(\mathfrak{p}) = A/\mathfrak{p}$. And $\mathfrak{p}=\{a\in A\,\mid\,\{n\mid a_n=0\}\in p\}$ for an ultrafilter $p$ on $\mathbb{N}$. If $p$ is principal, say $p=\{D\subseteq\mathbb{N}\mid n\in D\}$ for some $n\in\mathbb{N}$, then $\mathfrak{p}$ is the principal ideal of $A$ generated by $1-e_n$. In that case, $B \otimes_A \kappa(\mathfrak{p}) = B \otimes_A A/\mathfrak{p} \cong k$, since $e_n$ maps to $1$ in $A/\mathfrak{p}$ and $X$ becomes $0$. And if the ultrafilter $p$ is non-principal, so is the ideal $\mathfrak{p}$, and then $B \otimes_A \kappa(\mathfrak{p}) \cong K[X]$, where $K=A/\mathfrak{p}$ is the resulting ultraproduct field. Indeed, for such $p$ one has $\mathbb{N}-\{n\}\in p$, hence $e_n\in\mathfrak{p}$, for all $n$, hence $I\subseteq\mathfrak{p}[X]$.

It is easy to see that $\operatorname{min}(B)$ consists of the ideals $(1-e_n)B$ for $n\in\mathbb{N}$, plus the ideals $\mathfrak{p}[X]B$ where $\mathfrak{p}\in\operatorname{spec}(A)$ is non-principal. Now $\operatorname{min}(B)\subseteq D(X)\,\cup\,\bigcup_{n\in\mathbb{N}}D(e_n)$, where $D(b)$ denotes the basic-open set of $\operatorname{spec}(B)$ associated with $b\in B$ as usual. For if $X\in P\in \operatorname{min}(B)$, $P$ must be of the form $(1-e_n)B$ for some $n$, and hence $e_n\notin P$. This covering does not admit a finite subcovering.

Note: the $B$ that can arise in this context are necessarily reduced, and for a reduced ring $B$ compactness of $\operatorname{min}(B)$ is equivalent to $\prod_{P\in \operatorname{min}(B)}B_P$ being a flat $B$-module, or, alternatively, to the condition that for all $b\in B$ there is a finitely generated ideal $J$ of $B$ contained in $\operatorname{ann}_B(b)$ such that $\operatorname{ann}_B(bB+J)=0$. (Proposition 1.16 in Matlis, The minimal prime spectrum of a reduced ring). In the example above, the element $b:=X\in B$ clearly fails the latter condition.