Question: Let $A$ be a commutative von Neumann regular ring, and $B$ an $A$-algebra of finite presentation, i.e. $B = A[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$. Is $B$ a projective $A$-module?
In the special case that $B$ is $1$-generated, I know that the answer is affirmative. For then $B = A[x]/f$, and for each maximal ideal $\mathfrak{p}$ of $A$ we find that either (i) the coefficients of $f$ are in $\mathfrak{p}$, so that there exists an idempotent $e \in A \setminus \mathfrak{p}$ such that $B_e = A_e[x]$, or (ii) the coefficients of $f$ are not in $\mathfrak{p}$, so we find an idempotent $e \in A \setminus \mathfrak{p}$ such that $f$ is monic over $A_e$, hence $B_e$ is finitely presented and flat = projective over $A_e$. A finite set of these idempotents generate the unit ideal. Since being projective is a Zariski-local property, this shows that $B$ is projective.
Another special case is if $B$ has Krull dimension $0$, in which case it is already module-finite over $B$ (Zariski's lemma) and thus projective again.
This makes me wonder about some kind of inductive argument on the number of generators of $B$ or the dimension of $B$, but I am not sure how to make either of these work out. And I am also skeptical that there is a positive answer because I suspect it would be more of a "textbook" result.
It is perhaps convenient that we can invoke faithfully flat descent to reduce to the case that $A$ is an "algebraically closed" VNR, i.e. $A/\mathfrak{p}$ is an algebraically closed field for every $\mathfrak{p} \in \operatorname{Spec}(A)$. From there we can further reduce consideration to the case that there is a retract $B \rightarrow A$.
