# Are finitely presented algebras over VNRs projective?

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 $$B \rightarrow A$$.

Indeed, $$B$$ is trivially $$A$$-pure since for every prime (i.e. maximal) ideal $$p$$ of $$A$$, the henselization of $$A$$ at $$p$$ is just the residue field $$A/p$$.