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 $B \rightarrow A$.


1 Answer 1


This is a special case of Theorem 3.3.5 in

Raynaud, Michel; Gruson, Laurent, Critères de platitude et de projectivité. Techniques de ”platification” d’un module. (Criterial of flatness and projectivity. Technics of ”flatification of a module.), Invent. Math. 13, 1-89 (1971). ZBL0227.14010.

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 $A/p$.

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    $\begingroup$ Laurent, thank you very much for pointing me to this. I've never taken the time to understand relatively pure modules in the slightest so I wouldn't have made this connection. This will motivate me! I'd like to point out for documentation's sake that most of the seminal Raynaud-Gruson results have made it into the stacks project. The result you reference is [Proposition 05MD] in Stacks $\endgroup$ Commented May 12, 2021 at 2:11

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