I think the title says it all. If I have a finite map $p:X\to Y$ between schemes, and $F$ is a coherent sheaf on $X$ such that $p_*F$ is locally free, can I conclude that $F$ is locally free?

Assumptions I would be happy to make:

  1. The map $p$ is flat.
  2. $X$ and $Y$ are both $\mathbb{A}^n$.

I would be much more pleased with a reference than a proof, since I would like to use this result in a paper.


3 Answers 3


The reference is "Auslander-Buchsbaum formula". What matters is that $X$ and $Y$ are smooth of some common pure dimension $n$ (which forces flatness due to the quasi-finiteness, by the way), not that one has global affine spaces. Then every coherent sheaf on $X$ has stalks with finite projective dimension, and so all of projective dimension zero (= locally free) precisely when the depth of each stalk is equal to the dimension of its support on the local ring. A length-$n$ regular sequence at $p(x) \in Y$ exists by the local freeness on the base, and that's such a sequence at $x$ provided that the successive quotients arising in the definition of "regular sequence" really remain nonzero when localizing upstairs.

The completion of the $p(x)$-stalk of the pushforward is the product of the completions at the points of $p^{-1}(p(x))$, so each factor module is free and hence we're OK as long as such factor modules are all nonzero since we can use a regular sequence in the completed local ring at $p(x)$. So we have to rule out the possibility of a vanishing completed stalk, or equivalently a vanishing stalk, upstairs. In such cases, upon passing to connected=irreducible components, the coherent sheaf $F$ would vanish on a Zariski-dense open upstairs, and hence on some $p^{-1}(U)$ for a dense open $U$ in the base. Then $p_ {\ast}(F)$ would vanish on $U$ and hence vanish by local freeness, so $F = 0$. The case $F = 0$ is easy to handle directly.


Not without condition (2).

Our counter-example takes place inside the ring $k[t,u]$. Let $A$ be the subring $k[t^2,tu,u^2]$ and $B$ the subring $k[t^2, u^2]$ of $A$. Note that $B$ is a free $A$-module, with basis $(1, tu)$, so $\mathrm{Spec} \ A$ is flat over $\mathrm{Spec} \ B$. Let $M$ be the $A$-module of odd polynomials (those with $f(-t,-u) = - f(t,u)$).

Then $M$ is not locally free as an $A$-module. However, $M$ is free as a $B$-module, with basis $(t, u)$. Turning $M$ into a sheaf, we have a counter-example.

  • 3
    $\begingroup$ This nice example shows that one cannot remove the smoothness hypotheses (the natural version of condition (2)). $\endgroup$
    – Boyarsky
    Jun 28, 2010 at 15:10

Since you asked for reference, here are some references that may be helpful (the question is local):

1) (EDITED: Thanks to BCnrd for keeping me honest here!) If $(R,m)\to (S,n)$ is a map of Cohen-Macaulay local rings of same dimensions, $N$ a finite $S$-module which is $R$-free flat then $${\rm{depth}}_ S N ={\rm{depth}} R = {\rm{depth}} S.$$

This follows from Prop 1.2.16 and Theorem A.11 of Bruns-Herzog "Cohen-Macaulay rings". Namely, take $M=R$ in both results, one get from A.11 that $\dim_SN = \dim_RN + \dim_SN/mN$, so $\dim_SN/mN=0$, then 1.2.16 gives ${\rm{depth}}_SN = {\rm{depth}} R$.

2) If $S$ is regular, then f.g modules with maximal depth are free.

This is well-known. It follows from Auslander-Buchsbaum formula, as Boyarsky pointed out.

EDIT: Just to be clear, it is enought to assume: $f$ is finite, $X$ regular, and $Y$ Cohen-Macaulay scheme (certainly true if $Y$ also regular or smooth, but is a much weaker condition). For example the map induced by $k[x,y]/(xy) \to k[x]$ by killing $y$ works.

  • $\begingroup$ Hailong, in (1) it is restrictive to demand finiteness while also requiring $S$ to be local (rather than just semi-local). For example, in the chosen answer an extra argument with completions is needed to pass to the "finite local" case from the "finite semi-local" case. So for the CM case (in which case the trick of "connected=irreducible" is less relevant), it is better (for purpose of "global" applications) to take $S$ just semi-local and then need more argument to get to the local case. Does your CM reference apply verbatim for $R$-flat (?) semi-local $S$? $\endgroup$
    – BCnrd
    Jun 28, 2010 at 22:49
  • $\begingroup$ BCnrd: The reference does not require $S$ to be $R$-flat. If $S$ is semi-local, can't we just localize at each maximal ideal of $S$? $\endgroup$ Jun 28, 2010 at 23:02
  • $\begingroup$ @BCnrd: Also, may be EGA has some better reference? If you know, please share with us. Thanks. $\endgroup$ Jun 28, 2010 at 23:24
  • $\begingroup$ Hailong, I am surprised the CM reference does not require flatness. Anyway, if you just localize at top then you lose the finiteness hypothesis; that's why I asked about weakening to semi-local (and it is the reason completions -- or henselizations -- would have to intervene, as in the accepted answer, to bust apart semi-local into a direct product of locals). As for EGA ref., let me get back to you about that... $\endgroup$
    – BCnrd
    Jun 29, 2010 at 1:20
  • $\begingroup$ Hailong, assuming regular upstairs and CM downstairs, the proof in the accepted answer works without any need for flatness! As for relevant EGA references, not much to say. They prove Auslander-Buchsbaum only in the regular case (all we need) in $0_ {\rm{IV}}$, 17.3.4 (so a rare case when EGA missed optimal generality; for shame), and then 17.3.5(i) says the obvious conclusion that local freeness on top amounts to have full depth on stalks. But to verify that condition, one has to go through the argument in the accepted answer, adapted to the weaker hypotheses; so EGA is useless (for this). $\endgroup$
    – BCnrd
    Jun 29, 2010 at 2:42

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