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 $\mathbb{A}^n$.



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> I would be much more pleased with a reference than a proof, since I would like to use this result in a paper.
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Is it true that if the pushforward of a coherent sheaf is locally free, then the original sheaf is locally free?
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. 


