Suppose $X$ and $Y$ are two (smooth, affine) algebraic varieties. Let $\mathcal{F}$ be a locally free coherent sheaf over $X \times Y$, and let $\mathcal{G}$ be the pushforward of $\mathcal{F}$ to $X$. Is it true that $\mathcal{G}$ is a locally free quasicoherent sheaf?
5 Answers
The answer is "yes" (though I can't imagine a situation where one would really need this fact). More generally, if $A$ and $B$ are arbitrary commutative algebras over a field $k$ with $A$ noetherian and if $M$ is an $A \otimes_k B$module which is locally free as such (perhaps not finitely generated) then $M$ is locally free as an $A$module. Here, by "locally free" I meant relative to the Zariski topology.
Without loss of generality (since $A$ is noetherian), we may and do assume that ${\rm{Spec}}(A)$ is connected.
The first thing to observe is that $M$ is projective as an $A \otimes_k B$module. Indeed, projectivity is a Zariskilocal (even fpqclocal) property for modules over commutative rings, by 3.1.3 part II of RaynaudGruson (and the fact that faithfully flat ring maps satisfy their condition (C), using 3.1.4 part I of RaynaudGruson), so any locally free module over a commutative ring is projective. Thus, $M$ is a direct summand of a free $A \otimes_k B$module, which in turn is also free as an $A$module. Hence, $M$ is projective as a $A$module. If $M$ is modulefinite as such then it is certainly locally free (since $A$ is noetherian). But if it is not modulefinite then we're again done since $A$ is noetherian with connected spectrum, as then it follows that any projective $A$module that is not finitely generated is free! This is Bass' theorem "big projective modules are free"; see Corollary 4.5 in his paper with that title.

5$\begingroup$ It was suggested (in an earlier comment which was later deleted) that there must be something "wrong" with Bass' theorem, since for a nonfree finitely generated projective module P, surely an infinite direct sum of copies of P is not free. In case anyone else has doubts, it really is free (and Corollary 4.5 in Bass' paper is not wrong); that (among other things) is the marvelous surprise of Bass' theorem. This phenomenon can also be seen on a smaller scale: if $R$ is Dedekind and $I$ is a nonprincipal ideal then $I \oplus I \simeq R^2$ is free even though $I$ is not free. $\endgroup$– BCnrdMar 17, 2010 at 0:02

$\begingroup$ A fact similar to the last one you mention is also true for the Weyl algebra A_1 (differential operators on C[x] with polynomial coefficients). If M is a projective (left, say) A_1 module, $M \oplus M$ is free. $\endgroup$ Mar 17, 2010 at 6:15

$\begingroup$ Small typo: in my preceding comment, for the example at the end I meant to require that $I$ is also 2torsion in the class group. $\endgroup$– BCnrdMar 17, 2010 at 6:28
One can prove this also without Bass's theorem. Let $X= Spec A$ and $Y=Spec B$. The sheaf $\cal F$ comes from a finitely generated module $M$ over $C=A\otimes B$. Our first goal is to show that $M$, as an $A$module, is a direct sum of finitely generated, locally free modules. Since $A$ is noetherian, this is equivalent to $M$ being a direct sum of finitely generated projective $A$modules.
The fact that $M$ is locallyfree implies that $M$ is projective over $C$, further it is finitely generated, so there is a finitely generated $C$module $N$ such that $M\oplus N=\bigoplus_{i=1}^kC e_i$. Now each $e_i$ of this free basis can bewritten uniquely as $e_i=m_i+n_i$. Let $M_0$ be the $A$module generated by $m_1,\dots,m_k$. Let $(b_j)_{j\in J}$ be a basis of $B$ over the ground field, then $$ M=\bigoplus_{j\in J}b_j M_0. $$ Let $N_0$ be the $A$module generated by $n_1,\dots,n_k$. Then $M_0\oplus N_0= \bigoplus_i Ae_i$ is a free $A$module and so is Also $b_j(M_0\oplus N_0)=\bigoplus_iAb_je_i$. This means that we have written $M$ as a direct sum of finitely generated projective $A$modules as claimed.
Now to conclude remember that $A$ is noetherian, therefore for each point in $X$ there exists an open neighborhood, where all summands of $\cal G$ are free.
Hi. I don't know how to make a comment; this is on Torsten's example with elliptic curve minus zero point. When you take a line bundle corresponding to a point (for simplicity not a two torsion point) and add the line bundle corresponding to minus of that point then this rank=2 vector bundle restricted to the elliptic curve minus zero is (even globally) free.
Indeed, working on the elliptic curve twist with the degree one line bundle L corresponding to the zero point and take the canonical inclusion of the structure sheaf on both factors, the quotient must be isomorphic to L squared.
This is just to understand Bass' theorem on this nice example.
No. Take for example $X$ to be an elliptic curve, $Y = Pic^0(X) \cong X$ and $F = L$, the Poincare bundle (the universal bundle on $X \times Pic^0(X)$. Then the pushforward of $L$ onto $Pic^0(X)$ is the structure sheaf of the point, corresponding to the trivial line bundle.

3$\begingroup$ An elliptic curve is not affine. $\endgroup$– user1688Mar 16, 2010 at 20:51

Maybe I'm totally confused, but a locally free sheaf over an affine variety should be nothing else than a projective module over the ring of global functions. Push forward along the projection is then just restriction of scalars along the inclusion $$O_X\rightarrow O_X\otimes_k O_Y$$ Now $O_Y=\bigoplus k$ so $$O_X\otimes_k O_Y=O_X\otimes_k \bigoplus k=\bigoplus O_X$$ so free modules stay free and projective Modules=Summands of free modules stay projective.

3$\begingroup$ In the affine setting a locally free sheaf corresponds (under mild finiteness conditions) to a projective module, not to a free module. Also, your equality O_Y=\oplus k is incorrect. $\endgroup$ Mar 16, 2010 at 21:15

$\begingroup$ Thanks, I edited my post. O_Y=\oplus k is true since O_Y is a k algebra and any vectorspace is free. $\endgroup$ Mar 16, 2010 at 21:28

$\begingroup$ I think the question may be based on the fact that for nonfinitely generated modules projective and locally finite are not the same thing. Take for instance an elliptic curve (minus the origin to make it affine) and take the sum of all the line bundles (one for each point on the original elliptic curve). Then this sum is projective but there is no nonempty open subset of the spectrum over which it is free. $\endgroup$ Mar 16, 2010 at 22:04

$\begingroup$ By Bass' theorem, a nonfinitely generated projective module over any noetherian ring with connected spectrum is free. $\endgroup$– BCnrdMar 17, 2010 at 0:04

$\begingroup$ Mea culpa. My comment was wrong in almost any respect. I leave it both in order to leave the subsequent comment meaningful and as a reminder to myself. $\endgroup$ Mar 17, 2010 at 5:25