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?
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. 


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. 


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. 

