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Given two smooth projective surfaces $X$ and $Y$ over some algebraically closed field. Given a torsion free coherent sheaf $M$ on $X$. One has the projections $\pi_X$ and $\pi_Y$ from the product $X\times Y$. Then we have $\pi_X^{\*}M$ on $X\times Y$.

Question: Is $\pi_X^{\*}M$ flat over $Y$?

One has to show that for each $z\in X\times Y$ the $O_{X\times Y,z}$-module $\pi_X^{\*}M_{z}$ is a flat $O_{Y,\pi_Y(z)}$-module. Now $\pi_X^{\*}M_z=M_{\pi_X(z)} \otimes_{O_{X,\pi_X(z)}} O_{X\times Y,z}$. I don't see why this should be flat over $O_{Y,\pi_Y(z)}$. I just know that $O_{X\times Y}$ is flat over $O_X$ and $O_Y$, but that doesn't seem to help me.

Background: I'm reading www.imsc.res.in/library/pdf/shaves.pdf, page 144, Theorem 6.1.8. When they want to compute the relative $Ext$-sheaves explicitely, they choose a certain complex of sheaves for whose existence they cite an article by Banica, Putinar and Schumacher. Now i looked that article up, and the complex is constructed using that both sheaves are flat over the base scheme. In this proof the sheaf $\mathcal{E}$ is a quasi universal family, so it is flat by definition, but what about the pullback sheaf?

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Certainly it is flat, and you don't even have to assume that $M$ is torsion free. Indeed, the question is local, so you can assume that $X$ and $Y$ are affine, so the question is: given $k$-algebras $A$ and $B$ and an $A$-module $M$ check that $M\otimes_A(A\otimes_k B)$ is flat over $B$. But this is easy --- $M\otimes_A(A\otimes_k B) \cong M\otimes_k B$, so it is a free $B$-module.

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  • $\begingroup$ Ah, thanks. I still don't a feeling for if and when one can reduce to the affine case, which makes most problems, like in this case, very easy. $\endgroup$ – TonyS Nov 11 '10 at 21:39

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