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I am trying to get an interpretation of what means for a sheaf to be flat with respect to a base. The definition is that, given $f:X \rightarrow Y$ morphism of schemes, $\mathcal{F}$ is flat over $Y$ at $x \in X$ if $\mathcal{F}_x$ is a flat $\mathcal{O}_y$ module. It is said to be flat it if holds true on the whole $X$.

Given $\mathcal{F}$ flat over $Y$, does it imply that given a short exact sequence \begin{equation} 0 \rightarrow \mathcal{G}' \rightarrow \mathcal{G} \rightarrow \mathcal{G}'' \rightarrow 0 \end{equation} on $Y$, then \begin{equation} 0 \rightarrow f^*\mathcal{G}' \otimes \mathcal{F}\rightarrow f^*\mathcal{G}\otimes \mathcal{F} \rightarrow f^*\mathcal{G}'' \otimes \mathcal{F}\rightarrow 0 \end{equation} is exact on $X$? Is it true? I was thinking so, since the stalk of the pullback, say of $\mathcal{G}$, is $f^*\mathcal{G}_x=\mathcal{G}_{f(x)}\otimes_{\mathcal{O}_{f(x)}}\mathcal{O}_x$. Also $\mathcal{G}_{f(x)}\otimes_{\mathcal{O}_{f(x)}}\mathcal{O}_x \otimes_{\mathcal{O}_x} \mathcal{F}_x=\mathcal{G}_{f(x)}\otimes_{\mathcal{O}_y}\mathcal{F}_x$. So I basically have the exactness on stalks for the exact sequence I claim and then I could conclude.

Is my argument correct? If not, where am I mistaken? What is a nice interpretation for flatness over a base?

Thank you

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  • $\begingroup$ The proof is okay. What are your doubts concerning it? With basically the same argument, passing to stalks can be avoided by noting that $F$ is flat over $Y$ if and only if it is flat as $f^{-1}\mathcal{O}_Y$-module. $\endgroup$ – Ben Apr 11 '15 at 20:25
  • $\begingroup$ @Ben My concern is that it seems right to me, but I have not seen such an interpretation anywhere, so I started having a doubt about it! $\endgroup$ – Stefano Apr 11 '15 at 22:01
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It is true and the proof is fine. Here is a very similar one, based on the observation that, since $(f^{-1}\mathcal{O}_Y)_x = \mathcal{O}_{Y,f(x)}$ for all $x\in X$, $ F $ is flat over $Y$ if and only if it is flat as $f^{-1}\mathcal{O}_Y$-module. We don't work with the sheaf of rings $f^{-1}\mathcal{O}_Y$ very often, because it's not an $\mathcal{O}_X$-module, but $\mathcal{O}_X$ is an algebra over $f^{-1}\mathcal{O}_Y$. In this case, however it's the key.

The functor $G\mapsto f^*(G) \otimes_{\mathcal{O}_X} F$ in fact is exact. This follows from the obvious natural equivalence $$f^{-1}(-)\otimes_{f^{-1}\mathcal{O}_Y} F\Rightarrow f^{-1}(-)\otimes_{f^{-1}\mathcal{O}_Y}\mathcal{O}_X\otimes_{\mathcal{O}_X} F = f^*(-)\otimes_{\mathcal{O}_X} F$$ of functors from $\mathcal{O}_Y$-modules to $\mathcal{O}_X$- (or $f^{-1}\mathcal{O}_Y$-)modules, and the exactness of $f^{-1}(-)$. (Note that here it doesn't matter whether we regard $f^*(G) \otimes_{\mathcal{O}_X} F$ as $\mathcal{O}_X$- or $f^{-1}\mathcal{O}_Y$-module: all that matters is the underlying sheaf of groups.)

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