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Ravi gives a version of the claim below in exercise 24.4G in "Foundations...", but he adds the condition that $Y$ is reduced in order to conclude something additional (the equivalence of the condition $\pi_* O_X$ locally free to $\pi_* O_X$ locally constant rank).

However, I don't see where $Y$ reduced would be needed in the following claim.

Claim: If $ \pi : X \to Y$ is a finite morphism of locally Noetherian schemes. Then $\pi$ is flat iff $\pi_* O_X$ is locally free on $Y$.

Sketch of argument that pushforward locally free implies flat: We reduce to the affine case, since flatness can be checked on an open cover of the source and / or target. We further reduce to the case when these covers trivialize $\pi_* O_X$. But a free module is flat, so we can conclude.

I see no problem with this right now (and I did write out more details than what are above)... maybe I am missing something subtle or obvious? (Or maybe I am just being paranoid...)

Thank you!

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  • $\begingroup$ don't you also have to conclude the converse? It sounds like all you have done so far is prove free implies flat. That is, given that $\pi$ is flat, you have that $\pi_*\mathcal{O}_X$ is flat over $\mathcal{O}_Y$, but how do you know it is free? $\endgroup$ – John Martin Mar 30 '16 at 21:56
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    $\begingroup$ @JohnMartin For finitely generated modules over Noetherian local rings, flatness is equivalent to freeness. I don't think reduceness is needed there either, it is just an argument with the Nakayama lemma. But you are right - I only meant to ask about the direction I wrote the argument for. $\endgroup$ – Lorenzo Mar 30 '16 at 22:02
  • $\begingroup$ Sounds like you have it to me. You do need the reducedness to conclude constant rank though. The rank can jump at non-reduced points $\endgroup$ – John Martin Mar 30 '16 at 22:10
  • $\begingroup$ @JohnMartin Thanks! $\endgroup$ – Lorenzo Mar 31 '16 at 0:27
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Yes, you're correct that reducedness doesn't play a role. In fact, for any morphism $f:X\to S$, $f$ is finite locally free in the sense that $f$ is finite and $f_*\mathscr{O}_X$ is locally free on $S$ (necessarily of finite rank) if and only if $f$ is finite, flat, and locally of finite presentation. If $S$ is locally Noetherian, the final hypothesis is automatic, so indeed you just need the morphism to be finite and flat. For a reference, see http://stacks.math.columbia.edu/tag/02KB.

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