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In 'Fundamental Algebraic Geometry' by Fantechi there is a lemma in section 5.3.2, page 119:

LEMMA 5.5 Let $S$ be a noetherian scheme and let $F$ be a coherent sheaf on $\mathbb{P}^n_S$. Suppose there exist some integer $N$ such that for all $r \ge N$ the direct image $\pi_*F(r)= \pi_*(F \otimes O_{\mathbb{P}^n_S}(r))$ is locally free. ($\pi$ is the structure morphism $\mathbb{P}^n_S \to S$). Claim: Then $F$ is flat over $S$.

Now the author leaves as exercise to show that the converse(!) of the above lemma holds as well: if $F$ is flat over $S$ then $\pi_*F(r)$ is locally free for all sufficiently large $r$.

Does anybody know how the proof of the converse works or where I can find it?

About the choice of suitable $N$ I conjecture that Serre's vanishing theorem may play an important role in the proof:

Let $A$ a ring and $F$ coherent on $\mathbb{P}^n_A$. Then for all $i \ge 1$ there exist $m_0$ with $H^i(\mathbb{P}^n_A, F(m))$ for all $m \ge m_0$.

But that's just my suspucion, not more since I otherwise not know how a candidate for $N$ can be found.

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  • $\begingroup$ See for example Hartshorne, Theorem III.9.9. The idea is to use a Čech resolution and Serre vanishing. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jun 30, 2020 at 21:43
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    $\begingroup$ @R.vanDobbendeBruyn: You mean that one which starts with "Let $T$ be an integral noetherian scheme. Let $X \subset \mathbb{P}^n_T$ be a closed subscheme... And the claim is that $X$ is flat over $T$ iff Hilbert polynomial $P_t$ is independent of $t$. Here is required that $T$ is integral, especially reduced, the book seems to work without it but thats subtile. $\endgroup$
    – user267839
    Commented Jun 30, 2020 at 22:16
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    $\begingroup$ This can be deduced from semicontinuity theorem . For sufficiently large $N$, higher direct images of $F(N)$ vanish. Then the direct image is locally free. $\endgroup$
    – Mohan
    Commented Jun 30, 2020 at 23:03
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    $\begingroup$ You're right, the reference doesn't cover the precise statement, but as @Mohan points out Thm III.12.11 does (together with Thm III.8.8). $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jul 1, 2020 at 0:09
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    $\begingroup$ I also think the (very short) proof of Thm III.9.9 (i)$\Rightarrow$(ii) applies verbatim in this context, because that direction uses neither that $\mathscr F$ is the structure sheaf nor that $T$ is integral. In fact the first is not even stated as an assumption, and the second is only used for (iii)$\Rightarrow$(ii). $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jul 1, 2020 at 0:37

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