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If $f:X\rightarrow Y$ is a flat family of projective varieties over an algebraically closed field. Suppose that $X$ and $Y$ are irreducible. Let $F$ be a torsion-free coherent sheaf on $X$ of rank $r$.

1) Is there a non-empty open set $U$ of $Y$ such that the restriction of $F$ to every fiber $f^{-1}(y)$, $y\in Y$ is torsion-free?

2) Is it true that there is a nonempty open subset $U$ of $Y$ such that for all $y\in U$, the rank $F|_{f^{-1} (y)}=r$? I feel this is true because rank of fibers is upper semi continuous and rank of $F$ at the generic point of $X$ is $r$. Is this correct?

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    $\begingroup$ Under the additional hypothesis that the fibers of the support of $\mathcal{F}$ are (everywhere locally) pure dimensional, openness of the torsion-free locus (the $S_1$-locus) is proved in EGA IV_3, Theorem 12.1.1. EGA never quite gives a counterexample in case the fiber dimension is not pure dimensional, but that hypothesis is used in the proof. The rank is upper semicontinuous. $\endgroup$
    – Jason Starr
    Commented Mar 25, 2017 at 8:55
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    $\begingroup$ I just noticed that you are assuming that $X$ and $Y$ are irreducible. That settles the issue about pure dimensionality. I also noticed that you are not assuming that $\mathcal{F}$ is flat. However, by generic flatness, there does exist a dense open subset $U$ of $Y$ over which $\mathcal{F}$ is flat. After replacing $Y$ by $U$ and $X$ by $f^{-1}(U)$, you can apply Theorem 12.1.1. $\endgroup$
    – Jason Starr
    Commented Mar 25, 2017 at 9:13
  • $\begingroup$ @JasonStarr Prof. Starr, thank you for the reference. If I understand the comment correctly, the theorem proves that the torsion-free locus is open. Does it also mean it is non-empty? I asked this because in general for an open property, non-emptiness is not assured. $\endgroup$
    – user52991
    Commented Mar 25, 2017 at 10:00
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    $\begingroup$ The generic point, $\text{Spec}(\mathcal{O}_{Y,\eta_Y}),$ is contained in the maximal open subscheme over which $\mathcal{F}$ is flat and $S_1.$ $\endgroup$
    – Jason Starr
    Commented Mar 25, 2017 at 10:35
  • $\begingroup$ @JasonStarr, thank you. The restriction to the generic fiber $f^{-1}(\eta_Y)$ is torsion-free. This is because the generic point $\eta_X$ maps to $\eta_Y$ under $f$, thus $\eta_X$ is in the generic fiber. I hope this is right. $\endgroup$
    – user52991
    Commented Mar 25, 2017 at 10:40

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