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Mar 25, 2017 at 13:21 comment added user52991 @JasonStarr, Professor, I was also wondering if there is some characterization of the complement of the torsion-free locus in terms of codimension in $Y$
Mar 25, 2017 at 10:40 comment added user52991 @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.
Mar 25, 2017 at 10:35 comment added Jason Starr The generic point, $\text{Spec}(\mathcal{O}_{Y,\eta_Y}),$ is contained in the maximal open subscheme over which $\mathcal{F}$ is flat and $S_1.$
Mar 25, 2017 at 10:00 comment added user52991 @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.
Mar 25, 2017 at 9:13 comment added Jason Starr 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.
Mar 25, 2017 at 8:55 comment added Jason Starr 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.
Mar 25, 2017 at 7:45 history asked user52991 CC BY-SA 3.0