Let $k$ be an infinite field (not necessarily algebraically closed), $X$ a smooth, projective curve over $k$ and $F$ a pure, coherent sheaf on $X$. Let $F'$ be the maximal destabilizing sheaf of $F$. Is the quotient sheaf $F/F'$ locally free?
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1$\begingroup$ If $k=\mathbb C$, then you can argue the following: as $F'$ is saturated in $F$ (saturation increases slope), $F/F'$ is torsion free, so it is locally free on an open subset whose complement has codimension at least two. In particular, if the base is a smooth curve, $F/F'$ is locally free. For a reference, see for instance Kobayashi, Differential Geometry of Complex Vector Bundles, Corollary 5.15. $\endgroup$– HenriCommented Mar 10, 2017 at 0:28
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$\begingroup$ @Henri Thanks for the answer. I had a non-algebraically closed field in mind. I have specified it more clearly in the question now. Does the same argument still work for non-algebraically closed field? If I understand corrently, the Serre's conditions on normality is not preserved under base change, meaning isn't it possible that the quotient is torsion-free on $X$ but not geometrically torsion-free? $\endgroup$– user45397Commented Mar 10, 2017 at 8:59
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$\begingroup$ I don't think there is such a thing as geometrically (non-)torsion-free. $\endgroup$– Thomas PoguntkeCommented Mar 13, 2017 at 14:57
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