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Are there examples of projective varieties over a non-algebraically closed field such that every geometrically stable sheaf on the variety is simple? I see, for example in Huybrechts-Lehn and in some other mathoverflow posts, that any stable sheaf on a K3-surface is simple; however, it is not clear if the underlying field need be algebraically closed. Also, I have not seen a proof of this fact.

The part that I am most confused about is the following. Let $X$ be a (smooth) projective variety over a non-algebraically closed field $k$ and $E$ be a geometrically stable sheaf on $X$. Then, $$\overline{k} \cong \mathrm{End}(E_{\overline{k}}) \cong \mathrm{End}(E) \otimes_k {\overline{k}}. $$ But, Hom-functor commutes with flat base change, in particular with base field extension. This should imply that $\mathrm{End}(E) \cong k$. Is this argument wrong?

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  • $\begingroup$ Any geometrically stable sheaf $\mathscr F$ on a smooth projective $k$-variety $X$ is (geometrically) simple. Indeed, as you note, by flat base change it suffices to prove the result for $k = \bar k$. Then stability implies that a nonzero endomorphism $\phi \colon \mathscr F \to \mathscr F$ has to be surjective, since otherwise either $\operatorname{im} \phi$ or $\ker \phi$ has to be a proper subsheaf of larger slope. But a surjective endomorphism of a coherent sheaf is an isomorphism (see e.g. here). $\endgroup$
    – R. van Dobben de Bruyn
    Commented Aug 26, 2017 at 2:33
  • $\begingroup$ @R.vanDobbendeBruyn Thanks. By "(geometrically) simple", you mean simple as well as its pull-back to algebraic closure is simple? $\endgroup$
    – user43198
    Commented Aug 26, 2017 at 6:36
  • $\begingroup$ Yeah. Not sure if it's standard terminology. And depending on which definition of simple you use, it may or may not be automatic (does simple mean that $\operatorname{End}(\mathscr F)$ is a division algebra, or that it equals $k$?). $\endgroup$
    – R. van Dobben de Bruyn
    Commented Aug 26, 2017 at 12:52
  • $\begingroup$ @R.vanDobbendeBruyn For me, simple means that $\mathrm{End}(\mathcal{F})=k$. $\endgroup$
    – user43198
    Commented Aug 26, 2017 at 15:39

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