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Is every supersingular (thus unirational for ${\rm char }\ k = p\geq 5$, from Liedtke) $K3$ surface defined over a finite field? I guess this is true for Kummer surfaces, for example, since supersingular Kummer surfaces can be seen as the resolution of singularities of $(E\times E)/<-1>$, $E$ is a supersingular curve, so it is defined over a finite field but I don't know if there are examples of a supersingular surfaces defined over a function field over a finite field, say.

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    $\begingroup$ Doesn't the proof of unirationality involve studying the moduli space of supersingular K3 surfaces, which is ten-dimensional or something? A generic point of the moduli space will not be defined over any finite field. $\endgroup$
    – Will Sawin
    Commented Oct 31, 2017 at 12:35
  • $\begingroup$ You are probably thinking of the analogy with elliptic curves. A supersingular elliptic curve is defined over a finite field "because" the moduli space of supersingular elliptic curves is zero-dimensional. However, the moduli space of supersingular princ. pol'd abelian surfaces is positive dimensional. As in Will's comment, a generic point of (a positive-dimensional component of) this moduli space will not be defined over a finite field (although all supersingular abelian surfaces over an algebraically closed field $k$ are $k$-isogenous to a product of supersingular elliptic curves). $\endgroup$
    – Ariyan Javanpeykar
    Commented Oct 31, 2017 at 19:21
  • $\begingroup$ @AriyanJavanpeykar and WillSawin, Thanks for the comment. Ok, I see, but what I'd like to know is if there exists a known example of a elliptic surface over a function field that is K3 and supersingular (as a surface, $\rho = 22$). All papers I found treated special cases of supersingular k3 elliptic surfaces, but all are defined over a finite field.. $\endgroup$
    – Vinicius M.
    Commented Oct 31, 2017 at 21:18
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    $\begingroup$ @ViniciusM. Let $K$ be the function field of a one-dimensional subspace of the moduli $M_{ss}$ of supersingular (polarized) K3 surfaces. There is a finite field extension $K'/K$ and a "natural" morphism Spec $K'\to M_{ss}$. The universal family over $M_{ss}$ induces a polarized supersingular K3 surface $X$ over $K'$. Note that $K'$ is the function field of a curve over $\mathbb{F}_q$, and that $X$ can't be defined over a finite field. Finally, by construction, $X$ is supersingular, hence $\rho = 22$ and $X$ admits an elliptic fibration. Is this good enough? $\endgroup$
    – Ariyan Javanpeykar
    Commented Nov 1, 2017 at 9:46
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    $\begingroup$ @ViniciusM. If you want a more explicit example, find a supersingular abelian surface $A$ over $k(t)$ (with $k$ some finite field) which can't be defined over $\overline{\mathbb{F}_p}$, and then let $X$ be the associated Kummer surface. Probably, $X$ is a (supersingular) K3 surface which can't be defined over a finite field; see arxiv.org/pdf/1412.7107.pdf for examples of such abelian surfaces $\endgroup$
    – Ariyan Javanpeykar
    Commented Nov 1, 2017 at 9:51

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