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Let $A$ be a supersingular abelian surface over a sufficiently large finite field $\mathbb{F}_q$ of characteristic $2$ and let $K_A = A/(-1)$ be the Kummer surface. Shioda ("Kummer surfaces in characteristic 2") and Katsura ("On Kummer surfaces in characteristic 2") proved that $K_A$ is rational.

There is the open conjecture about $\mathbb{F}_q$-unirationality of a rational surface with a smooth $\mathbb{F}_q$-point. See, for example, "Rational varieties: algebra, geometry and arithmetic" by Manin and Tsfasman.

Let $A$ be $\mathbb{F}_q$-simple, it est, $A$ isn't isogenous a direct product of 2 elliptic curves over $\mathbb{F}_q$. Is $K_A$ unirational over $\mathbb{F}_q$ in this case? In particular, what is $\mathbb{F}_q$-minimal smooth model for $K_A$?

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  • $\begingroup$ Could you please clarify your question? Is the following interpretation correct: You have a simple abelian variety $A$ over a finite field $\mathbb{F}_q$. You are assuming that $A$ is isogenous to a product of supersingular elliptic curves after some finite extension of $\mathbb{F}_q$. Then your question is whether $K_A$ is unirational over $\mathbb{F}_q$? $\endgroup$ – Daniel Loughran Jul 7 '16 at 8:30
  • $\begingroup$ Yes, all right. $\endgroup$ – Dima Koshelev Jul 7 '16 at 8:41
  • $\begingroup$ This question looks difficult. For example, in the easier case of del Pezzo surfaces, this is currently unknown: Every del Pezzo surface $S$ of degree $1$ over a finite field $\mathbb{F}_q$ has a rational point, yet it is unknown whether every $S$ is actually unirational over $\mathbb{F}_q$. Such a surface $S$ even has an elliptic fibration, so it is quite similar to your case. $\endgroup$ – Daniel Loughran Jul 7 '16 at 9:02
  • $\begingroup$ Is $K_A$ birationally isomorphic over $\mathbb{F}_q$ to an elliptic surface? $\endgroup$ – Dima Koshelev Jul 13 '16 at 4:50
  • $\begingroup$ It certainly obtains an elliptic fibration after a finite field extension, given by projecting onto one of the elliptic curves. I'm not sure about over the ground field. $\endgroup$ – Daniel Loughran Jul 13 '16 at 7:35

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