# Is a supersingular Kummer surface $k$-unirational in characteristic 2?

Let $k$ be a perfect field of even characteristic. Consider the simplest example of a supersingular genus 2 curve, i.e., $$C\!: y^2 + y = x^5.$$

By the article of J. S. Müller "Explicit Kummer surface formulas for arbitrary characteristics" the Kummer surface of the Jacobian of $C$ is the quartic surface $$K\!: k_2^2k_4^2 + k_1^3k_4 + k_1^2k_2k_3 + k_1k_2^3 + k_3^4.$$ Let $k_1 := t$, $k_2 := 1$, $k_3 := x$, $k_4 := y$. Then we have $$K\!: y^2 + t^3y = x^4 +t^2x + t.$$ Associate with $K$ the curve $K_t$ over the function field $k(t)$.

What is the geometric genus of this curve? Unfortunately, the classical geometric genus formula for plane curve doesn't work, because $K_t$ has the cusp on the infinity.

If $g(K_t) = 0$, then the surface $K$ has the quasi-elliptic $k$-fibration and so $K$ is $k$-unirational.

I think the proof is still true for an arbitrary genus 2 supersingular curve over $k$.

• The change of variables $y=z+x^2$ changes your curve to a conic in $x,z$, so genus zero. – Felipe Voloch Aug 23 '16 at 20:35