# Is there a purely inseparable covering $\mathbb{A}^2 \to K$ of a Kummer surface $K$ over $\mathbb{F}_{p^2}$?

Let $E_i\!: y_i^2 = x_i^3 + a_4x_i + a_6$ be two copies ($i = 1$, $2$) of a supersingular elliptic curve over a finite field $\mathbb{F}_{p^2}$, for odd prime $p > 3$. Consider the Kummer surface $K = \mathrm{Kum}(A)$ for the superspecial abelian surface $A = E_1\times E_2$. Katsura and Schütt proved in the articles "Generalized Kummer surfaces and their unirationality in characteristic p (1987)" and "Zariski K3 surfaces (2017)" that the surface $K$ is a Zariski surface, i.e., there is a purely inseparable covering $\mathbb{A}^2 \to K$ over $\overline{\mathbb{F}_p}$, where $p \not\equiv 1$ (mod $12$). Is it still true over $\mathbb{F}_{p^2}$ at least for some supersingular elliptic curve and some $p$?

A positive answer to this question may have value for cryptography :) If $K$ is a Zariski surface over $\mathbb{F}_{p^2}$, then we have a method to compress a pair $(P_1, P_2) \in A(\mathbb{F}_{p^2})$ by computation of a map from $A$ to the affine plane $\mathbb{A}^2$ (through $K$) of separable degree 2. This is very compact and efficient method, because for decompression we need to solve only one quadratic equation. Computation of a preimage for a purely inseparable map is very fast. People usually take two projections on x-coordinates for $P_1$ and $P_2$ independently, hence they should solve two quadratic equations. The state of the art for this part of cryptography is represented, for example, in the article https://eprint.iacr.org/2017/1143.

• What's the cryptographic context? (Crypto or no, it looks like a natural enough question in arthmetic geometry) May 23, 2018 at 12:37
• I added cryptographic comments. May 23, 2018 at 16:45

A supersingular elliptic curve in characteristic $p>2$ will always be covered by the hyperelliptic curve $y^2=x^p-x$ as the Jacobian of the latter is isogenous to a product of supersingular elliptic curves. When $p \equiv 2 \pmod{3}$ this can be made very explicit for $y^2=x^3-1$, which is covered by $y^2=x^{p+1}-1$ in the obvious way and $y^2=x^{p+1}-1$ is isomorphic to $y^2=x^p-x$ by sending a Weierstrass point to infinity.
• How can we explicitly find a purely inseparable part $S \to K$ of the map $\mathbb{A}^2 \to K$, where $S$ is some "intermediate" surface? May 23, 2018 at 19:26
• @DimaKoshelev I missed the fact you wanted purely inseparable. But if $\mathbb{A}^2 \to X \to K$ and $X \to K$ is Galois, you can try to lift the Galois action to $\mathbb{A}^2$ and take the quotient, which will also be a rational surface. May 23, 2018 at 19:53
• Likewise for $p \equiv 3 \bmod 4$ by sending $y^2 = x^{p+1} - 1$ to $y^2 = X^4 - 1$ where $X = x^{(p+1)/4}$. May 23, 2018 at 20:45
• @DimaKoshelev Yes, a priori, only over the algebraic closure. I believe the cases of $p \equiv 2 \pmod{3}, p \equiv 3 \pmod{4}$ already mentioned will be quite explicit and can be checked directly. Shioda's paper Math. Ann. 211 (1974), 233–236, does the Fermat surface of degree $p+1$ explicitly and shows it's a Zariski surface over $\mathbb{F}_{p^2}$. May 24, 2018 at 6:59