Let $X$ be an irreducible surface such that $X \times \mathbb{P}^1$ is rational. Is it true that $X$ is rational?

If the field is not algebraically closed, the answer is no in general (see A. Beauville, J.-L. Colliot-Thélène, J.-J. Sansuc et Sir Peter Swinnerton-Dyer, Variétés stablement rationnelles non rationnelles, Ann. of Math. 121(1985) 283–318.).

If the field is algebraically closed of characteristic zero, the answer is yes.

**What happens when the field is algebraically closed, of positive characteristic?**

(one could ask the same for simply rationally connected surfaces).