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).



Are stably rational surfaces all rational?
The result is true in all characteristics. See O. Zariski, Illinois J. Math. 2(1958), 303-315.


