TITLE: Are stably rational surfaces all rational?
QUESTION [10 upvotes]: 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).

REPLY [13 votes]: The result is true in all characteristics. See O. Zariski, Illinois J. Math. 2(1958), 303-315.