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).

  • 1
    $\begingroup$ I imagine that this is a simple application of Castelnuovo's criterion, which is valid in all characteristics. $\endgroup$ – Daniel Loughran Mar 19 '19 at 9:41
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    $\begingroup$ At least to me, the question does not exactly fit with the title. The question "are stably rational surfaces rational" is rather whether $X\times\mathbf{P}^n$ rational (for some $n$) implies $X$ rational? $\endgroup$ – YCor Mar 19 '19 at 10:19

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

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