Are stably rational surfaces all rational?
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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).
ag.algebraic-geometry birational-geometry
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add a comment |
$begingroup$
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).
ag.algebraic-geometry birational-geometry
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$begingroup$
I imagine that this is a simple application of Castelnuovo's criterion, which is valid in all characteristics.
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– Daniel Loughran
2 days ago
2
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At least to me, the question does not exactly fit with the title. The question "are stably rational surfaces rational" is rather whether $Xtimesmathbf{P}^n$ rational (for some $n$) implies $X$ rational?
$endgroup$
– YCor
2 days ago
add a comment |
$begingroup$
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).
ag.algebraic-geometry birational-geometry
$endgroup$
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).
ag.algebraic-geometry birational-geometry
ag.algebraic-geometry birational-geometry
edited 2 days ago
Jérémy Blanc
asked 2 days ago
Jérémy BlancJérémy Blanc
4,19411536
4,19411536
$begingroup$
I imagine that this is a simple application of Castelnuovo's criterion, which is valid in all characteristics.
$endgroup$
– Daniel Loughran
2 days ago
2
$begingroup$
At least to me, the question does not exactly fit with the title. The question "are stably rational surfaces rational" is rather whether $Xtimesmathbf{P}^n$ rational (for some $n$) implies $X$ rational?
$endgroup$
– YCor
2 days ago
add a comment |
$begingroup$
I imagine that this is a simple application of Castelnuovo's criterion, which is valid in all characteristics.
$endgroup$
– Daniel Loughran
2 days ago
2
$begingroup$
At least to me, the question does not exactly fit with the title. The question "are stably rational surfaces rational" is rather whether $Xtimesmathbf{P}^n$ rational (for some $n$) implies $X$ rational?
$endgroup$
– YCor
2 days ago
$begingroup$
I imagine that this is a simple application of Castelnuovo's criterion, which is valid in all characteristics.
$endgroup$
– Daniel Loughran
2 days ago
$begingroup$
I imagine that this is a simple application of Castelnuovo's criterion, which is valid in all characteristics.
$endgroup$
– Daniel Loughran
2 days ago
2
2
$begingroup$
At least to me, the question does not exactly fit with the title. The question "are stably rational surfaces rational" is rather whether $Xtimesmathbf{P}^n$ rational (for some $n$) implies $X$ rational?
$endgroup$
– YCor
2 days ago
$begingroup$
At least to me, the question does not exactly fit with the title. The question "are stably rational surfaces rational" is rather whether $Xtimesmathbf{P}^n$ rational (for some $n$) implies $X$ rational?
$endgroup$
– YCor
2 days ago
add a comment |
1 Answer
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The result is true in all characteristics. See O. Zariski, Illinois J. Math. 2(1958), 303-315.
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
The result is true in all characteristics. See O. Zariski, Illinois J. Math. 2(1958), 303-315.
$endgroup$
add a comment |
$begingroup$
The result is true in all characteristics. See O. Zariski, Illinois J. Math. 2(1958), 303-315.
$endgroup$
add a comment |
$begingroup$
The result is true in all characteristics. See O. Zariski, Illinois J. Math. 2(1958), 303-315.
$endgroup$
The result is true in all characteristics. See O. Zariski, Illinois J. Math. 2(1958), 303-315.
answered 2 days ago
Laurent Moret-BaillyLaurent Moret-Bailly
14.4k14769
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$begingroup$
I imagine that this is a simple application of Castelnuovo's criterion, which is valid in all characteristics.
$endgroup$
– Daniel Loughran
2 days ago
2
$begingroup$
At least to me, the question does not exactly fit with the title. The question "are stably rational surfaces rational" is rather whether $Xtimesmathbf{P}^n$ rational (for some $n$) implies $X$ rational?
$endgroup$
– YCor
2 days ago