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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I imagine that this is a simple application of Castelnuovo's criterion, which is valid in all characteristics.
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– Daniel Loughran
Mar 19 at 9:41
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?
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– YCor
Mar 19 at 10:19
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 Mar 19 at 8:34
Jérémy Blanc
asked Mar 19 at 8:21
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
Mar 19 at 9:41
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
Mar 19 at 10:19
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
Mar 19 at 9:41
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
Mar 19 at 10:19
$begingroup$
I imagine that this is a simple application of Castelnuovo's criterion, which is valid in all characteristics.
$endgroup$
– Daniel Loughran
Mar 19 at 9:41
$begingroup$
I imagine that this is a simple application of Castelnuovo's criterion, which is valid in all characteristics.
$endgroup$
– Daniel Loughran
Mar 19 at 9:41
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
Mar 19 at 10:19
$begingroup$
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
Mar 19 at 10:19
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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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 Mar 19 at 9:14
Laurent Moret-BaillyLaurent Moret-Bailly
14.5k14769
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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
Mar 19 at 9:41
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
Mar 19 at 10:19