Are stably rational surfaces all rational?Del pezzo surfaces in positive characteristicNumerically negative exceptional divisor on a surface.Universal property of blowing downStabilisers of group actionsAre stably birational varieties birational?Methods of showing a variety is stably rationalConic bundles on quartic del Pezzo surfacesPurely inseparable $k$-rational dominant maps between an absolutely irreducible $k$-surface and $mathbbP^2$Is an open subscheme of a rationally connected variety, rationally connected?Is there an odd degree unirational parametrization of a cubic threefold?
Are stably rational surfaces all rational?
Del pezzo surfaces in positive characteristicNumerically negative exceptional divisor on a surface.Universal property of blowing downStabilisers of group actionsAre stably birational varieties birational?Methods of showing a variety is stably rationalConic bundles on quartic del Pezzo surfacesPurely inseparable $k$-rational dominant maps between an absolutely irreducible $k$-surface and $mathbbP^2$Is an open subscheme of a rationally connected variety, rationally connected?Is there an odd degree unirational parametrization of a cubic threefold?
$begingroup$
Let $X$ be an irreducible surface such that $X times mathbbP^1$ 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
asked Mar 19 at 8:21
Jérémy BlancJérémy Blanc
4,19411536
$endgroup$
add a comment |
$begingroup$
Let $X$ be an irreducible surface such that $X times mathbbP^1$ 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
asked Mar 19 at 8:21
Jérémy BlancJérémy Blanc
4,19411536
$endgroup$
$begingroup$
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 $XtimesmathbfP^n$ rational (for some $n$) implies $X$ rational?
$endgroup$
– YCor
Mar 19 at 10:19
add a comment |
$begingroup$
Let $X$ be an irreducible surface such that $X times mathbbP^1$ 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
asked Mar 19 at 8:21
Jérémy BlancJérémy Blanc
4,19411536
$endgroup$
Let $X$ be an irreducible surface such that $X times mathbbP^1$ 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
asked Mar 19 at 8:21
Jérémy BlancJérémy Blanc
4,19411536
asked Mar 19 at 8:21
Jérémy BlancJérémy Blanc
4,19411536
edited Mar 19 at 8:34
Jérémy Blanc
asked Mar 19 at 8:21
Jérémy BlancJérémy Blanc
4,19411536
asked Mar 19 at 8:21
Jérémy BlancJérémy Blanc
4,19411536
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 $XtimesmathbfP^n$ 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 $XtimesmathbfP^n$ 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 $XtimesmathbfP^n$ 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 $XtimesmathbfP^n$ rational (for some $n$) implies $X$ rational?
$endgroup$
– YCor
Mar 19 at 10:19
add a comment |
1 Answer
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$begingroup$
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
$endgroup$
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.
answered Mar 19 at 9:14
Laurent Moret-BaillyLaurent Moret-Bailly
14.5k14769
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
answered Mar 19 at 9:14
Laurent Moret-BaillyLaurent Moret-Bailly
14.5k14769
answered Mar 19 at 9:14
Laurent Moret-BaillyLaurent Moret-Bailly
14.5k14769
answered Mar 19 at 9:14
Laurent Moret-BaillyLaurent Moret-Bailly
14.5k14769
14.5k14769
add a comment |
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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 $XtimesmathbfP^n$ rational (for some $n$) implies $X$ rational?
$endgroup$
– YCor
Mar 19 at 10:19