irreducible smooth proper one-dimensional $mathbb{Z}$-schemes
$begingroup$
Is there an irreducible smooth proper one-dimensional $mathbb{Z}$-scheme not isomorphic to projective line?
algebraic-geometry
asked Jan 5 at 17:57
Aknazar KazhymuratAknazar Kazhymurat
1534
$endgroup$
add a comment |
$begingroup$
Is there an irreducible smooth proper one-dimensional $mathbb{Z}$-scheme not isomorphic to projective line?
algebraic-geometry
asked Jan 5 at 17:57
Aknazar KazhymuratAknazar Kazhymurat
1534
$endgroup$
2
$begingroup$
Do you mean relative dimension 1 or really dimension 1 ?
$endgroup$
– Roland
Jan 5 at 19:46
$begingroup$
@Roland lets say the structure morphism has relative dimension 1
$endgroup$
– Aknazar Kazhymurat
Jan 6 at 4:39
add a comment |
$begingroup$
Is there an irreducible smooth proper one-dimensional $mathbb{Z}$-scheme not isomorphic to projective line?
algebraic-geometry
asked Jan 5 at 17:57
Aknazar KazhymuratAknazar Kazhymurat
1534
$endgroup$
Is there an irreducible smooth proper one-dimensional $mathbb{Z}$-scheme not isomorphic to projective line?
algebraic-geometry
algebraic-geometry
asked Jan 5 at 17:57
Aknazar KazhymuratAknazar Kazhymurat
1534
asked Jan 5 at 17:57
Aknazar KazhymuratAknazar Kazhymurat
1534
asked Jan 5 at 17:57
Aknazar KazhymuratAknazar Kazhymurat
1534
asked Jan 5 at 17:57
Aknazar KazhymuratAknazar Kazhymurat
1534
asked Jan 5 at 17:57
Aknazar KazhymuratAknazar Kazhymurat
1534
1534
2
$begingroup$
Do you mean relative dimension 1 or really dimension 1 ?
$endgroup$
– Roland
Jan 5 at 19:46
$begingroup$
@Roland lets say the structure morphism has relative dimension 1
$endgroup$
– Aknazar Kazhymurat
Jan 6 at 4:39
add a comment |
2
$begingroup$
Do you mean relative dimension 1 or really dimension 1 ?
$endgroup$
– Roland
Jan 5 at 19:46
$begingroup$
@Roland lets say the structure morphism has relative dimension 1
$endgroup$
– Aknazar Kazhymurat
Jan 6 at 4:39
2
2
$begingroup$
Do you mean relative dimension 1 or really dimension 1 ?
$endgroup$
– Roland
Jan 5 at 19:46
$begingroup$
Do you mean relative dimension 1 or really dimension 1 ?
$endgroup$
– Roland
Jan 5 at 19:46
$begingroup$
@Roland lets say the structure morphism has relative dimension 1
$endgroup$
– Aknazar Kazhymurat
Jan 6 at 4:39
$begingroup$
@Roland lets say the structure morphism has relative dimension 1
$endgroup$
– Aknazar Kazhymurat
Jan 6 at 4:39
add a comment |
1 Answer
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The answer is no.
Let $X$ be an irreducible scheme and let $Xto mathrm{Spec} mathbb{Z} = S$ be a smooth proper morphism of schemes. Let $Xto Tto S$ be the Stein factorization of $Xto S$. Note that $T$ is irreducible and that $Tto S$ is smooth proper and finite. It follows from Hermite-Minkowski's theorem that $T =S$ (as $mathrm{Spec} mathbb{Z}$ is simply connected). Thus, $Xto S$ has geometrically connected fibres. Let $C := X_{mathbb{Q}}$ be the generic fibre of $Xto mathrm{Spec} mathbb{Z}$. Then $C$ is a smooth proper geometrically connected curve over $mathbb{Q}$ with good reduction everywhere. In particular, its Jacobian $Jac(C)$ has good reduction everywhere. By the theorem of Abrashkin-Fontaine (see https://link.springer.com/article/10.1007%2FBF01388584) it follows that $Jac(C_{mathbb{Q}})=0$, so that $C_{mathbb{Q}}$ is a smooth proper geometrically connected curve of genus zero. I leave it to you to conclude that $C_{mathbb{Q}} = mathbb{P}^1_{mathbb{Q}}$ and that $X = mathbb{P}^1_{mathbb{Z}}$. (Hint: use that $X(mathbb{F}_p)$ is non-empty for every $p$.)
answered Jan 8 at 13:58
Ariyan JavanpeykarAriyan Javanpeykar
1,317912
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add a comment |
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$begingroup$
The answer is no.
Let $X$ be an irreducible scheme and let $Xto mathrm{Spec} mathbb{Z} = S$ be a smooth proper morphism of schemes. Let $Xto Tto S$ be the Stein factorization of $Xto S$. Note that $T$ is irreducible and that $Tto S$ is smooth proper and finite. It follows from Hermite-Minkowski's theorem that $T =S$ (as $mathrm{Spec} mathbb{Z}$ is simply connected). Thus, $Xto S$ has geometrically connected fibres. Let $C := X_{mathbb{Q}}$ be the generic fibre of $Xto mathrm{Spec} mathbb{Z}$. Then $C$ is a smooth proper geometrically connected curve over $mathbb{Q}$ with good reduction everywhere. In particular, its Jacobian $Jac(C)$ has good reduction everywhere. By the theorem of Abrashkin-Fontaine (see https://link.springer.com/article/10.1007%2FBF01388584) it follows that $Jac(C_{mathbb{Q}})=0$, so that $C_{mathbb{Q}}$ is a smooth proper geometrically connected curve of genus zero. I leave it to you to conclude that $C_{mathbb{Q}} = mathbb{P}^1_{mathbb{Q}}$ and that $X = mathbb{P}^1_{mathbb{Z}}$. (Hint: use that $X(mathbb{F}_p)$ is non-empty for every $p$.)
answered Jan 8 at 13:58
Ariyan JavanpeykarAriyan Javanpeykar
1,317912
$endgroup$
add a comment |
$begingroup$
The answer is no.
Let $X$ be an irreducible scheme and let $Xto mathrm{Spec} mathbb{Z} = S$ be a smooth proper morphism of schemes. Let $Xto Tto S$ be the Stein factorization of $Xto S$. Note that $T$ is irreducible and that $Tto S$ is smooth proper and finite. It follows from Hermite-Minkowski's theorem that $T =S$ (as $mathrm{Spec} mathbb{Z}$ is simply connected). Thus, $Xto S$ has geometrically connected fibres. Let $C := X_{mathbb{Q}}$ be the generic fibre of $Xto mathrm{Spec} mathbb{Z}$. Then $C$ is a smooth proper geometrically connected curve over $mathbb{Q}$ with good reduction everywhere. In particular, its Jacobian $Jac(C)$ has good reduction everywhere. By the theorem of Abrashkin-Fontaine (see https://link.springer.com/article/10.1007%2FBF01388584) it follows that $Jac(C_{mathbb{Q}})=0$, so that $C_{mathbb{Q}}$ is a smooth proper geometrically connected curve of genus zero. I leave it to you to conclude that $C_{mathbb{Q}} = mathbb{P}^1_{mathbb{Q}}$ and that $X = mathbb{P}^1_{mathbb{Z}}$. (Hint: use that $X(mathbb{F}_p)$ is non-empty for every $p$.)
answered Jan 8 at 13:58
Ariyan JavanpeykarAriyan Javanpeykar
1,317912
$endgroup$
add a comment |
$begingroup$
The answer is no.
Let $X$ be an irreducible scheme and let $Xto mathrm{Spec} mathbb{Z} = S$ be a smooth proper morphism of schemes. Let $Xto Tto S$ be the Stein factorization of $Xto S$. Note that $T$ is irreducible and that $Tto S$ is smooth proper and finite. It follows from Hermite-Minkowski's theorem that $T =S$ (as $mathrm{Spec} mathbb{Z}$ is simply connected). Thus, $Xto S$ has geometrically connected fibres. Let $C := X_{mathbb{Q}}$ be the generic fibre of $Xto mathrm{Spec} mathbb{Z}$. Then $C$ is a smooth proper geometrically connected curve over $mathbb{Q}$ with good reduction everywhere. In particular, its Jacobian $Jac(C)$ has good reduction everywhere. By the theorem of Abrashkin-Fontaine (see https://link.springer.com/article/10.1007%2FBF01388584) it follows that $Jac(C_{mathbb{Q}})=0$, so that $C_{mathbb{Q}}$ is a smooth proper geometrically connected curve of genus zero. I leave it to you to conclude that $C_{mathbb{Q}} = mathbb{P}^1_{mathbb{Q}}$ and that $X = mathbb{P}^1_{mathbb{Z}}$. (Hint: use that $X(mathbb{F}_p)$ is non-empty for every $p$.)
answered Jan 8 at 13:58
Ariyan JavanpeykarAriyan Javanpeykar
1,317912
$endgroup$
The answer is no.
Let $X$ be an irreducible scheme and let $Xto mathrm{Spec} mathbb{Z} = S$ be a smooth proper morphism of schemes. Let $Xto Tto S$ be the Stein factorization of $Xto S$. Note that $T$ is irreducible and that $Tto S$ is smooth proper and finite. It follows from Hermite-Minkowski's theorem that $T =S$ (as $mathrm{Spec} mathbb{Z}$ is simply connected). Thus, $Xto S$ has geometrically connected fibres. Let $C := X_{mathbb{Q}}$ be the generic fibre of $Xto mathrm{Spec} mathbb{Z}$. Then $C$ is a smooth proper geometrically connected curve over $mathbb{Q}$ with good reduction everywhere. In particular, its Jacobian $Jac(C)$ has good reduction everywhere. By the theorem of Abrashkin-Fontaine (see https://link.springer.com/article/10.1007%2FBF01388584) it follows that $Jac(C_{mathbb{Q}})=0$, so that $C_{mathbb{Q}}$ is a smooth proper geometrically connected curve of genus zero. I leave it to you to conclude that $C_{mathbb{Q}} = mathbb{P}^1_{mathbb{Q}}$ and that $X = mathbb{P}^1_{mathbb{Z}}$. (Hint: use that $X(mathbb{F}_p)$ is non-empty for every $p$.)
answered Jan 8 at 13:58
Ariyan JavanpeykarAriyan Javanpeykar
1,317912
answered Jan 8 at 13:58
Ariyan JavanpeykarAriyan Javanpeykar
1,317912
answered Jan 8 at 13:58
Ariyan JavanpeykarAriyan Javanpeykar
1,317912
answered Jan 8 at 13:58
Ariyan JavanpeykarAriyan Javanpeykar
1,317912
1,317912
add a comment |
add a comment |
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$begingroup$
Do you mean relative dimension 1 or really dimension 1 ?
$endgroup$
– Roland
Jan 5 at 19:46
$begingroup$
@Roland lets say the structure morphism has relative dimension 1
$endgroup$
– Aknazar Kazhymurat
Jan 6 at 4:39