Proper Non Constant Morphism of Curves has Finite Fibers The Next CEO of Stack OverflowFibers of a proper birational morphismProperties of fibers of a morphism of varietiesMorphism whose fibers are finite and reduced is unramifiedEquivalence of two definitions of proper morphism for separated varietiesWhy are proper morphisms restricted to those of finite type?Schemes as closed subschemes in étale schemesProper Morphism closedValuation Criterion for Proper MorphismsFinite Morphism of Schemes is ProperExceptional Curves provided by Dominant Morphism
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Proper Non Constant Morphism of Curves has Finite Fibers
The Next CEO of Stack OverflowFibers of a proper birational morphismProperties of fibers of a morphism of varietiesMorphism whose fibers are finite and reduced is unramifiedEquivalence of two definitions of proper morphism for separated varietiesWhy are proper morphisms restricted to those of finite type?Schemes as closed subschemes in étale schemesProper Morphism closedValuation Criterion for Proper MorphismsFinite Morphism of Schemes is ProperExceptional Curves provided by Dominant Morphism
$begingroup$
Let $f: C to D$ be a proper, non constant scheme morphism of curves. (a curve is for me a $1$-dimensional, separated $k$-scheme of finite type).
Assume futhermore that $C$ and $D$ are irreducible and proper.
Let $b in D$ an arbitrary closed point of $D$. My question is how to see that the fiber $f^-1(b)$ is a finite set?
My considerations:
Since the property "finite type" is stable under base change we deduce that the scheme structure $C times_D kappa(b)$ of the fiber $f^-1(b)$ is a $kappa(b)$-scheme of finite type. Therefore the ring of grobal sections of $C times_D kappa(b)$ is Noetherian by Hilbert's Basissatz.
Futhermore $f^-1(b)$ is discrete and a union of closed points. But here I don't see why the beeing Noetherian property for the ring $O_C times_D kappa(b)(C times_D kappa(b))$ imply the Noether ascending property for $C times_D kappa(b)$ as topological space. The problem is that $C times_D kappa(b)$ isn't affine.
Another approach would be to show that every complement of an non empty open set in $C$ is finite but I'm not sure why it should here hold. I can only say that every open set of $C$ is dense but not more.
algebraic-geometry
asked 2 days ago
KarlPeterKarlPeter
5341316
$endgroup$
add a comment |
$begingroup$
Let $f: C to D$ be a proper, non constant scheme morphism of curves. (a curve is for me a $1$-dimensional, separated $k$-scheme of finite type).
Assume futhermore that $C$ and $D$ are irreducible and proper.
Let $b in D$ an arbitrary closed point of $D$. My question is how to see that the fiber $f^-1(b)$ is a finite set?
My considerations:
Since the property "finite type" is stable under base change we deduce that the scheme structure $C times_D kappa(b)$ of the fiber $f^-1(b)$ is a $kappa(b)$-scheme of finite type. Therefore the ring of grobal sections of $C times_D kappa(b)$ is Noetherian by Hilbert's Basissatz.
Futhermore $f^-1(b)$ is discrete and a union of closed points. But here I don't see why the beeing Noetherian property for the ring $O_C times_D kappa(b)(C times_D kappa(b))$ imply the Noether ascending property for $C times_D kappa(b)$ as topological space. The problem is that $C times_D kappa(b)$ isn't affine.
Another approach would be to show that every complement of an non empty open set in $C$ is finite but I'm not sure why it should here hold. I can only say that every open set of $C$ is dense but not more.
algebraic-geometry
asked 2 days ago
KarlPeterKarlPeter
5341316
$endgroup$
$begingroup$
The fiber is a closed subset of a noetherian topological space and is thus noetherian. Secondly, such a morphism will always be proper (see 01W6), so it doesn't really make sense to talk about replacing proper by surjective without further altering the question.
$endgroup$
– KReiser
2 days ago
$begingroup$
@KReiser: ok so the problem reduces the point to verify that a curve $C$ is a noetherian topological space. But here can only deduce that it is locally noetherian. Indeed, locally noetherian is clear since $C$ is $k$-scheme of finite type so we can find for each $c in C$ a wlog affine open neighborhood $U_c = Spec(R)$ such that $R = k[x_1,...,x_n]/I$ and by Hilbert $R$ is noetherian therefore $U_c$ is noetherian (especially as topological space). The proplem is that the argument $R$ noetherian $Leftrightarrow$ $Spec(R)$ noetherian works only for affine schemes.
$endgroup$
– KarlPeter
yesterday
$begingroup$
@KReiser:But $C$ is in general not affine so don't know how to show that $C$ is a noetherian space. The overkill argument would be to embedd it in a $mathbbP^n$. Do you see a more "elementary" argument?
$endgroup$
– KarlPeter
yesterday
add a comment |
$begingroup$
Let $f: C to D$ be a proper, non constant scheme morphism of curves. (a curve is for me a $1$-dimensional, separated $k$-scheme of finite type).
Assume futhermore that $C$ and $D$ are irreducible and proper.
Let $b in D$ an arbitrary closed point of $D$. My question is how to see that the fiber $f^-1(b)$ is a finite set?
My considerations:
Since the property "finite type" is stable under base change we deduce that the scheme structure $C times_D kappa(b)$ of the fiber $f^-1(b)$ is a $kappa(b)$-scheme of finite type. Therefore the ring of grobal sections of $C times_D kappa(b)$ is Noetherian by Hilbert's Basissatz.
Futhermore $f^-1(b)$ is discrete and a union of closed points. But here I don't see why the beeing Noetherian property for the ring $O_C times_D kappa(b)(C times_D kappa(b))$ imply the Noether ascending property for $C times_D kappa(b)$ as topological space. The problem is that $C times_D kappa(b)$ isn't affine.
Another approach would be to show that every complement of an non empty open set in $C$ is finite but I'm not sure why it should here hold. I can only say that every open set of $C$ is dense but not more.
algebraic-geometry
asked 2 days ago
KarlPeterKarlPeter
5341316
$endgroup$
Let $f: C to D$ be a proper, non constant scheme morphism of curves. (a curve is for me a $1$-dimensional, separated $k$-scheme of finite type).
Assume futhermore that $C$ and $D$ are irreducible and proper.
Let $b in D$ an arbitrary closed point of $D$. My question is how to see that the fiber $f^-1(b)$ is a finite set?
My considerations:
Since the property "finite type" is stable under base change we deduce that the scheme structure $C times_D kappa(b)$ of the fiber $f^-1(b)$ is a $kappa(b)$-scheme of finite type. Therefore the ring of grobal sections of $C times_D kappa(b)$ is Noetherian by Hilbert's Basissatz.
Futhermore $f^-1(b)$ is discrete and a union of closed points. But here I don't see why the beeing Noetherian property for the ring $O_C times_D kappa(b)(C times_D kappa(b))$ imply the Noether ascending property for $C times_D kappa(b)$ as topological space. The problem is that $C times_D kappa(b)$ isn't affine.
Another approach would be to show that every complement of an non empty open set in $C$ is finite but I'm not sure why it should here hold. I can only say that every open set of $C$ is dense but not more.
algebraic-geometry
algebraic-geometry
asked 2 days ago
KarlPeterKarlPeter
5341316
asked 2 days ago
KarlPeterKarlPeter
5341316
edited yesterday
KarlPeter
asked 2 days ago
KarlPeterKarlPeter
5341316
asked 2 days ago
KarlPeterKarlPeter
5341316
asked 2 days ago
KarlPeterKarlPeter
5341316
5341316
$begingroup$
The fiber is a closed subset of a noetherian topological space and is thus noetherian. Secondly, such a morphism will always be proper (see 01W6), so it doesn't really make sense to talk about replacing proper by surjective without further altering the question.
$endgroup$
– KReiser
2 days ago
$begingroup$
@KReiser: ok so the problem reduces the point to verify that a curve $C$ is a noetherian topological space. But here can only deduce that it is locally noetherian. Indeed, locally noetherian is clear since $C$ is $k$-scheme of finite type so we can find for each $c in C$ a wlog affine open neighborhood $U_c = Spec(R)$ such that $R = k[x_1,...,x_n]/I$ and by Hilbert $R$ is noetherian therefore $U_c$ is noetherian (especially as topological space). The proplem is that the argument $R$ noetherian $Leftrightarrow$ $Spec(R)$ noetherian works only for affine schemes.
$endgroup$
– KarlPeter
yesterday
$begingroup$
@KReiser:But $C$ is in general not affine so don't know how to show that $C$ is a noetherian space. The overkill argument would be to embedd it in a $mathbbP^n$. Do you see a more "elementary" argument?
$endgroup$
– KarlPeter
yesterday
add a comment |
$begingroup$
The fiber is a closed subset of a noetherian topological space and is thus noetherian. Secondly, such a morphism will always be proper (see 01W6), so it doesn't really make sense to talk about replacing proper by surjective without further altering the question.
$endgroup$
– KReiser
2 days ago
$begingroup$
@KReiser: ok so the problem reduces the point to verify that a curve $C$ is a noetherian topological space. But here can only deduce that it is locally noetherian. Indeed, locally noetherian is clear since $C$ is $k$-scheme of finite type so we can find for each $c in C$ a wlog affine open neighborhood $U_c = Spec(R)$ such that $R = k[x_1,...,x_n]/I$ and by Hilbert $R$ is noetherian therefore $U_c$ is noetherian (especially as topological space). The proplem is that the argument $R$ noetherian $Leftrightarrow$ $Spec(R)$ noetherian works only for affine schemes.
$endgroup$
– KarlPeter
yesterday
$begingroup$
@KReiser:But $C$ is in general not affine so don't know how to show that $C$ is a noetherian space. The overkill argument would be to embedd it in a $mathbbP^n$. Do you see a more "elementary" argument?
$endgroup$
– KarlPeter
yesterday
$begingroup$
The fiber is a closed subset of a noetherian topological space and is thus noetherian. Secondly, such a morphism will always be proper (see 01W6), so it doesn't really make sense to talk about replacing proper by surjective without further altering the question.
$endgroup$
– KReiser
2 days ago
$begingroup$
The fiber is a closed subset of a noetherian topological space and is thus noetherian. Secondly, such a morphism will always be proper (see 01W6), so it doesn't really make sense to talk about replacing proper by surjective without further altering the question.
$endgroup$
– KReiser
2 days ago
$begingroup$
@KReiser: ok so the problem reduces the point to verify that a curve $C$ is a noetherian topological space. But here can only deduce that it is locally noetherian. Indeed, locally noetherian is clear since $C$ is $k$-scheme of finite type so we can find for each $c in C$ a wlog affine open neighborhood $U_c = Spec(R)$ such that $R = k[x_1,...,x_n]/I$ and by Hilbert $R$ is noetherian therefore $U_c$ is noetherian (especially as topological space). The proplem is that the argument $R$ noetherian $Leftrightarrow$ $Spec(R)$ noetherian works only for affine schemes.
$endgroup$
– KarlPeter
yesterday
$begingroup$
@KReiser: ok so the problem reduces the point to verify that a curve $C$ is a noetherian topological space. But here can only deduce that it is locally noetherian. Indeed, locally noetherian is clear since $C$ is $k$-scheme of finite type so we can find for each $c in C$ a wlog affine open neighborhood $U_c = Spec(R)$ such that $R = k[x_1,...,x_n]/I$ and by Hilbert $R$ is noetherian therefore $U_c$ is noetherian (especially as topological space). The proplem is that the argument $R$ noetherian $Leftrightarrow$ $Spec(R)$ noetherian works only for affine schemes.
$endgroup$
– KarlPeter
yesterday
$begingroup$
@KReiser:But $C$ is in general not affine so don't know how to show that $C$ is a noetherian space. The overkill argument would be to embedd it in a $mathbbP^n$. Do you see a more "elementary" argument?
$endgroup$
– KarlPeter
yesterday
$begingroup$
@KReiser:But $C$ is in general not affine so don't know how to show that $C$ is a noetherian space. The overkill argument would be to embedd it in a $mathbbP^n$. Do you see a more "elementary" argument?
$endgroup$
– KarlPeter
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If there were an infinite fiber, then it would have an infinite irreducible component. That is, we have an infinite closed irreducible set inside of the curve $C$. This is impossible for dimension reasons unless that component is the whole curve. But the morphism is not constant, so the component can’t be the whole curve.
answered yesterday
Samir CanningSamir Canning
50039
$endgroup$
$begingroup$
Why this fiber would have an infinite irreducible component? Must it have a finite number of irreducible components?
$endgroup$
– KarlPeter
yesterday
$begingroup$
Seems that it also boils down to topological Noether property for the fiber.
$endgroup$
– KarlPeter
yesterday
$begingroup$
Curves are Noetherian because they’re locally Noetherian and finite type so in particular they are quasi compact. Subsets of Noetherian spaces are Noetherian. Therefore the fiber has only finitely many irreducible components.
$endgroup$
– Samir Canning
yesterday
$begingroup$
Do you know a reference /sketch of the argument that locally Noetherian + finite type imply Noetherian?
$endgroup$
– KarlPeter
yesterday
1
$begingroup$
Finite type by definition includes quasi compact. Locally Noetherian plus quasicompact by definition means Noetherian as a scheme. Noetherian schemes are Noetherian topological spaces.
$endgroup$
– Samir Canning
yesterday
add a comment |
Your Answer
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$begingroup$
If there were an infinite fiber, then it would have an infinite irreducible component. That is, we have an infinite closed irreducible set inside of the curve $C$. This is impossible for dimension reasons unless that component is the whole curve. But the morphism is not constant, so the component can’t be the whole curve.
answered yesterday
Samir CanningSamir Canning
50039
$endgroup$
$begingroup$
Why this fiber would have an infinite irreducible component? Must it have a finite number of irreducible components?
$endgroup$
– KarlPeter
yesterday
$begingroup$
Seems that it also boils down to topological Noether property for the fiber.
$endgroup$
– KarlPeter
yesterday
$begingroup$
Curves are Noetherian because they’re locally Noetherian and finite type so in particular they are quasi compact. Subsets of Noetherian spaces are Noetherian. Therefore the fiber has only finitely many irreducible components.
$endgroup$
– Samir Canning
yesterday
$begingroup$
Do you know a reference /sketch of the argument that locally Noetherian + finite type imply Noetherian?
$endgroup$
– KarlPeter
yesterday
1
$begingroup$
Finite type by definition includes quasi compact. Locally Noetherian plus quasicompact by definition means Noetherian as a scheme. Noetherian schemes are Noetherian topological spaces.
$endgroup$
– Samir Canning
yesterday
add a comment |
$begingroup$
If there were an infinite fiber, then it would have an infinite irreducible component. That is, we have an infinite closed irreducible set inside of the curve $C$. This is impossible for dimension reasons unless that component is the whole curve. But the morphism is not constant, so the component can’t be the whole curve.
answered yesterday
Samir CanningSamir Canning
50039
$endgroup$
$begingroup$
Why this fiber would have an infinite irreducible component? Must it have a finite number of irreducible components?
$endgroup$
– KarlPeter
yesterday
$begingroup$
Seems that it also boils down to topological Noether property for the fiber.
$endgroup$
– KarlPeter
yesterday
$begingroup$
Curves are Noetherian because they’re locally Noetherian and finite type so in particular they are quasi compact. Subsets of Noetherian spaces are Noetherian. Therefore the fiber has only finitely many irreducible components.
$endgroup$
– Samir Canning
yesterday
$begingroup$
Do you know a reference /sketch of the argument that locally Noetherian + finite type imply Noetherian?
$endgroup$
– KarlPeter
yesterday
1
$begingroup$
Finite type by definition includes quasi compact. Locally Noetherian plus quasicompact by definition means Noetherian as a scheme. Noetherian schemes are Noetherian topological spaces.
$endgroup$
– Samir Canning
yesterday
add a comment |
$begingroup$
If there were an infinite fiber, then it would have an infinite irreducible component. That is, we have an infinite closed irreducible set inside of the curve $C$. This is impossible for dimension reasons unless that component is the whole curve. But the morphism is not constant, so the component can’t be the whole curve.
answered yesterday
Samir CanningSamir Canning
50039
$endgroup$
If there were an infinite fiber, then it would have an infinite irreducible component. That is, we have an infinite closed irreducible set inside of the curve $C$. This is impossible for dimension reasons unless that component is the whole curve. But the morphism is not constant, so the component can’t be the whole curve.
answered yesterday
Samir CanningSamir Canning
50039
answered yesterday
Samir CanningSamir Canning
50039
answered yesterday
Samir CanningSamir Canning
50039
answered yesterday
Samir CanningSamir Canning
50039
50039
$begingroup$
Why this fiber would have an infinite irreducible component? Must it have a finite number of irreducible components?
$endgroup$
– KarlPeter
yesterday
$begingroup$
Seems that it also boils down to topological Noether property for the fiber.
$endgroup$
– KarlPeter
yesterday
$begingroup$
Curves are Noetherian because they’re locally Noetherian and finite type so in particular they are quasi compact. Subsets of Noetherian spaces are Noetherian. Therefore the fiber has only finitely many irreducible components.
$endgroup$
– Samir Canning
yesterday
$begingroup$
Do you know a reference /sketch of the argument that locally Noetherian + finite type imply Noetherian?
$endgroup$
– KarlPeter
yesterday
1
$begingroup$
Finite type by definition includes quasi compact. Locally Noetherian plus quasicompact by definition means Noetherian as a scheme. Noetherian schemes are Noetherian topological spaces.
$endgroup$
– Samir Canning
yesterday
add a comment |
$begingroup$
Why this fiber would have an infinite irreducible component? Must it have a finite number of irreducible components?
$endgroup$
– KarlPeter
yesterday
$begingroup$
Seems that it also boils down to topological Noether property for the fiber.
$endgroup$
– KarlPeter
yesterday
$begingroup$
Curves are Noetherian because they’re locally Noetherian and finite type so in particular they are quasi compact. Subsets of Noetherian spaces are Noetherian. Therefore the fiber has only finitely many irreducible components.
$endgroup$
– Samir Canning
yesterday
$begingroup$
Do you know a reference /sketch of the argument that locally Noetherian + finite type imply Noetherian?
$endgroup$
– KarlPeter
yesterday
1
$begingroup$
Finite type by definition includes quasi compact. Locally Noetherian plus quasicompact by definition means Noetherian as a scheme. Noetherian schemes are Noetherian topological spaces.
$endgroup$
– Samir Canning
yesterday
$begingroup$
Why this fiber would have an infinite irreducible component? Must it have a finite number of irreducible components?
$endgroup$
– KarlPeter
yesterday
$begingroup$
Why this fiber would have an infinite irreducible component? Must it have a finite number of irreducible components?
$endgroup$
– KarlPeter
yesterday
$begingroup$
Seems that it also boils down to topological Noether property for the fiber.
$endgroup$
– KarlPeter
yesterday
$begingroup$
Seems that it also boils down to topological Noether property for the fiber.
$endgroup$
– KarlPeter
yesterday
$begingroup$
Curves are Noetherian because they’re locally Noetherian and finite type so in particular they are quasi compact. Subsets of Noetherian spaces are Noetherian. Therefore the fiber has only finitely many irreducible components.
$endgroup$
– Samir Canning
yesterday
$begingroup$
Curves are Noetherian because they’re locally Noetherian and finite type so in particular they are quasi compact. Subsets of Noetherian spaces are Noetherian. Therefore the fiber has only finitely many irreducible components.
$endgroup$
– Samir Canning
yesterday
$begingroup$
Do you know a reference /sketch of the argument that locally Noetherian + finite type imply Noetherian?
$endgroup$
– KarlPeter
yesterday
$begingroup$
Do you know a reference /sketch of the argument that locally Noetherian + finite type imply Noetherian?
$endgroup$
– KarlPeter
yesterday
1
1
$begingroup$
Finite type by definition includes quasi compact. Locally Noetherian plus quasicompact by definition means Noetherian as a scheme. Noetherian schemes are Noetherian topological spaces.
$endgroup$
– Samir Canning
yesterday
$begingroup$
Finite type by definition includes quasi compact. Locally Noetherian plus quasicompact by definition means Noetherian as a scheme. Noetherian schemes are Noetherian topological spaces.
$endgroup$
– Samir Canning
yesterday
add a comment |
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$begingroup$
The fiber is a closed subset of a noetherian topological space and is thus noetherian. Secondly, such a morphism will always be proper (see 01W6), so it doesn't really make sense to talk about replacing proper by surjective without further altering the question.
$endgroup$
– KReiser
2 days ago
$begingroup$
@KReiser: ok so the problem reduces the point to verify that a curve $C$ is a noetherian topological space. But here can only deduce that it is locally noetherian. Indeed, locally noetherian is clear since $C$ is $k$-scheme of finite type so we can find for each $c in C$ a wlog affine open neighborhood $U_c = Spec(R)$ such that $R = k[x_1,...,x_n]/I$ and by Hilbert $R$ is noetherian therefore $U_c$ is noetherian (especially as topological space). The proplem is that the argument $R$ noetherian $Leftrightarrow$ $Spec(R)$ noetherian works only for affine schemes.
$endgroup$
– KarlPeter
yesterday
$begingroup$
@KReiser:But $C$ is in general not affine so don't know how to show that $C$ is a noetherian space. The overkill argument would be to embedd it in a $mathbbP^n$. Do you see a more "elementary" argument?
$endgroup$
– KarlPeter
yesterday