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Curve is a Noetherian Topological Space
The Next CEO of Stack OverflowSeparated and Finite Type Scheme over an Algebraically Closed FieldWhy is the disjoint union of a finite number of affine schemes an affine scheme?Why is every Noetherian zero-dimensional scheme finite discrete?Can we prove that a quasi-compact locally noetherian space is noetherian without Axiom of Choice?Does pullback of schemes by monomorphism produce topological pullback?Is a product of two Noetherian schemes over Spec $mathbb Z$ a Noetherian scheme?AG on non-Noetherian ringsFinite Morphism of Schemes is ProperGalois Action on underlying Topological space of a Group SchemeProper Non Constant Morphism of Curves has Finite Fibers
$begingroup$
Let $C$ be curve, so a $1$-dimensional, separated $k$-scheme of finite type).
My question is how to see that the underlying topological space of $C$ is a Noetherian space, so satisfies the descending chain condition for closed subsets.
My attempts:
Obviously $C$ is locally noetherian:
Because $C$ is $k$-scheme of finite type so we can find for each $c∈C$ a wlog affine open neighborhood $U_c=Spec(R)$ of $c$ 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 $⇔ Spec(R)$ noetherian works only for affine schemes.
Generally $C$ is not affine so I hornestly don't know how to show that $C$ is a noetherian space in satisfying way. One way to see it is would be to embedd it in a $mathbbP^n$. But I find it a bit overkill like so would like to prefer a more elementary argument basing on the definition of a curve as given above.
algebraic-geometry
asked 2 days ago
KarlPeterKarlPeter
5611316
$endgroup$
add a comment |
$begingroup$
Let $C$ be curve, so a $1$-dimensional, separated $k$-scheme of finite type).
My question is how to see that the underlying topological space of $C$ is a Noetherian space, so satisfies the descending chain condition for closed subsets.
My attempts:
Obviously $C$ is locally noetherian:
Because $C$ is $k$-scheme of finite type so we can find for each $c∈C$ a wlog affine open neighborhood $U_c=Spec(R)$ of $c$ 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 $⇔ Spec(R)$ noetherian works only for affine schemes.
Generally $C$ is not affine so I hornestly don't know how to show that $C$ is a noetherian space in satisfying way. One way to see it is would be to embedd it in a $mathbbP^n$. But I find it a bit overkill like so would like to prefer a more elementary argument basing on the definition of a curve as given above.
algebraic-geometry
asked 2 days ago
KarlPeterKarlPeter
5611316
$endgroup$
add a comment |
$begingroup$
Let $C$ be curve, so a $1$-dimensional, separated $k$-scheme of finite type).
My question is how to see that the underlying topological space of $C$ is a Noetherian space, so satisfies the descending chain condition for closed subsets.
My attempts:
Obviously $C$ is locally noetherian:
Because $C$ is $k$-scheme of finite type so we can find for each $c∈C$ a wlog affine open neighborhood $U_c=Spec(R)$ of $c$ 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 $⇔ Spec(R)$ noetherian works only for affine schemes.
Generally $C$ is not affine so I hornestly don't know how to show that $C$ is a noetherian space in satisfying way. One way to see it is would be to embedd it in a $mathbbP^n$. But I find it a bit overkill like so would like to prefer a more elementary argument basing on the definition of a curve as given above.
algebraic-geometry
asked 2 days ago
KarlPeterKarlPeter
5611316
$endgroup$
Let $C$ be curve, so a $1$-dimensional, separated $k$-scheme of finite type).
My question is how to see that the underlying topological space of $C$ is a Noetherian space, so satisfies the descending chain condition for closed subsets.
My attempts:
Obviously $C$ is locally noetherian:
Because $C$ is $k$-scheme of finite type so we can find for each $c∈C$ a wlog affine open neighborhood $U_c=Spec(R)$ of $c$ 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 $⇔ Spec(R)$ noetherian works only for affine schemes.
Generally $C$ is not affine so I hornestly don't know how to show that $C$ is a noetherian space in satisfying way. One way to see it is would be to embedd it in a $mathbbP^n$. But I find it a bit overkill like so would like to prefer a more elementary argument basing on the definition of a curve as given above.
algebraic-geometry
algebraic-geometry
asked 2 days ago
KarlPeterKarlPeter
5611316
asked 2 days ago
KarlPeterKarlPeter
5611316
asked 2 days ago
KarlPeterKarlPeter
5611316
asked 2 days ago
KarlPeterKarlPeter
5611316
asked 2 days ago
KarlPeterKarlPeter
5611316
5611316
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Since $C$ is finite type over $k$, it is quasicompact, so it is covered by finitely many affine open sets (each of which are Noetherian). It follows immediately that $C$ is Noetherian (given any descending sequence of closed sets, intersect it with each affine open set and it must eventually stabilize on each one).
answered 2 days ago
Eric WofseyEric Wofsey
191k14216349
$endgroup$
$begingroup$
The funny part is that an irreducible separated scheme locally of finite type and dimension 1 over a field is also quasi-compact (there was a MO post)
$endgroup$
– Aknazar Kazhymurat
yesterday
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Since $C$ is finite type over $k$, it is quasicompact, so it is covered by finitely many affine open sets (each of which are Noetherian). It follows immediately that $C$ is Noetherian (given any descending sequence of closed sets, intersect it with each affine open set and it must eventually stabilize on each one).
answered 2 days ago
Eric WofseyEric Wofsey
191k14216349
$endgroup$
$begingroup$
The funny part is that an irreducible separated scheme locally of finite type and dimension 1 over a field is also quasi-compact (there was a MO post)
$endgroup$
– Aknazar Kazhymurat
yesterday
add a comment |
$begingroup$
Since $C$ is finite type over $k$, it is quasicompact, so it is covered by finitely many affine open sets (each of which are Noetherian). It follows immediately that $C$ is Noetherian (given any descending sequence of closed sets, intersect it with each affine open set and it must eventually stabilize on each one).
answered 2 days ago
Eric WofseyEric Wofsey
191k14216349
$endgroup$
$begingroup$
The funny part is that an irreducible separated scheme locally of finite type and dimension 1 over a field is also quasi-compact (there was a MO post)
$endgroup$
– Aknazar Kazhymurat
yesterday
add a comment |
$begingroup$
Since $C$ is finite type over $k$, it is quasicompact, so it is covered by finitely many affine open sets (each of which are Noetherian). It follows immediately that $C$ is Noetherian (given any descending sequence of closed sets, intersect it with each affine open set and it must eventually stabilize on each one).
answered 2 days ago
Eric WofseyEric Wofsey
191k14216349
$endgroup$
Since $C$ is finite type over $k$, it is quasicompact, so it is covered by finitely many affine open sets (each of which are Noetherian). It follows immediately that $C$ is Noetherian (given any descending sequence of closed sets, intersect it with each affine open set and it must eventually stabilize on each one).
answered 2 days ago
Eric WofseyEric Wofsey
191k14216349
answered 2 days ago
Eric WofseyEric Wofsey
191k14216349
answered 2 days ago
Eric WofseyEric Wofsey
191k14216349
answered 2 days ago
Eric WofseyEric Wofsey
191k14216349
191k14216349
$begingroup$
The funny part is that an irreducible separated scheme locally of finite type and dimension 1 over a field is also quasi-compact (there was a MO post)
$endgroup$
– Aknazar Kazhymurat
yesterday
add a comment |
$begingroup$
The funny part is that an irreducible separated scheme locally of finite type and dimension 1 over a field is also quasi-compact (there was a MO post)
$endgroup$
– Aknazar Kazhymurat
yesterday
$begingroup$
The funny part is that an irreducible separated scheme locally of finite type and dimension 1 over a field is also quasi-compact (there was a MO post)
$endgroup$
– Aknazar Kazhymurat
yesterday
$begingroup$
The funny part is that an irreducible separated scheme locally of finite type and dimension 1 over a field is also quasi-compact (there was a MO post)
$endgroup$
– Aknazar Kazhymurat
yesterday
add a comment |
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