Irreducible of finite Krull dimension implies quasi-compact? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)A closed point in the closure of any point in the closure of any point of an irreducible schemeWhen is an irreducible scheme quasi-compact?not locally of finite type implies not universally closed?If $X,Y$ are regular and of finite type over $S$, can $Xtimes _S Y$ be embedded into a regular $S$-scheme? Krull dimension in equivariant cohomologyKrull dimension and Morley rankQuasi-finite morphisms of stacksQuasi-compactness of irreducible separated scheme locally of finite typeWhen does the image of a morphism of schemes support scheme structure?Sufficient condition for quasi-compactness of scheme morphismSchemes monomorphing into affine scheme of dimension 1
Irreducible of finite Krull dimension implies quasi-compact?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)A closed point in the closure of any point in the closure of any point of an irreducible schemeWhen is an irreducible scheme quasi-compact?not locally of finite type implies not universally closed?If $X,Y$ are regular and of finite type over $S$, can $Xtimes _S Y$ be embedded into a regular $S$-scheme? Krull dimension in equivariant cohomologyKrull dimension and Morley rankQuasi-finite morphisms of stacksQuasi-compactness of irreducible separated scheme locally of finite typeWhen does the image of a morphism of schemes support scheme structure?Sufficient condition for quasi-compactness of scheme morphismSchemes monomorphing into affine scheme of dimension 1
5
$begingroup$
Let $X$ be the underlying space of a scheme.
- If $X$ is irreducible of finite Krull dimension, is it necessarily
quasi-compact? - Is it necessarily Noetherian?
- What if we assume not
only that Krull dimension is finite but also that it is 1?
at.algebraic-topology gn.general-topology schemes
edited 11 hours ago
András Bátkai
3,84142342
asked 11 hours ago
schematic_ftmschematic_ftm
261
New contributor
schematic_ftm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
5
$begingroup$
Let $X$ be the underlying space of a scheme.
- If $X$ is irreducible of finite Krull dimension, is it necessarily
quasi-compact? - Is it necessarily Noetherian?
- What if we assume not
only that Krull dimension is finite but also that it is 1?
at.algebraic-topology gn.general-topology schemes
edited 11 hours ago
András Bátkai
3,84142342
asked 11 hours ago
schematic_ftmschematic_ftm
261
New contributor
schematic_ftm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
a separated 1d example: Enumerate the primes: $p_1=2,p_2,...$. Let $R_n$ be the ring of rational numbers whose denominator is not divisible by the first $n$ primes $p_1,...,p_n$. This has spectrum with 1 generic point and $n$ closed points with local rings $mathbb Z_(p_i)$. Each ring is a localization of the next: $R_n=R_n+1[p_n+1^-1]$, so its spectrum is open in the next. Then $X=bigcup_n mathrmSpecR_n$ is a noncompact scheme with global sections $mathbb Z$. Clearly not $mathrmSpecmathbb Z$!
$endgroup$
– Ben Wieland
4 hours ago
$begingroup$
A 2d example closer to practical geometry: Let $X_0=mathbb P^2$ and $p_0$ some point on it. Let $X_n+1$ be the blow-up of $X_n$ at $p_n$ and let $p_n+1in Dsubset X_n+1$ be some point of the exceptional divisor. Let $U_n=X_n-p_n$. Each is open in the next. So $U=bigcup U_n$ is a noncompact scheme... The inverse limit of the $X_n$ is probably a compact non-noetherian scheme with open set $U$... The sequence of $p_n$ specifies a 2d valuation ring, necessarily non-noetherian. It has compact spectrum, but the complement of the unique closed point is related to $U$, maybe even 1d.
$endgroup$
– Ben Wieland
4 hours ago
add a comment |
5
5
5
$begingroup$
Let $X$ be the underlying space of a scheme.
- If $X$ is irreducible of finite Krull dimension, is it necessarily
quasi-compact? - Is it necessarily Noetherian?
- What if we assume not
only that Krull dimension is finite but also that it is 1?
at.algebraic-topology gn.general-topology schemes
edited 11 hours ago
András Bátkai
3,84142342
asked 11 hours ago
schematic_ftmschematic_ftm
261
New contributor
schematic_ftm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let $X$ be the underlying space of a scheme.
- If $X$ is irreducible of finite Krull dimension, is it necessarily
quasi-compact? - Is it necessarily Noetherian?
- What if we assume not
only that Krull dimension is finite but also that it is 1?
at.algebraic-topology gn.general-topology schemes
at.algebraic-topology gn.general-topology schemes
edited 11 hours ago
András Bátkai
3,84142342
asked 11 hours ago
schematic_ftmschematic_ftm
261
New contributor
schematic_ftm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 11 hours ago
András Bátkai
3,84142342
asked 11 hours ago
schematic_ftmschematic_ftm
261
New contributor
schematic_ftm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 11 hours ago
András Bátkai
3,84142342
edited 11 hours ago
András Bátkai
3,84142342
edited 11 hours ago
András Bátkai
3,84142342
3,84142342
asked 11 hours ago
schematic_ftmschematic_ftm
261
New contributor
schematic_ftm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 11 hours ago
schematic_ftmschematic_ftm
261
asked 11 hours ago
schematic_ftmschematic_ftm
261
261
New contributor
schematic_ftm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
schematic_ftm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
schematic_ftm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
a separated 1d example: Enumerate the primes: $p_1=2,p_2,...$. Let $R_n$ be the ring of rational numbers whose denominator is not divisible by the first $n$ primes $p_1,...,p_n$. This has spectrum with 1 generic point and $n$ closed points with local rings $mathbb Z_(p_i)$. Each ring is a localization of the next: $R_n=R_n+1[p_n+1^-1]$, so its spectrum is open in the next. Then $X=bigcup_n mathrmSpecR_n$ is a noncompact scheme with global sections $mathbb Z$. Clearly not $mathrmSpecmathbb Z$!
$endgroup$
– Ben Wieland
4 hours ago
$begingroup$
A 2d example closer to practical geometry: Let $X_0=mathbb P^2$ and $p_0$ some point on it. Let $X_n+1$ be the blow-up of $X_n$ at $p_n$ and let $p_n+1in Dsubset X_n+1$ be some point of the exceptional divisor. Let $U_n=X_n-p_n$. Each is open in the next. So $U=bigcup U_n$ is a noncompact scheme... The inverse limit of the $X_n$ is probably a compact non-noetherian scheme with open set $U$... The sequence of $p_n$ specifies a 2d valuation ring, necessarily non-noetherian. It has compact spectrum, but the complement of the unique closed point is related to $U$, maybe even 1d.
$endgroup$
– Ben Wieland
4 hours ago
add a comment |
$begingroup$
a separated 1d example: Enumerate the primes: $p_1=2,p_2,...$. Let $R_n$ be the ring of rational numbers whose denominator is not divisible by the first $n$ primes $p_1,...,p_n$. This has spectrum with 1 generic point and $n$ closed points with local rings $mathbb Z_(p_i)$. Each ring is a localization of the next: $R_n=R_n+1[p_n+1^-1]$, so its spectrum is open in the next. Then $X=bigcup_n mathrmSpecR_n$ is a noncompact scheme with global sections $mathbb Z$. Clearly not $mathrmSpecmathbb Z$!
$endgroup$
– Ben Wieland
4 hours ago
$begingroup$
A 2d example closer to practical geometry: Let $X_0=mathbb P^2$ and $p_0$ some point on it. Let $X_n+1$ be the blow-up of $X_n$ at $p_n$ and let $p_n+1in Dsubset X_n+1$ be some point of the exceptional divisor. Let $U_n=X_n-p_n$. Each is open in the next. So $U=bigcup U_n$ is a noncompact scheme... The inverse limit of the $X_n$ is probably a compact non-noetherian scheme with open set $U$... The sequence of $p_n$ specifies a 2d valuation ring, necessarily non-noetherian. It has compact spectrum, but the complement of the unique closed point is related to $U$, maybe even 1d.
$endgroup$
– Ben Wieland
4 hours ago
$begingroup$
a separated 1d example: Enumerate the primes: $p_1=2,p_2,...$. Let $R_n$ be the ring of rational numbers whose denominator is not divisible by the first $n$ primes $p_1,...,p_n$. This has spectrum with 1 generic point and $n$ closed points with local rings $mathbb Z_(p_i)$. Each ring is a localization of the next: $R_n=R_n+1[p_n+1^-1]$, so its spectrum is open in the next. Then $X=bigcup_n mathrmSpecR_n$ is a noncompact scheme with global sections $mathbb Z$. Clearly not $mathrmSpecmathbb Z$!
$endgroup$
– Ben Wieland
4 hours ago
$begingroup$
a separated 1d example: Enumerate the primes: $p_1=2,p_2,...$. Let $R_n$ be the ring of rational numbers whose denominator is not divisible by the first $n$ primes $p_1,...,p_n$. This has spectrum with 1 generic point and $n$ closed points with local rings $mathbb Z_(p_i)$. Each ring is a localization of the next: $R_n=R_n+1[p_n+1^-1]$, so its spectrum is open in the next. Then $X=bigcup_n mathrmSpecR_n$ is a noncompact scheme with global sections $mathbb Z$. Clearly not $mathrmSpecmathbb Z$!
$endgroup$
– Ben Wieland
4 hours ago
$begingroup$
A 2d example closer to practical geometry: Let $X_0=mathbb P^2$ and $p_0$ some point on it. Let $X_n+1$ be the blow-up of $X_n$ at $p_n$ and let $p_n+1in Dsubset X_n+1$ be some point of the exceptional divisor. Let $U_n=X_n-p_n$. Each is open in the next. So $U=bigcup U_n$ is a noncompact scheme... The inverse limit of the $X_n$ is probably a compact non-noetherian scheme with open set $U$... The sequence of $p_n$ specifies a 2d valuation ring, necessarily non-noetherian. It has compact spectrum, but the complement of the unique closed point is related to $U$, maybe even 1d.
$endgroup$
– Ben Wieland
4 hours ago
$begingroup$
A 2d example closer to practical geometry: Let $X_0=mathbb P^2$ and $p_0$ some point on it. Let $X_n+1$ be the blow-up of $X_n$ at $p_n$ and let $p_n+1in Dsubset X_n+1$ be some point of the exceptional divisor. Let $U_n=X_n-p_n$. Each is open in the next. So $U=bigcup U_n$ is a noncompact scheme... The inverse limit of the $X_n$ is probably a compact non-noetherian scheme with open set $U$... The sequence of $p_n$ specifies a 2d valuation ring, necessarily non-noetherian. It has compact spectrum, but the complement of the unique closed point is related to $U$, maybe even 1d.
$endgroup$
– Ben Wieland
4 hours ago
add a comment |
1 Answer
1
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10
$begingroup$
The answer is no for all these questions. Take the line with infinite origins: the scheme obtained by gluing an infinite amount of copies of $mathbbA^1$ along the open subsets $mathbbG_m$. This has Krull dimension 1 (there are only closed points and the unique generic point) and it is irreducible (the only proper nonempty closed subsets are the closed points) but it is not quasi-compact (it contains an infinite discrete set), and so in particular not Noetherian.
answered 11 hours ago
Denis NardinDenis Nardin
9,17223565
$endgroup$
1
$begingroup$
Denis Nardin, I am not sure about the second paragraph. I thought nilpotent thickening never affects the underlying topological space. So while your nilpotent thickening gives a non-locally Noetherian scheme, does it give a non-locally Noetherian space (about which is the question)?
$endgroup$
– Stepan Banach
10 hours ago
1
$begingroup$
@StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it.
$endgroup$
– Denis Nardin
10 hours ago
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
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active
oldest
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10
$begingroup$
The answer is no for all these questions. Take the line with infinite origins: the scheme obtained by gluing an infinite amount of copies of $mathbbA^1$ along the open subsets $mathbbG_m$. This has Krull dimension 1 (there are only closed points and the unique generic point) and it is irreducible (the only proper nonempty closed subsets are the closed points) but it is not quasi-compact (it contains an infinite discrete set), and so in particular not Noetherian.
answered 11 hours ago
Denis NardinDenis Nardin
9,17223565
$endgroup$
1
$begingroup$
Denis Nardin, I am not sure about the second paragraph. I thought nilpotent thickening never affects the underlying topological space. So while your nilpotent thickening gives a non-locally Noetherian scheme, does it give a non-locally Noetherian space (about which is the question)?
$endgroup$
– Stepan Banach
10 hours ago
1
$begingroup$
@StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it.
$endgroup$
– Denis Nardin
10 hours ago
add a comment |
10
$begingroup$
The answer is no for all these questions. Take the line with infinite origins: the scheme obtained by gluing an infinite amount of copies of $mathbbA^1$ along the open subsets $mathbbG_m$. This has Krull dimension 1 (there are only closed points and the unique generic point) and it is irreducible (the only proper nonempty closed subsets are the closed points) but it is not quasi-compact (it contains an infinite discrete set), and so in particular not Noetherian.
answered 11 hours ago
Denis NardinDenis Nardin
9,17223565
$endgroup$
1
$begingroup$
Denis Nardin, I am not sure about the second paragraph. I thought nilpotent thickening never affects the underlying topological space. So while your nilpotent thickening gives a non-locally Noetherian scheme, does it give a non-locally Noetherian space (about which is the question)?
$endgroup$
– Stepan Banach
10 hours ago
1
$begingroup$
@StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it.
$endgroup$
– Denis Nardin
10 hours ago
add a comment |
10
10
10
$begingroup$
The answer is no for all these questions. Take the line with infinite origins: the scheme obtained by gluing an infinite amount of copies of $mathbbA^1$ along the open subsets $mathbbG_m$. This has Krull dimension 1 (there are only closed points and the unique generic point) and it is irreducible (the only proper nonempty closed subsets are the closed points) but it is not quasi-compact (it contains an infinite discrete set), and so in particular not Noetherian.
answered 11 hours ago
Denis NardinDenis Nardin
9,17223565
$endgroup$
The answer is no for all these questions. Take the line with infinite origins: the scheme obtained by gluing an infinite amount of copies of $mathbbA^1$ along the open subsets $mathbbG_m$. This has Krull dimension 1 (there are only closed points and the unique generic point) and it is irreducible (the only proper nonempty closed subsets are the closed points) but it is not quasi-compact (it contains an infinite discrete set), and so in particular not Noetherian.
answered 11 hours ago
Denis NardinDenis Nardin
9,17223565
edited 10 hours ago
answered 11 hours ago
Denis NardinDenis Nardin
9,17223565
answered 11 hours ago
Denis NardinDenis Nardin
9,17223565
answered 11 hours ago
Denis NardinDenis Nardin
9,17223565
9,17223565
1
$begingroup$
Denis Nardin, I am not sure about the second paragraph. I thought nilpotent thickening never affects the underlying topological space. So while your nilpotent thickening gives a non-locally Noetherian scheme, does it give a non-locally Noetherian space (about which is the question)?
$endgroup$
– Stepan Banach
10 hours ago
1
$begingroup$
@StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it.
$endgroup$
– Denis Nardin
10 hours ago
add a comment |
1
$begingroup$
Denis Nardin, I am not sure about the second paragraph. I thought nilpotent thickening never affects the underlying topological space. So while your nilpotent thickening gives a non-locally Noetherian scheme, does it give a non-locally Noetherian space (about which is the question)?
$endgroup$
– Stepan Banach
10 hours ago
1
$begingroup$
@StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it.
$endgroup$
– Denis Nardin
10 hours ago
1
1
$begingroup$
Denis Nardin, I am not sure about the second paragraph. I thought nilpotent thickening never affects the underlying topological space. So while your nilpotent thickening gives a non-locally Noetherian scheme, does it give a non-locally Noetherian space (about which is the question)?
$endgroup$
– Stepan Banach
10 hours ago
$begingroup$
Denis Nardin, I am not sure about the second paragraph. I thought nilpotent thickening never affects the underlying topological space. So while your nilpotent thickening gives a non-locally Noetherian scheme, does it give a non-locally Noetherian space (about which is the question)?
$endgroup$
– Stepan Banach
10 hours ago
1
1
$begingroup$
@StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it.
$endgroup$
– Denis Nardin
10 hours ago
$begingroup$
@StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it.
$endgroup$
– Denis Nardin
10 hours ago
add a comment |
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-at.algebraic-topology, gn.general-topology, schemes
$begingroup$
a separated 1d example: Enumerate the primes: $p_1=2,p_2,...$. Let $R_n$ be the ring of rational numbers whose denominator is not divisible by the first $n$ primes $p_1,...,p_n$. This has spectrum with 1 generic point and $n$ closed points with local rings $mathbb Z_(p_i)$. Each ring is a localization of the next: $R_n=R_n+1[p_n+1^-1]$, so its spectrum is open in the next. Then $X=bigcup_n mathrmSpecR_n$ is a noncompact scheme with global sections $mathbb Z$. Clearly not $mathrmSpecmathbb Z$!
$endgroup$
– Ben Wieland
4 hours ago
$begingroup$
A 2d example closer to practical geometry: Let $X_0=mathbb P^2$ and $p_0$ some point on it. Let $X_n+1$ be the blow-up of $X_n$ at $p_n$ and let $p_n+1in Dsubset X_n+1$ be some point of the exceptional divisor. Let $U_n=X_n-p_n$. Each is open in the next. So $U=bigcup U_n$ is a noncompact scheme... The inverse limit of the $X_n$ is probably a compact non-noetherian scheme with open set $U$... The sequence of $p_n$ specifies a 2d valuation ring, necessarily non-noetherian. It has compact spectrum, but the complement of the unique closed point is related to $U$, maybe even 1d.
$endgroup$
– Ben Wieland
4 hours ago