Irreducible of finite Krull dimension implies quasi-compact?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?A closed point in the closure of any point in the closure of any point of an irreducible schemeKrull dimension of the ring of global sections
Irreducible of finite Krull dimension implies quasi-compact?
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?A closed point in the closure of any point in the closure of any point of an irreducible schemeKrull dimension of the ring of global sections
6
$begingroup$
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 Apr 17 at 17:13
András Bátkai
3,9214 gold badges24 silver badges43 bronze badges
asked Apr 17 at 16:45
schematic_ftmschematic_ftm
311 bronze badge
$endgroup$
add a comment
|
6
$begingroup$
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 Apr 17 at 17:13
András Bátkai
3,9214 gold badges24 silver badges43 bronze badges
asked Apr 17 at 16:45
schematic_ftmschematic_ftm
311 bronze badge
$endgroup$
1
$begingroup$
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
Apr 18 at 0:11
1
$begingroup$
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
Apr 18 at 0:16
$begingroup$
@BenWieland I think that should be an answer, it's a better example than mine and it's a pity to leave it in the comments.
$endgroup$
– Denis Nardin
Apr 18 at 8:59
add a comment
|
6
6
6
$begingroup$
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 Apr 17 at 17:13
András Bátkai
3,9214 gold badges24 silver badges43 bronze badges
asked Apr 17 at 16:45
schematic_ftmschematic_ftm
311 bronze badge
$endgroup$
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 Apr 17 at 17:13
András Bátkai
3,9214 gold badges24 silver badges43 bronze badges
asked Apr 17 at 16:45
schematic_ftmschematic_ftm
311 bronze badge
edited Apr 17 at 17:13
András Bátkai
3,9214 gold badges24 silver badges43 bronze badges
asked Apr 17 at 16:45
schematic_ftmschematic_ftm
311 bronze badge
edited Apr 17 at 17:13
András Bátkai
3,9214 gold badges24 silver badges43 bronze badges
edited Apr 17 at 17:13
András Bátkai
3,9214 gold badges24 silver badges43 bronze badges
edited Apr 17 at 17:13
András Bátkai
3,9214 gold badges24 silver badges43 bronze badges
3,9214 gold badges24 silver badges43 bronze badges
asked Apr 17 at 16:45
schematic_ftmschematic_ftm
311 bronze badge
asked Apr 17 at 16:45
schematic_ftmschematic_ftm
311 bronze badge
asked Apr 17 at 16:45
schematic_ftmschematic_ftm
311 bronze badge
311 bronze badge
1
$begingroup$
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
Apr 18 at 0:11
1
$begingroup$
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
Apr 18 at 0:16
$begingroup$
@BenWieland I think that should be an answer, it's a better example than mine and it's a pity to leave it in the comments.
$endgroup$
– Denis Nardin
Apr 18 at 8:59
add a comment
|
1
$begingroup$
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
Apr 18 at 0:11
1
$begingroup$
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
Apr 18 at 0:16
$begingroup$
@BenWieland I think that should be an answer, it's a better example than mine and it's a pity to leave it in the comments.
$endgroup$
– Denis Nardin
Apr 18 at 8:59
1
1
$begingroup$
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
Apr 18 at 0:11
$begingroup$
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
Apr 18 at 0:11
1
1
$begingroup$
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
Apr 18 at 0:16
$begingroup$
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
Apr 18 at 0:16
$begingroup$
@BenWieland I think that should be an answer, it's a better example than mine and it's a pity to leave it in the comments.
$endgroup$
– Denis Nardin
Apr 18 at 8:59
$begingroup$
@BenWieland I think that should be an answer, it's a better example than mine and it's a pity to leave it in the comments.
$endgroup$
– Denis Nardin
Apr 18 at 8:59
add a comment
|
1 Answer
1
active
oldest
votes
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 Apr 17 at 17:19
Denis NardinDenis Nardin
10.5k2 gold badges42 silver badges74 bronze badges
$endgroup$
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)?
$endgroup$
– user137767
Apr 17 at 17:44
1
$begingroup$
@StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it.
$endgroup$
– Denis Nardin
Apr 17 at 18:13
add a comment
|
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "504"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/4.0/"u003ecc by-sa 4.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f328305%2firreducible-of-finite-krull-dimension-implies-quasi-compact%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 Apr 17 at 17:19
Denis NardinDenis Nardin
10.5k2 gold badges42 silver badges74 bronze badges
$endgroup$
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)?
$endgroup$
– user137767
Apr 17 at 17:44
1
$begingroup$
@StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it.
$endgroup$
– Denis Nardin
Apr 17 at 18:13
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 Apr 17 at 17:19
Denis NardinDenis Nardin
10.5k2 gold badges42 silver badges74 bronze badges
$endgroup$
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)?
$endgroup$
– user137767
Apr 17 at 17:44
1
$begingroup$
@StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it.
$endgroup$
– Denis Nardin
Apr 17 at 18:13
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 Apr 17 at 17:19
Denis NardinDenis Nardin
10.5k2 gold badges42 silver badges74 bronze badges
$endgroup$
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 Apr 17 at 17:19
Denis NardinDenis Nardin
10.5k2 gold badges42 silver badges74 bronze badges
edited Apr 17 at 18:13
answered Apr 17 at 17:19
Denis NardinDenis Nardin
10.5k2 gold badges42 silver badges74 bronze badges
answered Apr 17 at 17:19
Denis NardinDenis Nardin
10.5k2 gold badges42 silver badges74 bronze badges
answered Apr 17 at 17:19
Denis NardinDenis Nardin
10.5k2 gold badges42 silver badges74 bronze badges
10.5k2 gold badges42 silver badges74 bronze badges
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)?
$endgroup$
– user137767
Apr 17 at 17:44
1
$begingroup$
@StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it.
$endgroup$
– Denis Nardin
Apr 17 at 18:13
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)?
$endgroup$
– user137767
Apr 17 at 17:44
1
$begingroup$
@StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it.
$endgroup$
– Denis Nardin
Apr 17 at 18:13
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)?
$endgroup$
– user137767
Apr 17 at 17:44
$begingroup$
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)?
$endgroup$
– user137767
Apr 17 at 17:44
1
1
$begingroup$
@StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it.
$endgroup$
– Denis Nardin
Apr 17 at 18:13
$begingroup$
@StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it.
$endgroup$
– Denis Nardin
Apr 17 at 18:13
add a comment
|
draft saved
draft discarded
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f328305%2firreducible-of-finite-krull-dimension-implies-quasi-compact%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
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
Apr 18 at 0:11
1
$begingroup$
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
Apr 18 at 0:16
$begingroup$
@BenWieland I think that should be an answer, it's a better example than mine and it's a pity to leave it in the comments.
$endgroup$
– Denis Nardin
Apr 18 at 8:59