Strength of “every affine scheme is compact”
$begingroup$
Among the first results one usually sees, right after defining schemes, is that affine schemes are compact, however this statement is stronger than $mathsf{ZF}$: In $mathsf{ZFC}$ we have that $$operatorname{Spec}left(prod_{i=1}^inftyBbb F_2right)congbetaBbb N,$$the Stone-Čech compactification on $Bbb N$, but in $mathsf{ZF}$ the only ultrafilters we can prove to exist on $Bbb N$ are the principal ones, corresponding to $Bbb NsubseteqbetaBbb N$, which is a discrete subspace, hence not compact. So, if $betaBbb NsetminusBbb N=varnothing$ we have an affine scheme which is not compact.
Is the statement "every affine scheme is compact" equivalent to some more well-known statement over $mathsf{ZF}$? Or does it imply/is it implied by some more well known statements weaker than $mathsf{AC}$?
algebraic-geometry set-theory axiom-of-choice
asked Nov 26 '18 at 21:34
Alessandro CodenottiAlessandro Codenotti
3,76311539
$endgroup$
add a comment |
$begingroup$
Among the first results one usually sees, right after defining schemes, is that affine schemes are compact, however this statement is stronger than $mathsf{ZF}$: In $mathsf{ZFC}$ we have that $$operatorname{Spec}left(prod_{i=1}^inftyBbb F_2right)congbetaBbb N,$$the Stone-Čech compactification on $Bbb N$, but in $mathsf{ZF}$ the only ultrafilters we can prove to exist on $Bbb N$ are the principal ones, corresponding to $Bbb NsubseteqbetaBbb N$, which is a discrete subspace, hence not compact. So, if $betaBbb NsetminusBbb N=varnothing$ we have an affine scheme which is not compact.
Is the statement "every affine scheme is compact" equivalent to some more well-known statement over $mathsf{ZF}$? Or does it imply/is it implied by some more well known statements weaker than $mathsf{AC}$?
algebraic-geometry set-theory axiom-of-choice
asked Nov 26 '18 at 21:34
Alessandro CodenottiAlessandro Codenotti
3,76311539
$endgroup$
$begingroup$
Which step in proving that every affine scheme is compact is problematic for ZF?
$endgroup$
– Kenny Lau
Nov 26 '18 at 21:49
$begingroup$
That's a good question I was discussing with Mathei in chat, I'm not sure where the usual proof fails exactly though @KennyLau
$endgroup$
– Alessandro Codenotti
Nov 26 '18 at 21:50
$begingroup$
At some point you use that $V(I)=varnothing$ iff $I=(1)$ for any ideal $I$, which is not provable in ZF.
$endgroup$
– MatheinBoulomenos
Nov 26 '18 at 21:57
$begingroup$
Without looking it up, I'd guess the ultrafilter lemma is probably to blame.
$endgroup$
– Asaf Karagila♦
Nov 26 '18 at 22:14
add a comment |
$begingroup$
Among the first results one usually sees, right after defining schemes, is that affine schemes are compact, however this statement is stronger than $mathsf{ZF}$: In $mathsf{ZFC}$ we have that $$operatorname{Spec}left(prod_{i=1}^inftyBbb F_2right)congbetaBbb N,$$the Stone-Čech compactification on $Bbb N$, but in $mathsf{ZF}$ the only ultrafilters we can prove to exist on $Bbb N$ are the principal ones, corresponding to $Bbb NsubseteqbetaBbb N$, which is a discrete subspace, hence not compact. So, if $betaBbb NsetminusBbb N=varnothing$ we have an affine scheme which is not compact.
Is the statement "every affine scheme is compact" equivalent to some more well-known statement over $mathsf{ZF}$? Or does it imply/is it implied by some more well known statements weaker than $mathsf{AC}$?
algebraic-geometry set-theory axiom-of-choice
asked Nov 26 '18 at 21:34
Alessandro CodenottiAlessandro Codenotti
3,76311539
$endgroup$
Among the first results one usually sees, right after defining schemes, is that affine schemes are compact, however this statement is stronger than $mathsf{ZF}$: In $mathsf{ZFC}$ we have that $$operatorname{Spec}left(prod_{i=1}^inftyBbb F_2right)congbetaBbb N,$$the Stone-Čech compactification on $Bbb N$, but in $mathsf{ZF}$ the only ultrafilters we can prove to exist on $Bbb N$ are the principal ones, corresponding to $Bbb NsubseteqbetaBbb N$, which is a discrete subspace, hence not compact. So, if $betaBbb NsetminusBbb N=varnothing$ we have an affine scheme which is not compact.
Is the statement "every affine scheme is compact" equivalent to some more well-known statement over $mathsf{ZF}$? Or does it imply/is it implied by some more well known statements weaker than $mathsf{AC}$?
algebraic-geometry set-theory axiom-of-choice
algebraic-geometry set-theory axiom-of-choice
asked Nov 26 '18 at 21:34
Alessandro CodenottiAlessandro Codenotti
3,76311539
asked Nov 26 '18 at 21:34
Alessandro CodenottiAlessandro Codenotti
3,76311539
edited Nov 26 '18 at 21:48
Alessandro Codenotti
asked Nov 26 '18 at 21:34
Alessandro CodenottiAlessandro Codenotti
3,76311539
asked Nov 26 '18 at 21:34
Alessandro CodenottiAlessandro Codenotti
3,76311539
asked Nov 26 '18 at 21:34
Alessandro CodenottiAlessandro Codenotti
3,76311539
3,76311539
$begingroup$
Which step in proving that every affine scheme is compact is problematic for ZF?
$endgroup$
– Kenny Lau
Nov 26 '18 at 21:49
$begingroup$
That's a good question I was discussing with Mathei in chat, I'm not sure where the usual proof fails exactly though @KennyLau
$endgroup$
– Alessandro Codenotti
Nov 26 '18 at 21:50
$begingroup$
At some point you use that $V(I)=varnothing$ iff $I=(1)$ for any ideal $I$, which is not provable in ZF.
$endgroup$
– MatheinBoulomenos
Nov 26 '18 at 21:57
$begingroup$
Without looking it up, I'd guess the ultrafilter lemma is probably to blame.
$endgroup$
– Asaf Karagila♦
Nov 26 '18 at 22:14
add a comment |
$begingroup$
Which step in proving that every affine scheme is compact is problematic for ZF?
$endgroup$
– Kenny Lau
Nov 26 '18 at 21:49
$begingroup$
That's a good question I was discussing with Mathei in chat, I'm not sure where the usual proof fails exactly though @KennyLau
$endgroup$
– Alessandro Codenotti
Nov 26 '18 at 21:50
$begingroup$
At some point you use that $V(I)=varnothing$ iff $I=(1)$ for any ideal $I$, which is not provable in ZF.
$endgroup$
– MatheinBoulomenos
Nov 26 '18 at 21:57
$begingroup$
Without looking it up, I'd guess the ultrafilter lemma is probably to blame.
$endgroup$
– Asaf Karagila♦
Nov 26 '18 at 22:14
$begingroup$
Which step in proving that every affine scheme is compact is problematic for ZF?
$endgroup$
– Kenny Lau
Nov 26 '18 at 21:49
$begingroup$
Which step in proving that every affine scheme is compact is problematic for ZF?
$endgroup$
– Kenny Lau
Nov 26 '18 at 21:49
$begingroup$
That's a good question I was discussing with Mathei in chat, I'm not sure where the usual proof fails exactly though @KennyLau
$endgroup$
– Alessandro Codenotti
Nov 26 '18 at 21:50
$begingroup$
That's a good question I was discussing with Mathei in chat, I'm not sure where the usual proof fails exactly though @KennyLau
$endgroup$
– Alessandro Codenotti
Nov 26 '18 at 21:50
$begingroup$
At some point you use that $V(I)=varnothing$ iff $I=(1)$ for any ideal $I$, which is not provable in ZF.
$endgroup$
– MatheinBoulomenos
Nov 26 '18 at 21:57
$begingroup$
At some point you use that $V(I)=varnothing$ iff $I=(1)$ for any ideal $I$, which is not provable in ZF.
$endgroup$
– MatheinBoulomenos
Nov 26 '18 at 21:57
$begingroup$
Without looking it up, I'd guess the ultrafilter lemma is probably to blame.
$endgroup$
– Asaf Karagila♦
Nov 26 '18 at 22:14
$begingroup$
Without looking it up, I'd guess the ultrafilter lemma is probably to blame.
$endgroup$
– Asaf Karagila♦
Nov 26 '18 at 22:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The statement that every affine scheme is compact is equivalent to the Boolean prime ideal theorem, which is a well-known weak form of choice (strictly weaker than full AC) which is equivalent to many other common applications of choice (for instance, it is equivalent to the compactness theorem in first order logic, or Tychonoff's theorem for Hausdorff spaces).
Indeed, the usual proof that an affine scheme is compact uses the axiom of choice only to say that any nonzero commutative ring has a prime ideal, which is known to be equivalent to the Boolean prime ideal theorem. Conversely, in the special case of a ring $A=mathbb{F}_2^S$ for some set $S$, compactness of $operatorname{Spec} A$ is exactly the statement that any collection of subsets of $S$ with the finite intersection property is contained in an ultrafilter on $S$. This statement for arbitrary $S$ is equivalent to the Boolean prime ideal theorem.
answered Nov 26 '18 at 22:14
Eric WofseyEric Wofsey
183k13211338
$endgroup$
1
$begingroup$
Called it... :-)
$endgroup$
– Asaf Karagila♦
Nov 26 '18 at 22:18
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
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/3.0/"u003ecc by-sa 3.0 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
});
}
});
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%2fmath.stackexchange.com%2fquestions%2f3014962%2fstrength-of-every-affine-scheme-is-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
$begingroup$
The statement that every affine scheme is compact is equivalent to the Boolean prime ideal theorem, which is a well-known weak form of choice (strictly weaker than full AC) which is equivalent to many other common applications of choice (for instance, it is equivalent to the compactness theorem in first order logic, or Tychonoff's theorem for Hausdorff spaces).
Indeed, the usual proof that an affine scheme is compact uses the axiom of choice only to say that any nonzero commutative ring has a prime ideal, which is known to be equivalent to the Boolean prime ideal theorem. Conversely, in the special case of a ring $A=mathbb{F}_2^S$ for some set $S$, compactness of $operatorname{Spec} A$ is exactly the statement that any collection of subsets of $S$ with the finite intersection property is contained in an ultrafilter on $S$. This statement for arbitrary $S$ is equivalent to the Boolean prime ideal theorem.
answered Nov 26 '18 at 22:14
Eric WofseyEric Wofsey
183k13211338
$endgroup$
1
$begingroup$
Called it... :-)
$endgroup$
– Asaf Karagila♦
Nov 26 '18 at 22:18
add a comment |
$begingroup$
The statement that every affine scheme is compact is equivalent to the Boolean prime ideal theorem, which is a well-known weak form of choice (strictly weaker than full AC) which is equivalent to many other common applications of choice (for instance, it is equivalent to the compactness theorem in first order logic, or Tychonoff's theorem for Hausdorff spaces).
Indeed, the usual proof that an affine scheme is compact uses the axiom of choice only to say that any nonzero commutative ring has a prime ideal, which is known to be equivalent to the Boolean prime ideal theorem. Conversely, in the special case of a ring $A=mathbb{F}_2^S$ for some set $S$, compactness of $operatorname{Spec} A$ is exactly the statement that any collection of subsets of $S$ with the finite intersection property is contained in an ultrafilter on $S$. This statement for arbitrary $S$ is equivalent to the Boolean prime ideal theorem.
answered Nov 26 '18 at 22:14
Eric WofseyEric Wofsey
183k13211338
$endgroup$
1
$begingroup$
Called it... :-)
$endgroup$
– Asaf Karagila♦
Nov 26 '18 at 22:18
add a comment |
$begingroup$
The statement that every affine scheme is compact is equivalent to the Boolean prime ideal theorem, which is a well-known weak form of choice (strictly weaker than full AC) which is equivalent to many other common applications of choice (for instance, it is equivalent to the compactness theorem in first order logic, or Tychonoff's theorem for Hausdorff spaces).
Indeed, the usual proof that an affine scheme is compact uses the axiom of choice only to say that any nonzero commutative ring has a prime ideal, which is known to be equivalent to the Boolean prime ideal theorem. Conversely, in the special case of a ring $A=mathbb{F}_2^S$ for some set $S$, compactness of $operatorname{Spec} A$ is exactly the statement that any collection of subsets of $S$ with the finite intersection property is contained in an ultrafilter on $S$. This statement for arbitrary $S$ is equivalent to the Boolean prime ideal theorem.
answered Nov 26 '18 at 22:14
Eric WofseyEric Wofsey
183k13211338
$endgroup$
The statement that every affine scheme is compact is equivalent to the Boolean prime ideal theorem, which is a well-known weak form of choice (strictly weaker than full AC) which is equivalent to many other common applications of choice (for instance, it is equivalent to the compactness theorem in first order logic, or Tychonoff's theorem for Hausdorff spaces).
Indeed, the usual proof that an affine scheme is compact uses the axiom of choice only to say that any nonzero commutative ring has a prime ideal, which is known to be equivalent to the Boolean prime ideal theorem. Conversely, in the special case of a ring $A=mathbb{F}_2^S$ for some set $S$, compactness of $operatorname{Spec} A$ is exactly the statement that any collection of subsets of $S$ with the finite intersection property is contained in an ultrafilter on $S$. This statement for arbitrary $S$ is equivalent to the Boolean prime ideal theorem.
answered Nov 26 '18 at 22:14
Eric WofseyEric Wofsey
183k13211338
answered Nov 26 '18 at 22:14
Eric WofseyEric Wofsey
183k13211338
answered Nov 26 '18 at 22:14
Eric WofseyEric Wofsey
183k13211338
answered Nov 26 '18 at 22:14
Eric WofseyEric Wofsey
183k13211338
183k13211338
1
$begingroup$
Called it... :-)
$endgroup$
– Asaf Karagila♦
Nov 26 '18 at 22:18
add a comment |
1
$begingroup$
Called it... :-)
$endgroup$
– Asaf Karagila♦
Nov 26 '18 at 22:18
1
1
$begingroup$
Called it... :-)
$endgroup$
– Asaf Karagila♦
Nov 26 '18 at 22:18
$begingroup$
Called it... :-)
$endgroup$
– Asaf Karagila♦
Nov 26 '18 at 22:18
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- 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.
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%2fmath.stackexchange.com%2fquestions%2f3014962%2fstrength-of-every-affine-scheme-is-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
$begingroup$
Which step in proving that every affine scheme is compact is problematic for ZF?
$endgroup$
– Kenny Lau
Nov 26 '18 at 21:49
$begingroup$
That's a good question I was discussing with Mathei in chat, I'm not sure where the usual proof fails exactly though @KennyLau
$endgroup$
– Alessandro Codenotti
Nov 26 '18 at 21:50
$begingroup$
At some point you use that $V(I)=varnothing$ iff $I=(1)$ for any ideal $I$, which is not provable in ZF.
$endgroup$
– MatheinBoulomenos
Nov 26 '18 at 21:57
$begingroup$
Without looking it up, I'd guess the ultrafilter lemma is probably to blame.
$endgroup$
– Asaf Karagila♦
Nov 26 '18 at 22:14