Strength of “every affine scheme is compact”












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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







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  • $begingroup$
    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
















5












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














5












5








5





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 '18 at 21:48







Alessandro Codenotti

















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


















  • $begingroup$
    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










1 Answer
1






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$begingroup$

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.






share|cite|improve this answer









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    $begingroup$
    Called it... :-)
    $endgroup$
    – Asaf Karagila♦
    Nov 26 '18 at 22:18











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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









6












$begingroup$

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.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Called it... :-)
    $endgroup$
    – Asaf Karagila♦
    Nov 26 '18 at 22:18
















6












$begingroup$

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.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Called it... :-)
    $endgroup$
    – Asaf Karagila♦
    Nov 26 '18 at 22:18














6












6








6





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 26 '18 at 22:14









Eric WofseyEric Wofsey

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183k13211338








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    $begingroup$
    Called it... :-)
    $endgroup$
    – Asaf Karagila♦
    Nov 26 '18 at 22:18














  • 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


















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