Mixing
algebraic geometry for specific fields means no axiom of choice?
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
I'm beginning to study commutative algebra and algebraic geometry, and something confuses me.
Many proofs, e.g. of the Nullstellensatz, require the axiom of choice (to get the right results about ideals, I suppose). Is there any way to avoid that axiom, if you start with an algebraically closed field that is countable? What about the complex field?
I've been looking for something on this topic on the Internet, with no luck. Thanks for any help!
algebraic-geometry commutative-algebra axiom-of-choice
asked Aug 3 at 17:08
Chris Sanders
1,138216
add a comment |Â
up vote
3
down vote
favorite
I'm beginning to study commutative algebra and algebraic geometry, and something confuses me.
Many proofs, e.g. of the Nullstellensatz, require the axiom of choice (to get the right results about ideals, I suppose). Is there any way to avoid that axiom, if you start with an algebraically closed field that is countable? What about the complex field?
I've been looking for something on this topic on the Internet, with no luck. Thanks for any help!
algebraic-geometry commutative-algebra axiom-of-choice
asked Aug 3 at 17:08
Chris Sanders
1,138216
1
If your field is countable, you can probably get away with most things. Do note, however, that it is consistent that $Bbb Q$ has two non-isomorphic algebraic closures (but only one of them is countable). So even that's to be taken with a pinch of salt.
– Asaf Karagila
Aug 3 at 17:16
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'm beginning to study commutative algebra and algebraic geometry, and something confuses me.
Many proofs, e.g. of the Nullstellensatz, require the axiom of choice (to get the right results about ideals, I suppose). Is there any way to avoid that axiom, if you start with an algebraically closed field that is countable? What about the complex field?
I've been looking for something on this topic on the Internet, with no luck. Thanks for any help!
algebraic-geometry commutative-algebra axiom-of-choice
asked Aug 3 at 17:08
Chris Sanders
1,138216
I'm beginning to study commutative algebra and algebraic geometry, and something confuses me.
Many proofs, e.g. of the Nullstellensatz, require the axiom of choice (to get the right results about ideals, I suppose). Is there any way to avoid that axiom, if you start with an algebraically closed field that is countable? What about the complex field?
I've been looking for something on this topic on the Internet, with no luck. Thanks for any help!
algebraic-geometry commutative-algebra axiom-of-choice
asked Aug 3 at 17:08
Chris Sanders
1,138216
asked Aug 3 at 17:08
Chris Sanders
1,138216
asked Aug 3 at 17:08
Chris Sanders
1,138216
asked Aug 3 at 17:08
Chris Sanders
1,138216
1,138216
1
If your field is countable, you can probably get away with most things. Do note, however, that it is consistent that $Bbb Q$ has two non-isomorphic algebraic closures (but only one of them is countable). So even that's to be taken with a pinch of salt.
– Asaf Karagila
Aug 3 at 17:16
add a comment |Â
1
If your field is countable, you can probably get away with most things. Do note, however, that it is consistent that $Bbb Q$ has two non-isomorphic algebraic closures (but only one of them is countable). So even that's to be taken with a pinch of salt.
– Asaf Karagila
Aug 3 at 17:16
1
1
If your field is countable, you can probably get away with most things. Do note, however, that it is consistent that $Bbb Q$ has two non-isomorphic algebraic closures (but only one of them is countable). So even that's to be taken with a pinch of salt.
– Asaf Karagila
Aug 3 at 17:16
If your field is countable, you can probably get away with most things. Do note, however, that it is consistent that $Bbb Q$ has two non-isomorphic algebraic closures (but only one of them is countable). So even that's to be taken with a pinch of salt.
– Asaf Karagila
Aug 3 at 17:16
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
The axiom of choice is pretty much never necessary for doing any sort of commutative algebra or algebraic geometry which involves only finitely generated algebras over a field. In particular, any arguments that involve the existence of maximal ideals (or ideals which are maximal with a given property) can be carried out in these rings without invoking the axiom of choice, since they are Noetherian.
To be more precise, without assuming the axiom of choice, you can prove that any polynomial ring in finitely many variables over a field (and therefore also any quotient of such a ring) is Noetherian in the strong sense that any set of ideals contains a maximal element. The proof is a minor modification of the usual proof of the Hilbert basis theorem; you can find it written out in full detail at Existence of a prime ideal in an integral domain of finite type over a field without Axiom of Choice.
answered Aug 3 at 21:30
Eric Wofsey
161k12188297
Hrm... define Noetherian without choice. :-)
– Asaf Karagila
Aug 3 at 22:20
Right, there are multiple inequivalent definitions. That's why I clarified which definition I was using (namely, the strongest one).
– Eric Wofsey
Aug 3 at 22:24
Right, yeah, I should have finished reading the answer before I posted the comment. :-P
– Asaf Karagila
Aug 3 at 22:24
add a comment |Â
up vote
2
down vote
Many applications of choice can be removed when we restrict attention to well-orderable - or even better, countable - fields and other objects. As Asaf says, however, there are nonetheless some facts which really do require the axiom of choice. At the same time, these often have slight weakenings that (a) don't require choice and (b) do everything we really need (e.g. that there is only one countable algebraic closure of $mathbbQ$ up to isomorphism).
You might be concerned at this point about results in "concrete" mathematics which go through principles with no apparent choice-free proof. E.g. is it possible that the Riemann hypothesis could be proved, but only using choice due to its reliance on some "choicey" piece of algebraic geometry? It turns out that there is a very powerful metatheorem which says that the answer is "no:" Shoenfield absoluteness. In particular, this implies that if ZFC proves the Riemann hypothesis, or Goldbach, or Fermat's last theorem, or basically any fact from number theory or "low-level" algebraic geometry, then so does ZF. In fact it says much more: it says e.g. that we can also through arithmetic axioms like GCH onto the pile while still not changing the provability of the Riemann hypothesis. The motto I would give here is that Shoenfield absoluteness shows that number theory doesn't rely on choice.
answered Aug 3 at 18:02
Noah Schweber
110k9138259
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The axiom of choice is pretty much never necessary for doing any sort of commutative algebra or algebraic geometry which involves only finitely generated algebras over a field. In particular, any arguments that involve the existence of maximal ideals (or ideals which are maximal with a given property) can be carried out in these rings without invoking the axiom of choice, since they are Noetherian.
To be more precise, without assuming the axiom of choice, you can prove that any polynomial ring in finitely many variables over a field (and therefore also any quotient of such a ring) is Noetherian in the strong sense that any set of ideals contains a maximal element. The proof is a minor modification of the usual proof of the Hilbert basis theorem; you can find it written out in full detail at Existence of a prime ideal in an integral domain of finite type over a field without Axiom of Choice.
answered Aug 3 at 21:30
Eric Wofsey
161k12188297
Hrm... define Noetherian without choice. :-)
– Asaf Karagila
Aug 3 at 22:20
Right, there are multiple inequivalent definitions. That's why I clarified which definition I was using (namely, the strongest one).
– Eric Wofsey
Aug 3 at 22:24
Right, yeah, I should have finished reading the answer before I posted the comment. :-P
– Asaf Karagila
Aug 3 at 22:24
add a comment |Â
up vote
3
down vote
accepted
The axiom of choice is pretty much never necessary for doing any sort of commutative algebra or algebraic geometry which involves only finitely generated algebras over a field. In particular, any arguments that involve the existence of maximal ideals (or ideals which are maximal with a given property) can be carried out in these rings without invoking the axiom of choice, since they are Noetherian.
To be more precise, without assuming the axiom of choice, you can prove that any polynomial ring in finitely many variables over a field (and therefore also any quotient of such a ring) is Noetherian in the strong sense that any set of ideals contains a maximal element. The proof is a minor modification of the usual proof of the Hilbert basis theorem; you can find it written out in full detail at Existence of a prime ideal in an integral domain of finite type over a field without Axiom of Choice.
answered Aug 3 at 21:30
Eric Wofsey
161k12188297
Hrm... define Noetherian without choice. :-)
– Asaf Karagila
Aug 3 at 22:20
Right, there are multiple inequivalent definitions. That's why I clarified which definition I was using (namely, the strongest one).
– Eric Wofsey
Aug 3 at 22:24
Right, yeah, I should have finished reading the answer before I posted the comment. :-P
– Asaf Karagila
Aug 3 at 22:24
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The axiom of choice is pretty much never necessary for doing any sort of commutative algebra or algebraic geometry which involves only finitely generated algebras over a field. In particular, any arguments that involve the existence of maximal ideals (or ideals which are maximal with a given property) can be carried out in these rings without invoking the axiom of choice, since they are Noetherian.
To be more precise, without assuming the axiom of choice, you can prove that any polynomial ring in finitely many variables over a field (and therefore also any quotient of such a ring) is Noetherian in the strong sense that any set of ideals contains a maximal element. The proof is a minor modification of the usual proof of the Hilbert basis theorem; you can find it written out in full detail at Existence of a prime ideal in an integral domain of finite type over a field without Axiom of Choice.
answered Aug 3 at 21:30
Eric Wofsey
161k12188297
The axiom of choice is pretty much never necessary for doing any sort of commutative algebra or algebraic geometry which involves only finitely generated algebras over a field. In particular, any arguments that involve the existence of maximal ideals (or ideals which are maximal with a given property) can be carried out in these rings without invoking the axiom of choice, since they are Noetherian.
To be more precise, without assuming the axiom of choice, you can prove that any polynomial ring in finitely many variables over a field (and therefore also any quotient of such a ring) is Noetherian in the strong sense that any set of ideals contains a maximal element. The proof is a minor modification of the usual proof of the Hilbert basis theorem; you can find it written out in full detail at Existence of a prime ideal in an integral domain of finite type over a field without Axiom of Choice.
answered Aug 3 at 21:30
Eric Wofsey
161k12188297
answered Aug 3 at 21:30
Eric Wofsey
161k12188297
answered Aug 3 at 21:30
Eric Wofsey
161k12188297
answered Aug 3 at 21:30
Eric Wofsey
161k12188297
161k12188297
Hrm... define Noetherian without choice. :-)
– Asaf Karagila
Aug 3 at 22:20
Right, there are multiple inequivalent definitions. That's why I clarified which definition I was using (namely, the strongest one).
– Eric Wofsey
Aug 3 at 22:24
Right, yeah, I should have finished reading the answer before I posted the comment. :-P
– Asaf Karagila
Aug 3 at 22:24
add a comment |Â
Hrm... define Noetherian without choice. :-)
– Asaf Karagila
Aug 3 at 22:20
Right, there are multiple inequivalent definitions. That's why I clarified which definition I was using (namely, the strongest one).
– Eric Wofsey
Aug 3 at 22:24
Right, yeah, I should have finished reading the answer before I posted the comment. :-P
– Asaf Karagila
Aug 3 at 22:24
Hrm... define Noetherian without choice. :-)
– Asaf Karagila
Aug 3 at 22:20
Hrm... define Noetherian without choice. :-)
– Asaf Karagila
Aug 3 at 22:20
Right, there are multiple inequivalent definitions. That's why I clarified which definition I was using (namely, the strongest one).
– Eric Wofsey
Aug 3 at 22:24
Right, there are multiple inequivalent definitions. That's why I clarified which definition I was using (namely, the strongest one).
– Eric Wofsey
Aug 3 at 22:24
Right, yeah, I should have finished reading the answer before I posted the comment. :-P
– Asaf Karagila
Aug 3 at 22:24
Right, yeah, I should have finished reading the answer before I posted the comment. :-P
– Asaf Karagila
Aug 3 at 22:24
add a comment |Â
up vote
2
down vote
Many applications of choice can be removed when we restrict attention to well-orderable - or even better, countable - fields and other objects. As Asaf says, however, there are nonetheless some facts which really do require the axiom of choice. At the same time, these often have slight weakenings that (a) don't require choice and (b) do everything we really need (e.g. that there is only one countable algebraic closure of $mathbbQ$ up to isomorphism).
You might be concerned at this point about results in "concrete" mathematics which go through principles with no apparent choice-free proof. E.g. is it possible that the Riemann hypothesis could be proved, but only using choice due to its reliance on some "choicey" piece of algebraic geometry? It turns out that there is a very powerful metatheorem which says that the answer is "no:" Shoenfield absoluteness. In particular, this implies that if ZFC proves the Riemann hypothesis, or Goldbach, or Fermat's last theorem, or basically any fact from number theory or "low-level" algebraic geometry, then so does ZF. In fact it says much more: it says e.g. that we can also through arithmetic axioms like GCH onto the pile while still not changing the provability of the Riemann hypothesis. The motto I would give here is that Shoenfield absoluteness shows that number theory doesn't rely on choice.
answered Aug 3 at 18:02
Noah Schweber
110k9138259
add a comment |Â
up vote
2
down vote
Many applications of choice can be removed when we restrict attention to well-orderable - or even better, countable - fields and other objects. As Asaf says, however, there are nonetheless some facts which really do require the axiom of choice. At the same time, these often have slight weakenings that (a) don't require choice and (b) do everything we really need (e.g. that there is only one countable algebraic closure of $mathbbQ$ up to isomorphism).
You might be concerned at this point about results in "concrete" mathematics which go through principles with no apparent choice-free proof. E.g. is it possible that the Riemann hypothesis could be proved, but only using choice due to its reliance on some "choicey" piece of algebraic geometry? It turns out that there is a very powerful metatheorem which says that the answer is "no:" Shoenfield absoluteness. In particular, this implies that if ZFC proves the Riemann hypothesis, or Goldbach, or Fermat's last theorem, or basically any fact from number theory or "low-level" algebraic geometry, then so does ZF. In fact it says much more: it says e.g. that we can also through arithmetic axioms like GCH onto the pile while still not changing the provability of the Riemann hypothesis. The motto I would give here is that Shoenfield absoluteness shows that number theory doesn't rely on choice.
answered Aug 3 at 18:02
Noah Schweber
110k9138259
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Many applications of choice can be removed when we restrict attention to well-orderable - or even better, countable - fields and other objects. As Asaf says, however, there are nonetheless some facts which really do require the axiom of choice. At the same time, these often have slight weakenings that (a) don't require choice and (b) do everything we really need (e.g. that there is only one countable algebraic closure of $mathbbQ$ up to isomorphism).
You might be concerned at this point about results in "concrete" mathematics which go through principles with no apparent choice-free proof. E.g. is it possible that the Riemann hypothesis could be proved, but only using choice due to its reliance on some "choicey" piece of algebraic geometry? It turns out that there is a very powerful metatheorem which says that the answer is "no:" Shoenfield absoluteness. In particular, this implies that if ZFC proves the Riemann hypothesis, or Goldbach, or Fermat's last theorem, or basically any fact from number theory or "low-level" algebraic geometry, then so does ZF. In fact it says much more: it says e.g. that we can also through arithmetic axioms like GCH onto the pile while still not changing the provability of the Riemann hypothesis. The motto I would give here is that Shoenfield absoluteness shows that number theory doesn't rely on choice.
answered Aug 3 at 18:02
Noah Schweber
110k9138259
Many applications of choice can be removed when we restrict attention to well-orderable - or even better, countable - fields and other objects. As Asaf says, however, there are nonetheless some facts which really do require the axiom of choice. At the same time, these often have slight weakenings that (a) don't require choice and (b) do everything we really need (e.g. that there is only one countable algebraic closure of $mathbbQ$ up to isomorphism).
You might be concerned at this point about results in "concrete" mathematics which go through principles with no apparent choice-free proof. E.g. is it possible that the Riemann hypothesis could be proved, but only using choice due to its reliance on some "choicey" piece of algebraic geometry? It turns out that there is a very powerful metatheorem which says that the answer is "no:" Shoenfield absoluteness. In particular, this implies that if ZFC proves the Riemann hypothesis, or Goldbach, or Fermat's last theorem, or basically any fact from number theory or "low-level" algebraic geometry, then so does ZF. In fact it says much more: it says e.g. that we can also through arithmetic axioms like GCH onto the pile while still not changing the provability of the Riemann hypothesis. The motto I would give here is that Shoenfield absoluteness shows that number theory doesn't rely on choice.
answered Aug 3 at 18:02
Noah Schweber
110k9138259
answered Aug 3 at 18:02
Noah Schweber
110k9138259
answered Aug 3 at 18:02
Noah Schweber
110k9138259
answered Aug 3 at 18:02
Noah Schweber
110k9138259
110k9138259
add a comment |Â
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2871301%2falgebraic-geometry-for-specific-fields-means-no-axiom-of-choice%23new-answer', 'question_page');
);
Post as a guest
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
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
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
1
If your field is countable, you can probably get away with most things. Do note, however, that it is consistent that $Bbb Q$ has two non-isomorphic algebraic closures (but only one of them is countable). So even that's to be taken with a pinch of salt.
– Asaf Karagila
Aug 3 at 17:16