$text{Gal}(bar{mathbb{Q}}/mathbb{Q})$ without choice
$begingroup$
My understand (from this question) is that we can construct a countable algebraic closure of $mathbb{Q}$ without using the axiom of choice (i.e. we can construct $overline{mathbb{Q}} subseteq mathbb{C}$), although perhaps this oversimplifying thing.
What can we say about $text{Gal}(bar{mathbb{Q}}/mathbb{Q})$ without using the axiom of choice? I think the infinite Galois correspondence requires some form of choice, so maybe this the automorphism group of $bar{mathbb{Q}}$ doesn't deserve to be called a Galois group without choice. Can it be shown that it is uncountable? Or that it has more than $2$ elements?
Motivation: people often say that it is impossible to write down any non-trivial element of $text{Gal}(bar{mathbb{Q}}/mathbb{Q})$. If it is consistent with ZF that $text{Gal}(bar{mathbb{Q}}/mathbb{Q})$ has size $2$, then this would be a way to formalize that statement.
set-theory galois-theory
asked Dec 16 '18 at 23:34
vukovvukov
788414
$endgroup$
|
show 5 more comments
$begingroup$
My understand (from this question) is that we can construct a countable algebraic closure of $mathbb{Q}$ without using the axiom of choice (i.e. we can construct $overline{mathbb{Q}} subseteq mathbb{C}$), although perhaps this oversimplifying thing.
What can we say about $text{Gal}(bar{mathbb{Q}}/mathbb{Q})$ without using the axiom of choice? I think the infinite Galois correspondence requires some form of choice, so maybe this the automorphism group of $bar{mathbb{Q}}$ doesn't deserve to be called a Galois group without choice. Can it be shown that it is uncountable? Or that it has more than $2$ elements?
Motivation: people often say that it is impossible to write down any non-trivial element of $text{Gal}(bar{mathbb{Q}}/mathbb{Q})$. If it is consistent with ZF that $text{Gal}(bar{mathbb{Q}}/mathbb{Q})$ has size $2$, then this would be a way to formalize that statement.
set-theory galois-theory
asked Dec 16 '18 at 23:34
vukovvukov
788414
$endgroup$
$begingroup$
See this very similar question, which went unanswered.
$endgroup$
– Alex Kruckman
Dec 16 '18 at 23:39
$begingroup$
To be clear, do you mean for $bar{mathbb{Q}}$ to refer specifically to a countable algebraic closure of $mathbb{Q}$, or to an arbitrary algebraic closure? Not all algebraic closures are necessarily countable without choice.
$endgroup$
– Eric Wofsey
Dec 17 '18 at 0:11
$begingroup$
@EricWofsey I was thinking about the "standard algebraic closure" of $mathbb{Q}$, i.e. the set of algebraic numbers in $mathbb{C}$. If it isn't possible to make sense of that without choice, then I would be interested if it was possible to show that there exists an algebraic closure of $mathbb{Q}$ with more than $2$ elements in its automorphism group.
$endgroup$
– vukov
Dec 17 '18 at 0:17
1
$begingroup$
I don't think people usually say it's impossible to write down a non-trivial element of $text{Gal}(bar{mathbb{Q}}/mathbb{Q})$ (there's just not any simple formula for one). Rather, it's $text{Gal}(mathbb{C}/mathbb{Q})$ which you can't write down any nontrivial elements of.
$endgroup$
– Eric Wofsey
Dec 17 '18 at 0:39
1
$begingroup$
@JyrkiLahtonen: Well, every automorphism of $overline{mathbb{Q}}$ of order $2$ is conjugate to complex conjugation (and of course, if you don't have a chosen embedding in $mathbb{C}$, you can only hope to describe it up to conjugation). See math.stackexchange.com/questions/2563758/….
$endgroup$
– Eric Wofsey
Dec 17 '18 at 18:47
|
show 5 more comments
$begingroup$
My understand (from this question) is that we can construct a countable algebraic closure of $mathbb{Q}$ without using the axiom of choice (i.e. we can construct $overline{mathbb{Q}} subseteq mathbb{C}$), although perhaps this oversimplifying thing.
What can we say about $text{Gal}(bar{mathbb{Q}}/mathbb{Q})$ without using the axiom of choice? I think the infinite Galois correspondence requires some form of choice, so maybe this the automorphism group of $bar{mathbb{Q}}$ doesn't deserve to be called a Galois group without choice. Can it be shown that it is uncountable? Or that it has more than $2$ elements?
Motivation: people often say that it is impossible to write down any non-trivial element of $text{Gal}(bar{mathbb{Q}}/mathbb{Q})$. If it is consistent with ZF that $text{Gal}(bar{mathbb{Q}}/mathbb{Q})$ has size $2$, then this would be a way to formalize that statement.
set-theory galois-theory
asked Dec 16 '18 at 23:34
vukovvukov
788414
$endgroup$
My understand (from this question) is that we can construct a countable algebraic closure of $mathbb{Q}$ without using the axiom of choice (i.e. we can construct $overline{mathbb{Q}} subseteq mathbb{C}$), although perhaps this oversimplifying thing.
What can we say about $text{Gal}(bar{mathbb{Q}}/mathbb{Q})$ without using the axiom of choice? I think the infinite Galois correspondence requires some form of choice, so maybe this the automorphism group of $bar{mathbb{Q}}$ doesn't deserve to be called a Galois group without choice. Can it be shown that it is uncountable? Or that it has more than $2$ elements?
Motivation: people often say that it is impossible to write down any non-trivial element of $text{Gal}(bar{mathbb{Q}}/mathbb{Q})$. If it is consistent with ZF that $text{Gal}(bar{mathbb{Q}}/mathbb{Q})$ has size $2$, then this would be a way to formalize that statement.
set-theory galois-theory
set-theory galois-theory
asked Dec 16 '18 at 23:34
vukovvukov
788414
asked Dec 16 '18 at 23:34
vukovvukov
788414
edited Dec 17 '18 at 0:29
vukov
asked Dec 16 '18 at 23:34
vukovvukov
788414
asked Dec 16 '18 at 23:34
vukovvukov
788414
asked Dec 16 '18 at 23:34
vukovvukov
788414
788414
$begingroup$
See this very similar question, which went unanswered.
$endgroup$
– Alex Kruckman
Dec 16 '18 at 23:39
$begingroup$
To be clear, do you mean for $bar{mathbb{Q}}$ to refer specifically to a countable algebraic closure of $mathbb{Q}$, or to an arbitrary algebraic closure? Not all algebraic closures are necessarily countable without choice.
$endgroup$
– Eric Wofsey
Dec 17 '18 at 0:11
$begingroup$
@EricWofsey I was thinking about the "standard algebraic closure" of $mathbb{Q}$, i.e. the set of algebraic numbers in $mathbb{C}$. If it isn't possible to make sense of that without choice, then I would be interested if it was possible to show that there exists an algebraic closure of $mathbb{Q}$ with more than $2$ elements in its automorphism group.
$endgroup$
– vukov
Dec 17 '18 at 0:17
1
$begingroup$
I don't think people usually say it's impossible to write down a non-trivial element of $text{Gal}(bar{mathbb{Q}}/mathbb{Q})$ (there's just not any simple formula for one). Rather, it's $text{Gal}(mathbb{C}/mathbb{Q})$ which you can't write down any nontrivial elements of.
$endgroup$
– Eric Wofsey
Dec 17 '18 at 0:39
1
$begingroup$
@JyrkiLahtonen: Well, every automorphism of $overline{mathbb{Q}}$ of order $2$ is conjugate to complex conjugation (and of course, if you don't have a chosen embedding in $mathbb{C}$, you can only hope to describe it up to conjugation). See math.stackexchange.com/questions/2563758/….
$endgroup$
– Eric Wofsey
Dec 17 '18 at 18:47
|
show 5 more comments
$begingroup$
See this very similar question, which went unanswered.
$endgroup$
– Alex Kruckman
Dec 16 '18 at 23:39
$begingroup$
To be clear, do you mean for $bar{mathbb{Q}}$ to refer specifically to a countable algebraic closure of $mathbb{Q}$, or to an arbitrary algebraic closure? Not all algebraic closures are necessarily countable without choice.
$endgroup$
– Eric Wofsey
Dec 17 '18 at 0:11
$begingroup$
@EricWofsey I was thinking about the "standard algebraic closure" of $mathbb{Q}$, i.e. the set of algebraic numbers in $mathbb{C}$. If it isn't possible to make sense of that without choice, then I would be interested if it was possible to show that there exists an algebraic closure of $mathbb{Q}$ with more than $2$ elements in its automorphism group.
$endgroup$
– vukov
Dec 17 '18 at 0:17
1
$begingroup$
I don't think people usually say it's impossible to write down a non-trivial element of $text{Gal}(bar{mathbb{Q}}/mathbb{Q})$ (there's just not any simple formula for one). Rather, it's $text{Gal}(mathbb{C}/mathbb{Q})$ which you can't write down any nontrivial elements of.
$endgroup$
– Eric Wofsey
Dec 17 '18 at 0:39
1
$begingroup$
@JyrkiLahtonen: Well, every automorphism of $overline{mathbb{Q}}$ of order $2$ is conjugate to complex conjugation (and of course, if you don't have a chosen embedding in $mathbb{C}$, you can only hope to describe it up to conjugation). See math.stackexchange.com/questions/2563758/….
$endgroup$
– Eric Wofsey
Dec 17 '18 at 18:47
$begingroup$
See this very similar question, which went unanswered.
$endgroup$
– Alex Kruckman
Dec 16 '18 at 23:39
$begingroup$
See this very similar question, which went unanswered.
$endgroup$
– Alex Kruckman
Dec 16 '18 at 23:39
$begingroup$
To be clear, do you mean for $bar{mathbb{Q}}$ to refer specifically to a countable algebraic closure of $mathbb{Q}$, or to an arbitrary algebraic closure? Not all algebraic closures are necessarily countable without choice.
$endgroup$
– Eric Wofsey
Dec 17 '18 at 0:11
$begingroup$
To be clear, do you mean for $bar{mathbb{Q}}$ to refer specifically to a countable algebraic closure of $mathbb{Q}$, or to an arbitrary algebraic closure? Not all algebraic closures are necessarily countable without choice.
$endgroup$
– Eric Wofsey
Dec 17 '18 at 0:11
$begingroup$
@EricWofsey I was thinking about the "standard algebraic closure" of $mathbb{Q}$, i.e. the set of algebraic numbers in $mathbb{C}$. If it isn't possible to make sense of that without choice, then I would be interested if it was possible to show that there exists an algebraic closure of $mathbb{Q}$ with more than $2$ elements in its automorphism group.
$endgroup$
– vukov
Dec 17 '18 at 0:17
$begingroup$
@EricWofsey I was thinking about the "standard algebraic closure" of $mathbb{Q}$, i.e. the set of algebraic numbers in $mathbb{C}$. If it isn't possible to make sense of that without choice, then I would be interested if it was possible to show that there exists an algebraic closure of $mathbb{Q}$ with more than $2$ elements in its automorphism group.
$endgroup$
– vukov
Dec 17 '18 at 0:17
1
1
$begingroup$
I don't think people usually say it's impossible to write down a non-trivial element of $text{Gal}(bar{mathbb{Q}}/mathbb{Q})$ (there's just not any simple formula for one). Rather, it's $text{Gal}(mathbb{C}/mathbb{Q})$ which you can't write down any nontrivial elements of.
$endgroup$
– Eric Wofsey
Dec 17 '18 at 0:39
$begingroup$
I don't think people usually say it's impossible to write down a non-trivial element of $text{Gal}(bar{mathbb{Q}}/mathbb{Q})$ (there's just not any simple formula for one). Rather, it's $text{Gal}(mathbb{C}/mathbb{Q})$ which you can't write down any nontrivial elements of.
$endgroup$
– Eric Wofsey
Dec 17 '18 at 0:39
1
1
$begingroup$
@JyrkiLahtonen: Well, every automorphism of $overline{mathbb{Q}}$ of order $2$ is conjugate to complex conjugation (and of course, if you don't have a chosen embedding in $mathbb{C}$, you can only hope to describe it up to conjugation). See math.stackexchange.com/questions/2563758/….
$endgroup$
– Eric Wofsey
Dec 17 '18 at 18:47
$begingroup$
@JyrkiLahtonen: Well, every automorphism of $overline{mathbb{Q}}$ of order $2$ is conjugate to complex conjugation (and of course, if you don't have a chosen embedding in $mathbb{C}$, you can only hope to describe it up to conjugation). See math.stackexchange.com/questions/2563758/….
$endgroup$
– Eric Wofsey
Dec 17 '18 at 18:47
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
If $overline{mathbb{Q}}$ is a countable algebraic closure of $mathbb{Q}$, the axiom of choice is hardly if ever needed to study $text{Gal}(overline{mathbb{Q}}/mathbb{Q})$. All of the usual arguments using the axiom of choice work without it, by just explicitly using an enumeration of $overline{mathbb{Q}}$ in place of the well-ordering that would be used via the axiom of choice. For instance, you can construct elements of $text{Gal}(overline{mathbb{Q}}/mathbb{Q})$ by one-by-one extending automorphisms of subfields of $overline{mathbb{Q}}$ which are finite and Galois over $mathbb{Q}$ (no choice is ever needed since you have a well-ordering of $overline{mathbb{Q}}$ which you can use to make all your choices). The usual diagonalization argument goes through to show there are uncountably many such elements. Similarly, you can prove that any two countable algebraic closures of $mathbb{Q}$ are isomorphic.
More generally, pretty much the same can be said for any algebraically closed field $K$ that is well-orderable. The only way the usual theory of algebraically closed fields (and more generally of infinite field extensions) uses the axiom of choice is to have a well-ordering on the field so that you can construct homomorphisms out of (and/or into) it by transfinite recursion. So, if you already know that such a well-ordering exists, you don't need the axiom of choice.
answered Dec 17 '18 at 0:34
Eric WofseyEric Wofsey
186k14214341
$endgroup$
$begingroup$
As far as I can see, the set of algebraic numbers in $mathbb C$ is countable, even without choice. The point is that (1) we can enumerate all the nonconstant polynomials over $mathbb Q$, (2) each polynomial has only finitely many roots, and (3) each finite subset of $mathbb C$ has a canonical ordering, namely the lexicographic ordering according to real parts and then imaginary parts.
$endgroup$
– Andreas Blass
Dec 17 '18 at 3:32
1
$begingroup$
Perhaps a better formulation of my preceding comment is that, if an algebraic extension of a countable field is linearly orderable as a set, then it is countable.
$endgroup$
– Andreas Blass
Dec 17 '18 at 3:33
$begingroup$
The "...to show there are uncountable many such elements" part seems to be, at least, confusing if not completely wrong. There are only countable many algebraic elements over the rationals, and thus $;|overline{Bbb Q}|=|Bbb Q|=aleph_0;$ . Nevertheless, it seems to be the "uncountable elements" there refers to the elements of the full Galois group, though I can't see how the diagonal argument of Cantor can be applied in a meaningful way there.
$endgroup$
– DonAntonio
Dec 17 '18 at 8:11
add a comment |
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$begingroup$
If $overline{mathbb{Q}}$ is a countable algebraic closure of $mathbb{Q}$, the axiom of choice is hardly if ever needed to study $text{Gal}(overline{mathbb{Q}}/mathbb{Q})$. All of the usual arguments using the axiom of choice work without it, by just explicitly using an enumeration of $overline{mathbb{Q}}$ in place of the well-ordering that would be used via the axiom of choice. For instance, you can construct elements of $text{Gal}(overline{mathbb{Q}}/mathbb{Q})$ by one-by-one extending automorphisms of subfields of $overline{mathbb{Q}}$ which are finite and Galois over $mathbb{Q}$ (no choice is ever needed since you have a well-ordering of $overline{mathbb{Q}}$ which you can use to make all your choices). The usual diagonalization argument goes through to show there are uncountably many such elements. Similarly, you can prove that any two countable algebraic closures of $mathbb{Q}$ are isomorphic.
More generally, pretty much the same can be said for any algebraically closed field $K$ that is well-orderable. The only way the usual theory of algebraically closed fields (and more generally of infinite field extensions) uses the axiom of choice is to have a well-ordering on the field so that you can construct homomorphisms out of (and/or into) it by transfinite recursion. So, if you already know that such a well-ordering exists, you don't need the axiom of choice.
answered Dec 17 '18 at 0:34
Eric WofseyEric Wofsey
186k14214341
$endgroup$
$begingroup$
As far as I can see, the set of algebraic numbers in $mathbb C$ is countable, even without choice. The point is that (1) we can enumerate all the nonconstant polynomials over $mathbb Q$, (2) each polynomial has only finitely many roots, and (3) each finite subset of $mathbb C$ has a canonical ordering, namely the lexicographic ordering according to real parts and then imaginary parts.
$endgroup$
– Andreas Blass
Dec 17 '18 at 3:32
1
$begingroup$
Perhaps a better formulation of my preceding comment is that, if an algebraic extension of a countable field is linearly orderable as a set, then it is countable.
$endgroup$
– Andreas Blass
Dec 17 '18 at 3:33
$begingroup$
The "...to show there are uncountable many such elements" part seems to be, at least, confusing if not completely wrong. There are only countable many algebraic elements over the rationals, and thus $;|overline{Bbb Q}|=|Bbb Q|=aleph_0;$ . Nevertheless, it seems to be the "uncountable elements" there refers to the elements of the full Galois group, though I can't see how the diagonal argument of Cantor can be applied in a meaningful way there.
$endgroup$
– DonAntonio
Dec 17 '18 at 8:11
add a comment |
$begingroup$
If $overline{mathbb{Q}}$ is a countable algebraic closure of $mathbb{Q}$, the axiom of choice is hardly if ever needed to study $text{Gal}(overline{mathbb{Q}}/mathbb{Q})$. All of the usual arguments using the axiom of choice work without it, by just explicitly using an enumeration of $overline{mathbb{Q}}$ in place of the well-ordering that would be used via the axiom of choice. For instance, you can construct elements of $text{Gal}(overline{mathbb{Q}}/mathbb{Q})$ by one-by-one extending automorphisms of subfields of $overline{mathbb{Q}}$ which are finite and Galois over $mathbb{Q}$ (no choice is ever needed since you have a well-ordering of $overline{mathbb{Q}}$ which you can use to make all your choices). The usual diagonalization argument goes through to show there are uncountably many such elements. Similarly, you can prove that any two countable algebraic closures of $mathbb{Q}$ are isomorphic.
More generally, pretty much the same can be said for any algebraically closed field $K$ that is well-orderable. The only way the usual theory of algebraically closed fields (and more generally of infinite field extensions) uses the axiom of choice is to have a well-ordering on the field so that you can construct homomorphisms out of (and/or into) it by transfinite recursion. So, if you already know that such a well-ordering exists, you don't need the axiom of choice.
answered Dec 17 '18 at 0:34
Eric WofseyEric Wofsey
186k14214341
$endgroup$
$begingroup$
As far as I can see, the set of algebraic numbers in $mathbb C$ is countable, even without choice. The point is that (1) we can enumerate all the nonconstant polynomials over $mathbb Q$, (2) each polynomial has only finitely many roots, and (3) each finite subset of $mathbb C$ has a canonical ordering, namely the lexicographic ordering according to real parts and then imaginary parts.
$endgroup$
– Andreas Blass
Dec 17 '18 at 3:32
1
$begingroup$
Perhaps a better formulation of my preceding comment is that, if an algebraic extension of a countable field is linearly orderable as a set, then it is countable.
$endgroup$
– Andreas Blass
Dec 17 '18 at 3:33
$begingroup$
The "...to show there are uncountable many such elements" part seems to be, at least, confusing if not completely wrong. There are only countable many algebraic elements over the rationals, and thus $;|overline{Bbb Q}|=|Bbb Q|=aleph_0;$ . Nevertheless, it seems to be the "uncountable elements" there refers to the elements of the full Galois group, though I can't see how the diagonal argument of Cantor can be applied in a meaningful way there.
$endgroup$
– DonAntonio
Dec 17 '18 at 8:11
add a comment |
$begingroup$
If $overline{mathbb{Q}}$ is a countable algebraic closure of $mathbb{Q}$, the axiom of choice is hardly if ever needed to study $text{Gal}(overline{mathbb{Q}}/mathbb{Q})$. All of the usual arguments using the axiom of choice work without it, by just explicitly using an enumeration of $overline{mathbb{Q}}$ in place of the well-ordering that would be used via the axiom of choice. For instance, you can construct elements of $text{Gal}(overline{mathbb{Q}}/mathbb{Q})$ by one-by-one extending automorphisms of subfields of $overline{mathbb{Q}}$ which are finite and Galois over $mathbb{Q}$ (no choice is ever needed since you have a well-ordering of $overline{mathbb{Q}}$ which you can use to make all your choices). The usual diagonalization argument goes through to show there are uncountably many such elements. Similarly, you can prove that any two countable algebraic closures of $mathbb{Q}$ are isomorphic.
More generally, pretty much the same can be said for any algebraically closed field $K$ that is well-orderable. The only way the usual theory of algebraically closed fields (and more generally of infinite field extensions) uses the axiom of choice is to have a well-ordering on the field so that you can construct homomorphisms out of (and/or into) it by transfinite recursion. So, if you already know that such a well-ordering exists, you don't need the axiom of choice.
answered Dec 17 '18 at 0:34
Eric WofseyEric Wofsey
186k14214341
$endgroup$
If $overline{mathbb{Q}}$ is a countable algebraic closure of $mathbb{Q}$, the axiom of choice is hardly if ever needed to study $text{Gal}(overline{mathbb{Q}}/mathbb{Q})$. All of the usual arguments using the axiom of choice work without it, by just explicitly using an enumeration of $overline{mathbb{Q}}$ in place of the well-ordering that would be used via the axiom of choice. For instance, you can construct elements of $text{Gal}(overline{mathbb{Q}}/mathbb{Q})$ by one-by-one extending automorphisms of subfields of $overline{mathbb{Q}}$ which are finite and Galois over $mathbb{Q}$ (no choice is ever needed since you have a well-ordering of $overline{mathbb{Q}}$ which you can use to make all your choices). The usual diagonalization argument goes through to show there are uncountably many such elements. Similarly, you can prove that any two countable algebraic closures of $mathbb{Q}$ are isomorphic.
More generally, pretty much the same can be said for any algebraically closed field $K$ that is well-orderable. The only way the usual theory of algebraically closed fields (and more generally of infinite field extensions) uses the axiom of choice is to have a well-ordering on the field so that you can construct homomorphisms out of (and/or into) it by transfinite recursion. So, if you already know that such a well-ordering exists, you don't need the axiom of choice.
answered Dec 17 '18 at 0:34
Eric WofseyEric Wofsey
186k14214341
edited Dec 17 '18 at 0:42
answered Dec 17 '18 at 0:34
Eric WofseyEric Wofsey
186k14214341
answered Dec 17 '18 at 0:34
Eric WofseyEric Wofsey
186k14214341
answered Dec 17 '18 at 0:34
Eric WofseyEric Wofsey
186k14214341
186k14214341
$begingroup$
As far as I can see, the set of algebraic numbers in $mathbb C$ is countable, even without choice. The point is that (1) we can enumerate all the nonconstant polynomials over $mathbb Q$, (2) each polynomial has only finitely many roots, and (3) each finite subset of $mathbb C$ has a canonical ordering, namely the lexicographic ordering according to real parts and then imaginary parts.
$endgroup$
– Andreas Blass
Dec 17 '18 at 3:32
1
$begingroup$
Perhaps a better formulation of my preceding comment is that, if an algebraic extension of a countable field is linearly orderable as a set, then it is countable.
$endgroup$
– Andreas Blass
Dec 17 '18 at 3:33
$begingroup$
The "...to show there are uncountable many such elements" part seems to be, at least, confusing if not completely wrong. There are only countable many algebraic elements over the rationals, and thus $;|overline{Bbb Q}|=|Bbb Q|=aleph_0;$ . Nevertheless, it seems to be the "uncountable elements" there refers to the elements of the full Galois group, though I can't see how the diagonal argument of Cantor can be applied in a meaningful way there.
$endgroup$
– DonAntonio
Dec 17 '18 at 8:11
add a comment |
$begingroup$
As far as I can see, the set of algebraic numbers in $mathbb C$ is countable, even without choice. The point is that (1) we can enumerate all the nonconstant polynomials over $mathbb Q$, (2) each polynomial has only finitely many roots, and (3) each finite subset of $mathbb C$ has a canonical ordering, namely the lexicographic ordering according to real parts and then imaginary parts.
$endgroup$
– Andreas Blass
Dec 17 '18 at 3:32
1
$begingroup$
Perhaps a better formulation of my preceding comment is that, if an algebraic extension of a countable field is linearly orderable as a set, then it is countable.
$endgroup$
– Andreas Blass
Dec 17 '18 at 3:33
$begingroup$
The "...to show there are uncountable many such elements" part seems to be, at least, confusing if not completely wrong. There are only countable many algebraic elements over the rationals, and thus $;|overline{Bbb Q}|=|Bbb Q|=aleph_0;$ . Nevertheless, it seems to be the "uncountable elements" there refers to the elements of the full Galois group, though I can't see how the diagonal argument of Cantor can be applied in a meaningful way there.
$endgroup$
– DonAntonio
Dec 17 '18 at 8:11
$begingroup$
As far as I can see, the set of algebraic numbers in $mathbb C$ is countable, even without choice. The point is that (1) we can enumerate all the nonconstant polynomials over $mathbb Q$, (2) each polynomial has only finitely many roots, and (3) each finite subset of $mathbb C$ has a canonical ordering, namely the lexicographic ordering according to real parts and then imaginary parts.
$endgroup$
– Andreas Blass
Dec 17 '18 at 3:32
$begingroup$
As far as I can see, the set of algebraic numbers in $mathbb C$ is countable, even without choice. The point is that (1) we can enumerate all the nonconstant polynomials over $mathbb Q$, (2) each polynomial has only finitely many roots, and (3) each finite subset of $mathbb C$ has a canonical ordering, namely the lexicographic ordering according to real parts and then imaginary parts.
$endgroup$
– Andreas Blass
Dec 17 '18 at 3:32
1
1
$begingroup$
Perhaps a better formulation of my preceding comment is that, if an algebraic extension of a countable field is linearly orderable as a set, then it is countable.
$endgroup$
– Andreas Blass
Dec 17 '18 at 3:33
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Perhaps a better formulation of my preceding comment is that, if an algebraic extension of a countable field is linearly orderable as a set, then it is countable.
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– Andreas Blass
Dec 17 '18 at 3:33
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The "...to show there are uncountable many such elements" part seems to be, at least, confusing if not completely wrong. There are only countable many algebraic elements over the rationals, and thus $;|overline{Bbb Q}|=|Bbb Q|=aleph_0;$ . Nevertheless, it seems to be the "uncountable elements" there refers to the elements of the full Galois group, though I can't see how the diagonal argument of Cantor can be applied in a meaningful way there.
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– DonAntonio
Dec 17 '18 at 8:11
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The "...to show there are uncountable many such elements" part seems to be, at least, confusing if not completely wrong. There are only countable many algebraic elements over the rationals, and thus $;|overline{Bbb Q}|=|Bbb Q|=aleph_0;$ . Nevertheless, it seems to be the "uncountable elements" there refers to the elements of the full Galois group, though I can't see how the diagonal argument of Cantor can be applied in a meaningful way there.
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– DonAntonio
Dec 17 '18 at 8:11
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See this very similar question, which went unanswered.
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– Alex Kruckman
Dec 16 '18 at 23:39
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To be clear, do you mean for $bar{mathbb{Q}}$ to refer specifically to a countable algebraic closure of $mathbb{Q}$, or to an arbitrary algebraic closure? Not all algebraic closures are necessarily countable without choice.
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– Eric Wofsey
Dec 17 '18 at 0:11
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@EricWofsey I was thinking about the "standard algebraic closure" of $mathbb{Q}$, i.e. the set of algebraic numbers in $mathbb{C}$. If it isn't possible to make sense of that without choice, then I would be interested if it was possible to show that there exists an algebraic closure of $mathbb{Q}$ with more than $2$ elements in its automorphism group.
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– vukov
Dec 17 '18 at 0:17
1
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I don't think people usually say it's impossible to write down a non-trivial element of $text{Gal}(bar{mathbb{Q}}/mathbb{Q})$ (there's just not any simple formula for one). Rather, it's $text{Gal}(mathbb{C}/mathbb{Q})$ which you can't write down any nontrivial elements of.
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– Eric Wofsey
Dec 17 '18 at 0:39
1
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@JyrkiLahtonen: Well, every automorphism of $overline{mathbb{Q}}$ of order $2$ is conjugate to complex conjugation (and of course, if you don't have a chosen embedding in $mathbb{C}$, you can only hope to describe it up to conjugation). See math.stackexchange.com/questions/2563758/….
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– Eric Wofsey
Dec 17 '18 at 18:47