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I want to show that, if $K$ is a maximal subfield of $\mathbb C$ without $\sqrt{2}$ in it, then $\mathbb C$ is of countable degree over $K$. Is it the case that an algebraic extension of an uncountable field always has countable degree? if so, I can try to show $K$ is uncountable.

Nishant
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1 Answers1

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Yes. Here's an example.

Suppose that $B$ is an uncountable set of algebraically independent (over $\Bbb Q$) numbers from $\Bbb C$, and consider $F$ to be the smallest field containing $B$ and $\Bbb Q$.

Now consider $K=F[\{\sqrt x\mid x\in B\}]$. Since the elements of $B$ are algebraically independent it follows that none of the square roots is in $F$. But this is certainly an algebraic extension.

Asaf Karagila
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  • Thanks. So, do you have any hints as to how to solve my original problem? – Nishant Jun 17 '14 at 07:55
  • My immediate guess is to appeal to the Artin-Schreier theorem and show that there's no such field. – Asaf Karagila Jun 17 '14 at 07:58
  • I think Zorn's Lemma implies that such a field exists. That theorem seems to say that the degree of $\mathbb C$ over it is infinite, but how do I show it's not uncountable? – Nishant Jun 17 '14 at 07:59
  • Oh, yeah. You're right, it's not that $K[\sqrt2]=\Bbb C$. To show that $\Bbb C$ cannot be an uncountable degree away, I'd try to show that $\Bbb C$ can be obtained by adding only algebraic numbers. – Asaf Karagila Jun 17 '14 at 08:05
  • But if $K$ is uncountable, then there may be uncountably many algebraic numbers, since there are uncountably many polynomials... – Nishant Jun 17 '14 at 15:53
  • I meant algebraic over $\Bbb Q$. (Otherwise I'd have said "algebraic over $K$".) – Asaf Karagila Jun 17 '14 at 15:54
  • Oh, I see. I guess everything transcendental has already been added, so adding all other algebraic numbers should get all of $\mathbb C$?

    Hmm, now for the nonfiniteness part...I'm assuming the intended solution has something to do with the fact that the Galois group of any extension between $K$ and $\mathbb C$ has to be a cyclic 2-group. Do you have any ideas as to how to proceed from that direction?

     – Nishant Jun 17 '14 at 16:00
  • Well, to show it's infinite is quite easy. $\sqrt2$ not there, so $\sqrt[4]2$ is not there, and so on and so forth. Also, it's not true that every transcendental number has been added, if $\pi$ is in $K$ then $\pi\sqrt2$ is not; and vice versa. But you can say that given a transcendental number, it is algebraically dependent with one from $K$ (which is the correct argument, I suppose). – Asaf Karagila Jun 17 '14 at 16:09
  • Maybe, but how do you know that by adjoining $\sqrt{2}$, you don't also get $\sqrt[4]{2}$?

    Yeah, that's a good point. But the argument does still work, I think.

     – Nishant Jun 17 '14 at 16:20
  • What is the degree of $K[\sqrt2]$ over $K$? What is the degree of $K[\sqrt[4]2]$ over $K$? – Asaf Karagila Jun 17 '14 at 16:26
  • Well, they seem to be $2$ and $4$...if $x^4-2$ is reducible, then it follows that $\sqrt{2}\in K$. And I suppose this can be extended to all of the roots of $2$? – Nishant Jun 17 '14 at 16:32
  • And why shouldn't it? – Asaf Karagila Jun 17 '14 at 16:34
  • Also, I have a doubt about the countable argument: even if a transcendental has a polynomial relation with algebraic numbers and elements of $K$, how do we know that we can use that relation to "solve for" the transcendental as a rational function of elements of $K\overline{\mathbb Q}$? – Nishant Jun 17 '14 at 16:34
  • Well, suppose there was something we couldn't get from $K\overline{\Bbb Q}$, $x$, is it possible that $\sqrt2\in K[x]$? – Asaf Karagila Jun 17 '14 at 16:38
  • Well, $x$ is algebraic over $K$, just by squaring. So some polynomial $p\in K[x]$ satisfies $p(x)=\sqrt{2}$, but I don't see that this can be used to solve for $x$... – Nishant Jun 17 '14 at 16:43
  • I think this would be the place to appeal to the maximality of $K$. – Asaf Karagila Jun 17 '14 at 16:45
  • Didn't I do that to write $\sqrt{2}$ as a polynomial over $K[x]$? Or do you mean in a different way? – Nishant Jun 17 '14 at 16:47
  • No, we assume towards contradiction that $\sqrt2\in K[x]$. Now it's time to use maximality to have that $x\in K\overline{\Bbb Q}$. I should perhaps say that we've longed crossed the limits of my knowledge in field theory. I'm just suggesting ideas that I'd find natural to pursue, had I been trying to prove this statement. – Asaf Karagila Jun 17 '14 at 16:49
  • Hmm, so does this means that $K(x)$ is contained in $K\overline{\mathbb Q}$? If so, then $x\in K\overline{\mathbb Q}$... – Nishant Jun 17 '14 at 16:55
  • The point is, I think, that if $K[x]$ is not a subset of $K\overline{\Bbb Q}$ then there is some $y\in K[x]\setminus K$ such that $\sqrt2\notin K[y]$. This will contradict the maximality of $K$. – Asaf Karagila Jun 17 '14 at 17:05
  • How do we construct such a $y$? If we assume for contradiction that $x\notin K\overline{\mathbb Q}$, then wouldn't $y=x$? or is there a special way to pick $y$? – Nishant Jun 17 '14 at 17:31
  • Well, clearly if $\sqrt2\notin K[x]$ then $x\in K$ already, by maximality. So $\sqrt2\in K[x]$. Now using some standard Galois-theoretic (which I truly can't recall) you find some $y$ such that $K[x]=K[y][\sqrt2]=K[\sqrt2][y]$, or something like that (if I recall correctly, you may assume $K[x]$ is a Galois extension, then you can take the subfield which is fixed by some particular subgroup, and pick a generator for it, or something like that). – Asaf Karagila Jun 17 '14 at 17:55
  • Hmm, so then $y\notin K(\sqrt{2})$, but why does that imply that $\sqrt{2}\notin K(y)$? – Nishant Jun 17 '14 at 18:35
  • If $K[x]=K[y][\sqrt2]$ then $\sqrt2\notin K[y]$. – Asaf Karagila Jun 17 '14 at 18:38
  • Why not? Why can't $K[\sqrt{2}][y]=K[y]$? – Nishant Jun 17 '14 at 18:40
  • No, this is due to the choice of $y$ (which should exist due to some Galois-theoretic considerations that I don't fully remember at the moment). – Asaf Karagila Jun 17 '14 at 18:43