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I want to prove that the degree of $\mathbb C$ over $K$ is countably infinite. I think it's infinite because the polynomials $x^{2^n}-2$ are all irreducible over $K$ (can someone confirm this?), and it was suggested here (Can an algebraic extension of an uncountable field be of uncountable degree?) that adjoining all elements of $\overline{\mathbb Q}$ would give $\mathbb C$. Is this the case? I have had a long discussion in the comments of that post, but I'm not quite able to get the proof.

If it helps, I've shown that $\mathbb C$ is algebraic over $K$ and that any finite extension of $K$ within $\mathbb C$ is Galois and has cyclic Galois group of order $2^k$.

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Nishant
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  • Regarding the irreducibility of $f_n(x) = x^{2^n}-2$, for each $n \geq 1$ consider the monic polynomial $g_n(x) = f_n(x-2)$. The constant term of $g_n(x)$ is $g_n(0) = (-2)^{2^n} - 2 = 2(2^{2^n-1}-1)$, which is divisible by $2$ but not by $4$. Use the binomial theorem to see that $g_n$ is irreducible by Eisenstein's criterion. Hence so is $f_n$. In particular, this confirms that $K({\sqrt[n]{2}})$ is a countably infinite intermediate extension of $\mathbb{C}/K$, hence $\mathbb{C}/K$ is also at least countably infinite. I'll have to think on how to answer your other questions. – Dan Jun 17 '14 at 23:37
  • I think you can just apply Eisenstein directly to $x^{2^n}-2$, but I don't think Eisenstein applies here since we're talking about irreducibility over $K$, not $\mathbb Q$. – Nishant Jun 18 '14 at 04:29
  • I apparently forgot where I was going with that comment. You're absolutely right that Eisenstein's criterion is inadequate to show irreducibility. I think you can get around this by arguing more directly. If $\zeta \sqrt[2^k]{2} \in K$ for some positive $k$ and some primitive $2^k$-th root of unity $\zeta$, then since $K$ is a field $(\zeta\sqrt[2^k]{2})^{2^{k-1}} \in {\pm \sqrt{2}}$ and also in $K$. This contradicts the definition of $K$. Therefore the $2^n$-th roots of $2$ lie outside $K$, for all $n \in\mathbb{N}$. – Dan Jun 18 '14 at 19:18
  • Yes, I agree, but the polynomial $x^{2^n}-2$ can be reducible without having roots, right? – Nishant Jun 18 '14 at 19:35
  • It can. Do you need it to be irreducible (as opposed to factor into a product of non-linear irreducibles)? – Dan Jun 18 '14 at 20:09
  • Yeah, I'm trying to show that $\mathbb C$ is of infinite degree over $K$, so I'm trying to show that the splitting fields of $x^{2^n}-2$ produce extensions of $K$ of arbitrarily large degree. – Nishant Jun 18 '14 at 20:10

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