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If we let $K$ be a maximal subfield of $\mathbb C$ that doesn't contain $\sqrt{2}$, then if $L$ is any finite extension of $K$, then its Galois with a cyclic 2-group. I want to use this to conclude that $[\mathbb C:K]$ is countable and not finite. I know that $\mathbb C$ is algebraic over $K$.

I'm not quite sure how to do this. I sort of have an idea for the not finite part: if it were finite, then $\mathbb C$ would be a cyclic extension of $K$, and so it would have a unique index 2 subfield. If I can prove that the only index $2$ subfield of $\mathbb C$ is $\mathbb R$, then I think I can finish the argument, since $K$ "should" contain imaginary elements (though I'm not too sure how to prove this either).

For countable, does the fact that every complex number is a root of a polynomial in $K[x]$ and the fact that $[K[x]:K]$ is countable work?

Martin Sleziak
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Nishant
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1 Answers1

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See Algebraically closed fields with proper maximal subfields at MathOverflow for a succinct answer (shorter than the one I wrote, until I recognized it as one I had read over there, looked up the reference, and deleted my answer).

Community
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Eric Towers
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  • Hmm, so it seems that $\mathbb R$ is not the only index $2$ subfield...can I still modify my argument to work? or is there a simpler way to conclude that it's not finite? – Nishant Jun 17 '14 at 05:25
  • @Nishant: Troublingly, the accepted answer shows that $[\mathbb{C}:K]$ (your $K$, not Laurent Moret-Bailly's $K$) is finite (in fact, $2$). – Eric Towers Jun 17 '14 at 05:38
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    Wait, really?! This is an exercise from Dummit and Foote, and this is not on the errata... – Nishant Jun 17 '14 at 05:42
  • Seems relevant: http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/artinschreier.pdf – Eric Towers Jun 17 '14 at 05:58
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    Wait, that just says that if $K$ WERE of finite index, it would be of index $2$. i guess that's why it's NOT of finite index...? – Nishant Jun 17 '14 at 06:00
  • Which part of Laurent Moret-Bailly's argument do you think does not apply to your problem? – Eric Towers Jun 17 '14 at 06:03
  • So if $[\mathbb C:K]$ were finite, it would be of index $2$, but I'm not sure how to derive a contradiction from that... – Nishant Jun 17 '14 at 06:06
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    The question you linked to deals with maximal subfields. In particular, it must be that $K(\alpha)=\mathbb{C}$ for any $\alpha \in \mathbb{C} - K$. But for this question, $K$ is only maximal with respect to the condition $\sqrt{2} \notin K$. There are many extensions of $K$ which are not $\mathbb{C}$: $K(\sqrt{2})$ (does not contain $\sqrt[4]{2}$), $K(\sqrt[4]{2})$, etc. – Jacob Bond Dec 28 '14 at 20:54