The maximal subfield of $\mathbb C$ not containing $\sqrt2$

Question

He said,

"...fixed field is an extension of $K$ which doesn't contain $\sqrt{2}$, and thus must be $K$ itself. Thus, the Galois group is cyclic..."

Why is that so?

Could you give a reason or hint?

Thank you!

There's not any "the" maximal subfield not containing $\sqrt 2$. Take for example $\mathbb Q(\pi)$ and $\mathbb Q(\sqrt 2\pi)$. We can extend each of these to a maximal subfield not containing $\sqrt 2$ (by Zorn's lemma), but those maximal subfields cannot be the same. — hmakholm left over Monica, Jan 03 '16 at 16:42
@HenningMakholm Thanks for your comment. Can you give some advice about What I asked? — nicksohn, Jan 03 '16 at 16:52

score 0 · Answer 1 · edited Apr 13 '17 at 12:19

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This boils down to the group theoretic proposition that any 2-group with unique subgroup of index $2$ is cyclic. A proof is here:Abelian $p$-group with unique subgroup of index $p$

To see the reduction let $L$ be any alg. ext. of $K$ and $E$ its normal closure. Let $G$ be the Galois group of $E/K$ then $G$ is a $2$-group since if the $2$-sylow subgroup has index, which must be odd, greater than $1$ then you get an odd extension of $K$. Further any extension contains $K(\sqrt{2})$ and so there is a unique subgroup of index $2$. It then follows that $G$ is cyclic and all intermediate fields are also Galois and cyclic.

edited Apr 13 '17 at 12:19

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answered Jan 03 '16 at 18:34

Rene Schipperus

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Thnks a lot! What if we replace $\sqrt2$ by any irrational number which is in a algebraic closure of Rationals. Simialr Can work? I don't see... – nicksohn Jan 03 '16 at 21:01
If it has prime order then al least I think its clear. – Rene Schipperus Jan 03 '16 at 22:07
I think it reduces to this case by an easy argument so... yes. – Rene Schipperus Jan 03 '16 at 22:25

The maximal subfield of $\mathbb C$ not containing $\sqrt2$

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