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Let $F$ be a subextension of $\mathbb{C}$ maximal with respect to not containing $\sqrt2$. Let $K/F$ be a finite extension with $K\subset\mathbb{C}$. Then $K/F$ is of Galois and $[K:F]$ is a power of a prime. Could you help me solve this exercise?

Following the idea of Olivier:

Let $L$ be the normal closure of $K/F$. If $K\neq F$, then $\sqrt2\in L$ and so $2|[L:F]=|\mathrm{Gal}(L/F)|=G$, let $H<G$ be a 2-Sylow. I want to prove $L^H=F$. If $L^H/F$ is proper then $\sqrt2\in L^H$, as in the answer of Olivier we prove that every element of $G$ fixes $\sqrt2$ and so $\sqrt2\in L^G=F$ contradiction, so $L^H=F$ and so $G=H$, so $[L:F]$ is a power of 2 and so it is $[K:F]$. What I'm missing is how to prove that $L=K$, i.e. that $K/F$ is normal.

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Let $L/F$ be the normal closure of $K/F$. Let $G=Gal(L/F)$, and let $P$ be a Sylow 2-subgroup of $G$. Then $M=\mathrm{Inv}(P)$ is an odd degree extension field of $F$, so it cannot contain $\sqrt2$ either, and we conclude by maximality that $M=F$ and that $G$ is a 2-group.

Let $H=Gal(L/K)$. As you observed, if we can show that $H$ is a normal subgroup of $G$, then we are done. Let $G_0=Gal(L/F[\sqrt2])$. As $F[\sqrt2]$ is contained in all the finite extensions of $F$, we see that $G_0$ must be the unique maximal subgroup of $G$.

The only abelian $2$-groups with a unique maximal subgroup are the cyclic ones. Using this we can prove the following.

Claim. A 2-group $G$ with a unique maximal subgroup $G_0$ is abelian.

Proof. We know that $G$ is nilpotent by virtue of being a 2-group. Consider the lower central series: $G_1=[G,G]$, $G_2=[G,G_1]$. The derived subgroup of the quotient $\overline{G}=G/G_2$ is then obviously $\tilde{G}=[\overline{G},\overline{G}]=G_1/G_2$. Because $[\tilde{G},\overline{G}]$ is trivial, we have $\tilde{G}\subseteq Z(\overline{G})$. The maximal subgroups of the abelian group $G/[G,G]$ are in 1-1 correspondence of maximal subgroups of $G$ containing $[G,G]$. As $G_0$ is the only such subgroup of $G$, we must have that $G_0/[G,G]$ is the only maximal subgroup of $G/[G,G]$. Therefore $G/[G,G]$ must be cyclic. We are now in a position to apply the well-known lemma: $$G/Z(G)\ \text{cyclic}\implies G\ \text{abelian}$$ to the group $\overline{G}$. Therefore $[\overline{G},\overline{G}]$ is trivial, and we thus have $G_2=G_1$. The lower central series of a nilpotent group can stall only at the trivial subgroup, so we have shown that $G_1=\{1_G\}$. Q.E.D.

Of course, this settles the original question as all the subgroups of an abelian group are normal. At this point we can also conclude that $G$ must be cyclic.

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I am tacitly using the fact that $\mathbf{C}$ is algebraically closed. This implies that any finite extension $E$ of $F$ is also contained in $\mathbf{C}$. Only with this "extra bit" does the maximality of $F$ to imply that $\sqrt{2}\in E$. –  Jyrki Lahtonen May 28 '12 at 13:52
    
In the second paragraph I think you want $G_0=Gal(L/F(\sqrt2))$, am I right? –  Berry May 28 '12 at 19:43
    
Absolutely, @Berry! Sorry about that. –  Jyrki Lahtonen May 28 '12 at 19:53
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I think I might have a solution to this but since I know very little actual field theory there might be flaws.

It hinges on the following "fact" that I can more or less prove: if $K/k$ is a finite extension and $\sigma$ is an automorphism of $k$, then it extends to an automorphism of $K$. Indeed, suppose $x_1,\dots,x_n$ generate the extension, then define extensions step by step $k\subset k(x_1)\subset\cdots\subset k(x_1,\dots,x_n)=K$ by saying something like the extension $k_i\subset k_i(x_{i+1})=k_{i+1}$ is simple and if $\sigma_i$ is already defined, just extend it by only changing the coefficients in $k_i$ for polynomial expressions in $x_{i+1}$.

this is the step I am not so sure about... If this cannot be done, the following will not suffice.

In any case, if $L$ is a field and $H$ a finite group of field automorphisms then the extension $L/L^H$ is Galois and has $Gal(L/L^H)=H$ and $[L:L^H]=|H|$.

Set $G=Gal(K/F)$ and consider a $2$-Sylow $H$. Then the extension $K/K^H$ is Galois of degree a power of $2$ and Galois group $H$. We would be done were $K^H$ equal to $F$.

Suppose by contradiction $L=K^H/F$ is a proper extension. Then it contains $x=$ the square root of $2$. Thus all automorphisms in $H$ stabilize it. If $\sigma$ is any automorphism in $G$, then it sends $x$ to itself or to its opposite. If the order of $\sigma$ is $2^a(2b+1)$, then $\sigma^{2b+1}$ belongs to some $2$-Sylow subgroup of $G$, and must thus stabilize $x$ (this would need some elaboration, but I believe it works, take some other $2$-Sylow, and by degree considerations it cannot fix only $F$ so it fixes $x$ aswell). Thus ($\sigma^{2b+1}$ and thus) $\sigma$ actually sends $x$ to itself. This means that every automorphism in $G$ stabilizes the square root of $2$, i.e. $x$.

But we have the extension $F\subset F(x)\subset K$ and an automorphism of $F\subset F(x)$ sending $x$ to its opposite $-x$ and by the first paragraph we get at least one element in $G$ that doesn't stabilize $x$ which is a contradiction. So $x\notin K^H$, and $F=K^H$ and we are done showing the extension is Galois.

We then have $G=Gal(K\F)=Gal(K/K^H)=H$ is a $2$-group.

Edit There is a flaw in the proof of the "fact" : if we want to extend an automorphism $\sigma$ of a field $L$ to an automorphism of some simple algebraic extension $L(y)$, we need to send the adjoined element $y$ to a root in $L$ of $Q=P^{\sigma}$ where $P$ is the minimal polynomial of $y$ and $Q$ is $P$ except all coefficients are replaced by their image under $\sigma$...

All we really need to make the previous proof work is to show that there is some automorphism of $K$ in $G$ that sends $x$ to its opposite.

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About the step you are not sure, I would observe that the fact holds (trivially) for any extension. Simply consider $K$ as a vector space over $k$ and let the extended automorphism act only on the "scalars". Unfortunately I don't see where are you using the fact! –  Giovanni De Gaetano May 9 '12 at 14:10
    
@Giovanni de Gaetano I don't think so unless I misunderstand your comment, we want the extension to remain a field automorphism. –  Olivier Bégassat May 9 '12 at 14:14
    
I use the "fact" at the very end to obtain somebody in $G$ that doesn't fix $x$ the square root of $2$, thus producing a contradiction. –  Olivier Bégassat May 9 '12 at 14:26
    
I'm very sorry! You are totally right! I thought that, according to the real meaning of your statement, the condition you want is equivalent to $K/k$ being Galois. But I do NOT have a proof for this! So, I apologize again for the comment above and I try not to add other dumb observations. :) –  Giovanni De Gaetano May 9 '12 at 14:28
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A simpler counterexample to the "fact" at the top is the extension ${\mathbf Q}(\sqrt[4]{2})/{\mathbf Q}(\sqrt{2})$. The automorphism of the base field $a + b\sqrt{2} \rightarrow a - b\sqrt{2}$ can't extend to an automorphism of ${\mathbf Q}(\sqrt[4]{2})$. –  KCd May 9 '12 at 17:55
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