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Let $K=\Bbb Q(X)$ where $X=\{\sqrt p: p$ is prime$\}$. Then $K$ is galois over $\Bbb Q$. If $\sigma \in Gal(K/ \Bbb Q)$, let $Y_{\sigma}=\{\sqrt p: \sigma(\sqrt p)=-\sqrt p\}$. Then how to prove

a) $Y_{\sigma}=Y_{\tau}$ then $\sigma=\tau$

b) If $Y \subseteq X$ then there is a $\sigma \in Gal(K/ \Bbb Q) $ with $Y_{\sigma}=Y$

c)If $\mathcal P(X)$ is the power set of $X$, show that $|Gal(K/\Bbb Q|=|\mathcal P(X)|$ and that $|X|=[K:\Bbb Q]$ then how to conclude that $|Gal(K/\Bbb Q)|>[K:\Bbb Q]$.

I am adding the proof of a) Here each $\sigma \in Gal(K/ \Bbb Q)$ will send $\sqrt p$ to a root of $x^2-p$. Now if $\sigma \neq \tau$ then $\exists q$ prime s.t $\sigma(\sqrt q)=\sqrt q$ but $\tau(\sqrt q)=-\sqrt q$. Hence $Y_\sigma \neq Y_\tau$

b) it is done also the same arguement as of a) that for each $\sigma \in Gal(K/ \Bbb Q)$ will send $\sqrt p$ to a root of $x^2-p$. But as it is infinite set I think that Zorn's lemma will be needed here.

c) For a single part of c) i.e |Gal(K/\Bbb Q)|=| \mathcal P(X)|$ follows from b).

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    $\begingroup$ Isn't $[K:\mathbb Q]=\infty$ in this case? $\endgroup$ – Gregory Grant Nov 3 '15 at 20:29
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    $\begingroup$ $[K:\mathbb{Q}]=2^{|X|}=|Gal(K/\mathbb{Q})|=\infty$. $\endgroup$ – Dietrich Burde Nov 3 '15 at 20:32
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    $\begingroup$ The dimension of $K$ over $\mathbb{Q}$ is countably iinfinite. The Galois group is uncountable, we can freely choose whether to map $\sqrt{p_n}$ to itself or to $-\sqrt{p_n}$. $\endgroup$ – André Nicolas Nov 3 '15 at 20:36
  • $\begingroup$ @user: See math.stackexchange.com/questions/1512 $\endgroup$ – Watson Aug 20 '16 at 16:04

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