# Let $K=\Bbb Q(X)$ where $X=\{\sqrt p: p$ is prime$\}$. How to conclude that $|Gal(K/\Bbb Q)|>[K:\Bbb Q]$

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)|$follows from b). • Isn't$[K:\mathbb Q]=\infty$in this case? – Gregory Grant Nov 3 '15 at 20:29 •$[K:\mathbb{Q}]=2^{|X|}=|Gal(K/\mathbb{Q})|=\infty$. – Dietrich Burde Nov 3 '15 at 20:32 • The dimension of$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}\$. – André Nicolas Nov 3 '15 at 20:36
• @user: See math.stackexchange.com/questions/1512 – Watson Aug 20 '16 at 16:04