TITLE: Need help determining the Galois group of an extension
QUESTION [5 upvotes]: In an assignment I got I have been asked to try to determine when $E/F$ is a Galois Extension, and determine the Galois group of such an extension.
$\textbf{Context:}$ $F$ is any field of characteristic zero and $E = F(\sqrt{c},\sqrt{a + b\sqrt{c}})$, where $a,b,c \in F$, $c$ is not a square in $F$ and $a  + b\sqrt{c}$ is not a square in $F(\sqrt{c})$. Now assume that $b \neq 0$; I have determined that this is a Galois extension iff
$$\sqrt{ a -b\sqrt{c}}$$ is in $E$. This holds iff $a - b\sqrt{c}$ is the square of an element in $F(\sqrt{c})$, or $(a + b\sqrt{c})(a - b\sqrt{c} ) = a^2 - b^2c$ is the square of an element in $F(\sqrt{c})$. Now the first case cannot hold because it contradicts our assumption that $a + b\sqrt{c}$ is not a square in $F(\sqrt{c})$. So the second case holds, that is when $a^2 -b^2c = (h + g\sqrt{c})^2$ for some $h,g \in F$.
Expanding this out and comparing coefficients, it must be the case that either $h = 0$ or (exclusively) $g = 0$. The first case gives that 
$$a^2 - b^2c = g^2c$$ for some $g \in F$, or (exclusively) 
$$a^2 - b^2c = h^2$$ for some $h \in F$. 
$\textbf{Where I'm stuck:}$ Now the problem for me comes in determining the Galois group of $E/F$. Firstly $E$ can be written as $F(\sqrt{a  + b\sqrt{c}}, \sqrt{a - b\sqrt{c}})$ or even just $F(\sqrt{a + b\sqrt{c}})$.
If we take $E  = F(\sqrt{a  + b\sqrt{c}}, \sqrt{a - b\sqrt{c}})$ then noticing that $E$ is the splitting field of $(x^2 - a)^2 - b^2c$ over $F$, my guess is that $\sigma \in \operatorname{Gal}(E/F)$ can only take $\sqrt{a + b\sqrt{c}}$ to $- \sqrt{a + b\sqrt{c}}$, it can't take $\sqrt{a  + b\sqrt{c}}$ to say $\sqrt{a  - b\sqrt{c}}$. I am guessing this based on looking at the polynomial $(x^2 - a)^2 - b^2c$  which can be written as 
$$(x^2 - (a + b\sqrt{c}))(x^2 - (a - b\sqrt{c})).$$
However it seems to me that I am not taking advantage of the conditions found for $a^2 - b^2c$ above. In addition how can I determine whether given some $\sigma$ a permutation of the roots, that it is actually a valid member of the Galois group $\operatorname{Gal}(E/F)$?
Thanks. 
$\textbf{Edit:}$ I think the second case where $a^2 -b^2c = h^2$ cannot hold because this would contradict the fact that the degree of $E = F(\sqrt{a  +b\sqrt{c}},\sqrt{a - b\sqrt{c}})$ over $F$ is 4.

REPLY [3 votes]: Since you have already established that if $a^2 - b^2c = h^2$ for rational $h$, that $\mathrm{Gal}(E/F) \cong V$, I will only address the "other case" where $a^2 - b^2c = g^2c$ for some rational $g$.
It suffices to show that $\sigma:\sqrt{a+b\sqrt{c}} \mapsto \sqrt{a-b\sqrt{c}}$ is of order $4$. Let $\alpha = \sqrt{a+b\sqrt{c}}$ and $\beta = \sqrt{a-b\sqrt{c}}$. Then $\sigma$ is of order $4$ iff $\sigma(\beta) \neq \alpha$.
Note that $\alpha\beta = (g/b)(\alpha^2 - a)$, so $\beta = (g/b)(\alpha - (a/\alpha))$. Hence:
$\sigma(\beta) = (g/b)(\beta - (a/\beta)) = (g/b)((\beta^2 - a)/\beta) = (g/b)(-b\sqrt{c}/\beta) = (-g\sqrt{c}/\beta)$
$ = (-\alpha\beta/\beta) = -\alpha$,
so $\sigma$ is of order 4, so in this case $\mathrm{Gal}(E/F) \cong C_4$.