Properties of Iterated Quadratic extensions

Question

Let $L=F(\sqrt{c})$ and $E=L(\sqrt{a+b\sqrt{c}})$ for some $a,b,c\in F$, where $F$ is a field of characteristic other than $2$. Let $E/F$ be non-Galois. Let $M=E(a-b\sqrt{c})$. Then, can we show that $M$ is Galois over $F$? and if it is Galois, what is the structure of the Galois group $G(M/F)$? I think we have to use the Quadratic formula here and the Fundamental theorem of Galois theory. Any ideas. Thanks beforehand.

Answer 1

As pointed out in the comments, I assume that you mean $M=E(\sqrt{a-b\sqrt{c}})$, otherwise the question does not make much sense.

It's easy to see that the degree of all field extensions that are involved will be a power of two, so as $\operatorname{char} F \neq 2$, we get seperability for free.

So we're concerned only about normality. Let $\alpha = \sqrt{a+b\sqrt{c}}$, $\beta = \sqrt{a-b\sqrt{c}}$, so that $M = F(\alpha,\beta)$. Consider $P(x) = (x^2-a)^2-b^2c$.

We have $\alpha^2= a -b \sqrt{c} \Rightarrow \alpha^2-a=-b\sqrt{c} \Rightarrow (\alpha^2-a)^2=b^2c \Rightarrow P(\alpha)=0$ Similiarly $P(\beta)=0$. Due to $P(x)=P(-x)$, we also have $P(-\alpha)= P(-\beta) = 0 $. As $\operatorname{char} F \neq 2$, we have found four distinct roots of $P$, thus $M$ is a splitting field of $P$, hence normal.

About the Galois group, we know that it is a subgroup of $S_4$, whose order is a power of two which has a non-normal subgroup (hence cannot be abelian), the only group which satisfies these requierements is $D_4$