Suppose $K$ is a field of characteristic $\neq 2$ and suppose $c \in K\setminus K^2$ and $F=K(\sqrt{c})$. Suppose $\alpha=a+b\sqrt{c}$ with $a,b\in K$ so that $\alpha \notin K^2$ and $E=F(\sqrt{\alpha})$. Prove the following equivalences(prove they imply each other):
-$E/K$ is a Galois extension
-$E=F(\sqrt{\alpha'})$ with $\alpha'=a-b\sqrt{c}$
-$\alpha\alpha'=a^2-b^2c\in K^2$ or $c\alpha\alpha'\in K^2$
I am sorry about the lay out, I have no clue how to do it. I am following a beginner course in Galois theory and I really don't know how to start this exercise. Any help is appreciated! Thanks
To understand when $E/K$ might be Galois, it helps to try to write down the minimal polynomial for the generator $\sqrt{\alpha}$ over $K$. Set $\gamma = \sqrt{\alpha}$. Also, set $\alpha' = a-b\sqrt{c}$ and pick $\gamma'$ to be a square root of $\alpha'$. One can guess that when $E/K$ is Galois, then the conjugates of $\gamma$ are $-\gamma$, $\gamma'$, and $-\gamma'$. In fact,
$$ (x - \gamma) (x+ \gamma) (x-\gamma') (x+\gamma') = \left(x^2 - (a+b\sqrt{c}) \right) \left( x^2 - (a-b\sqrt{c}) \right)\\ = x^4 - 2a x^2 + (a^2 - b^2 c) $$
Thus, if this polynomial splits in $E/K$, then $E/K$ will be Galois. Conversely, if $E/K$ is Galois, then some automorphism of $E$ will take $\alpha$ to $\alpha'$ and hence $\gamma$ to $\gamma'$, and hence the polynomial splits.
To summarize, $E/K$ is Galois if and only if $\gamma' = \sqrt{\alpha'}$ is in $E$. The equivalence of the first two statements follows almost immediately from this.
For the equivalence with the third statement, I suggest trying to show that if $E/K$ is Galois, then $\alpha\alpha'$ is in $\left(F^{\times} \right)^2 \cap K$ by considering the action of various authomorphisms on $\sqrt{\alpha}$ and $\sqrt{\alpha'}$. Then try to show that every element of this intersection is either in $\left(K^{\times} \right)^2$ or is $c$ times such an element. The converse can be shown by assuming that $\alpha \alpha'$ is in $\left(F^{\times} \right)^2 \cap K$ and showing that $\sqrt{\alpha'} \in E$.