Galois Theory Problem (Fundamental theorem of Galois) Let $k$ be a field of characteristic$>2$. Let $c\in k$, $c\notin k^2$. Let $F=k(\sqrt{c})$. Let $\alpha=a+b\sqrt{c}$ with $a,b\in k$ and not both $a,b=0$. Ket $E=F(\sqrt{\alpha})$. Prove that the following conditions are equivalent.
(1) $E$ is Galois over $k$.
(2) $E=F(\sqrt{\alpha'})$ where $\alpha'=a-b\sqrt{c}$
(3) Either $\alpha\alpha'=a^2-cb^2\in k^2$ or $c\alpha\alpha'\in k^2$.
Show that when these conditions are satisfied, then $E$ is cyclic over $k$ of degree $4$ iff $c\alpha\alpha'\in k^2$.
This is a problem from Serge Lang. I proved 1 is equivalent to 2. But I do not know what to do afterward. Do I have to prove 2 equivalent to 3 by proof by contradiction?
 A: If you've shown $(1)\Longleftrightarrow(2)$, I'll show $(1)\Longrightarrow(3)$ and leave the converse for you (or you can prove $(3)\Longrightarrow (2)$, whichever suits you better).
Suppose $E$ is Galois over $k$. Then we'll show that $\sqrt{\alpha\alpha'}\in F$. Now, if $f(x)=(x^2-a)^2-cb^2$ is irreducible, then $\sqrt{\alpha'}$ is a conjugate of $\sqrt{\alpha}$, and thus belong to $E=F(\sqrt{\alpha})$ since $E$ is Galois. Thus, we have $\sqrt{\alpha'}=x\sqrt{\alpha}+y$ for some $x,y\in F$. By an easy calculation, we have $2xy\sqrt{\alpha}=\alpha'-x^2\alpha-y^2$. The right-hand side is in $F$, so we have three possible cases: $\sqrt{\alpha}\in F$, $x=0$, or $y=0$. Whatever the case, we see that $\sqrt{\alpha\alpha'}\in F$.
If $f$ is reducible, then either $\sqrt{\alpha\alpha'}\in k$ or $\sqrt{\alpha^2}=\alpha\in k$, which implies $b=0$. Both cases gives us $\sqrt{\alpha\alpha'}\in F$, so we have $$a^2-cb^2=\alpha\alpha'=(s+t\sqrt{c})^2=s^2+t^2c+2st\sqrt{c}$$for some $s,t\in k$. Thus $s=0$ or $t=0$, which implies $c\alpha\alpha'\in k^2$ or $\alpha\alpha'\in k^2$.
