# $\alpha \alpha' \in K^2$ if and only if $Gal(E/K) \cong C_2 \times C_2$

K= field of characteristic p $$\neq$$ 2, $$c \in K-K^2$$ and $$F=K(\sqrt{c})$$. Let $$\alpha = a+b\sqrt{c}$$ such that $$\alpha \notin F^2$$ and $$E=F(\sqrt\alpha)$$. Define $$\alpha'=a-b\sqrt{c}$$.

How can I prove the next statement:

$$\alpha \alpha'= a^2-b^2c \in K^2$$ if and only if $$Gal(E/K) \cong C_2 \times C_2$$.

Let me give a more « conceptual» Galois answer. First , fix an integer m and consider a field K of characteristic not dividing m, containing a primitive m-th root of unity. Let F/K be a finite Galois extension, and let $E = F (\sqrt[m]{\alpha})$, with $\alpha \in K*$ . By Kummer theory, E only depends on the class $[\alpha]$ of $\alpha$ modulo $F*^m$ , and E/F is cyclic of degree dividing $m$ . Question : when is E/K Galois ? Answer : iff $[\alpha]$ is invariant under the action of Gal(F/K), in other words, for any s in Gal(F/K), $s(\alpha)/\alpha$ $\in$ $F*^m$ (this is a classical exercise in Galois theory).
Coming back to your problem, take $m = 2$ , F/K quadratic , s a generator of its Galois group. According to the criterion above, for $E = F(\sqrt{\alpha})$ , E/K is Galois (of degree 4) iff $s(\alpha)/\alpha$ is a square in F*, equivalently iff $\alpha . s(\alpha)$ is a square in F* . It is important to stress « in F*, not necessarily in K* », because $\alpha . s(\alpha) = N(\alpha)$, where N is the norm map from F* to K*, and $N(\alpha)$ $\in$ K* . Suppose the criterion is met, then Gal(E/K) is a group of order 4, hence abelian, isomorphic to $C_4$ or $C_2 × C_2$ . Question : how to distinguish between the two cases, cyclic and biquadratic ? We must go back to the idea of proof in the above Galois exercise. In our situation, the original expression $s(\alpha) = \alpha . x^2$, $x$ $\in$ F*, implies that N(x) = $\pm$ 1. How to distinguish between the two signs ? By Hilbert 90 (or mere calculation), N(x) = 1 iff x is of the form $s(y)/y$, y$\in$ F*, hence $s(\alpha)/\alpha$ = $(s(y)/y)^2$, or equivalently, $\alpha = a. y^2$, a$\in$ K*. The latter equality means that E = F($\sqrt a$), or equivalently, writing F = K($\sqrt c$) in your original notation, that F is the biquadratic field K($\sqrt a, \sqrt c$), a and c $\in$ K* .
This solution may seem complicated, but could be useful in more elaborate associated problems, for instance : describe all the quadratic extensions L of K($\sqrt a, \sqrt c$) which are Galois but non abelian (i.e. which are diedral or quaternionic) over K .
Let $\beta = \alpha'$ for clarity.
If $\sqrt{\alpha\beta} \in K$, then $\sqrt{\beta} \in E$, thus $E$ is the splitting field of $$f(x) = (x - \sqrt{\alpha})(x + \sqrt{\alpha})(x - \sqrt{\beta})(x + \sqrt{\beta}).$$ Now $[E : K] = 4$, so $G = \textrm{Gal}(E/K)$ is isomorphic to $C_4$ or $C_2 \times C_2$. Since $\sqrt{\alpha\beta} \in K$, for any $\sigma \in G$, we have $$\sigma(\sqrt{\alpha\beta}) = \sqrt{\alpha\beta}.$$ Label $\alpha,-\alpha,\beta,-\beta$ as $1,2,3,4$, respectively. Try to use the above equation to show that the elements of $G$ are the identity and $$(12)(34),(13)(24),(14)(23),$$hence $G \cong C_2 \times C_2$.
Now suppose $G \cong C_2 \times C_2$. If $E/K$ is Galois, we must have $\sqrt{\beta} \in E$. Try to conclude that no non-identity element of $G$ fixes any of the roots of $f(x)$. You can then conclude that $G$ is the group by the above permutations of the roots and then conclude that $\sqrt{\alpha\beta}$ is fixed by $G$, hence $\sqrt{\alpha\beta} \in K$.