4
$\begingroup$

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 $Gal(E/K) \cong C_2 \times C_2$.

$\endgroup$
2
$\begingroup$

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 .

$\endgroup$
2
$\begingroup$

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 $\sqrt{\alpha\beta} \in K$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.