There are fields $\mathbb{Q}$, $L$ and $M$ $s.t.$ $\mathbb{Q} \subset L \subset M$
Here the $\alpha \in L$ and $[L ; \mathbb{Q}] = 2$
The field $M(= L(\sqrt \alpha) )$ is a finite normal extension over $\mathbb{Q}$
Let the Galois group, $G= G(M/\mathbb{Q})$ and it is cyclic group with ordr 4
Show the existence of the $\beta \in L$ $s.t.$ $\Pi_{\sigma \in H } \sigma(\beta)=-1$ for $H = G(L/\mathbb{Q})$
Solving the above question, I tried to find the $\beta$ but failed.
Plus I already knew the fact that $G(L/\mathbb{Q}) \simeq \mathbb{Z}_2$ by Galois thm
Any help would be appreciated.
That it is a tower of quadratic extensions means $$L=\Bbb{Q}(\sqrt{d}),\quad M=\Bbb{Q}(\sqrt{d},r), \qquad r=\sqrt{u+v\sqrt{d}}$$ Let $r^*=\sqrt{u-v\sqrt{d}}$, that $M/\Bbb{Q}$ is Galois means $r^* \in M$. That it is not biquadratic means $v\ne 0, r\ne r^*$. Let $\phi(r)=-r,\phi(\sqrt{d})=\sqrt{d}\in Aut(M/L)$. That $M/\Bbb{Q}$ is cyclic means $\phi$ is the only element of order $2$.
The Galois group acts transitively so there is $\sigma\in Aut(M/\Bbb{Q}),\sigma(r)=r^*$. Then $\sigma(u+v\sqrt{d})=\sigma(r^2)=(r^*)^2=u-v\sqrt{d}$. Thus $\sigma(\sqrt{d})=-\sqrt{d}$ and $\sigma^2(\sqrt{d})=\sqrt{d}$ means that $\sigma^2(r)=\pm r$. The sign has to be $-$ since otherwise $\sigma$ would have order $2$.
And hence the Galois group is generated by $\sigma$.
Since $\sigma^2(rr^*)=rr^*$ then $rr^*$ is in the subfield fixed by $\phi=\sigma^2$ which is $L$. $$rr^*=a+b\sqrt{d}$$ $$N_{L/\Bbb{Q}}(\frac{a+b\sqrt{d}}{u+v\sqrt{d}})=\frac{rr^*\sigma(rr^*)}{u^2-v^2d}=\frac{-r^2 (r^*)^2}{r^2 (r^*)^2}=-1$$