How to solve the following question?!
Let $K$ be an imaginary, quadratic field and let $L/K$ be a Galois extension. If $\tau$ is complex conjugation, show that:
(a) $L/\Bbb Q$ is Galois iff $\tau(L)=L$
(b) If $L/\Bbb Q$ is Galois, then
(i) $[L\cap\Bbb R:\Bbb Q]=[L:K]$
(ii) For $\alpha\in L\cap \Bbb R,\,L\cap\Bbb R=\Bbb Q(\alpha)\iff L=K(\alpha)$
Thanks in advance
This is an exercise in group theory mostly. Clearly if $L/\Bbb Q$ is Galois, then $\tau(L)=L$ since Galois fields are equal to their conjugates for any automorphism of the algebraic closure. For the converse, note that the only automorphism of $K/\Bbb Q$ is complex conjugation, and the extension theorem gives an extension to $L/\Bbb Q$, and any automorphism of $L/\Bbb Q$ differs from one of $L/K$ by $\tau$ if at all. But if $\tau$ doesn't conjugate $L$ to another field, it must be that $L$ is already a Galois extension of the rationals.
For the other parts, we use (a) to characterize things. If $\tau(L)=L$ then $L\cap \Bbb R=L^{\tau}$. Since $K\subseteq L$, we have that $L\cap \Bbb R\subset L$ is proper, and since the order of complex conjugation is $2$, we know $[L:L\cap \Bbb R]=2$, but then
$$2[L\cap\Bbb R:\Bbb Q]=[L:L\cap\Bbb R][L\cap\Bbb R:\Bbb Q]=[L:K][K:\Bbb Q]=[L:\Bbb Q]=2[L:K].$$
which proves (i).
Finally, we note that
$$L\cap\Bbb R=\Bbb Q(\alpha)\iff L\cap\Bbb R=\Bbb Q[x]/m_\alpha(x)$$
since $L$ is Galois. But then We note that $m_\alpha(x)$ remains irreducible in $K[x]$ since $\alpha$ is totally real, hence $\alpha$ retains its degree in $K$, so that
$$2[K(\alpha):K]=[L:\Bbb Q].$$
Since clearly $K(\alpha)\subseteq L$ we have equality. For the converse we note that if $L=K(\alpha)$ then alpha must be fixed by complex conjugation and have the right degree, hence $L\cap\Bbb R=\Bbb Q(\alpha)$.