Question about the Galois extension of a given field extension Let $K=\mathbb{Q}(\omega)$ be given, where $\omega^3=1$. I want to know:
(1) Whether there is a Galois extension $L/\mathbb{Q}$ containing $K$ such that $\mathrm{Gal}(L/\mathbb{Q})\cong\mathbb{Z}_4$?
(2) Whether there is a Galois extension $L/\mathbb{Q}$ containing $K$ such that $\mathrm{Gal}(L/\mathbb{Q})\cong Q$, where $Q$ is the quaternion group with 8 elements.
I have no idea how to solve such problem.
 A: The answer to part 1 is that such a field $L$ cannot exist. This is seen as follows. Assume that $L=\mathbb{Q}(\alpha)$ were such a field (the extension is known to be simple). Then the complex conjugate $\overline{\alpha}$ would also be an element of $L$ as is shares the minimal polynomial (over the rationals!) with $\alpha$. The argument can be repeated for all the elements of $L$, so we have shown that $L$ is stable under complex conjugation. 
But then the fixed field of complex conjugation, $\mathbb{R}\cap L$, is necessarily a quadratic extension of the rationals. Therefore $L$ contains at least two intermediate quadratic fields contradicting the fact that a cyclic group contains a unique subgroup of each possible order.

The non-existence of the field in part 2 is a bit different. As in part 1 we see that $L$ has to be stable under complex conjugation. The restriction of complex conjgation to $L$ is clearly of order two (can't be identity, when $K\subset L$). But the quaternion group $Q$ has a unique element of order two ($-1=i^2=j^2=k^2$), so the field $M=\mathbb{R}\cap L$
is the only quartic subfield of $L$. The other subgroups of $Q$ are all cyclic of order four, these are $\langle i\rangle$, $\langle j\rangle$ and $\langle k\rangle$. Therefore $L$ has exactly three subfields that are quadratic extensions of the rationals. The catch is that Galois correspondence forces all these to be subfields of $M$, and thus real. Hence $K$ cannot be one of them.
