Suppose that $K/\mathbb{Q}$ is an abelian Galois extension with Galois group $H$. Let $m= |H|=[K \colon \mathbb{Q}]$. Given an abelian group $G$ with $H \leq G$ and $|G|=km$, is it possible to find an extension $L/K/\mathbb{Q}$ with Galois group $G$? That is, given an abelian Galois extension, is it possible to extend this to another Galois field of desired degree so that the original Galois group is a specified subgroup?
For example, if $[K \colon \mathbb{Q}]=2$ and $G= \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, one can just take the $L$ to be the compositum of $K$ and another degree 2 extension.
But is this sort of idea always possible for 'nice' extensions, i.e. abelian Galois extensions? It is always easy to create an extension $L$ of degree $km$ or usually not difficult create a Galois field extension having $H$ as a subgroup, but to do both has proved a difficult task.
Best Answer
Not always. This is the extension problem. As an example, consider $\Bbb Q(i)/\Bbb Q$. There is no extension $L/\Bbb Q(i)$ such that $L/\Bbb Q$ is Galois over $\Bbb Q$ with cyclic Galois group of order $4$.
To see this, observe that complex conjugation must induce an order two automorphism of $L$, so must be the square of the generator. But $\Bbb Q(i)$ is the fixed field of the square of the generator (by the Galois correspondence) but its elements are not fixed by complex conjugation.