# Constructing Galois Extensions of Desired Degree

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.

• I don't think your question is phrased correctly. If $L/\mathbb{Q}$ is an abelian Galois extension with Galois group $G = \mathrm{Gal}(L/\mathbb{Q})$ and $H$ is a subgroup of $G$, then the corresponding fixed field, say $K$, of $H$ is Galois over $\mathbb{Q}$. $L/K$ is Galois regardless of whether the extension is abelian or not. We then have $\mathrm{Gal}(K/\mathbb{Q}) = G/H$ and $\mathrm{Gal}(L/K) = H$. So if $|G| = km$ and $|H|=m$, then $K/F$ is an extension of degree $k$ and $L/K$ is of degree $m$. Jul 25, 2019 at 3:55
• Edit: I meant $K/\mathbb{Q}$ and not $K/F$. Jul 25, 2019 at 4:06
• @ParthivBasu Yes, I did incorrectly state the degrees in the $|G|=km$ because of the inclusion-reversing nature of the correspondence. But as to the rest, yes, $K$ is Galois because $L$ is but the content is that you start with $K$ and you want to extend to an abelian Galois extension $L/\mathbb{Q}$ of a desired degree with $H \leq G$ in a specified way. Jul 25, 2019 at 4:26

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.
• Thanks! The extension problem has never crossed my path so I was not aware. Do you know where I could find more or even for which types of fields this does not work for, especially in low degree extensions like a fixed cubic or quintic Galois field or given a Galois groups like $\mathbb{Z}/6\mathbb{Z}$, $\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}$. Jul 25, 2019 at 4:03