Why is this Galois group abelian?

Consider the field extension $\mathbb Q(\zeta_3,\sqrt2,\zeta_8)/\mathbb Q(\zeta_3)$, with intermediate fields $\mathbb Q(\zeta_3,\sqrt2)$ and $\mathbb Q(\zeta_3,\zeta_8)$.

Denote

$L:=\mathbb Q(\zeta_3,\sqrt2,\zeta_8)$, $K=\mathbb Q(\zeta_3)$, $M_1=\mathbb Q(\zeta_3,\sqrt2)$, $M_2=\mathbb Q(\zeta_3,\zeta_8)$

Now I know that the extension $L/K$ is galois. I think that all other subextensions are galois too (?)

How do I see from this, that the Galois group of $L/K$ is abelian?

• I assume that $\zeta_8$ is an eighth-root of 1 but which one? – Piquito Jun 18 '15 at 15:09
• How do you know that those are the only intermediate fields? The ones you have listed are all Galois over $K$ since they are all splitting fields. I would calculate the order of the Galois group. Note that the finite nonabelian groups that have all subgroups normal must have order divisible by $8$. – Taylor Jun 18 '15 at 15:32
• I'm not sure what your'e looking for here, exactly. It's not true that $M_1$ and $M_2$ are the only intermediate fields -- there's also $\mathbb{Q}(\zeta_3,i)$, $\mathbb{Q}(\zeta_3,\sqrt{2})$, $\mathbb{Q}(\zeta_3,i\sqrt{2})$, $\mathbb{Q}(\zeta_3,i\sqrt{2})$, $\mathbb{Q}(\zeta_3,\sqrt{2})$, and $\mathbb{Q}(\zeta_3,i\sqrt{2})$. Do you just want to know what the Galois group is? – Jim Belk Jun 18 '15 at 18:09
• I didn't mean to claim that they are the only ones. I just think that the intermediate fields help to split the Galois group into two parts – Dan Jun 19 '15 at 11:40

Basic facts about linearly disjoint extensions and how they mesh with extensions being Galois give you everything you want. If you have not seen the concept before you can use the notes I prepared for a local seminar as a first aid. Pete L. Clark has published more extensive and polished lecture notes on his web page.

Anyway:

1. $M_1/K$ is a Galois extension and cyclic of degree three because it is a root extension of the type, where the base field contains the appropriate roots of unity.
2. $M_2/K$ is a Galois extension, of degree four with the Galois group isomorphic to the Klein four group. This is because $M_2$ is the 24th cyclotomic field.
3. Because $[M_1:K]=3$ and $[M_2:K]=4$ are coprime we have $M_1\cap M_2=K$.
4. For two Galois extensions (of the same field $K$) the trivial intersection condition of item 3 is equivalent to the extensions being linearly disjoint, so (see Proposition 7 in my notes) the compositum $L=M_1M_2$ is Galois over $K$, and the Galois group $$Gal(L/K)\cong Gal(M_1/K)\times Gal(M_2/K)$$ is thus abelian as a direct product of two abelian groups.

The relevant results can surely also be found in many textbooks:

• Two Galois extensions are linearly disjoint, iff their intersection is trivial.
• The compositum of two linearly disjoint Galois extensions is itself Galois, and the corresponding Galois group is the direct product of the Galois groups of the two smaller extensions.
• One answer-lesson that I liked – Piquito Jun 18 '15 at 19:15

I don’t think this is so hard. Your ground field is $K=\Bbb Q(\zeta_3)$, the field of cube roots of unity. You have one extension gotten by adjoining to $K$ the cube root of $2$, and this is abelian, in fact cyclic of order $3$. The other extension of $K$ is gotten by adjoining the eighth roots of unity, and that’s abelian, since it’s already abelian over $\Bbb Q$. So you have a compositum of two abelian extensions of $K$, and this extension is consequently abelian.