4
$\begingroup$

I'm trying to do the following exercise:

find the Galois group $ G(E/\mathbb{Q}) $, where $ E $ is the splitting field of $ x^3 - 7 $, all its subgroups and the intermediate subfields $ E^H $ (subfields of $ E $ which are fixed by $ H \subset G $).

Of course $ E = \mathbb{Q}(\sqrt[3]{7}, \varepsilon_3)$, where $ \varepsilon_3 $ denotes the primitive root of unity of degree $ 3 $

The degree of this extension is $ 6 $ and I am able to find two automorphisms of $ E/\mathbb{Q} $ which do not commute, hence $ G \simeq S_3 $, since there are only two groups of order $6 $.

I know the subgroup structure of $ S_3 $ - all its proper subgroups are cyclic.

Is there a more elegant way to find all the intermediate subfields other than just checking which elements are fixed by every automorphism?

I'd appreciate some help with that.

$\endgroup$
1
  • $\begingroup$ You can show that the Galois group G is $S_3$. It is easy to see that it has degree 6($Q(\sqrt[3]7)$ is real) and $G$ embeds into $S_3$ since it permutes the $3$ roots.. $\endgroup$
    – Asvin
    Jun 1, 2015 at 20:58

1 Answer 1

1
$\begingroup$

You know that $x^{3}-7$ is irreducible over $\mathbf{Q}$ so the Galois group acts transitively on the set of roots, so $\mathbf{Q}(\sqrt[3]{7})$,$\mathbf{Q}(\zeta_{3}\sqrt[3]{7})$, $\mathbf{Q}(\zeta_{3}^{2}\sqrt[3]{7})$ are subfields of degree 3. And you know that these correspond to the subgroups of order 2 of $S_{3}$, and $S_{3}$ has exactly three such subgroups. Now $\mathbf{Q}(\zeta_{3})$ is a subfield of degree 2, and hence corresponds to a subgroup of order 3 in $S_{3}$, and $S_{3}$ has a unique such subgroup.

$\endgroup$
2
  • $\begingroup$ Thank you. I'm wondering what happens if I substitute $ \mathbb{Q} $ with $ \mathbb{Q}(i) $. Then the polynomial $ f $ is still irreducible and I think that this (my and your) argument still holds. Am I right or does something crucial change? $\endgroup$
    – Jytug
    Jun 1, 2015 at 21:24
  • $\begingroup$ Jup, I still think so, but the subfields will be $\mathbf{Q}(i)(\zeta_{3}) \cdots$ $\endgroup$
    – mich95
    Jun 1, 2015 at 21:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .