1

I'm trying to determine the Galois group of $\mathbb{Q}(\sqrt[5]{3},\zeta):\mathbb{Q}$, where $\zeta=e^{2\pi i/5}$. I have the following information:

  • $\mathbb{Q}(\sqrt[5]{3},\zeta)$ is the splitting field for $x^5-3$, whose roots are $\sqrt[5]{3}, \zeta \sqrt[5]{3}, \zeta^2 \sqrt[5]{3}, \zeta^3 \sqrt[5]{3},$ and $\zeta^4 \sqrt[5]{3}$.
  • $\mathbb{Q}(\sqrt[5]{3},\zeta):\mathbb{Q}$ has degree 20.
  • If $\sigma$ is an element of the Galois group, then $\sigma$ must send a root of $x^5-3$ to a root of $x^5-3$. So the Galois group is isomorphic to a subgroup of $S_5$ with order 20.
  • $\sigma$ is completely determined by where it sends $\sqrt[5]{3}$ and $\zeta$ (the basis elements for $\mathbb{Q}(\sqrt[5]{3},\zeta):\mathbb{Q}$).

There is a unique subgroup of $S_5$ of order 20, which is $\langle (1\,2\,3\,4\,5),(1\,2\,4\,3)\rangle$, but without using this, how can I determine the Galois group structure?

Also, how can I show it's not abelian?

user500144
  • 927
  • To show it is not abelian, note that you have one automorphism given by complex conjugation (fixes $\sqrt[5]{3}$, and exchanges $\zeta\sqrt[5]{3}$ with $\zeta^4\sqrt[5]{3}$ and $\zeta^2\sqrt[5]{3}$ with $\zeta^3\sqrt[5]{3}$), and one that fixes $\zeta$ and sends $\sqrt[5]{3}$ to $\zeta\sqrt[5]{3}$. Check what happens when you compose them in the two different orders. – Arturo Magidin Apr 08 '19 at 04:37
  • 1
    To realize it as a subgroup of $S_5$ explicitly, think about what it does to the roots of $x^5-3$; verify that you can just decide where to map $\sqrt[5]{3}$ and $\zeta\sqrt[5]{3}$ (as long as they are not mapped to the same thing), and then number the five roots and see what happens to them under the maps. – Arturo Magidin Apr 08 '19 at 04:40
  • Did you mean $\zeta^4$ where you wrote $\zeta^5$? – J. W. Tanner Apr 08 '19 at 04:41
  • @J.W.Tanner Yes; it's fixed – user500144 Apr 08 '19 at 05:00
  • @ArturoMagidin For your second comment, deciding where $\sqrt[5]{3}$ and $\zeta \sqrt[5]{3}$ are sent to determines where the basis elements $\sqrt[5]{3}$ and $\zeta$ are sent to, which then completely determines the automorphism, correct? And there are $5\cdot 4=20$ possible choices for mapping those two elements. – user500144 Apr 08 '19 at 05:19
  • @ArturoMagidin To see the group structure, do I need to explicitly write out each of the 20 elements and figure out the subgroups? – user500144 Apr 08 '19 at 05:21
  • @ArturoMagidin Or I guess now I've found 2 elements, one corresponding to $(1,2,3,4,5)$ and another corresponding to $(2,3,5,4)$. Sorry I'm rusty on group theory - how does this determine the whole group structure; or how do I get all the subgroups from this? – user500144 Apr 08 '19 at 05:29
  • What I’m saying is that instead of thinking about $\sqrt[5]{3}$ and $\zeta$ separately, you can think about what is happening to the roots of $x^5-3$. Which has the added benefit of having some geometry to it, since they happen to be the vertices of a nice regular pentagon on the complex plane...And you can try to figure out what happens to that pentagon as a result of the automorphisms.... – Arturo Magidin Apr 08 '19 at 05:40

1 Answers1

2

This is the Galois group of $x^5-3$. In general, the Galois group of $x^n-a$ is explicitly known, see here:

Computing the Galois group of polynomials $x^n-a \in \mathbb{Q}[x]$

Proposition [Jacobson, Velez]: One has $G\cong \mathbb{Z}/n \rtimes (\mathbb{Z}/n)^{\times}$ if and only if $n$ is odd, or $n$ is even and $\sqrt{a}\not\in \mathbb{Q}(\zeta_n)$.

Here $n=5$ is odd, so we know that $G\cong \Bbb Z/5\rtimes \Bbb Z/4$ is a solvable, non-abelian group of order $20$.

Dietrich Burde
  • 130,978