# Galois group of the splitting field of $x^3-2$

I want to find $Gal(E/\mathbb{Q})$ where $E$ is the splitting field of $f(x)=x^3-2$.

I started out finding the zeros, which is $2^{1/3},2^{1/3}\omega, 2^{1/3}\omega^2$, where $\omega={-1+i\sqrt3\over 2}$, and $\omega^2={-1-i\sqrt3\over 2}$.

So I was able to conclude that my splitting field $E$ is $\mathbb{Q}(2^{1/3}, \omega)$.

My question is, since $2^{1/3},$and $\omega, \omega^2$ are the zeros and their minimal polynomial is $x^3-2$, and $x^2+x+1$, which have power 2 and 3, may I conclude that $Gal(E/\mathbb{Q})$ is isomorphic to $\mathbb{Z_6}$ or $\mathbb{Z_2\times Z_3}$?

• You are forgetting the other group of order $6$, which happens to be the right one. – Alex Becker Apr 15 '14 at 16:01
• – lhf Apr 15 '14 at 16:07

The Galois group is actually $S_3$. As you correctly note, the fact that a cubic and a quadratic which are irreducible over $\mathbb Q$ split over $E$ implies that $E/\mathbb Q$ has subextensions of degree $2$ and $3$, so must have degree at least $\mathrm{lcm}(2,3)=6$. Thus $|\mathrm{Gal}(E/\mathbb Q)|\ge 6$. On the other hand, since the Galois group of the splitting field of $f(x)$ acts faithfully on the roots of $f$ (i.e. it permutes the roots, and the only element which fixes every root is the trivial automorphism), we have that $\mathrm{Gal}(E/\mathbb Q)$ embeds in $S_3$. But since $|S_3|=6$, $\mathrm{Gal}(E/\mathbb Q)$ must be the whole group so $\mathrm{Gal}(E/\mathbb Q)\cong S_3$.
• So if I am to list out the elements of $Gal(E/\mathbb{Q})$, what are they? Will they be different automorphisms that map $2^{1/3}, \omega 2^{1/3}$, and $\omega^2 2^{1/3}$ to each other? – Akaichan Apr 17 '14 at 14:19
• @Akaichan Yes, they will be all permutations of the set $\{\sqrt2,\omega\sqrt2,\omega^2\sqrt2\}$, extended to automorphisms of $\mathbb Q(\sqrt2,\omega)$. – Alex Becker Apr 17 '14 at 16:44