# Why is every element of a Galois group identifiable with a permutation

Why is every element of $Gal(\mathbb Q(\sqrt[3]{2}, i\sqrt{3}):\mathbb Q)$ may be identified with a permutation of the three cube roots of 2?

I am new to Galois groups. I would think the permutation would come from the automorphism. So is every element written as the roots times a coefficient that's in our field?

EDIT: Since I am concurrently trying to solve that $Gal(\mathbb Q(\sqrt[3]{2}, i\sqrt{3}):\mathbb Q)\cong S_3$, I have found some information and wanted to see if I now understand more clearly. So for Galois group $x^3-2$ we need to look at the automorphism of this splitting field over $\mathbb Q$. It is completely determined by where the three roots of $x^3−2$ are sent (this would be the permutations as referenced above, I believe); more precisely, the Galois group embeds in $S_3$. Since the extension has of degree, we are done.

• Are you certain that that extension is Galois? – Arthur Dec 8 '17 at 16:17
• I am wanting to understand this so I can then prove $Gal(\mathbb Q(\sqrt[3]{2}, i\sqrt{3}):\mathbb Q)\cong S_3$. So I would assume that it is. – K Math Dec 8 '17 at 16:19

The general idea here is that the Galois group is the group of automorphisms of the extension field that fix $\mathbb Q$. Since each member of the Galois group fixes $\mathbb Q$, it will leave any polynomial expression whose co-efficients are in $\mathbb Q$ unchanged. But if this polynomial has all of its roots in the extension field then it can be expressed as the product of linear terms of the form $x-\alpha_i$ where $\alpha_i$ is an element of the extension field. Since a Galois group member leaves the polynomial unchanged, it can only change the order of these terms i.e. it must permute the roots $\alpha_i$.
In general if $f \in \mathbb{Q}[x]$ factorizes as $f(x) = \prod_{j=1}^d (x-\alpha_j)$ and $K = \mathbb{Q}(\alpha_1,\ldots,\alpha_d)$, then $K/\mathbb{Q}$ is Galois, any automorphism $\sigma \in Aut(K)$ is defined by its action on the $\alpha_j$, and $f(\alpha_j) = 0 \implies \sigma(f(\alpha_j)) =f(\sigma(\alpha_j))= 0$, thus $\sigma(\alpha_j)$ is another root of $f$, thus $\sigma$ acts as a permutation of the $\alpha_j$.
In your example $f(x) = (x^3-2)(x^2+x+1) = \prod_{l=1}^3 (x-\zeta_3^l \sqrt[3]{2})(x-\zeta_3)(x-\zeta_3^2)$ where $\zeta_3 = \frac{-1+i \sqrt{3}}{2}$.