Elements of the Galois Group of the splitting field of $x^3-5$

Question

So as in the title, I am trying to find what the elements of the Galois Group are. First of all we do the extension $Q(\sqrt[3]{5})$ which is of degree 3 using the polynomial $x^3-5$. Now we extend again to get $Q(\sqrt[3]{5},\omega\sqrt[3]{5})=E$ which is an extension of degree two. Hence we know that $|Gal(E/Q)|=6$. Now i want to find the elements of this group by looking at the automorphisms. We know we have to map $\sqrt[3]{5}$ to the other zeroes of $x^3-5$ and then map $\omega\sqrt[3]{5}$ to $\omega^2\sqrt[3]{5}$ or itself. However, we already know where the root of 5 maps so all we consider is where $\omega$ maps to. This is sort of obvious simply because $Q(\sqrt[3]{5},\omega\sqrt[3]{5})=Q(\omega,\sqrt[3]{5})$. However, i am having trouble finding a polynomial in $Q(\sqrt[3]{5})$ whose zeroes are $\omega$ and $\omega^2$. How do I find such a polynomial? Is it cyclotomic?

Answer 1

The automorphisms are induced by all permutations of $\sqrt[3]{5}$, $\omega\sqrt[3]{5}$, and $\omega^2\sqrt[3]{5}$. It seems like you want to know where each of these permutations takes $\omega$. Well, the 3-cycles leave it alone since they have fixed field $\Bbb{Q}(\omega)$ by the Galois correspondence. (Or you can note that $\omega=\omega\sqrt[3]{5}/\sqrt[3]{5}$ will be taken to, say, $\omega^2\sqrt[3]{5}/\omega\sqrt[3]{5}=\omega$ by $\sqrt[3]{5}\rightarrow\omega\sqrt[3]{5}\rightarrow\omega^2\sqrt[3]{5}$.)

And the transpositions take $\omega$ to $\omega^2$, because half of the elements need to swap $\omega$ with its conjugate $\omega^2$ (or by looking, e.g., at what $\sqrt[3]{5}\rightarrow\omega\sqrt[3]{5}\rightarrow\sqrt[3]{5}$ does to $\omega=\omega\sqrt[3]{5}/\sqrt[3]{5}$).