TITLE: Calculate $\mathrm{Gal}(\mathbb{Q}(\sqrt[5]{3})/\mathbb{Q})$
QUESTION [5 upvotes]: I'm attempting some of my first problems in solving for Galois Groups, and this one has stumped me.
What I've done so far is found that $\mathbb{Q}(\sqrt[5]{3})$ is not a normal extension, because the minimum polynomial of $\sqrt[5]{3}$ is $x^5-3$, and $\mathbb{Q}(\sqrt[5]{3})$ is not the splitting field of this polynomial.
So $\mathbb{Q}(\sqrt[5]{3})$ is not normal, thus not a Galois Extension. I know if it was a Galois extension you can determine the order of the Galois Group since it equals the degree of the extension over $\mathbb{Q}$. But for an extension that is not Galois I can only get that the degree of the extension is an upper bound on the order of the Galois group. And from here I don't know where to go...
Any help would be greatly appreciated!

REPLY [7 votes]: Step 1: Show that any automorphism of $\mathbb Q(\sqrt[5]{3})$ is determined by its action on $\sqrt[5]{3}$.
Step 2: If $\sigma$ is an automorphism of $\mathbb Q(\sqrt[5]{3})$ show that $\sigma(\sqrt[5]{3})=\sqrt[5]{3}$.