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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!

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Is your definition of the Galois group the group of automorphisms of the extension? – JSchlather Feb 25 '13 at 4:36
Yes that's the definition I am using! – Guest Feb 25 '13 at 4:40
Consider writing $\text{Aut}(E/F)$ rather than $\text{Gal}(E/F)$ when it's not a Galois extension, so that it's obvious what you mean. – Hurkyl Feb 25 '13 at 5:03
@Hurkyl: I agree that making a notational distinction is better (I am a late convert, I persisted in always writing $\mathrm{Gal}$ until much later than my first exposure to Galois theory). – Zev Chonoles Feb 25 '13 at 5:10
Noted, thanks for the pointer, much appreciated! – Guest Feb 26 '13 at 0:59

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 $\sigma(\sqrt[5]{3})=\sqrt[5]{3}$.

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