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I'm trying to calculate the Galois group of $\mathbb{Q}(\sqrt[5]{3}):\mathbb{Q}$. I'm not sure how to proceed since this is not a normal extension.

The normal closure of this is $\mathbb{Q}(\sqrt[5]{3},\zeta)$ where $\zeta=e^{2\pi i/5}$, and $\mathbb{Q}(\sqrt[5]{3},\zeta):\mathbb{Q}$ has degree 20. There is a unique subgroup of $S_5$ of order 20, so Gal$(\mathbb{Q}(\sqrt[5]{3},\zeta):\mathbb{Q})$ is isomorphic to that group. I don't know if this is useful since $\mathbb{Q}(\sqrt[5]{3}):\mathbb{Q}$ is not normal. How do I calculate the Galois group?

Thanks for any help in advance.

user500144
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    Is your definition of “Galois group” the group of automorphisms that fix the base field? If so, note that $f\colon\mathbb{Q}(\sqrt[5]{3})\to\mathbb{Q}(\sqrt[5]{3})$ is completely determined by what happens to $\sqrt[5]{3}$, and that whatever it is that happens, it must map to a root of $x^5-3$ which lies in $\mathbb{Q}(\sqrt[5]{3})$. – Arturo Magidin Apr 08 '19 at 00:23

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The Galois group $G$ of $X^5-3$ is the Galois group of the extension $\Bbb Q(\sqrt[5]{3},\zeta_5)$ over $\Bbb Q$, which has degree $20$ and is a transitive subgroup of $S_5$. The only such group is the Frobenius group $Fr_{20}$, see here. This is the Galois group you are looking for, since the extension $\mathbb{Q}(\sqrt[5]{3})$ over $\Bbb Q$ is not normal.

References:

Galois group of solvable quintic is subgroup of Fr20

Galois group of $\mathbb{Q}(\sqrt[5]{3},\zeta):\mathbb{Q}$

Dietrich Burde
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