Calculations gave me that the Galois group of $\mathbb{Q}(\sqrt[5]{3},\zeta):\mathbb{Q}$, where $\zeta=e^{2\pi i/5}$, has order 20 and is isomorphic to the unique subgroup of $S_5$ with order 20. Also, the set of $\mathbb{Q}$-automorphisms of $\mathbb{Q}(\sqrt[5]{3})$ is the trivial group, since the only automorphism allowed is the identity map. So the subgroup of the Galois group of $\mathbb{Q}(\sqrt[5]{3},\zeta):\mathbb{Q}$ whose fixed field is $\mathbb{Q}(\sqrt[5]{3})$ is the trivial group?

But, I thought the 1-1 correspondence between the subgroups of the Galois group and the intermediate fields between $\mathbb{Q}$ and $\mathbb{Q}(\sqrt[5]{3},\zeta)$ gave the correspondence between the trivial subgroup and the entire field $\mathbb{Q}(\sqrt[5]{3},\zeta)$. What am I missing?

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    $\begingroup$ The subgroup whose fixed field is $\mathbb{Q}(\sqrt[5]{3})$ gives you the Galois group of $\mathbb{Q}(\sqrt[5]{3},\zeta)\colon\mathbb{Q}(\sqrt[5]{3})$; not the Galois group of $\mathbb{Q}(\sqrt[5]{3})\colon\mathbb{Q}$. $\endgroup$ – Arturo Magidin Apr 8 at 2:37
  • $\begingroup$ To put it another way, think about a map that takes $\zeta$ to $\zeta^2$, while fixing $\root5\of3$. $\endgroup$ – Gerry Myerson Apr 8 at 3:38

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