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?
• 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 $\mathbb{Q}(\sqrt[5]{3})\colon\mathbb{Q}$. – Arturo Magidin Apr 8 at 2:37
• To put it another way, think about a map that takes $\zeta$ to $\zeta^2$, while fixing $\root5\of3$. – Gerry Myerson Apr 8 at 3:38