Let $\xi_5$ be a primitive fifth root of unity in $\mathbb{C}$, then we know the extension $\mathbb{Q}(\xi_5)\supseteq\mathbb{Q}$ has degree 4 so the Galois group is of order 4. I am trying to find all intermediate extensions and it turns out the only one is $\mathbb{Q}(\sqrt{5})\supseteq\mathbb{Q}$. It is not hard to find this, but how to show that there are no more intermediate extensions?
[Math] Field extension, primitive root of unity
abstract-algebrafield-theorygalois-theory
Best Answer
The Galois group is of order $4$ because the degree of the extension is $4$, but more is true. It's canonically isomorphic to $(\mathbf{Z}/5\mathbf{Z})^\times\cong\mathbf{Z}/4\mathbf{Z}$, i.e. it is cyclic of order $4$. Galois theory gives a bijective correspondence between intermediate fields and subgroups of the Galois group, so, since $\mathbf{Z}/4\mathbf{Z}$ only has one proper, non-trivial subgroup, $\mathbf{Q}(\xi_5)/\mathbf{Q}$ has a unique non-trivial intermediate extension.