Let $K/F$ be a cyclic extension of degree $n$, and suppose that $F$ contains a primitive $n$th root of unity. If $K=F(\sqrt[n]{a})$ with $a\in F$, then any intermediate field of $K/F$ is of the form $F(\sqrt[m]{a})$ for some divisor $m$ of $n$.
In the proof of it, $\sigma$ is assumed to be the generator of $Gal(K/F)$. Then any subgroup of $Gal(K/F)$ is of the form $<\sigma^t>$ for some divisor of $n$. So the intermediate fields will be the fixed field of $\sigma^t$. If $t$ is a divisor of $n$, write $n=tm$ and let $\alpha = \sqrt[n]{a}$. Then $\sigma^t(\alpha^m)=(\omega^t\alpha)^m=\alpha^m$, so $\alpha^m$ is fixed by $\sigma^t$.
My doubt:
- $\alpha^m$ is fixed by $\sigma^t \implies$ $F(\alpha^m)=F(\sqrt[t]{a})$ is the fixed field,
OR,
- $\alpha^m$ is fixed by $\sigma^t \implies$ the fixed field of $\sigma^t$ has degree $m$ over $F$, so $F(\sqrt[m]{a})$ is the fixed field.
I cannot understand. Can someone please clear my doubt?
Thank you
If $F$ contains a $n$-root of unity $e_n$, $ e_n^p\sqrt[n]a,p=1,2,...,n$ are roots of $X^n-a$ so $K$ is the splitting field of $X^n-a$ and is separable, so the extension is Galoisian. You can apply the fundamental theorem of the Galois theory.
An intermediate field $L$ if the set of elements fixed by $\sigma^t$ and if $n=mt$, $\sigma^t(\alpha^m)=\alpha^m$, we deduce that $\alpha^m\in L$, $[L:F]=t$ and the minimal polynomial of $\alpha^m$ is $X^t-a$. We deduce that $[F(\alpha^m):F]=t$ and $F(\alpha^m)=L$.
https://en.wikipedia.org/wiki/Fundamental_theorem_of_Galois_theory