Answering that recent stackoverflow question, I encountered the following related problem :
Let $n,m,p\geq 2$ be integers, and let $K$ be a subfield of $\mathbb C$ containing all $nmp$-th roots of unity.
Consider a radical extension $K(a)$ of $K$ with degree $n$, so that $a^n\in K$ (and $X^n-a^n$ is irreducible over $K$). Consider also a radical extension $K(a)(b)$ of $K(a)$ with degree $m$, so that $b^m\in K(a)$ (and $X^m-b^m$ is irreducible over $K(a)$).
Of course, the extension $K(b)/K$ is not radical in general. It will be, however, if $b^m$ is a monomial in $a$, $b^m=ca^i$ with $c\in K, i\in{\mathbb N}$. Is the converse also true ? If $K(b)/K$ is radical, must $b^m$ be a monomial in $a$ ?
Perhaps this is a “stupid”, very easy question for Kummer theorists.
Yes, that is true. If $K(b)/K$ is radical then $b^{mn}\in K$, so $b^{mn}$ is fixed by any generator $\sigma$ of $\text{Gal}(K(a)/K)$. Since $b^m\in K(a)$, it follows that $\sigma(b^m)=\gamma b^m$ where $\gamma^n=1$. Pick a primitive $n$-th root of unity $\zeta$, and let $\sigma$ be the automorphism of $K(a)$ which fixes every element of $K$ and maps $a\mapsto \zeta a$. Since $b^m\in K(a)$, we can write $$ b^m = c_{n-1} a^{n-1} + c_{n-2} a^{n-2} + ... + c_0 $$ with every $c_i\in K$. Now we compute $$ \sigma(b^m) = c_{n-1} \sigma(a)^{n-1} + c_{n-2} \sigma(a)^{n-2} + ... + c_0 = c_{n-1} \zeta{n-1} a^{n-1} + c_{n-2}\zeta^{n-2} a^{n-2} + ... + c_0, $$ so that if $\sigma(b^m)=\gamma b^m$ with $\gamma\in K$ then for each $i$ with $0\le i\le n-1$ we must have $c_i\zeta^i a_i = \gamma c_i a^i$, whence either $c_i=0$ or $\gamma=\zeta^i$. Since $\zeta^0,\zeta^1,\dots,\zeta^{n-1}$ are pairwise distinct, it follows that $c_i$ is nonzero for at most one value $i$. Therefore $\sigma(b^m)=c_i a^i$ for some $i$.