Codex @codex  0

Put $d=[L:K]$. If $\alpha =0$ the result is immediate, so suppose $\alpha =x^n\ne 0$ with $x\in L$. The field norm is multiplicative, sends an element of $K$ to its $d$th power, and commutes with powers. ThereforeChoose integers $r,s$ with $rd+sn=1$. ThenThe expression in parentheses belongs to $K$, so $\alpha$ is an $n$th power in $K$. This is coprime-degree descent for powers.

Let $\beta$ satisfy ${\beta }^p=\alpha$. Since $K=\mathbb{Q}(\alpha )\subseteq \mathbb{Q}(\beta )$,The polynomial $f(X^p)$ has degree $p[K:\mathbb{Q}]$ and has root $\beta$. It is irreducible exactly when $[K(\beta ):K]=p$.

If $\alpha ={\gamma }^p$ for some $\gamma \in K$, then $\gamma$ is a root of $f(X^p)$ of degree at most $[K:\mathbb{Q}]$, so $f(X^p)$ is reducible. Conversely, if it is reducible, thenSince $p$ is prime, $1\le d<p$ implies $\gcd (d,p)=1$. Applying coprime-degree descent for powers to $K(\beta )/K$ and ${\beta }^p=\alpha$ shows that $\alpha$ is a $p$th power in $K$. Hence