= Solution
Let $\beta$ satisfy $\beta^p=\alpha$. Since $K=\mathbb Q(\alpha)\subseteq\mathbb Q(\beta)$,
$$
[\mathbb Q(\beta):\mathbb Q]=[K(\beta):K][K:\mathbb Q].
$$
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, then
$$
d=[K(\beta):K]<p.
$$
Since $p$ is prime, $1\leq d<p$ 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
$$
\boxed{f(X^p)\text{ is irreducible}\quad\Longleftrightarrow\quad
\alpha\notin K^p.}
$$
Back to article page
