For $p(x)$ and $q(x)$ in $\overline{\mathbb{Q}}[x]$, if $p(x)$ is coprime to $q(x)$, and $m\leq n$, then is $p(x)^n$ coprime to $q(x)^m$?

Question

Let $m,n\in\mathbb{N}_+$ with $m\le n$,
$p(x),q(x)\in \overline{\mathbb{Q}}[x]$,
and let $\perp$ denote the coprimality over $\overline{\mathbb{Q}}$.

Does

$$p(x)\perp q(x)\ \implies\ p(x)^n\perp q(x)^m$$

hold?

If yes, how can one show that?

Answer 1

The proof is simpler than you might expect. We do not use that $\overline{\mathbb{Q}}$ is algebraically closed. In fact, we only use that $k[x]$ is a PID and therefore a UFD, for each field $k$.

Indeed, $\overline{\mathbb{Q}}[x]$ is a UFD, since it is a PID. That way, we can factor $p$ and $q$ into irreducibles, and note that they have disjoint prime factorizations (none of the primes occuring in one factorization occur in the other) (that is equivalent to being coprime). The powers of $p$ and $q$ respectively have the same prime factorization (if the prime factorizations are $\prod a_i^{e_i}$ and $\prod b_j^{f_j}$, then the prime factorization of the powers is $\prod a_i^{e_i n}$ and $\prod b_j^{f_j m}$. Therefore the powers are coprime.

As mentioned, this works in any UFD.