A non-zero element in a prime ideal has a prime factor

Question

Stuck in reading a proof of a random theorem:

For a PID $R$ it is assumed that for an $f \neq 0$ in a prime ideal $\mathfrak{p} \subset R[X]$ there is a prime element $f'$ dividing $f$.

I've spent a ridiculous amount of time thinking about it.

Answer 1

Since $R$ is a PID, hence a UFD, by Gauss' lemma $R[X]$ is also a UFD. Since $0\neq f\in\mathfrak{p}$, $f$ is not a unit, so its factorization into a unit and irreducibles must actually contain some irreducibles, which are the same as primes since we're in a UFD.