This problem is from a Ph.D Qualifying Exam in algebra.
Let $R$ be a PID and $I$ be a nonzero prime ideal of the polynomial ring $R[x]$ such that $R\cap I=0$. Prove that if $R$ has infinitely many prime elements, then $I$ is not maximal.
My solution: I just figured out that $I=(f)$ for some irreducible and primitive $f\in R[x]$. Since $R\cap I=0$, one can find a prime element $p \in R$ such that $p \notin I$. Then $J=\langle p,f \rangle$ is an ideal properly containing $I$, and by infinitude of primes, there is another prime $p'\in R\backslash J$. Therefore, $I$ is not maximal.
Is my solution correct?
What I'm worried is that even if $R$ has only finitely many primes, my solution seems to make sense.(i.e. the infinitude of primes of $R$ is too strong.) Is there any solution to this problem fully exploiting the infinitude of primes?
Any comments or corrections are welcome!