Nonprincipal prime ideals contain two relatively prime elements

Question

Let $R$ be a principal ideal domain and let $P$ be a nonprincipal prime ideal of $R[x]$. I'm having trouble seeing why $P$ must contain two elements with no common divisor.

Can anyone help me? Thanks.

Answer 1

I found the answer in Pete L. Clark's notes.

Since $R[x]$ is a UFD (this must be shown separately) the nonzero prime ideal $P$ contains a prime (hence irreducible) element $f_1$ (take nonzero $x\in P$ and factor into primes; since $P$ is prime, one of these must be in $P$). Since $P$ is nonprincipal there exists $f_2\in P\setminus (f_1)$. Now if $g$ is a common factor of $f_1$ and $f_2$ then in particular $g$ is a factor of $f_1$. Since $f_1$ is irreducible this implies that $g$ is a unit or associate to $f_1$. But if $g$ is associate to $f_1$ then the fact that $g$ divides $f_2$ means that $f_1$ divides $f_2$, contradiction. Hence $f_1$ and $f_2$ have no common factor.

Answer 2

If $p \ne 0$ is a prime element in $R$, consider $(p, x) \subseteq R[x]$. (By the way, it is not true that every prime ideal in $R[x]$ is nonprincipal - for example $(x)$ is a principal prime ideal - but the maximal ideals are guaranteed to be nonprincipal).