Prime ideal in R[x] is either principal or $\mathfrak p = (q,f)$ 
$R$ is a PID and $\mathfrak p$ a prime ideal of $R[x]$. Show that $\mathfrak p$ is principal or $\mathfrak p = (q,f)$ for some $q\in R$ prime and $f \in R[x]$ monic.

I can't figure out this problem. Any hint is appreciated! Thanks

Note that $(q) := \mathfrak p \cap R$ is a prime ideal in $R$. Either $q=0$ or $q$ is a prime element of $R$.
Consider two cases: 


*

*$q=0$, hence there is no constant in $\mathfrak p$. Let $m \in \mathfrak p$ be of minimal degree with leading coefficient $r \in R$. We want to show $\mathfrak p=(m)$. Let $f \in \mathfrak p$ arbitrary. We can do division with remainder in $Q(R)[x]$ and get $f=am+b$ with $\deg b < \deg m$. Note that the denominators in $a$ and $b$ are all of the form $r^s$ and we can pick the same $s$ for all denominators, hence we get the equality $$r^sf=r^sam+r^sb$$ in $R[x]$. We deduce $r^sb \in \mathfrak p$. By assumption we have $r^s \notin \mathfrak p$, thus we get $b \in \mathfrak p$, which - by the minimality of $m$ - yields $b=0$, i.e. the desired $f \in (m)$.

*$q$ is a non-zero prime element of $R$. We have $(q) \subset \mathfrak p$, i.e. $\mathfrak p/(q)$ is a prime ideal of $R[x]/(q) = (R/(q))[x]$. The latter is a principal ideal domain, hence we get $\mathfrak p/(q)=(\overline f)$ for some $\overline f \in (R/(q))[x]$. This easily yields $\mathfrak p=(q,f)$ (Note that $f$ might be zero, then we have $(q)=\mathfrak p$, which is of course possible).
