Prime ideals in $R[x]$, $R$ a PID 
Let $R$ be a PID. Show that if $\mathfrak p \subset R[x]$ is a prime ideal, $(r) = \left\{h(0) \colon h(x) \in \mathfrak p \right\}$, and $$\mathfrak p = (r, f(x), g(x)),$$ where $f(x), g(x) \in R[x]$ are nonconstant irreducible polynomials, then $f(x)$ and $g(x)$ are associates. So $\mathfrak p = (r, f(x))$.  


As stated, the claim being made in the post is false, as user26857 makes clear in the comments. To reiterate, a counterexample is $$\mathfrak{p} = (2, x, x+2) \subset \mathbf{Z}[x].$$
