 Solution

Suppose $f$ is prime in $F[X]$ and divides $gh$ in $R[X]$. It divides one factor in $F[X]$, say $g=fq$ for $q\in F[X]$. Write $q=a{q}_0$ with $q_0\in R[X]$ primitive and $a\in F$. The product $f{q}_0$ is primitive. Sincehas all coefficients in $R$, a denominator of $a$ would divide every coefficient of the primitive polynomial $f{q}_0$; it must therefore be a unit. Thus $a\in R$, so $q\in R[X]$ and $f$ divides $g$ in $R[X]$. Hence $f$ is prime in $R[X]$.