I want to prove Lemma 4. Source: http://homepages.math.uic.edu/~radford/math516f06/RRXUFD.pdf
Proposition 1: Every primitive polynomial of positive degree in $R[X]$ is the product of (primitive) irreducible polynomials of positive degree in $R[X]$.
Lemma 4: Let $p(X) \in R[X]$ be a primitive of positive degree. If $p(X)$ is an irreducible polynomial of $F[X]$ where $F$ is the field of quotients of $R$, then $p(X)$ is an irreducible polynomial of $R[X]$.
Proof: Suppose $p(X) \in F[X]$ is an irreducible polynomial of $F[X]$ and $p(X)=q(X)r(X)$, where $q(X), r(X) \in R[X]$. By proposition 1, $q(X),r(X) \in R[X]$ are irreducible. So, by definition, that means $q(X)$ and $r(X)$ are not invertible in $R[X]$ and both are only divisible by unit elements of $R$.
I don't know how to proceed because in order to show $p(X)$ is irreducible in $R[X]$, either $q(X)$ or $r(X)$ must be invertible. However, this leads to contradiction...? What am I missing?