Situating definitions.
Definitions: Let $R$ be a UFD. Let $f \in R[x]$, the content of $f$ is denoted $C(f)$ and is the principal ideal generated by a greatest common divisor of the coefficients of $f$. We say that $f$ is primitive if $C(f) = R$ or equivalently if $1$ is a gcd of the coefficients of $f$.
Gauss's Lemma: Let $f,g \in R[x]$ be primitive. Then $fg$ is also primitive.
The proof proceeds by the following. We want to show that for an arbitrary irreducible element $p$ of $R$, that $p$ does not divide at least one coefficient of $fg$.
Why is this sufficient prove the statement that $fg$ is primitive? Is the following the justification?
Suppose that $p \in R$ was irreducible and divided all the coefficients of $fg$ and let $d$ be a gcd of those coefficients. If $C(fg) =1$ then $(d) = R$ so $1 \in (d)$. Also by definition of gcd $p \mid d$ and do $d = pr$ for some $r \in R$. But if $1 \in (d)$ then $d$ has a multiplicative inverse in $R$ so $dd^{-1} = prd^{-1} \implies 1 = p(rd^{-1})$ i.e $r \in R^\times$ contradicting that $r$ is irreducible, so it must be that $C(fg) \neq R$.
This seems slightly long and annoying for something which was stated immediately with no justification, and I haven't even shown the converse statement to justify that not being primitive is equivalent to having an irreducible element which divides all coefficients. What obvious fact am I missing? Thanks in advance for the clarification.