I am trying to prove the following result:
Let $R$ be a UFD and let $S \subset R$ be a multiplicative set. Then if $q \in R$ is irreducible, either $q/1$ is a unit or is irreducible in $S^{-1}R$.
I thought I had figured out a proof, but then I realized that I was implicitly assuming that all fractions $a/b \in S^{-1}R$ could be taken in ``reduced form'', i.e., $\gcd(a, b) = 1$. Although this is certainly true for, say, the field of fractions over $R$, I do not see why it would generally true for an arbitrary localization. It requires the property that if $a \in R$ and $b \in S$, then $b / \gcd(a, b) \in S$, and I am not convinced that this needs to be true.
My question: Does anybody know a proof of this result that does not assume reduced forms exist in $S^{-1}R$ (this would be the most desirable answer). Alternatively, is there some proof that reduced forms always exist (I find this doubtful)? Or, another possible alternative, does the result fail to hold for some $R$ and $S$ such that reduced forms do not exist?
Suppose $x$ is irreducible in $R$. Then if $x$ divides some element of $S$, say $xy = s$ for $s\in S$, then $x$ clearly becomes a unit in $S^{-1}R$ since $$\frac{x}{1}\cdot\frac{y}{s} = \frac{xy}{s} = 1.$$ in $S^{-1}R$. Alternatively, if $x$ does not divide an element of $S$, then suppose that $x$ factors in $S^{-1}R$, say $x = \frac{y}{s'}\cdot\frac{z}{s''}$. Then $xs's'' = yz$. By unique factorization, $x$ must divide either $y$ or $z$. It cannot divide both since there is only one factor of $x$ on the LHS. Suppose it divides $y$. Then all irreducibles in the factorization of $z$ are divisors of $s'$ or $s''$ and thus divide some element of $S$. So $\frac{z}{s''}$ must be a unit (by the first case) and thus $x$ is irreducible.