Here is my attempt to prove the following:
Let $A$ be a unique factorization domain and $S$ a multiplicative subset such that $0 \notin S$. Show that $R = S^{-1}A$ is also a UFD, and that the prime elements of $R$ are those primes $p$ in $A$ such that $(p) \cap S = \emptyset$.
To prove the first part, observe that a non-invertible element $\frac{a}{s} \in R$ has a factorization into primes $\frac{b}{s} \prod_{i}$ $\frac{p_i}{1}$. To prove uniqueness, suppose there are two prime factorization $\frac{p_1 … p_t}{s} = \frac{q_1 … q_r}{s’}$ which implies $p_1 … p_t s’ = q_1 … q_r s$. Then $p_1$ divides either $q_1 … q_t$ or $s$. If it divides $s$ then $\frac{p_1}{1} \frac{d}{s} = \frac{1}{1}$ where $d$ is $s$ without one of the $p_1$ factors, and so $\frac{p_1}{1}$ is invertible contrary to assumption. Thus $p_1$ divides some $q_i$ and so repeating this idea proves uniqueness.
I will (attempt to) prove the second part by contraposition. If $(p) \cap S$ contains some element s, then $px = s$ for some element $x \in A$. So $p$ is not prime in $R$ and by contraposition we have $\frac{p}{1}$ prime implies $(p) \cap S = \emptyset$. Conversely, if $\frac{p}{1}$ is not prime then it has some inverse $\frac{a}{s}$. Thus $pa = s$ and $(p) \cap S$ is non-empty and again by contraposition we get the other implication.
