This result generalizes to the fractions of any ufd. Once reduced, the numerator and denominator have no primes in common, each has a unique factorization, and these can be combined to give a prime factorization for the fraction.
We can extend this result to dedekind domains. Every fractional ideal is a unique product of prime ideals, although some of the exponents may be negative.
Let's begin by proving every prime ideal is invertible. Start off with a lemma.
The product of finitely many fractional ideals is invertible iff each component is invertible. If each fractional ideal is invertible, multiply their inverses together to get the inverse of the product. Conversely, let J*(H1*H2*H3*…*Hn) = R. Each factor Hi is multiplied by something else to get R, hence each Hi is invertible.
Now let P be a prime ideal and let x be a nonzero element in P. Write x*R as a product of prime ideals Q1Q2Q3… If P contains a product of ideals it contains at least one of them. Suppose P contains Q1. Yet one prime ideal cannot properly contain another, hence P = Q1. Thus P is one of the primes in the factorization of x*R. Since x*R is invertible, so is P.
An immediate corollary is that all nontrivial proper ideals are invertible, as they are products of prime ideals, which are invertible.
Now, let H be a fractional ideal with denominator d. Thus d*H is an ideal in R, with a unique decomposition. The ideal generated by d also has a unique decomposition. Since every prime ideal is invertible, divide through by the ideals of d*R to find a factorization of H. Some other factorization, when multiplied by the prime ideals in d*R, would produce a different factorization for d*H, which is a contradiction. Therefore H is uniquely represented as a product of prime ideals or their inverses.
As an immediate corollary, every nontrivial fractional ideal is invertible.