Let F be a field, so that the polynomials F[x] form a pid. In what follows, fractions are always assumed to be proper. In other words, the polynomial on top has degree strictly less than the polynomial downstairs. (Given an improper fraction, use synthetic division to create a polynomial plus a proper fraction.)
Let w1 w2 w3 … wk be monic irreducible polynomials over F. Fix an exponent ei for each wi, and let vi = wi raised to the ei. Let q be the product over vi. Thus q has been factored.
Let p be another polynomial whose degree is less than the degree of q, so that p/q is a proper fraction. We would like to express p/q as a sum of fractions whose denominators are v1 through vk. In other words, p/q can be separated into partial fractions.
Two irreducible polynomials have a gcd of 1, and by bezout's identity, they span the ring. Thus wi and wj span the ring. Being powers of wi and wj, vi and vj are also coprime. Thus we can apply the chinese remainder theorem. Polynomials mod q correspond 1-1 with the direct product of polynomials mod vi. If another polynomial, s(x), is equal to p(x) mod vi, for each i, and if s has degree less than the degree of q, then s = p.
As rational expressions, consider the sum of ai/vi, where the degree of ai is less than the degree of vi. Build the common denominator q, and call the numerator s. The first term in s is a1 times v2*v3*…*vk. Reduce this mod vi, for i > 1, and get 0. Reduce this mod v1, and all the other terms of s go away. Select a1 so that a1 times v2 through vk = b1, where b1 = p mod v1. This can be done because v2 through vk are invertible mod v1. Select a2 through ak in a similar fashion. Of course each ai is reduced mod vi, and has degree less than that of vi. Each partial fraction is proper. Now s and p are equal mod vi, for each i, hence s = p. Therefore, p/q is represented as a sum of partial fractions.
If another sum of partial fractions is equal to p, reason as above to show a1 = b1, and so on through ak = bk. Therefore the representation is unique.