First we claim that K(x) consists of the elements g(x) with g K[X]. For if g(x) 0, then g is not divisible by f as f is irreducible. Thus, there are polynomials a, b in K[X] with af + bg = 1. Substitution of x for X yields: g(x)^-1 = b(x). Therefore, the inverse of g(x) also belongs to K[x]. Thus, the expressions of the form g(x), where g K[X] form a subfield of K(x) containing x. As K(x) is the smallest field containing x, the claim follows.

Consider the subsitution map K[X] -> K(x),    g -> g(x).

It is easily seen to be a morphism. By the first part of the proof, it is surjective. Its kernel is (f), since x is a zero of f and the latter is irreducible. The First isomorphism theorem then gives that there is an isomorphism as required.

The proof for is similar.

By the previous part, the composition · ^-1 is an isomorphism K(x) -> K(y).