Let k be any field, and let A := k[[x]]. Let K := k((x)), and let L := K(x^{\frac{1}{m}} , m ∈ ℕ). Let B denote the integral closure of A in L. By construction, B is integrally closed. Proposition I.5.6 shows that B has dimension 1. The domain B is clearly not a principal ideal domain since x = (x^{\frac{1}{m}})^{m} shows that x cannot be factored into a product of finitely many prime elements. Let us show that B is local. Let M_{1} and M_{2} be any maximal ideals of B. If M_{1} ≠ M_{2}, then let z ∈ M_{1} \ M_{2}. Consider a finite extension F/K of the form F = K(x^{\frac{1}{n}}) that contains K(z). Let B_{F} denote the integral closure of A in F. Then M_{1} ∩ B_{F} ≠ M_{2} ∩ B_{F}. This is a contradiction since B_{F} is local; indeed, F/K is an Eisenstein extension (III.7.7).