Let A be a local principal ideal domain, with maximal ideal (π) = P. Let f(y) = ∑_{i=0}^{n} a_{i}y^{i} ∈ A[y], with a_{n} = 1. Assume that ord_{M}(a_{0}) = m ≥ 1, with gcd(m, n) = 1. Let α ∈ \overline{K} be a root of f(y). Let L = K(α). In the proof below, we need to use the fact that the integral closure B of A in L is a Dedekind domain. Since we have only proved this fact so far when L/K is separable, we will assume that L/K is separable. This always happen, for instance, when n < p (with char(K) = p > 0). This condition shows that [L : K] is not divisible by p and thus L/K is separable.

Theorem 1 Let L = K(α) and assume that L/K is separable. Assume also that ord_{M}(a_{i}) ≥ m\frac{n−i}{n}, for all i = 0, …, n. Then the integral closure of A in L is a local principal ideal domain B with maximal ideal M. The degree of L/K in n, so that f(y) is irreducible in K[y]. Moreover, πB = M^{n}.

Proof: By hypothesis, B is a Dedekind domain. Let M be any maximal ideal of B containing (π). Let d := ord_{M}(α) and e := ord_{M}(π). Consider the relation in B:

α^{n} + a_{n−1}α^{n−1} + … + a_{1}α + a_{0} = 0.

Then

ord_{M}(α^{n}) = nd,
ord_{M}(a_{i}α^{i}) = ord_{M}(a_{i}) + id ≥ me\frac{n − i}{n} + id = me + (d − \frac{me}{n}) i,
ord_{M}(a_{0}) = me.

Assume nd ≠ me. Then d − \frac{me}{n} ≠ 0, and the set \{ord_{M}(a_{i}α^{i}), i = 0, …, n\} has a single minimal element. Thus ord_{M}(α^{n} + … + a_{0}) = min_{i}{ord_{M}(a_{i}α^{i})}. This is a contradiction. It follows that nd = me. Since gcd(m, n) = 1, we find that d = m and e = n. Thus

πB = M^{n} · M_{2}^{e_{2}} … M_{s}^{e_{s}}.

Since ∑_{i=1}^{s} e_{i}f_{i} = [L; K], we find that [L : K] ≥ n. So [L : K] = n and f(x) is irreducible, and πB = M^{n}. Since B has only one maximal ideal that contains (π) and since all maximal ideals of B contain π (the unique maximal ideal of A), we find that B is a local ring. Hence, since B is a Dedekind domain with finitely many maximal ideals, it is a principal ideal domain. Another way of proving this fact in our context is to argue as follows. Note that M ∩ A[α] = (α, π). If m > 1, then (α, π)B = M^{min(m,n)} is not maximal in B. Write um + vn = 1, and let β := α^{u}π^{v}. Then ord_{M}(β) = 1 and M = (β). Since B/A is integral, dim(B) = 1. So B is a local principal ideal domain.

Exercise 1 Show that f(y) ∈ A[y] is irreducible when A is any Dedekind domain and P ∈ Max(A).

Exercise 2 Show that the mininal polynomial of β over K is a classical P - Eisenstein polynomial of degree n (i.e., its constant term is in P \ P^{2}.)