By Antoine Chambert-Loir

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Let K ⊂ F be a finite normal extension of fields. Let A be a subring of K which is integrally closed in K and let B be the integral closure of A in F. Let G be the group of automorphisms of F which restrict to identity on K. a) For every σ ∈ G, one has σ(B) = B; b) For every point x ∈ Spec(A), the group G acts transitively on the fiber a ( φ)−1 (x) in Spec(B). Proof. — a) Let b ∈ B. Then σ(b) belongs to F and is integral over A. One thus has σ(b) ∈ B. This shows that σ(B) ⊂ B. Similarly, one has σ −1 (B) ⊂ B, hence B ⊂ σ(B).

Let F′ be a finite extension of F which is normal over K, let B′ be the integral closure of A in F′ . 4), there exists a chain q′0 ⊂ ⋅ ⋅ ⋅ ⊂ q′n of prime ideals of B′ such that pi = q′i ∩ A for every i. Let ̃ qn be a prime ideal of B′ such that ̃ qn ∩ B = qn . 2, there exists an automorphism σ of F′ such that σ∣K = id and σ(q′n ) = ̃ qn . For every integer i such that 0 ⩽ i ⩽ n − 1, let qi = σ(q′i ) ∩ B. Then q0 ⊂ ⋅ ⋅ ⋅ ⊂ qn is a chain of prime ideals of B. For every integer i, one has qi ∩ A = σ(q′i ) ∩ B ∩ A = σ(q′i ∩ A) = σ(pi ) = pi , hence the theorem.

Let a ∈ A be such that x n = ax n+1 . Since x ≠ 0 and A is an integral domain, we may simplify by x n , hence ax = 1. This shows that x is invertible. b) Let p be a prime ideal of A. Then, A/p is an artinian ring which is an integral domain. By part a), it is a field, hence p is a maximal ideal. c) Since every prime ideal of A is maximal, every point of Spec(A) is closed. If Spec(A) were infinite, there would exist an infinite sequence (x n ) of pairwise distinct points in Spec(A). The infinite sequence ∅ ⊂ {x1 } ⊂ {x1 , x2 } ⊂ .