Bourbaki

Exercise 13, § 1

Commutative Algebra, V, 357

Bourbaki marks this one as harder than the rest.

Let A be a local integral domain and K its field of fractions. Consider the two following properties: (i) A is Henselian (Chapter 111, § 4, Exercise 3). (ii) For every finite extension L of K, the integral closure A' of A in L is a local ring. Show that (i) implies (ii) and that the converse is true if A is integrally closed. (To see that (i) implies (ii), note that A' is the union of a right directed family of finite A-algebras and use the definition of a Henselian ring. To see that (ii) implies (i) if A is integrally closed, show that, if P∈A[X]P \in A[X] is an irreducible monic polynomial and f:A→kf : A \to k is the canonical homomorphism of A onto its residue field kk, f(P)f(P) is irreducible in K[X]K[X] and consider the field L=K[X]/(f)L = K[X]/(f).) If condition (ii) is fulfilled, the integral closure A of A in any extension of K is a local ring. Consider the case of complete Hausdorff local rings; if A is a complete Hausdorff Noetherian local ring, every A-algebra contained in A' and which is a finitely generated A-module is a complete Hausdorff local ring.

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