Bourbaki

Exercise 9, § 3

Commutative Algebra, III, 247

Bourbaki marks this one as harder than the rest.

(a) Let A be a Zariski ring, m a defining ideal of the topology on A and B a ring such that A⊂B⊂A^A \subset B \subset \hat{A}, which is a finitely generated A-module; show that, if mB is open in B under the topology induced by the topology on A^\hat{A}, of necessity B=AB = A (use Nakayama's Lemma). (b) Let A be a commutative Noetherian ring and m an ideal of A. Show that, if the Hausdorff completion A^\hat{A} of A with respect to the m-adic topology is a finitely generated A-module, A is complete with the m-adic topology (reduce it to the case where A is Hausdorff; use Chapter V, § 2, no. 1, Proposition 1 and Theorem 1 to show that A is a Zariski ring with the m-adic topology and conclude with the aid of (a)). \*

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