Bourbaki

Exercise 8, § 2

Commutative Algebra, VII, 557

Bourbaki marks this one as harder than the rest.

Let A\mathbf{A} be an integral domain in which every prime ideal ≠0\neq 0 is invertible. Show that A\mathbf{A} is a Dedekind domain. (Prove first that every prime ideal q′≠0q' \neq 0 of A\mathbf{A} is maximal, noting that, if pp is a prime ideal such that q≠p′q \neq p' and p⊃p′p \supset p', then p′:p=p′p': p = p' (Chapter 11, § 1, Exercises 8 (b)) and on the other hand that p′:p=p′p−1p': p = p'p^{-1}, whence a contradiction is deduced. Deduce that A\mathbf{A} is Noetherian (Chapter II, § 1, Exercise 6 (b)) and finally apply Chapter II, § 5, no. 6, Theorem 4 to show that Am\mathbf{A}_m is a field or a discrete valuation ring for every maximal ideal mm of A\mathbf{A}.)

Solution

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Transcribed from the printed volume. Not translated.