Bourbaki

Proposition 11

Tag 04ZE, permanent. Cite it as [Bourbaki, Tag 04ZE].

Commutative Algebra, chapter VII, § 1. Krull domains

Let A\mathbf{A} be a local Krull domain (in particular an integrally closed local Noetherian domain) and mm its maximal ideal. The following conditions are equivalent: (a) A\mathbf{A} is discrete valuation ring; (b) mm is invertible; (c) A:m≠A\mathbf{A}:m \neq \mathbf{A}; (d) mm is divisorial; (e) mm is the only non-zero prime ideal of A\mathbf{A}.

As every non-zero ideal of a discrete valuation ring is principal (Chapter VI, § 3, no. 6, Proposition 9), it is invertible and hence (a) implies (b). If mm is invertible, its inverse is A:m\mathbf{A}:m (Chapter II, § 5, no. 6, Proposition 10) and hence A:m≠A\mathbf{A}:m \neq \mathbf{A}; hence (b) implies (c). If A:m≠A\mathbf{A}:m \neq \mathbf{A}, then A:(A:m)≠A\mathbf{A}:(\mathbf{A}:m) \neq \mathbf{A}; now m⊂A:(A:m)m \subset \mathbf{A}:(\mathbf{A}:m); hence m=A:(A:m)m = \mathbf{A}:(\mathbf{A}:m) since mm is maximal, so that mm is divisorial (no. 1, Proposition 1 (c)); thus (c) implies (d). The fact that (d) implies (e) follows from Theorem 3 of no. 6, Finally, if mm is the only non-zero prime ideal of A\mathbf{A}, it is of height 1 and hence Am\mathbf{A}_m is a discrete valuation ring (no. 6, Theorem 4); as A\mathbf{A} is local, Am=A\mathbf{A}_m = \mathbf{A}, which shows that (e) implies (a).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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