Lemma 1

Tag 0DBM, permanent. Cite it as [Bourbaki, Tag 0DBM].

Commutative Algebra, chapter X, § 9. Modules dualisants

Let B be a local noetherian integral ring of dimension 1, and M a finitely generated torsion-free B-module. Suppose that one has $\mathrm{long}_B(M/xM) = \mathrm{long}_B(B/xB)$ for every nonzero element x of B. Then the B-module M is of rank 1.

For let $r$ be the rank of M; there exists a free submodule L of M of rank $r$ such that $M/L$ is a torsion module (VII, § 4, No. 1, corollary of proposition 1), hence of finite length (VII, § 2, No. 5, lemma 1). The annihilator of $M/L$ is not reduced to 0, and therefore contains a nonzero element $x$ of $m_B$ . Consider the commutative diagram

\begin{array}{cccccccc} 0 & \to & L & \longrightarrow & M & \longrightarrow & M/L & \to 0 \\ & & \downarrow_{x_L} & & \downarrow_{x_M} & & \downarrow_0 & \\ 0 & \to & L & \longrightarrow & M & \longrightarrow & M/L & \to 0 . \end{array}

By the snake lemma (A, X, p. 4, proposition 2), one deduces from it an exact sequence

0 \to M/L \longrightarrow L/xL \longrightarrow M/xM \longrightarrow M/L \to 0 ,

whence $\operatorname{long}(M/xM) = \operatorname{long}(L/xL)$ . Since $\operatorname{long}(M/xM) = \operatorname{long}(B/xB)$ by hypothesis and $\operatorname{long}(L/xL) = r \operatorname{long}(B/xB)$ , one deduces that $r = 1$ .

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

Cited by

Nothing.

Cites

Nothing.