Corollary 3

Tag 045X, permanent. Cite it as [Bourbaki, Tag 045X].

Commutative Algebra, chapter II, § 2. Rings and modules of fractions

Let M be a finitely generated A-module. For an ideal $a \& A$ to be such that $aM = M$ , it is necessary and sufficient that there exist $a \in a$ such that $(1 + a)M = 0$ .

Clearly the relation $(1 + a)M = 0$ implies $M = aM$ . To prove the converse we use the following lemma:

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

Cited by

Exercise 15, § 4
Proposition 5, § 3

Cites

Nothing.