Bourbaki

Exercise 15, § 1

Commutative Algebra, IV, 288

Let A be a Noetherian ring, E a finitely generated A-module, F a submodule of E and m an ideal of A. For F to be closed in E under the m-adic topology, it is necessary and sufficient that p+m≠Ap + m \neq A for every ideal p∈Ass⁡(E/F)p \in \operatorname{Ass}(E/F). (Reduce it to the case where F = 0, use Chapter 111, § 3, no. 2, Corollary to Proposition 5 and apply Corollary 2 to Proposition 2 of no. 1 to an element 1+m1 + m, where m∈mm \in m.)

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