Proposition 16

Tag 05L3, permanent. Cite it as [Bourbaki, Tag 05L3].

Algebra, chapter II, § 1. Modules

Let $M$ be an $A$ -module and $N$ a submodule of $M$ ; for $M$ to be of finite length, it is necessary and sufficient that $N$ and $M/N$ be so, and then

\operatorname{long}(N) + \operatorname{long}(M/N) = \operatorname{long}(M).

The proof has been given in I, § 4, no. 7, Proposition 10.

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

Cited by

Corollary 3, § 1
Exercise 12, § 4
Proposition 11, § 4
Lemma 3, § 1
Proposition 1, § 1
Example 1, § 11

Cites

I, § 4, no. 7, Proposition 10