Proposition 1

An A-module M has finite length (II, §1, No. 10, p. 212) if and only if it is both Artinian and Noetherian.

Suppose that M has finite length $d$. Then every strictly increasing or strictly decreasing sequence of submodules of M has at most $d+ 1$ terms (I, §4, No. 7, p. 44). Consequently, M is Artinian and Noetherian.

Conversely, suppose that M is Artinian and Noetherian. Let $\mathscr{S}$ be the set of submodules of M of finite length. The zero submodule is an element of $\mathscr{S}$, and since M is Noetherian, $\mathscr{S}$ has a maximal element N. Let us give a proof by contradiction and suppose that $M\not= N$. The set of submodules of M distinct from N and containing N then has a minimal element P because M is Artinian. The module $P/N$ has length 1, and since N is a module of finite length, the same holds for P (II, §1, No. 10, p. 212, Proposition 16). This contradicts the definition of N.