Bourbaki

Theorem 1

Tag 04JN, permanent. Cite it as [Bourbaki, Tag 04JN].

Commutative Algebra, chapter IV, § 1. Prime ideals associated with a module

Let A be a Noetherian ring and M a finitely generated A-module. There exists a composition series (Mi)0≤i≤n(M_i)_{0 \leq i \leq n} of M such that, for 0≤i≤n−10 \leq i \leq n - 1, Mi/Mi+1M_i / M_{i+1} is isomorphic to A/piA / \mathfrak{p}_i, where pi\mathfrak{p}_i is a prime ideal of A.

Let G\mathscr{G} be the set of submodules of M which have a composition series with the property of the statement. As G\mathscr{G} is non-empty (for {0}\{0\} belongs to G\mathscr{G}) and M is Noetherian, G\mathscr{G} has a maximal element N. If M≠NM \neq N, then M/N≠0M/N \neq 0 and hence Ass⁡(M/N)≠∅\operatorname{Ass}(M/N) \neq \varnothing (no. 1, Corollary 1 to Proposition 2); M/NM/N therefore contains a submodule N′/NN'/N isomorphic to an A-module of the form A/pA/\mathfrak{p}, where p\mathfrak{p} is prime; then by definition N′∈GN' \in \mathscr{G}, which contradicts the maximal character of N. Then necessarily N=MN = M.

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

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Transcribed from the printed volume. Not translated.