Bourbaki

Theorem 2

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

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

Let M be a finitely generated module over a Noetherian ring A and (Mi)0≤i≤n(M_i)_{0 \leq i \leq n} a composition series 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. Then

(4)

Ass⁡(M)⊂{p0,…,pn−1}⊂Supp⁡(M); \operatorname{Ass}(M) \subset \{\mathfrak{p}_0, \ldots, \mathfrak{p}_{n-1}\} \subset \operatorname{Supp}(M);

the minimal elements of these three sets are the same and coincide with the minimal elements of the set of prime ideals containing Ann(M).

The inclusion Ass(M) ⊂{p0,…,pn−1}\subset \{p_0, \ldots, p_{n-1}\} follows immediately from Propositions 1 and 3 of no. 1. For 0≤i≤n−10 \leq i \leq n - 1,

pi∈Supp⁡(A/pi)=Supp⁡(Mi/Mi+1) p_i \in \operatorname{Supp}(A/p_i) = \operatorname{Supp}(M_i/M_{i+1})

(Chapter II, § 4, no. 4, Example), whence pi∈Supp⁡(Mi)⊂Supp⁡(M)p_i \in \operatorname{Supp}(M_i) \subset \operatorname{Supp}(M) (Chapter II, § 4, no. 4, Proposition 16), which shows the inclusion

{p0,…,pn−1}⊂Supp⁡(M). \{p_0, \ldots, p_{n-1}\} \subset \operatorname{Supp}(M).

Corollary 1 to Proposition 7 of no. 3 shows that Ass(M) and Supp(M) have the same minimal elements and (4) shows that these are just the minimal elements of {p0,…,pn−1}\{p_0, \ldots, p_{n-1}\}. The last assertion then follows from Chapter II, § 4, no. 4, Proposition 17.

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

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