Bourbaki

Theorem 1

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

Commutative Algebra, chapter IV, § 2. Primary decomposition

Let MM be a finitely generated module over a Noetherian ring and let NN be a submodule of MM. There exists a primary decomposition of NN of the form

N=⋂p∈Ass⁡(M/N)Q(p) N = \bigcap_{p \in \operatorname{Ass}(M/N)} Q(p)
where, for all p∈Ass⁡(M/N)p \in \operatorname{Ass}(M/N), Q(p)Q(p) is pp-primary with respect to MM.

We may replace MM by M/NM/N and therefore suppose that N=0N = 0. By § 1, no. 4, Corollary to Theorem 2, Ass⁡(M)\operatorname{Ass}(M) is finite; by § 1, no. 1, Proposition 4, there exists, for each p∈Ass⁡(M)p \in \operatorname{Ass}(M), a submodule Q(p)Q(p) of MM such that Ass⁡(M/Q(p))={p}\operatorname{Ass}(M/Q(p)) = \{p\} and Ass⁡(Q(p))=Ass⁡(M)−{p}\operatorname{Ass}(Q(p)) = \operatorname{Ass}(M) - \{p\}. Let us write P=⋂p∈Ass⁡(M)Q(p)P = \bigcap_{p \in \operatorname{Ass}(M)} Q(p); for all p∈Ass⁡(M)p \in \operatorname{Ass}(M), Ass⁡(P)⊂Ass⁡(Q(p))\operatorname{Ass}(P) \subset \operatorname{Ass}(Q(p)) and hence Ass⁡(P)=∅\operatorname{Ass}(P) = \varnothing, which implies P=0P = 0 (§ 1, no. 1, Corollary 1 to Proposition 2) and therefore proves the theorem.

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

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