Bourbaki

Definition 1

Tag 04J7, permanent. Cite it as [Bourbaki, Tag 04J7].

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

Let M be a module over a ring A. A prime ideal p is said to be associated with M if there exists x ∈ M such that p is equal to the annihilator & x. The set of prime ideals associated with M is denoted by Ass_A(M), or simply Ass(M).

\* Example. Let a be an ideal in the polynomial ring A = \mathbf{C}[X_1, ..., X_n], V the corresponding affine algebraic variety and V_1, ..., V_p the irreducible components of V. If M is taken to be the ring A/a of functions which are regular on V, the set of prime ideals associated with M consists of the ideals of V_1, ..., V_p and in general other prime ideals each of which contains one of the ideals of the V_i. \*

As the annihilator of 0 is A, an element x ∈ M whose annihilator is a prime ideal is necessarily ≠ 0. To say that a prime idea p is associated with M

(*) The results of this chapter depend only on Books I to VI and Chapters I to III of this Book, excluding Chapter I, § 4 and Chapter III, § 5.

amounts to saying that MM contains a submodule isomorphic to A/pA/p (namely AxAx, for all x∈Mx \in M whose annihilator is pp).

If an AA-module MM is the union of a family (Mi)i∈I(M_i)_{i \in I} of submodules, then clearly

Ass⁡(M)=⋃i∈IAss⁡(Mi). \operatorname{Ass}(M) = \bigcup_{i \in I} \operatorname{Ass}(M_i).

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.