Bourbaki

Theorem 1

Tag 0D2R, permanent. Cite it as [Bourbaki, Tag 0D2R].

Commutative Algebra, chapter X, § 2. Modules et anneaux macaulayens

Let A be a noetherian local ring, M a nonzero finitely generated A-module, d the dimension of M, x = (x₁, ..., x_d) a sequence of elements of mA secant for M, and J the ideal generated by it. The following conditions are equivalent:

(i) the A-module M is Macaulay ;

(ii) the sequence xx is M-regular ; (iii) the sequence xx is completely secant for MM (A, X, p. 157, Def. 2) ; (iv) the multiplicity (VIII, § 7, No. 1, Def. 1) eJ(M)e_J(M) of MM relative to the ideal JJ is equal to the length of the A-module M/JMM/JM ; (v) for each integer ii such that 1≤i≤d1 \leq i \leq d, the A-module M/(x1M+…+xi−1M)M/(x_1M + \ldots + x_{i-1}M) is pure.

The equivalence of (ii) and (iii) follows from A, X, p. 160, Cor. 1 to Th. 1. Since the A-module M/JMM/JM is of finite length (VIII, § 3, No. 2, Th. 1), the equivalence of (iii) and (iv) follows from VIII, § 4, No. 3, Prop. 4 and No. 4, Th. 3. It remains to prove the equivalence of (i), (ii), and (v).

(i) ⇒\Rightarrow (v) : if MM is Macaulay, each of the modules M/(x1M+…+xi−1M)M/(x_1M + \ldots + x_{i-1}M) is Macaulay (Prop. 4), hence pure.

(v) ⇒\Rightarrow (ii) : this follows from Lemma 1 applied to each of the modules M/(x1M+…+xi−1M)M/(x_1M + \ldots + x_{i-1}M).

(ii) ⇒\Rightarrow (i) : this follows from Prop. 4, since M/JMM/JM is of finite length, hence Macaulay.

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

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