Proposition 6

Let A be a noetherian ring, M a finitely generated Macaulay A-module, and S a finite subset of A strongly secant for M. Then the A-module M/SM is Macaulay.

For every maximal ideal $m \in \mathrm{Supp}(M/SM)$, the image of S in $A_m$ is secant for $M_m$ (Prop. 5). Since $M_m$ is a Macaulay $A_m$-module, the same is true of $(M/SM)_m$ (Prop. 4), whence the proposition.

Theorem 2 (Macaulay-Cohen).— Let A be a noetherian ring and M a finitely generated A-module. The following conditions are equivalent:

(i) the A-module M is Macaulay;

(ii) for every ideal J of A generated by an M-regular sequence of elements of A, the A-module M/JM has no embedded associated prime ideals;

(iii) for every finite subset S of A strongly secant for M, the A-module M/SM has no embedded associated prime ideals.

(i) $\Rightarrow$ (iii) : Let S be a finite subset of A strongly secant for M. The A-module M/SM is Macaulay (Prop. 6) and in particular has no embedded associated prime ideals (No. 2, Prop. 2, a)).

(iii) $\Rightarrow$ (ii) : This follows from the cor. to Prop. 5.

(ii) $\Rightarrow$ (i): Let $p \in \mathrm{Supp}(M)$; let us prove that the $A_p$-module $M_p$ is Macaulay. We argue by induction on the integer $\dim(M_p)$. If $\dim(M_p)$ is zero, $M_p$ is an $A_p$-module of finite length, hence Macaulay (Example 1, No. 1). Suppose that $\dim(M_p)$ is nonzero, that is to say that $p$ is not a minimal element of $\mathrm{Supp}(M)$ (VIII, § 1, No. 4, Prop. 9 a)). Since M has no embedded associated prime ideals, $p$ is contained in no associated prime ideal of M, and there exists an element $x$ of $p$ such that the homothety $x_M$ is injective (§ 1, Remark 2). The homothety $x_{M_p}$ is then injective, and one has $\dim(M_p/xM_p) < \dim(M_p)$ (VIII, § 3, No. 2, Prop. 3). By the induction hypothesis applied to the A-module $M/xM$ and to the prime ideal $p$ of $\mathrm{Supp}(M/xM)$, the $A_p$-module $M_p/xM_p$ is Macaulay, which implies that the $A_p$-module $M_p$ is Macaulay (No. 3, Prop. 4).