Proposition 7

Let $A$ be a noetherian ring. The following conditions are equivalent:

(i) $A$ is a Macaulay ring; (ii) for every closed subset $F$ of $\mathrm{Spec}(A)$, one has $\mathrm{prof}_F(A) = \mathrm{codim}(F)$; (iii) every ideal $J$ of $A$ contains an $A$-regular sequence of length $\mathrm{ht}(J)$; (iii') every maximal ideal $m$ of $A$ contains an $A$-regular sequence of length $\mathrm{ht}(m)$; (iv) for every ideal $J$ of $A$, one has $\mathrm{Ext}_A^i(A/J, A) = 0$ for $i < \mathrm{ht}(J)$; (iv') for every maximal ideal $m$ of $A$, one has $\mathrm{Ext}_A^i(A/m, A) = 0$ for $i < \mathrm{ht}(m)$; (v) for every prime ideal $p$ of $A$ and every ideal $J$ of $A_p$ generated by a maximal secant sequence for $A_p$, one has $e_J(A_p) = \mathrm{long}(A_p/ JA_p)$; (v') for every maximal ideal $m$ of $A$, there exists an ideal $J$ of $A_m$, generated by a maximal secant sequence for $A_m$, satisfying $e_J(A_m) = \mathrm{long}(A_m/ JA_m)$; (vi) (Macaulay-Cohen criterion) for every finite subset $S$ of $A$ such that the ideal $\mathfrak{S}$ generated by $S$ is of height $\mathrm{Card}(S)$, the $A$-module $A/\mathfrak{S}$ has no embedded associated prime ideals.

The equivalence of (i) and (ii) follows from No. 1, cor. to Prop. 1. By Th. 2 of § 1, No. 4, and the definition of depth, conditions (iii) and (iv) (resp. (iii') and (iv')) mean that one has $\mathrm{prof}_A(J; A) \geq \mathrm{ht}(J)$ for every ideal (resp. every maximal ideal) $J$ of $A$. Hence one has

But (iv') implies, for every maximal ideal $m$ of $A$, $\mathrm{Ext}_{A_m}^i(\kappa(m), A_m) = 0$ for $i < \dim(A_m)$, whence $\mathrm{prof}(A_m) \geq \dim(A_m)$, so that $A$ is a Macaulay ring.

The equivalence of (i), (v) and (v’) follows from Th. 1 of No. 3, and that of (i) and (vi) from Th. 2 of No. 4.