Proposition 1

Let $A$ be a noetherian ring and $\Omega$ a dualizing $A$-module.

a) $A$ is a Macaulay ring, and the $A$-module $\Omega$ is Macaulay.

b) One has $\mathrm{di}_A(\Omega) = \dim(\Omega) = \dim(A)$.

Suppose first that the ring $A$ is local, and let $d$ denote its dimension. Prop. 6 of § 3, No. 3 implies $\mathrm{di}_A(\Omega) = d$, hence $\mathrm{prof}(A) = d$ by prop. 9 of § 3, No. 6, so that $A$ is a Macaulay ring. Moreover, one has $\mathrm{prof}(\Omega) = d$ by definition of depth; since one has $\mathrm{prof}(\Omega) \leq \dim(\Omega) \leq d$, one deduces the proposition in this case.

In the general case, the $A_m$-module $\Omega_m$ is dualizing for every maximal ideal $m$ of $A$, hence $A_m$ is a Macaulay ring and $\Omega_m$ an $A_m$-module which is Macaulay by what precedes, which implies a). Moreover one has $\mathrm{di}_{A_m}(\Omega_m) = \dim(\Omega_m) = \dim(A_m)$ for every maximal ideal $m$, whence b) by passing to the least upper bound (§ 3, No. 2, prop. 3).