Proposition 2

Let A be a noetherian ring, $\Omega$ a dualizing A-module. For every prime ideal $p$ of A, the $A_p$-module $\Omega_p$ is dualizing.

Consider a saturated chain $p \subset p_1 \subset \ldots \subset p_r$ of prime ideals of A such that the ideal $p_r$ is maximal. Reasoning by induction on $r$, one may suppose that the $A_{p_1}$-module $\Omega_{p_1}$ is dualizing. Replacing A by $A_{p_1}$ and $p$ by $pA_{p_1}$, one is reduced to the case where the ring A is local and where the chain $p \subset m_A$ is saturated.

Put then $d = \dim(A) = \mathrm{ht}(m_A)$. One has $\dim(A_p) = \mathrm{ht}(p) = d - 1$ since A is a Macaulay ring (§ 2, No. 2, cor. of prop. 2). For every integer $i$, the $A_p$-module $\mathrm{Ext}^i_{A_p}(\kappa(p), \Omega_p)$ is isomorphic to $\mathrm{Ext}^i_A(A/p, \Omega)_p$ (§ 3, No. 2, prop. 2); it is therefore enough to prove that the $A/p$-module $\mathrm{Ext}^i_A(A/p, \Omega)$ vanishes for $i \neq d - 1$ and is of rank one for $i = d - 1$.

Let $x$ be an element of $m_A - p$, and $\overline{x}$ its class in $A/p$. Consider the exact sequence of A-modules

The A-module $A/(p + xA)$ is of finite length since its support is reduced to $m_A$; since the A-module $\Omega$ is dualizing, one has $\mathrm{Ext}^i_A(A/(p + xA), \Omega) = 0$ for $i \neq d$ ($§ 8$, No. 5, example 3). It then follows from the exact sequence of extension modules associated with the sequence above and with $\Omega$ that the homothety of ratio $x$ in the A-module $\mathrm{Ext}^i_A(A/p, \Omega)$ is surjective for $i \neq d - 1$, which implies that this module is zero (Nakayama lemma). In particular $\mathrm{Ext}^d_A(A/p, \Omega)$ is zero, and one obtains an exact sequence

One has $\mathrm{long}_A(\mathrm{Ext}^i_A(A/(p + xA), \Omega)) = \mathrm{long}_A(A/(p + xA))$ (loc. cit.); the proposition then follows from the following lemma, applied to the ring $B = A/p$ and the B-module $M = \mathrm{Ext}^{d-1}_A(A/p, \Omega)$: