Proposition 4

*Let $A$ be a noetherian ring, $J$ an ideal of $A$ generated by an $A$-regular sequence $x$, and $\Omega$ a finitely generated $A$-module.

a) If the $A$-module $\Omega$ is dualizing, the sequence $x$ is $\Omega$-regular and the $A/J$-module $\Omega/J\Omega$ is dualizing;

b) If the $A/J$-module $\Omega/J\Omega$ is dualizing, if $J$ is contained in the radical of $A$, and if the sequence $x$ is $\Omega$-regular, the $A$-module $\Omega$ is dualizing.*

Reasoning by induction on the length of the sequence $x$, we are reduced to the case where the latter consists of a single element $x$. Suppose that the $A$-module $\Omega$ is dualizing. For every maximal ideal $m$ of $A$ containing $x$, one has $\dim(A_m/xA_m) = \dim(A_m) - 1$

Let \overline{A} denote the ring A/xA; let m be a maximal ideal of A containing x, and let \overline{m} be its image in \overline{A}. The A-module A/m is annihilated by x, and is identified with \overline{A}/\overline{m} ; hence for every integer i \geqslant 1 there is an isomorphism Ext^i_A(A/m, \Omega) \longrightarrow \operatorname{Ext}^{i-1}_{\overline{A}}(\overline{A}/\overline{m}, \Omega/x\Omega) (§ 3, n° 4, prop. 7). One has

(VIII, § 3, n° 1, cor. 2, a)). Now the maximal ideals of \overline{A} are the ideals \overline{m}, where m is a maximal ideal of A containing x; and if moreover x belongs to the radical of A, every maximal ideal of A contains x. The proposition follows.