Proposition 7

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

a) One has $\mathrm{Ext}_A^i(\Omega, \Omega) = 0$ for $i > 0$.

b) The canonical homomorphism $\gamma : A \to \mathrm{End}_A(\Omega)$ is bijective.

c) Every dualizing $A$-module is of the form $\Omega \otimes_A L$ where $L$ is a projective $A$-module of rank 1.

A) Let us first treat the case where the ring $A$ is local. In this case condition c) means simply that two dualizing modules are isomorphic.

Let $\Omega'$ be a dualizing $A$-module. One has $\mathrm{prof}_A(\Omega') = \dim_A(\Omega') = \dim(A)$ (No. 1, prop. 1), therefore $\mathrm{Ext}_A^i(\Omega', \Omega) = 0$ for $i \neq 0$ (No. 1, prop. 3, c)), whence a).

Let us prove b) and c) by induction on the integer $\dim(A)$ (equal to $\mathrm{prof}(A)$). If it is zero, the ring $A$ is artinian, $\Omega'$ and $\Omega$ are Matlis $A$-modules (No. 1, Example 1); hence they are isomorphic ($\S$ 8, No. 1, Prop. 1) and the canonical mapping $A \to \mathrm{End}_A(\Omega)$ is bijective ($\S$ 8, No. 2, Prop. 3, c)). Suppose $\dim(A) > 0$ and let $x$ be a cancellable element of $\mathfrak{m}_A$. The homothety $x_\Omega$ is injective (No. 2, Prop. 4), and we have an exact sequence

Since $\mathrm{Ext}_A^1(\Omega', \Omega)$ is zero and $\mathrm{Hom}_A(\Omega', \Omega/x\Omega)$ is identified with $\mathrm{Hom}_{A/xA}(\Omega'/x\Omega', \Omega/x\Omega)$, we deduce an exact sequence

where $p$ is the canonical mapping. By Prop. 4, the $A/xA$-modules $\Omega/x\Omega$ and $\Omega'/x\Omega'$ are dualizing, hence isomorphic by the induction hypothesis. Let $\overline{u}$ be an isomorphism of $\Omega'/x\Omega'$ onto $\Omega/x\Omega$. In view of the exact sequence (1), there exists an $A$-homomorphism $u : \Omega' \to \Omega$ such that $p(u) = \overline{u}$; by Lemma 2, $u$ is bijective, which proves c). By the induction hypothesis, the canonical homomorphism $A/xA \longrightarrow \mathrm{End}_{A/xA}(\Omega/x\Omega)$ is bijective. In view of the exact sequence (1), this homomorphism is identified with the homomorphism $\overline{\gamma} : A/xA \longrightarrow \mathrm{End}_A(\Omega)/x\mathrm{End}_A(\Omega)$ induced by $\gamma$; it then follows from Lemma 2 that $\gamma$ is bijective, whence b).

B) Let us pass to the general case. For every maximal ideal $m$ of $A$ and every integer $i > 0$, we have $\mathrm{Ext}^i_{A_m}(\Omega_m, \Omega_m) = 0$ by what precedes, hence $\mathrm{Ext}^i_A(\Omega, \Omega)_m = 0$ ($§ 3$, No. 2, Prop. 2), which implies $\mathrm{Ext}^i_A(\Omega, \Omega) = 0$ (II, $§ 3$, No. 3, Cor. 2 of Th. 1). Analogously, the homomorphism $\gamma_m : A_m \to \mathrm{End}_A(\Omega)_m$ is bijective for every maximal ideal $m$ of $A$, hence $\gamma$ is bijective (loc. cit., Th. 1).

Let us finally prove c). Let $\Omega'$ be a dualizing $A$-module. Denote by $L$ the $A$-module $\mathrm{Hom}_A(\Omega', \Omega)$, and by $v : \Omega' \otimes_A L \to \Omega$ the homomorphism such that $v(x \otimes f) = f(x)$ for $x \in \Omega'$, $f \in L$. Let $m$ be a maximal ideal of $A$. The $A_m$-module $L_m$ is identified with $\mathrm{Hom}_{A_m}(\Omega'_m, \Omega_m)$; by the case already treated it is free of rank one, and every isomorphism $h : \Omega'_m \to \Omega_m$ is a generator of it. When one identifies $L_m$ with $A_m$ by means of the generator $h$, the homomorphism $v_m : \Omega'_m \otimes_{A_m} L_m \to \Omega_m$ is identified with $h$, hence is bijective. Since this holds for every maximal ideal $m$ of $A$, the $A$-module $L$ is projective of rank one (II, $§ 5$, No. 3, Theorem 2), and the homomorphism $v$ is bijective (II, $§ 3$, No. 3, Theorem 1).