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).

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 $M = \mathrm{Ext}^{d-1}_A(A/p, \Omega)$:

For let $r$ be the rank of M; there exists a free submodule L of M of rank $r$ such that $M/L$ is a torsion module (VII, § 4, No. 1, corollary of proposition 1), hence of finite length (VII, § 2, No. 5, lemma 1). The annihilator of $M/L$ is not reduced to 0, and therefore contains a nonzero element $x$ of $m_B$. Consider the commutative diagram

By the snake lemma (A, X, p. 4, proposition 2), one deduces from it an exact sequence

whence $\operatorname{long}(M/xM) = \operatorname{long}(L/xL)$. Since $\operatorname{long}(M/xM) = \operatorname{long}(B/xB)$ by hypothesis and $\operatorname{long}(L/xL) = r \operatorname{long}(B/xB)$, one deduces that $r = 1$.

Indeed a dualizing module over a local ring is nonzero by definition.

The first assertion follows from A, X, p. 108, corollary. Let $\mathfrak{p}$ be a prime ideal of the support of $\operatorname{Ext}_A^i(M, \Omega)$. One has $\operatorname{Ext}_A^i(M, \Omega)_\mathfrak{p} \neq 0$, hence $\operatorname{Ext}_{A_\mathfrak{p}}^i(M_\mathfrak{p}, \Omega_\mathfrak{p}) \neq 0$ (§ 3, No. 2, proposition 2), which implies $\operatorname{di}_{A_\mathfrak{p}}(\Omega_\mathfrak{p}) \geqslant i$. Since $\Omega_\mathfrak{p}$ is a dualizing $A_\mathfrak{p}$-module (proposition 2), one has $\operatorname{di}_{A_\mathfrak{p}}(\Omega_\mathfrak{p}) = \dim(A_\mathfrak{p})$ (proposition 1), whence the corollary.

b) Put $c = \dim(A) - \dim_A(M)$. If M is nonzero, the A-module $\operatorname{Ext}_A^c(M, \Omega)$ is nonzero.

c) One has $\operatorname{Ext}_A^i(M, \Omega) = 0$ for $i > \dim(A) - \operatorname{prof}_A(M)$.

Assume M nonzero and denote by F its support. By proposition 9 of § 1, No. 5, the conjunction of assertions a) and b) is equivalent to $\operatorname{prof}_F(\Omega) = c$. But since $\Omega$ is Cohen-Macaulay and its support is equal to Spec(A) (proposition 1 and corollary 2 of proposition 2), one has

(§ 2, No. 1, corollary of proposition 1 and No. 2, corollary of proposition 2).

Let us prove c) by induction on the depth of M. If $\operatorname{prof}_A(M) = 0$, one indeed has $\operatorname{Ext}_A^i(M, \Omega) = 0$ for $i > \dim(A)$, since $\operatorname{di}_A(\Omega) = \dim(A)$ (proposition 1). Assume $\operatorname{prof}_A(M) > 0$; then there exists an element $x$ of $\mathfrak{m}_A$ such that the homothety of ratio $x$ is injective in M. One has $\operatorname{prof}_A(M/xM) = \operatorname{prof}_A(M) - 1$ (§ 1, No. 4, proposition 7).

Consider the exact sequence of extension modules

associated with the exact sequence

For $i > \dim(A) - \operatorname{prof}_A(M)$, the $A$-module $\operatorname{Ext}_A^{i+1}(M/xM, \Omega)$ is zero by the induction hypothesis; therefore the homothety with ratio $x$ is surjective in $\operatorname{Ext}_A^i(M, \Omega)$, which implies that this $A$-module is zero (Nakayama's lemma). This proves c).

The first assertion follows from Prop. 3, a) and c). Let $p \in \operatorname{Supp}(M)$; by Prop. 1 of § 2, No. 1, applied to $M$ and to $A$, we have

since the $A_p$-module $\Omega_p$ is dualizing (Prop. 2), it follows from Prop. 3, b) that the $A_p$-module $\operatorname{Ext}_A^c(M_p, \Omega_p)$ is not zero. Hence the support of $\operatorname{Ext}_A^c(M, \Omega)$ is equal to that of $M$.

Finally, let us prove, by induction on $\dim(M)$, that the $A$-module $\operatorname{Ext}_A^c(M, \Omega)$ is Macaulay. The assertion is true when $\dim(M) = 0$, since every module of finite length is Macaulay. Suppose $\dim(M) > 0$ and choose an element $x$ of $\mathfrak{m}_A$ such that the homothety $x_M$ is injective. The $A$-module $M/xM$ is Macaulay ($§ 2$, No. 3, Prop. 4), of dimension $\dim(M) - 1$. In view of the foregoing, the exact sequence of extension modules associated with the exact sequence $0 \to M \xrightarrow{x} M \longrightarrow M/xM \to 0$ reduces to

Prop. 4 of § 2, No. 3 and the induction hypothesis then imply that $\operatorname{Ext}_A^c(M, \Omega)$ is Macaulay, whence the corollary.

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.

Let \Omega be a dualizing A-module; it is torsion-free by prop. 4. Let K be the field of fractions of A; the K-vector space K \otimes_A \Omega is dualizing (n° 1, prop. 2), hence of dimension 1.

(i) the A-module \Omega is dualizing ; (ii) the A-module \Omega is a Macaulay module of dimension equal to \dim(A), and \Omega/J\Omega is an indecomposable injective module over the artinian local ring A/J ; (iii) the sequence x is \Omega-regular and \Omega/J\Omega is an indecomposable injective module over the artinian local ring A/J ; (iv) the sequence x is \Omega-regular, one has \operatorname{long}_A(\Omega/J\Omega) = \operatorname{long}_A(A/J) and the \kappa_A-vector space \operatorname{Hom}_A(\kappa_A, \Omega/J\Omega) is of dimension 1.

(i) \Rightarrow (ii) : if \Omega is dualizing, it is Macaulay and of dimension \dim(A) (n° 1, prop. 1). The sequence x is A-regular since A is a Macaulay ring; by prop. 4, the A/J-module \Omega/J\Omega is dualizing, and hence is a Matlis A/J-module (n° 1, example 1).

(ii) \Rightarrow (iii) : under assumption (ii), one has \dim(\Omega) = \dim(A) and \dim(\Omega/J\Omega) = \dim(A/J) = 0, so that the sequence x is secant for \Omega, hence \Omega-regular (§ 2, n° 3, th. 1).

(iii) \Rightarrow (i) : under the assumptions of (iii), the A/J-module \Omega/J\Omega is a Matlis A-module, and hence is dualizing (n° 1, example 1) ; by prop. 4 the A-module \Omega is dualizing.

(iii) \Leftrightarrow (iv) : this follows from the remark in § 8, n° 3.

Let $n$ be a maximal ideal of $B$, and $p$ its inverse image in $A$. The $A_p$-module $B_n$ is flat, the $A_p$-module $\Omega_p$ is dualizing, $\Omega_{(B)} \otimes_B B_n$ is identified with $\Omega_p \otimes_{A_p} B_n$ and $\kappa_{A_p} \otimes_{A_p} B_n$, which is identified with a ring of fractions of $\kappa(p) \otimes_A B$, is a Gorenstein ring. It is therefore enough to prove the proposition when $\rho$ is a local homomorphism of local rings, which we shall henceforth assume.

Let us first treat the case where the rings $A$ and $B$ are artinian. Put $C = B/\mathfrak{m}_A B$. Since $B$ is flat over $A$, the $B$-module $\mathrm{Hom}_B(C, \Omega_{(B)})$ is isomorphic to $\mathrm{Hom}_A(\kappa_A, \Omega) \otimes_A B$ (I, § 2, n° 10, prop. 11), hence to $\mathrm{Hom}_A(\kappa_A, \Omega) \otimes_{\kappa_A} C$. One deduces a sequence of isomorphisms

The $\kappa_A$-vector space $\mathrm{Hom}_A(\kappa_A, \Omega)$ is of dimension 1 since $\Omega$ is dualizing, and the same holds for the $\kappa_C$-vector space $\mathrm{Hom}_C(\kappa_C, C)$ since $C$ is a Gorenstein ring; consequently the $\kappa_B$-vector space $\mathrm{Hom}_B(\kappa_B, \Omega_{(B)})$ is of dimension 1.

Let $M$ be a $B$-module of finite length; let us prove by induction on $\mathrm{long}_B(M)$ that one has $\mathrm{long}_B(\mathrm{Hom}_B(M, \Omega_{(B)})) \leqslant \mathrm{long}_B(M)$. The assertion is clear if $M = 0$, and it follows from the preceding if $M = \kappa_B$. Suppose $\mathrm{long}_B(M) \geqslant 2$. There exists an exact sequence of $B$-modules

with $\mathrm{long}_B(M') < \mathrm{long}_B(M)$. One deduces from this an exact sequence and one concludes by applying the induction hypothesis to $M'$.

Let $N$ be the kernel of the canonical surjection of $\kappa_A \otimes_A B$ onto $\kappa_B$. Put $m = \mathrm{long}_B(\kappa_A \otimes_A B)$; one has $\mathrm{long}_B(N) = m - 1$. Consider the exact sequence of $B$-modules

The $B$-modules $\mathrm{Hom}_B(\kappa_A \otimes_A B, \Omega_{(B)})$ and $\mathrm{Ext}_B^1(\kappa_A \otimes_A B, \Omega_{(B)})$ are respectively isomorphic to $\mathrm{Hom}_A(\kappa_A, \Omega) \otimes_A B$ and $\mathrm{Ext}_A^1(\kappa_A, \Omega) \otimes_A B$, that is to say to $\kappa_A \otimes_A B$ and to 0. The lengths of the $B$-modules $\mathrm{Hom}_B(\kappa_B, \Omega_{(B)})$ and $\mathrm{Hom}_B(\kappa_A \otimes_A B, \Omega_{(B)})$ are 1 and $m$, and that of $\mathrm{Hom}_B(N, \Omega_{(B)})$) is $\leqslant m - 1$; one deduces from this that the $B$-module $\mathrm{Ext}_B^1(\kappa_B, \Omega_{(B)})$ is zero. By Prop. 6 of § 3, No. 3, the $B$-module $\Omega_{(B)}$ is injective; consequently it is a dualizing module (No. 1, Example 1).

Let us pass to the general case. Put $C = \kappa_A \otimes_A B$; by hypothesis this is a Gorenstein ring, hence a Macaulay ring ($§ 3, n° 7$, prop. 10). By prop. 1 of No. 1, $A$ is a Macaulay ring, and the $A$-module $\Omega$ is Macaulay. Hence $B$ is a Macaulay ring, and the $B$-module $\Omega_{(B)}$ is Macaulay ($§ 2, n° 7$, cor. 1 of prop. 9). Put $r = \dim(A)$, $s = \dim(C)$. There exists a sequence $(x_1, \ldots, x_r)$ of elements of $\mathfrak{m}_A$ regular for the $A$-modules $A$ and $\Omega$, and a sequence $(y_1, \ldots, y_s)$ of elements of $\mathfrak{m}_B$ regular for the $B$-module $C$; let $x$ denote the ideal of $A$ and $\mathfrak{g}$ the ideal of $B$ generated by them respectively. The sequence $(y_1, \ldots, y_s, \rho(x_1), \ldots, \rho(x_r))$ is regular for the $B$-modules $B$ and $\Omega_{(B)}$ ($§ 1, n° 6$, prop. 11), and the $A$-module $B/\mathfrak{g}$ is flat (loc. cit., prop. 10). Put $A' = A/x$, $B' = B/(xB + \mathfrak{g})$ and let $\rho' : A' \to B'$ denote the homomorphism deduced from $\rho$ by passing to quotients. The rings $A'$ and $B'$ are artinian, the $A'$-module $B'$ is flat, the ring $\kappa_{A'} \otimes_{A'} B'$, which is identified with $C/\mathfrak{g}$, is a Gorenstein ring ($§ 3, n° 7$, example 2) and the $A'$-module $\Omega_{(A')}$ is dualizing (No. 2, prop. 4). By the first part of the proof, the $B'$-module $\Omega_{(B')}$ is dualizing. It then follows from loc. cit. that the $B$-module $\Omega_{(B)}$ is dualizing.

Indeed, for every prime ideal $\mathfrak{p}$ of $A$, the ring $\kappa(\mathfrak{p}) \otimes_A A[X]$ is identified with $\kappa(\mathfrak{p})[X]$, which is regular, hence Gorenstein.

We have $\dim(B) = \dim_A(B) \leq \dim(A)$ (VIII, § 2, No. 3, Theorem 1 c)); let $c = \dim(A) - \dim(B)$. We have $\mathrm{Ext}_A^i(B, \Omega) = 0$ for $i \neq c$ since the $A$-module $B$ is Macaulay (No. 1, Corollary to Proposition 3). Let us prove that the $B$-module $\mathrm{Ext}_A^c(B, \Omega)$ is dualizing.

Suppose first that $\dim(B) = 0$. The spectrum $X$ of $B$ is finite and formed of maximal ideals (IV, § 2, No. 5, Proposition 9); the canonical mapping $B \to \prod_{n \in X} B_n$ is an isomorphism (loc. cit., Corollary 1). The $B$-module $\Omega' = \mathrm{Ext}_A^c(B, \Omega)$ is therefore the direct sum of the modules $\mathrm{Ext}_A^c(B_n, \Omega)$; since $\mathrm{Ext}_A^c(B_n, \Omega)$ has support in $\{n\}$, it identifies with $\Omega'_n$. We have $\dim(B_n) = 0$ for every $n$; to prove that the $B$-module $\Omega'$ is dualizing, it is therefore enough to prove that this is so for the $B_n$-module $\mathrm{Ext}_A^c(B_n, \Omega)$ for every $n \in X$, which brings us back to the case where the ring $B$ is local. In this case, by Example 6 of § 8, No. 5, the $B$-module $\mathrm{Ext}_A^c(B, \Omega)$ is isomorphic to $\mathrm{Hom}_A(B, I)$, where $I$ is a Matlis $A$-module; it is consequently a Matlis $B$-module (§ 8, No. 6, Corollary to Proposition 5), hence a dualizing $B$-module (No. 1, Example 1).

Now suppose that $\dim(B) > 0$ and reason by induction on $\dim(B)$. We have $\mathrm{prof}_A(B) = \dim_A(B) = \dim(B)$, whence $\mathrm{prof}_A(B) > 0$; on the other hand we have $\mathrm{prof}(A) = \dim(A) > 0$ (No. 1, Proposition 1), and consequently $\mathrm{prof}_A(A \oplus B) > 0$. There therefore exists an element $x$ of $\mathfrak{m}_A$ such that the homotheties $x_A$ and $x_B$ are injective.

Consider the exact sequence of extension modules associated with the exact sequence $0 \to B \xrightarrow{x_B} B \longrightarrow B/xB \to 0$ and with the $A$-module $\Omega$. The $A$-module $B/xB$ is Macaulay ($§ 2$, No. 1, Example 3), of dimension $\dim(B) - 1$ (VIII, $§ 3$, No. 2, Proposition 3) ; we therefore have $\mathrm{Ext}_A^i(B/xB, \Omega) = 0$ for $i \neq c + 1$ (No. 1, Corollary to Proposition 3). Since we have $\mathrm{Ext}_A^i(B, \Omega) = 0$ for $i \neq c$, we obtain an exact sequence of $B$-modules

By the induction hypothesis, the $B/xB$-module $\mathrm{Ext}_A^{c+1}(B/xB, \Omega)$ is dualizing. Since the $A$-algebra $B$ is finite, the image of $m_A$ in $B$ is contained in the radical of $B$ (V, $§ 2$, No. 1, Proposition 1) ; by Proposition 4 of No. 2, the $B$-module $\mathrm{Ext}_A^c(B, \Omega)$ is dualizing.

Let $\Omega'$ denote the $B$-module $\mathrm{Ext}_A(B, \Omega)$. Let $n$ be a maximal ideal of $B$; its inverse image in $A$ is a maximal ideal $m$ (V, $§ 2$, No. 1, prop. 1). The $A_m$-algebra $B_m = A_m \otimes_A B$ is finite, and is a Macaulay $A_m$-module; by the proposition, the $B_m$-module $\Omega'_m$, which identifies with $\mathrm{Ext}_{A_m}(B_m, \Omega_m)$ ($§ 3$, No. 2, prop. 2), is dualizing. Since $B_n$ is a ring of fractions of $B_m$, the $B_n$-module $\Omega'_n$ is dualizing, whence the corollary.

This results from cor. 1 and the cor. of prop. 5.

For let indeed $R$ be a regular ring and $A$ a Macaulay ring quotient of $R$. The $R$-module $A$ is a Macaulay module ($§ 2$, No. 5, example 5), and $R$ has a dualizing module (No. 1, example 2); hence the same is true of $A$ by cor. 1. Moreover, it has already been observed that a complete local noetherian ring is presentable ($§ 4$, No. 4, prop. 6, c)).

More generally, every Macaulay ring quotient of a Gorenstein ring has a dualizing module. Conversely, one can prove that a local Macaulay ring which has a dualizing module is a quotient of a Gorenstein local ring (exercise 1).

The assertion concerning the surjectivity of $u$ results from Nakayama's lemma (II, § 3, No. 2, cor. 1 of prop. 4), without any assumption on $x_N$. Consider the commutative diagram with exact rows

by means of the snake lemma (I, § 1, No. 4, prop. 2), one deduces from it an exact sequence $\mathrm{Ker}\, u \xrightarrow{x} \mathrm{Ker}\, u \longrightarrow \mathrm{Ker}\, \overline{u}$. If $\overline{u}$ is injective, the homothety with ratio $x$ is surjective in $\mathrm{Ker}\, u$, which implies $\mathrm{Ker}\, u = 0$ by Nakayama's lemma.

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).