Proposition 8

The A-module $V/\Omega$ is a Matlis module.

Consider the exact sequence

the A-module V is injective (A, X, p. 18, example 1), and one has $\mathrm{di}_A(\Omega) = 1$ (No. 1, prop. 1). One deduces on the one hand that $V/\Omega$ is injective (§ 3, No. 1, prop. 1), on the other hand that, for every maximal ideal $m$ of A, the $A/m$-vector space $\mathrm{Hom}_A(A/m, V/\Omega)$ is isomorphic to $\mathrm{Ext}_A^1(A/m, \Omega)$, hence of dimension 1. Since $V/\Omega$ is a torsion module, its associated prime ideals are maximal; this proves the proposition (§ 8, No. 4).

Let M be an A-module; in accordance with loc. cit., we shall denote by D(M) the A-module $\mathrm{Hom}_A(M, V/\Omega)$. One can apply the constructions of No. 5 by taking for I the complex

where V is placed in degree 0, and p denotes the canonical surjection. The complex $\mathbf{D}(M)$ is

with $p_M = \mathrm{Hom}(1_M, p)$. One has a canonical isomorphism of graded A-modules $\varphi(M, I) : \mathrm{H}(\mathbf{D}(M)) \to \mathrm{Ext}_A(M, \Omega)$ (A, X, p. 100, th. 1).

When M is a torsion module, the module $\mathbf{D}(M)^0 = \mathrm{Hom}_A(M, V)$ is zero, and $\varphi(M, I)$ is an isomorphism of $\mathbf{D}(M)$ onto $\mathrm{Ext}_A^1(M, \Omega)$; the morphism $\alpha_M : M \to \mathbf{D}(\mathbf{D}(M))$ is none other than the canonical homomorphism of A-modules

defined in § 8, No. 3, which is an isomorphism when M is of finite type (that is to say, of finite length). In this case one recovers the situation of loc. cit.

Let us return to the general case, and suppose the A-module M to be of finite type. Then $\mathbf{D}(M)$ is a torsion module, hence the A-module $\mathbf{D}(\mathbf{D}(M))^{-1} = \mathrm{Hom}_A(\mathbf{D}(M), V)$ is zero. On the other hand the A-module $\mathrm{Hom}_A(\mathbf{D}(M)^0, V) = \mathrm{Hom}_A(\mathrm{Hom}_A(M, V), V)$ identifies naturally with $K \otimes_A M$, in such a way that the homomorphism

identifies with the mapping

such that $j(\lambda \otimes m)(f) = p(\lambda f(m))$ for $\lambda \in K,\ m \in M,\ f \in \mathrm{Hom}_A(M, V)$. Theorem 1 of No. 5 therefore translates into the exactness of the sequence

where $i$ denotes the canonical mapping of $M$ into $K \otimes_A M$. The kernel of $i$ identifies with the torsion submodule $T(M)$ of $M$, and its cokernel with $(K/A) \otimes_A M$. Consider the commutative diagram with exact rows

where the second row is obtained by Matlis duality from the exact sequence

Theorem 1 then means that the homomorphisms of $A$-modules

deduced from $\alpha_M$ and $j$ respectively, are bijective. Since the $A$-module $T(M)$ is of finite length, the $A$-module $\mathrm{Ext}_A^1(M, \Omega)$ is of finite length and identifies with the Matlis dual of $D(T(M))$, and one has $[\mathrm{Ext}_A^1(M, \Omega)] = [T(M)]$ in the group $Z_0(A)$ and $\mathrm{long}_A(\mathrm{Ext}_A^1(M, \Omega)) = \mathrm{long}_A(T(M))$ (§ 8, No. 4, Proposition 4). On the other hand, when one takes $M = A$, one obtains a canonical isomorphism $\gamma^1(A) : K/A \to D(\Omega)$.

Let $B$ be a subring of $K$ containing $A$, finite over $A$. For every maximal ideal $m$ of $A$, one has $\mathrm{prof}_{A_m}(B_m) = \dim_{A_m}(B_m) = 1$ (§ 1, No. 1, Remark 2 and VIII, § 2, No. 3, Theorem 1), so that $B$ is a Macaulay A-module. Consequently the $B$-module $\Omega_B = \mathrm{Hom}_A(B, \Omega)$ is dualizing (No. 3, Remark). The canonical mapping of $\Omega_B = \mathrm{Hom}_A(B, \Omega)$ into $\Omega = \mathrm{Hom}_A(A, \Omega)$ is injective; its image is formed by the elements $\omega$ of $\Omega$ such that the B-submodule $B\omega$ of $V$ is contained in $\Omega$. Thus $\Omega_B$ identifies with the greatest B-submodule of $\Omega$. The $A$-module $B/A$ is of finite length; the exact sequence

makes it possible to identify $\Omega/\Omega_B$ with $\mathrm{Ext}_A^1(B/A, \Omega)$, hence from what precedes with $D(B/A)$. In particular, one has $[B/A] = [\Omega/\Omega_B]$ in $Z_0(A)$ and $\mathrm{Ann}_A(B/A) = \mathrm{Ann}_A(\Omega/\Omega_B)$ (§ 8, No. 4, prop. 4).

The ideal $c = \mathrm{Ann}_A(B/A)$ is the transporter $A : B$, that is to say (VII, § 1, No. 1) the set of elements $x$ of $K$ such that $xB \subset A$. It is a (nonzero) ideal of

A and of B; it is in fact the largest ideal of B contained in A. Since $\Omega_B : \Omega \subset \Omega : \Omega = A$ (No. 4, prop. 7, b)), one has $\operatorname{Ann}_A(\Omega/\Omega_B) = \Omega_B : \Omega$, whence finally

Since $\Omega_B$ is a B-module, the relation $x \Omega \subset \Omega_B$ is equivalent to $xB\Omega \subset \Omega_B$, so that one also has $c = \Omega_B : B\Omega$.

We shall specialise the preceding discussion to the case where B is the integral closure of A; the assumption that B be a finitely generated A-module is satisfied when the ring A is Japanese (IX, § 4, No. 1, def. 1), which is the case when it is local and complete (loc. cit., No. 2, th. 2), or when it is essentially of finite type over a field (loc. cit., No. 1, remark 2 and example). The ring B is then a Dedekind ring (VII, § 2, No. 2, th. 1), and the torsion-free B-modules $\Omega_B$, $B\Omega$ and $c$ are projective of rank 1 (VII, § 4, No. 10, prop. 22). The relation $c = \Omega_B : B\Omega$ then signifies that the linear mapping $c \otimes_B B\Omega \to \Omega_B$ deduced from the action of K on V is an isomorphism (II, § 5, No. 6, prop. 11). In particular one has $\Omega_B = c(B\Omega) = c\Omega$.