Proposition 9

Let B be the integral closure of A, and $c = A : B$. Suppose that B is a finitely generated A-module. One has the inequality $[B/c] \leq 2[B/A]$ in $Z_0(A)$. For there to be equality, it is necessary and sufficient that A be a Gorenstein ring.

One has $[B/c] = [B/A] + [A/c]$, so that the inequality under consideration is equivalent to $[A/c] \leq [B/A]$.

A) For every maximal ideal $m$ of A, the integral closure of $A_m$ is $B_m$ (V, § 1, No. 5, cor. 1), and one has $c_m = A_m : B_m$. Moreover, by definition, one has $[B/c] = \sum_m \operatorname{long}_{A_m}(B_m/c_m)[m]$ and $[B/A] = \sum_m \operatorname{long}_{A_m}(B_m/A_m)[m]$. This reduces us to proving the proposition when the ring A is local, which we shall henceforth assume. In this case the ordered group $Z_0(A)$ identifies canonically with $\mathbf{Z}$, in such a way that the class of a module of finite length is its length. The ring B is semilocal and the B-module $B\Omega$ is free of rank 1 (II, § 5, No. 3, prop. 5).

B) If A is a Gorenstein ring, the A-module $\Omega$ is free of rank 1 (No. 4, cor. 1 of prop. 7); let us choose a generator $\omega$ of $\Omega$. One has $\Omega = A\omega$ and $\Omega_B = c\Omega = c\omega$, and consequently $\operatorname{long}(A/c) = \operatorname{long}(\Omega/\Omega_B) = \operatorname{long}(B/A)$.

C) Suppose the residue field $\kappa_A$ is infinite. For every maximal ideal $n$ of B, let $L(n)$ denote the $B/n$-vector space (of dimension 1) $B\Omega/nB\Omega$, and $\operatorname{pr}_n$ the canonical projection of $\bigoplus_n L(n)$ onto $L(n)$. Let $\varphi : \Omega \longrightarrow \bigoplus_n L(n)$ be the restriction to $\Omega$ of the canonical homomorphism $B\Omega \longrightarrow \bigoplus_n L(n)$. The image of $\varphi$ is a $\kappa_A$-vector subspace of $\bigoplus_n L(n)$; it is not contained in $\operatorname{Ker} \operatorname{pr}_n$, for otherwise one would have $\Omega \subset nB\Omega$ and consequently $B\Omega \subset nB\Omega$, which is contradictory. Thus the image of $\varphi$ is not contained in the union of the $\operatorname{Ker} \operatorname{pr}_n$ (A, V, p. 40, lemma 1); there therefore exists an element $\omega$ of $\Omega$ whose image in $B\Omega/nB\Omega$ is non-zero for every $n$, which implies that $\omega$ generates the B-module $B\Omega$ (II, § 3, No. 3, prop. 11).

Let $a \in A$; if $a\omega$ belong to $\Omega_B$, one has $aB\omega \subset \Omega_B$, hence $a\Omega \subset \Omega_B$, which implies that $a \in c$. The mapping $a \mapsto a\omega$ therefore induces an injection of $A/c$ into $\Omega/\Omega_B$; consequently one has $\operatorname{long}(A/c) \leq \operatorname{long}(\Omega/\Omega_B) = \operatorname{long}(B/A)$.

If $\operatorname{long}(A/c) = \operatorname{long}(B/A)$, one has $A\omega + \Omega_B = \Omega$. One may suppose that the ideal $c$ is contained in $m_A$ (otherwise $A$ is equal to $B$, hence is a Gorenstein ring). Since $\Omega_B = c\Omega$ is contained in $m_A\Omega$, it follows from Nakayama's lemma that $\omega$ generates $\Omega$. Thus the $A$-module $\Omega$ is monogenous, hence free of rank 1, which means that $A$ is a Gorenstein ring (No. 4, cor. 1 of prop. 7).

D) Let us treat the general case. Let us denote by $A'$ the ring $A[X]$, that is to say (IX, App., No. 2) the local ring of the polynomial ring $A[X]$ at the prime ideal $m_A A[X]$; it is a flat, integral $A$-algebra of dimension 1, whose residue field $\kappa_{A'}$ is identified with $\kappa_A(X)$ and whose field of fractions is identified with $K(X)$ (loc. cit.). By the corollary to proposition 5 of No. 3, the $A'$-module $A' \otimes_A \Omega$ is dualizing. Put $B' = A' \otimes_A B$; this is the integral closure of $A'$ in $K(X)$ (V, § 1, No. 3, proposition 13 and No. 5, proposition 16). The conductor $c' = A' : B'$ is equal to $cA'$ (I, § 2, No. 10, formula (11)). For every $A$-module $M$ of finite length, one has $\operatorname{long}_{A'}(A' \otimes_A M) = \operatorname{long}_A(M)$: in fact, since the $A$-algebra $A'$ is flat, it is enough to prove this relation when $M$ is simple, that is to say isomorphic to $\kappa_A$; but in this case $A' \otimes_A \kappa_A$ is identified with $\kappa_{A'}$, whence our assertion. Thus one has

The ring $A'$ satisfies the hypotheses of the proposition, and its residue field is infinite. By part C) of the proof, one has $\operatorname{long}_{A'}(B'/c') \leq 2 \operatorname{long}_{A'}(B'/A')$, and equality implies that $A'$ is a Gorenstein ring; but this last condition implies that $A$ is a Gorenstein ring (§ 3, No. 8, corollary 1 to proposition 12).