Lemma 5

Let $M_1, M$, be two finitely generated $A$-modules, and $f : M_1 \to M_2$ an injective homomorphism. Then the kernel of $f \otimes 1 ; M_1 \otimes_A B \to M_2 \otimes_A B$ is pseudo-zero.

Let $\mathfrak{p}$ be a prime ideal of $A$ of height $\leq 1$. Then $(M_i \otimes_A B)_\mathfrak{p} = (M_i)_\mathfrak{p} \otimes_{A_\mathfrak{p}} B_\mathfrak{p}$ ($i = 1, 2$) (Chapter 11, § 2, no. 7, Proposition 18) and $(f \otimes 1_B)_\mathfrak{p} = f_\mathfrak{p} \otimes 1_{B_\mathfrak{p}}$; the hypothesis that $f$ is injective implies that so is $f$, (Chapter II, § 2, no. 4, Theorem 1); on the other hand, in view of the choice of $\mathfrak{p}$, $A_\mathfrak{p}$ is a principal ideal domain and $B_\mathfrak{p}$ a finitely generated torsion-free $A_\mathfrak{p}$-module and hence free; we conclude that $f_\mathfrak{p} \otimes 1_{B_\mathfrak{p}}$ is itself injective. If $I = \mathrm{Ker}(f \otimes 1)$, then $I_\mathfrak{p} = \mathrm{Ker}((f \otimes 1)_\mathfrak{p})$ (Chapter 11, § 2, no. 4, Theorem 1); therefore $I_\mathfrak{p} = 0$, whence a fortiori $I_\mathfrak{p} = (I_\mathfrak{p})_\mathfrak{p} = 0$ for $\mathfrak{p}|p$, which proves the lemma (no. 4, Definition 2).

We return now to the proof of (ii). For every finitely generated torsion $A$-module $M$, write $\phi(M) = \chi_B(M \otimes_A B)$; it follows from (i) that, if $M$ is pseudo-zero, $\phi(M) = 0$. On the other hand, consider an exact sequence of finitely generated torsion $A$-modules :

It follows from Lemma 4 that there is an exact sequence of $B$-modules: where $I$ is pseudo-zero. Using no. 5, Corollary to Proposition 10, we therefore obtain $\phi(M_2) = \phi(M_1) + \phi(M_3)$. We therefore conclude from Proposition 11 of no. 5 that there exists a homomorphism $\theta : D(A) \to D(B)$ such that $\phi(M) = \theta(\chi_A(M))$ for every finitely generated torsion $A$-module $M$. To prove that $\theta = i$, it suffices to show that $\phi(A/p) = i(p)$ for all $p \in P(A)$; now $(A/p) \otimes_A B = B/pB$ and, for all $\mathfrak{p} \in P(B)$, $(B/pB)_{\mathfrak{p}} = B_{\mathfrak{p}}/pB_{\mathfrak{p}}$; the last module is 0 if $\mathfrak{p}$ does not lie above $p$; if on the contrary $\mathfrak{p}|p$, $B_{\mathfrak{p}}/pB_{\mathfrak{p}}$ is a $B_{\mathfrak{p}}$-module of length $e(\mathfrak{p}/p)$ by definition of the ramification index (Chapter VI, § 8, no. 1); therefore $\chi_B(B/pB) = \sum_{\mathfrak{p}|p} e_{\mathfrak{p}/p} . 13 = i(p)$, which completes the proof of (ii).

Formula (24) is immediate, for

and the rank of $(M \otimes_A K) \otimes_K L$ over $L$ is equal to the rank of $M \otimes_A K$ over $K$. To show (23), consider a free submodule $H$ of $M$ such that $Q = M/H$ is a torsion $A$-module. Applying Lemma 4 as above, we obtain an exact sequence of $B$-modules:

where $I$ is pseudo-zero. It therefore follows from no. 7, Proposition 16(ii) and (v) and Corollary 1 to Proposition 16 that

by virtue of (ii); but by definition of the homomorphism $i : C(A) \to C(B)$, $c_B(i(\chi_A(Q))) = i(c_A(\chi_A(Q))) = -i(c_A(M))$, which completes the proof of (23).

Remarks (1) If $M$ is a reflexive $A$-module, $M \otimes_A B$ is not necessarily reflexive (exercise 6). However it is so when $B$ is a flat $A$-module (no. 2, Proposition 8). (2) Let $C$ be a third integrally closed Noetherian domain such that $B \subset C$ and $C$ is a finitely generated $B$-module (and hence also a finitely generated $A$-module). Then we have the transitivity equations:

(25)

(26)

which follow immediately from the transitivity equations for the ramification indices and the residue class degrees (Chapter VI, § 8, no. 1, Lemma 1).