Proposition 18

(i) For $R$ to be a pseudo-zero $B$-module, it is necessary and sufficient that the $A$-module $R_{[A]}$ be pseudo-zero.

(ii) For $R$ to be finitely generated torsion $B$-module, it is necessary and sufficient that $R_{[A]}$ be a finitely generated torsion $A$-module and then:

(19)

(iii) For $R$ to be finitely generated $B$-module, it is necessary and sufficient that $R_{[A]}$ be a finitely generated $A$-module and then:

(20)

(21)

As $B$ is a finitely generated $A$-module, for $R$ to be a finitely generated $B$-module, it is necessary and sufficient that $R_{[A]}$ be a finitely generated $A$-module. Moreover, if $b$ is the annihilator of $R$, $b \cap A = a$ is the annihilator of $R_{[A]}$; as $B$ is integral over $A$, there is no ideal other than 0 lying above the ideal 0 of $A$ (Chapter V, § 2, no. 1, Corollary 1 to Proposition 1) and hence it amounts to the same to say that $a \neq 0$ or that $b \neq 0$.

(i) By virtue of this last remark, we may confine our attention to the case where $R$ is a torsion $B$-module. If $b$ is contained in a prime ideal $\mathfrak{p} \in P(B)$, $a$ is contained in $\mathfrak{p} \cap A = p$, which is of height 1. Conversely, if $a$ is contained in a prime ideal $p \in P(A)$, there exists a prime ideal $\mathfrak{p}$ of $B$ which contains $b$ and lies above $p$ (Chapter V, § 2, no. 1, Corollary 2 to Theorem 1). Assertion (i) follows from these remarks and no. 4, Definition 2.

(ii) For every finitely generated torsion $B$-module $R$, we write

clearly (for finitely generated torsion $B$-modules) $\phi$ satisfies conditions (1) and (2) of Proposition II of no. 5 (taking account of (i)). There therefore exists a homomorphism $\theta : D(B) \to D(A)$ such that $\phi(R) = \theta(\chi_B(R))$ for every finitely generated torsion $B$-module $R$. The homomorphism $\theta$ is determined by its value for every $B$-module of the form $B/\mathfrak{p}$ where $\mathfrak{p} \in P(B)$, since $\chi_B(B/\mathfrak{p}) = \mathfrak{p}$. Now, for every prime ideal $q \neq p = \mathfrak{p} \cap A$ in $P(A)$, $p \notin q$ and hence $(B/\mathfrak{p})_q = 0$. On the other hand, if we set $S = A - p$, $\mathfrak{p}.S^{-1}B$ is a maximal ideal of $S^{-1}B$ and $(B/\mathfrak{p})_p = S^{-1}B/\mathfrak{p}$. $S^{-1}B$ is isomorphic to the field of fractions of $B/\mathfrak{p}$ (Chapter II, § 2, no. 5, Proposition 11), that is to the residue field of $v_{\mathfrak{p}}$; its length as an $A$-module is therefore $f_{\mathfrak{p}/p}$; which proves that $\theta = N$ (no. 5, Definition 4).

(iii) If $T$ is the torsion submodule of $R$, $T_{[A]}$ is the torsion submodule of $R_{[A]}$ and $(R/T)_{[A]} = R_{[A]}/T_{[A]}$; to prove (21) we may therefore confine our attention to the case where $R$ is torsion-free. Then $R$ is identified with a sub-$B$-module of $R_{(L)}$ and contains a basis $(e_i)_1 < < m$ of $R_{(L)}$ over $L$. If $(b_j)_1 s < n$ is a basis of $L$ over $K$ consisting of elements of $B$, the $b_j e_i$ form a basis of $R_{(L)}$ over $K$ consisting of elements of $R$, whence (21). On the other hand let $M$ be a free sub-$B$-module of $R$ such that $R/M$ is a torsion $B$-module; as $M_{[A]}$ is a direct sum of $r_B(R)$ $A$-modules isomorphic to $B$, by Proposition 16 (i)

Moreover, $c_A((R/M)_{[A]}) = -c_A(N(\chi_B(R/M)))$ by virtue of (19); but by definition of the homomorphism $N : C(B) \to C(A)$, $c_A(N(d)) = N(c_B(d))$ for all $d \in D(B)$ and, as $c_B(\chi(R/M)) = -c_B(R)$ by definition, finally

then it suffices to apply Proposition 16 (i) to obtain (20).