Lemma 3

Let $A$ be a Noetherian domain such that every non-zero prime ideal $\& A$ is maximal, $M$ a torsion-free $A$-module of finite rank $r$ and a non-zero element of $A$. Then $A/Aa$ is an $A$-module of finite length and:

Lemma 1 shows that $\operatorname{long}_A(A/Aa)$ is finite. We show (6) first in the case where $M$ is finitely generated. As $M$ is torsion-free and of rank $r$, there exists a submodule $L$ of $M$ which is isomorphic to $A'$ and such that $Q = M/L$ is a finitely generated torsion $A$-module and hence of finite length (Lemma 1). For every integer $n \geq 1$, the kernel of the canonical surjection $M/a^nM \to Q/a^nQ$ is equal to $(L + a^nM)/a^nM$ and isomorphic to $L/(a^nM \cap L)$; as therefore Now, since $M$ is torsion-free, multiplication by $a$ defines an isomorphism of

M/aM onto aA/a^2M; similarly for L; whence, by induction on n, the formulae:

Taking account of (7) we deduce:

for all $n > 0$; as L is isomorphic to A', $\text{long}_A(L/aL) = r \text{long}_A(A/Aa)$; whence (6) by letting $n$ tend to infinity in (9).

We now pass to the general case. Let $(M_i)$ be the family of finitely generated submodules of M. The module $T = M/aM$ is the union of the submodules $T_i = (M_i + aM)/aM = M_i/(M_i \cap aM)$. Now, T, is isomorphic to a quotient of $M_i/aM_i$ and hence

by what we have just proved. Whence

by Lemma 2.