# Lemma for the Krull-Akizuki Theorem

This is from Matsumura's Commutative Ring Theory (Lemma for Theorem 11.7)

Lemma for the Krull-Akizuki Theorem Let $A$ and $K$ be as in the theorem, and let $M$ be a torision-free $A$-module of rank $r < \infty$. Then for $0 \not= a \in A$ we have $$l(M/aM) \leq r \cdot l(A/aA).$$

Proof    I understand the proof for the finite case, but I'm having trouble with the infinite case. So we have, $$l(\sum A\bar\omega_i) \leq r \cdot l(A/aA)$$ and the right-hand side is independent of $\bar{N}$. So this means that any finitely generated submodule $\bar{N}$ has finite length. I'm not really sure how this implies that $\bar{M}$ is finitely generated.

Let $M$ be an $A$-module and $(M_i)$ the family of all finitely generated submodules of $M$. If $\sup_i l_A(M_i)<\infty$, then $M=M_i$ for some $i$.
Set $t= \sup_i l_A(M_i)$, and let $i_0$ be such that $t=l_A(M_{i_0})$. If $M\ne M_{i_0}$, then choose an element $x\in M-M_{i_0}$ and consider $M'=xA+M_{i_0}$ in order to get a contradiction.