1
$\begingroup$

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

enter image description here enter image description here enter image description here enter image description here

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.