关于Artinian模有限长度简化证明的有效性验证请求

Hi everyone,

I'm working through Lemma 6.4.10 from Antoine Chambert-Loir's Mostly Commutative Algebra and wanted to check if a simplified proof I came up with holds water. First, here's the lemma for context:

Lemma 6.4.10. Let $A$ be a ring and let $n$ be a positive integer. Let $M$ be an artinian $A$-module. Assume that $M$ is the sum of its submodules of length $\le n$. Then $M$ has finite length.

The textbook proof uses induction on $n$, which is solid, but I wondered if there was a more direct approach. Here's my attempt:

Suppose $M = \sum_{i\in I} M_i$, where each $M_i$ is an $A$-submodule of $M$ with length at most $n$. First, I can remove redundant terms from the sum—meaning we can assume that for every $i\in I$, $M_i \not\subseteq \sum_{j\ne i} M_j$ (no single submodule is contained in the sum of the others).

We already know that a finite sum of submodules with finite length also has finite length, so the key is to show $I$ must be finite. Suppose for contradiction that $I$ is infinite. We can pick an infinite sequence of distinct elements $(i_k)$ from $I$, then consider this chain of submodules:
$$\sum_{k=1}^\infty M_{i_k} \supset \sum_{k=2}^\infty M_{i_k} \supset \sum_{k=3}^\infty M_{i_k} \supset \cdots$$
This is a strictly descending infinite chain, which directly contradicts the definition of an Artinian module (since Artinian modules cannot have infinite strictly descending chains of submodules). That would mean our assumption that $I$ is infinite is wrong, so $I$ is finite, and thus $M$ has finite length. Q.E.D.

I'm hoping to get some feedback—are there any gaps in this reasoning, or unstated assumptions I'm relying on? Any corrections or suggestions would be really appreciated!

备注：内容来源于stack exchange，提问作者Bernard Pan