If an Artinian module $M$ is a sum of submodules of length $\le n$, then it is also of finite length

Question

Here is Lemma 6.4.10 in (Mostly) Commutative Algebra by Antoine Chambert-Loir:

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 proof there uses induction on $n$, which looks pretty good. But I want to see if there were any simpler way to prove it. Here are my attempts:

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Suppose that $M=\sum_{i\in I}{M_i}$, where each $M_i$ is an $R$-submodule of $M$ of length at most $n$. By dropping extra terms, we may assume that $M_i\not\subseteq\sum_{j\ne i}{M_j}$ for all $i\in I$.

We know that a finite sum of submodules of finite length also has finite length, so it suffices to show that $I$ is finite in this case: If $I$ is infinite, we can find an infinite sequence $(i_k)$ of distinct elements in $I$. Then $$\sum_{k=1}^\infty{M_{i_k}}\supset\sum_{k=2}^\infty{M_{i_k}}\supset\sum_{k=3}^\infty{M_{i_k}}\supset\cdots$$ is a strictly descending chain in $M$, contradicting the fact that $M$ is Artinian. Q.E.D.

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I wonder if my arguments above are valid. Any suggestions or corrections are highly welcomed!

Answer 1

As thinking about this question for several days, I realize that the proof presented above is false. The key problem is the assumption "$M_j\not\subseteq\sum_{j\ne i}{M_j}$."

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Counterexample. Consider the Prüfer group $\mathbb{Z}(p^\infty)$, which is Artinian but not Noetherian over $\mathbb{Z}$. The complete list of its subgroups is given by $$0<G_1:=(p^{-1}\mathbb{Z})/\mathbb{Z}<G_2:=(p^{-2}\mathbb{Z})/\mathbb{Z}<\cdots<\mathbb{Z}(p^\infty).$$ Now we have $\mathbb{Z}(p^\infty)=\sum_{n=1}^\infty{G_n}$. Suppose we want to restrict the index set $\mathbb{N}$ for the sum to some nonempty subset $I$.

• If $I$ is finite, then $\sum_{i\in I}{G_i}=G_k<\mathbb{Z}(p^\infty)$, where $k=\max(I)$. Such case is thus not possible.

• However, let $I\subseteq\mathbb{N}$ be infinite. Then for every $i\in I$, there always exists $j\in I$ such that $j>i$. Then $G_i<G_j\le\sum_{j\ne i}{G_j}$.

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Here I explain how I found such counterexample. I noticed that such argument does not rely on the fact that all $M_i$'s have length at most $n$, but only on the Artinian property of $M$. If it is valid, it should work for all Artinian modules.

However, I recalled that

Theorem. An $R$-module $M$ is finitely generated if and only if for every family $(M_i)_{i\in I}$ of submodules of $M$, $$M=\sum_{i\in I}{M_i}\implies\exists~\text{finite}~J\subseteq I:M=\sum_{i\in J}{M_i}.$$

Consequently, if the previous arguments are valid, we would see that every $R$-submodule of the Artinian module $M$ is also finitely generated, hence $M$ must be Noetherian then! Thus I searched for an example of Artinian but not Noetherian module and formed up the counterexample above.