Noetherian module does not contain a submodule $N$ which is a direct sum of $n$ simple modules

Question

The question:

Let $R$ be a ring and $M$ be a Noetherian module. Prove there is $k \in \mathbb{N}$ such for all $n > k$, $M$ does not contain a submodule $N$ which is a direct sum of $n$ simple modules.

My thoughts: Since M is Noetherian, this means any submodule and factor module of M is Noetherian. I considered a submodule, $N$. Since N is noetherian, N satisfies the ascending chain condition. Is this where I pick my k? I am not sure about other properties on Noetherian modules.

Answer 1

Here's a sketch of proof (to be honest I thought it would be simpler):

(i) There's a finite number of isomorphism classes of simple modules that appear as submodules of $M$ (because $M$ is noetherian)

(ii) Given a simple module $N$, the natural numbers $n$ such that $N^n$ is (isomorphic to) a submodule of $M$ are bounded : this is the "hard" bit : consider an increasing sequence $n_i$ of such $n$'s, let $N^{n_i}\simeq L_i \subset M$ and consider $\displaystyle\sum_i L_i$, which is finitely generated, therefore there is a finite amount of $L_i$'s such that any $L_j$ is in their sum. Take $n_j > \sum_i n_i$ where the sum runs over this finite amount of $i$'s

Now by induction prove that any finite sum of submodules of $M$ isomorphic to $N$ is isomorphic to a single $N^p$ for some $p$ smaller than the number of submodules; thus for this $j$ we have an injection $N^{n_j}\to N^p$ for some $p<n_j$; and that's not possible (indeed it is represented by an $n_j\times p$ matrix over the division ring $\hom_R(N,N)$ and such a matrix necessarily has a nonzero vector $(f_1,...,f_{n_j})$ in its kernel by dimension theory, thus if you apply it to some $n\in N$ such that $(f_1(n),...,f_{n_j}(n))\neq 0$ you get a nonzero element of the kernel of $N^{n_j}\to N^p$)

(iii) You can then find $k$ to be the sum over the finite number of isomorphism classes of simple modules of the bound given by (ii)

Answer 2

Let $N$ be the sum of all simple submodules of $M$. Since $M$ is Noetherian, $N$ is finitely generated, so it is in fact the sum of finitely many simple submodules $S_1,\dots,S_k$. It follows that $N$ is a module of finite length (specifically, length at most $k$). Now any submodule of $M$ which is a direct sum of $n$ simple modules is a submodule of $N$ of length $n$, so we must have $n\leq k$.