If $M$ is either Noetherian or Artinian and $M^{(n)}\cong M^{(m)}$, then $m=n$.

Question

The question is:

Prove that if $_RM$ is either artinian or noetherian and if $m,n\in\mathbb{N}$ with $M^{(m)}\cong M^{(n)}$, then $m=n$. Here, $M^{(m)}$ denotes the direct sum of $m$ times $M$.

I had tried to solve using induction over $n$ and an isomorphism between $M^{(n)}/M$ and $M^{(n-1)}$, but I have many troubles with that.

There is a suggestion (and I can't imagine how to use): Use the next lemma

Let $M$ be a module and let $f$ be an endomorphism of $M$.

$(1)$ If $M$ is Artinian, then $\operatorname{Im}\, f^n + \ker\, f^n=n$ for some $n$, whence $f$ is an automorphism if and only if $f$ is monic;

$(2)$ If $M$ is Noetherian, then $\operatorname{Im}\, f^n\cap \ker\, f^n=0$ for some $n$, whence $f$ is an automorphism if and only if $f$ is epic.

Answer 1

If $M^{(m)} \cong M^{(n)}$ via the map $g$ and we assume $n<m$, then if $f: M^{(n)} \to M^{(m)}$ is an inclusion map, we have $g \circ f$ is an monomorphism of $M^{(n)}$. But then if $M$ and therefore $M^{(n)}$ is Artinian, this means that $g \circ f$ is an automorphism. But this is impossible because then $f$ is an isomorphism, and we assumed it was just an inclusion map.

Does this help?