Power of simple object cannot be a proper subobject of itself

Question

I am trying to prove that in every abelian category and for every simple object $A$, for distinct natural numbers $m\neq n$ we have $A^m \ncong A^n$. I am not completely sure whether this is true for all abelian categories; or whether this might even hold for more general objects $A$ (for instance, if it holds for simple $A$, it should also hold for non-zero, semisimple $A$ composed of finitely many direct summands). But let's just assume that $A$ is simple, if this helps. And if assumptions have to be introduced on the category, feel free to do so.

The best instrument available seems to be the fact that $A^n \hookrightarrow A^m$ is a proper subobject for $n <m$ (in many different ways). It therefore suffices to show that $A^n$ cannot be a proper subobject of itself. If it is, we have a short exact sequence $$ 0 \to A^n \overset{f}{\to} A^n \to B \to 0$$ for some $B \cong A^n / f(A^n) \neq 0$. This might not bring us much further... But maybe there is a way to use that $A$ is simple, somewhere?

If we had, for instance, a faithful, exact functor to the category of finite dimensional vector spaces, we could apply it to the exact sequence to get a contradiction (exact implies that the sequence stays exact; faithful implies that $B$ is not sent to $0$). But I don't expect every abelian category to have such a functor.

Remark: maybe one could just use that $\mathrm{End}(A)$ is a division ring, $\mathrm{Hom}(A^n,A^m) \cong \mathrm{End}(A)^{m \times n}$, and non-square matrices cannot be invertible, hence $A^m \ncong A^n$. But this really sounds a bit sloppy, and I am convinced that there must be a more elegant argument.

Answer 1

Suppose $i:A^n\to A^n$ were monic but not epic. Then $i,i^2,i^3,\dots$ would be an infinite strictly descending sequence of subobjects of $A^n$. But $A$ is simple, and in particular Artinian, so $A^n$ is also Artinian, so this is impossible. (More generally, this argument shows an Artinian object cannot be a proper subobject of itself, and a Noetherian object cannot be a proper quotient of itself.)

Separately, let me remark that a much more general statement is true: the Jordan-Hölder theorem is true in any abelian category. That is, if an object has a composition series (a finite sequence of subobjects where the successive quotients are all simple), then any two composition series have the same length and the same simple quotients up to a permutation. The proof is essentially the same as for modules.

Answer 2

How about $\newcommand\Hom{\operatorname{Hom}}\newcommand\End{\operatorname{End}}\Hom(A,A^m)\cong \End(A)^m$ and $\Hom(A,A^n)\cong \End(A)^n$, so as long as $\End(A)$ has invariant basis number (IBN), then the claim holds. However, $\End(A)$ having IBN is essentially the same as non-square matrices over $\End(A)$ not being invertible, so this is essentially the same argument as your remark. (Note that in particular, if $A$ is simple, it is a division ring and division rings have IBN).

That said, I'm skeptical that a proof that doesn't somehow rely on division rings having invariant basis number exists, since the proof above goes backwards.

That is, if $\End(A)$ doesn't have IBN, then $\End(A)^n\cong \End(A)^m$ for some $n\ne m$, and matrices $B\in M^{n\times m}(\End(A))$ and $C\in M^{m\times n}(\End(A))$ such that $BC=I$ and $CB=I$. Then $B$ and $C$ also induce inverse isomorphisms between $A^n$ and $A^m$.