Let $R$ be a commutative ring. Let $M_1$ and $M_2$ be two indecomposable $R$-modules with $\ell(M_1)=\ell(M_2)$, where >$\ell$ denotes the length of a module.
Let $f: M_1 \rightarrow M_2$ be a non bijective homomorphism of $R$-modules.
Is it true that $f$ can't be surjective?
Best Answer
Suppose $f$ is onto. We have $$0\rightarrow \operatorname{Ker}f\rightarrow M_1\rightarrow \operatorname{M_2}\rightarrow 0$$ is a short exact sequence. Since $\ell$ is additive, we get $\ell(\operatorname{Ker}f)0\implies \operatorname{Ker}f=0$
So $f$ is injective as well.
Note: I am assuming that in your question both $M_1$ and $M_2$ have finite length.