I'm redoing some tasks and came across the following.

Let $M$ be an artinian $R$-module. Let $\phi : M \rightarrow M$ be an injective homomorphism. Show that $\phi$ is surjective.

My idea was that I might be able to show this by a short exact sequence (I already have a solution for this problem, but I would prefer this one a lot if it is possible to do it). Consider $0 \rightarrow M \stackrel{\phi}\rightarrow M \stackrel{\pi}\rightarrow M/im(\phi) \rightarrow 0$ with $\phi$ as given above from M to M and $\pi$ being the standard homomorphism from $M$ to $M/im(\phi)$. Then I know that $M \cong M \oplus M/im(\phi)$. Is there a way to conclude $M=im(\phi)$? If so I think this proof would be pretty elegant. Thank you in advance for your help!


It's for the same reason that an injective linear transformation of finite dimensional vector spaces is surjective. Look at the decreasing chain of submodules $\phi(M) \supset \phi^2(M) \supset \phi^3(M) \cdots$.

I don't know if the way you were trying it will work. Of course

$$0 \rightarrow M \xrightarrow{\phi} M \rightarrow M/\phi(M) \rightarrow 0$$

will split because $M/\phi(M)$ will be zero, but how can you show this before proving what you want?

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