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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!

user26857
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eu271828
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1 Answers1

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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?

D_S
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