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I am having a conceptual problem with the first isomomorphism theorem, probaly the question is silly, but I don't know how to precede.

Supose that we have an homomorphism of Modules $f : M \to M$, then by the isomorphism theorem, $M/kerf$ is isomorphic to $f(M)$ so, if $f$ is injective, $M$ is isomorphic to $f(M)$ and then $f$ is surjective. But this is not true for every module, (in fact the question was motivated by the problem to show that this is tru for artinian modules), for example, if we consider $\mathbb{Z}$ as a $\mathbb{Z}$-module, $f:$ $\mathbb{Z} \to $ $\mathbb{Z}$ such that $f(a) = 2a$ is monomorphism but not surjective.

I don't see what I am missing here. I'd appreciate any help.

Thank you

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$f(M)\neq M$ necessarily (though there is an isomorphism), so $f$ is not surjective necessarily.

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  • $\begingroup$ Ah than you so much! I see it now! The homomorphism in the question is not a homorphism of rings. Thank you again! $\endgroup$ – User43029 Feb 2 '15 at 15:47
  • $\begingroup$ you are welcome $\endgroup$ – user 1 Feb 2 '15 at 15:51
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The problem is, that you have only $M \cong f(M)$ and you want to deduce $M = f(M)$ (as subsets of $M$). This is not valid.

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