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Could someone suggest a simple $\phi\in $End$_R(A)$ where $A$ is a finitely generated module over ring $R$ where $\phi$ is injective but not surjective? I have a hunch that it exists but I can't construct an explicit example. Thanks.

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    It doesn't have to exist for every ring and every module.2012-03-16
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    They also had a hunch back in Victor Hugo's time: it is to take $R=A=\mathbb Z$ and $\phi(z)=2z$.2012-03-16
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    I am very sorry, I have forgotten to include the condition that $A$ has to be finitely generated.2012-03-16
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    But Georges' $A$ is.2012-03-16
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    Don't worry, Teenager, it was quasi modo implicit.2012-03-16
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    @GeorgesElencwajg: Thanks!2012-03-16
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    @ymar: indeed :) I just thought I should point out my edit2012-03-16
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    Dear @Georges: $+1$ for your mathematical comment, and $+$ the power of the continuum for your puns!2012-03-16
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    Dear @Pierre-Yves, thanks a lot: I really appreciate your kind comment.2012-03-16

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Let $R=K$ be a field, and let $A=K[x]$ be the polynomial ring in one variable over $K$ (with the module structure coming from multiplication). Then let $\phi(f)=xf$. It is injective, but has image $xK[x]\ne K[x]$.

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    Thank you! I am very sorry, I have forgotten to include the condition that $A$ has to be finitely generated.2012-03-16
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    No problem! I would have guessed that there wasn't one if $A$ was finitely generated, so thanks to Georges for setting me straight on that!2012-03-16
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    Dear Matt, +1 for your fine answer and even more for your honest admission, which most wouldn't have made.2012-03-16
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Consider the morphism of $\mathbb{R}$-modules:

$$ \varphi : \mathbb{R}^\infty \longrightarrow \mathbb{R}^\infty $$

defined by

$$ \varphi (x_1, x_2, \dots , x_n, \dots ) = (0, x_1, x_2 , \dots , x_n , \dots ) \ . $$

This example is not possible with finite dimension vector spaces, because then, with endomorphisms, you have

$$ \text{isomorphism} \quad \Longleftrightarrow \quad \text{monomorphism} \quad \Longleftrightarrow \quad \text{epimorphism} \ . $$

EDIT. Now I see you've added the finitely generated condition. So, this example doesn't apply any more obviously.

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    Thank you, Agusti. I am very sorry about the edit.2012-03-16
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    Don't worry. But you've got your example in the comment by Georges Elencwajg. (Maybe he should make it an answer, so you could chose as your accepted one.)2012-03-16
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    Dear Agustí: +1 since you perfectly answered the original question. And thanks but no, I won't write an answer. I only made the comment because I couldn't resist the temptation to make a pun in English (which isn't my mother tongue) !2012-03-16