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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. –  user23211 Mar 16 '12 at 11:50
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They also had a hunch back in Victor Hugo's time: it is to take $R=A=\mathbb Z$ and $\phi(z)=2z$. –  Georges Elencwajg Mar 16 '12 at 11:50
    
I am very sorry, I have forgotten to include the condition that $A$ has to be finitely generated. –  Teenager Mar 16 '12 at 11:53
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But Georges' $A$ is. –  user23211 Mar 16 '12 at 11:55
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Don't worry, Teenager, it was quasi modo implicit. –  Georges Elencwajg Mar 16 '12 at 11:56

2 Answers 2

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. –  Teenager Mar 16 '12 at 11:54
    
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! –  Matt Pressland Mar 16 '12 at 12:01
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Dear Matt, +1 for your fine answer and even more for your honest admission, which most wouldn't have made. –  Georges Elencwajg Mar 16 '12 at 12:38

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. –  Teenager Mar 16 '12 at 11:58
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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.) –  a.r. Mar 16 '12 at 12:01
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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) ! –  Georges Elencwajg Mar 16 '12 at 12:48

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