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.
An example of an endomorphism
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0Dear @Pierre-Yves, thanks a lot: I really appreciate your kind comment. – 2012-03-16
2 Answers
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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2Dear Matt, +1 for your fine answer and even more for your honest admission, which most wouldn't have made. – 2012-03-16
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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2Dear 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