0
$\begingroup$

Given $R$ a ring (commutative and unitary) and $M$ a Noetherian $R$-module, does there exist an endomorphism $f\in End(M)$ which is injective but not surjective?

I already prooved that if $f$ is surjective then it has to be injective (by taking the chain of $Ker(f)\subseteq Ker(f^2) \subseteq ...$ and using the ascending chain condition) but I'm stuck with this part...

$\endgroup$
1
$\begingroup$

Plenty. The endomorphism ring of $R$ as a module over itself is isomorphic to $R$; any nonzero divisor induces an injective endomorphism by multiplication. Is a nonzero divisor necessarily invertible?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.