If $A$ is Noetherian, the surjectivity of a homomorphism from $A$ to itself is necessarily injective.

Thus, it's clear that if $\phi: A \to A$ is surjective, but not injective, the ring in question would have to be non-Noetherian.

Reformulating the question using the first isomorphism theorem. Given $\phi:A \to A$, we know that $A/ \ker \phi \cong \textrm{im} \phi \cong A$. When is it the case that the kernel is nontrivial?

Related Question: If every surjective homomorphism $A \to A$ is an isomorphism, does it follow that $A$ is Noetherian?

  • $\begingroup$ Have you tried to use Search before posting? $\endgroup$ – user26857 Sep 19 '16 at 21:29
  • $\begingroup$ Indeed, I found one answer that gave the negative answer provided by Corvus, and another that supplied a proof for my first sentence. $\endgroup$ – Andres Mejia Sep 19 '16 at 21:59

Here is a large class of examples. A continuous map $f : X \to Y$ between compact Hausdorff spaces is injective iff the corresponding pullback map $f^{\ast} : C(Y) \to C(X)$ on spaces of continuous functions to $\mathbb{C}$ is surjective, and dually with injective and surjective switched; this is an aspect of the commutative Gelfand-Naimark theorem. So it suffices to find an endomorphism $f : X \to X$ of a compact Hausdorff space which is injective but not surjective. An easy example to visualize is an inclusion of a closed interval into a bigger closed interval.

  • $\begingroup$ Took a little bit of googling, but this is a really nice answer. As an aside, I enjoy a lot of your answers here and elsewhere. Thank you. $\endgroup$ – Andres Mejia Sep 22 '16 at 22:13

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.