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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?

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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