Ascending chain of ideals

Question

Let $R$ be a commutative ring with identity such that every ascending chain of ideals terminate. Let $f:R \to R$ be a surjective homomorphism. Prove that it is an isomorphism.

Answer 1

Expanding on Daniel Fischer's comment, since $f$ is surjective all we need to show is injectivity, and the way to do that is to show that $\ker f=0$. How can we use the ascending chain condition? One way is to consider the kernels of $f^2, f^3, \dots$, which form an ascending chain (why?): $$ \ker(f) \subseteq \ker(f^2) \subseteq \ker(f^3) \subseteq\cdots $$ What can we conclude about this chain? And how might we use that conclusion to show that $\ker f=0$?

Incidentally, this proposition has a dual statement which I would encourage you to prove as an exercise.

Let $R$ be a commutative ring with identity such that every descending chain of ideals terminates. Let $f : R \to R$ be an injective homomorphism. Prove that $f$ is an isomorphism.