# Ascending chain of ideals [duplicate]

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.

## marked as duplicate by rschwieb, Claude Leibovici, Davide Giraudo, Yiyuan Lee, Sujaan KunalanMar 23 '14 at 15:46

• Let $\mathfrak{a}_n = \ker f^n$. – Daniel Fischer Mar 22 '14 at 19:31
• That was the first closest duplicate I could find, but I'm pretty sure there are one or two closer ones. – rschwieb Mar 23 '14 at 14:25

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 $\ker f=0$?
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.
• Kernels of ring homomorphisms behave like kernels of linear maps in the sense that a ring homomorphism $f:R\to S$ is injective precisely if $\ker f=0$. (That follows from the corresponding fact for abelian groups.) I may not have addressed your answer at the right level because I was unsure of your background. Does that clear things up? – user134824 Mar 22 '14 at 20:13