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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.

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

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

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  • $\begingroup$ Am I correct if I say that these kernels behave much like that of linear transformations ? $\endgroup$ – user119065 Mar 22 '14 at 20:02
  • $\begingroup$ 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? $\endgroup$ – user134824 Mar 22 '14 at 20:13
  • $\begingroup$ Yep it does. And thanks for answering. For the dual part, I was thinking of considering any r in R then I need to create an ideal which has decreasing inclusion and thus will terminate giving an x such that f(x) = r. Am I on the correct path ? $\endgroup$ – user119065 Mar 22 '14 at 20:16
  • $\begingroup$ That should do it. You'll find the injectivity hypothesis very useful. Let me know if you hit a snag. $\endgroup$ – user134824 Mar 22 '14 at 20:24
  • $\begingroup$ I'm not being able to think of that ideal. The only thing that is popping in my head is the map f(x) - r, but its not even a homomorphism. $\endgroup$ – user119065 Mar 22 '14 at 20:40

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