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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.

Brian Fitzpatrick
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user119065
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1 Answers1

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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.

user134824
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  • Am I correct if I say that these kernels behave much like that of linear transformations ? – user119065 Mar 22 '14 at 20:02
  • 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
  • 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 ? – user119065 Mar 22 '14 at 20:16
  • That should do it. You'll find the injectivity hypothesis very useful. Let me know if you hit a snag. – user134824 Mar 22 '14 at 20:24
  • 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. – user119065 Mar 22 '14 at 20:40
  • Pick $r\in R$. We want to show $r$ is in the image of $f$, right? A natural ideal to pick is $(r)$. By analogy with the solution to the previous problem, see what happens when you iterate $f$ on $(r)$. – user134824 Mar 22 '14 at 20:43
  • Their images have the descending chain property. So for some n f^n(x) = r so f(y) = r where y is in R. Thus f is surjective. Is this correct ? – user119065 Mar 22 '14 at 21:02
  • Almost. All we can conclude from the descending chain property is that for some $n$, $f^n((a))=f^{n+1}((a))=\cdots$. But this is enough. Do you see why? – user134824 Mar 22 '14 at 21:11
  • Due to injective property of f I suppose ? – user119065 Mar 22 '14 at 21:14
  • We want to show that $a$ has a preimage under $f$, and we know that $f^n(a)=f^{n+1}(r)$ for some $r\in R$ since $f^n$ and $f^{n+1}$ have the same image. Now use injectivity. – user134824 Mar 22 '14 at 21:24
  • Thanks man! Cleared all my doubt. You are great ! – user119065 Mar 22 '14 at 21:32
  • If you're interested, rings that satisfy the ascending chain condition are called Noetherian. Noetherian rings are very important in commutative algebra -- they're "finite-dimensional" in a certain sense and have many nice properties. Rings that satisfy the descending chain condition are called Artinian. It turns out that every Artinian ring is Noetherian, although this isn't obvious from the definitions. – user134824 Mar 23 '14 at 00:27