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I have this problem:

Let $R$ a PID and $J\subseteq R$ a non-zero ideal. Prove that $R/J$ has a finite number of ideals.

I think that I must use the biyective correspondance between the ideals of $R$ that contain $J$ and the ideals of $R/J$, but I still don't see why I have a finite number of ideals that contain $J$. Can anyone give me a hint or I'm going in the wrong way?

iam_agf
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  • Also see the questions linked to the duplicate for more help. Also please search for your question before you post: you could find all of these just by searching for "PID finitely many ideals" and so on. – rschwieb Jan 25 '17 at 11:58
  • I tried to avoid duplicating a post, but I didn't try with that title to see. Now I see why it has finite divisors. – iam_agf Jan 25 '17 at 11:59
  • If you follow the links between questions, you'll hit this one which clarifies the application of the more general result. – rschwieb Jan 25 '17 at 12:01
  • Does exist any post about integral domains with prime finite ideals? I don't find it as posted before. – iam_agf Jan 25 '17 at 12:14
  • Isn't the link in my previous comment such a post? (the link again) – rschwieb Jan 25 '17 at 12:27

1 Answers1

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Since $R$ is a PID, any ideal in $R$ is principal. Let $J=(x)$ and let $I=(y)$ be another ideal. Then $J\subseteq I$ if and only if $y$ divides $x$.

Can you show that there are (up to multiplication by units) only finitely many elements of $R$ that divide $x$?

John Gowers
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