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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?

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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. – 2017-01-25
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    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. – 2017-01-25
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    If you follow the links between questions, you'll hit [this one](http://math.stackexchange.com/q/2055846/29335) which clarifies the application of the more general result. – 2017-01-25
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    Does exist any post about integral domains with prime finite ideals? I don't find it as posted before. – 2017-01-25
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    Isn't the link in my previous comment such a post? ([the link again](http://math.stackexchange.com/q/2055846/29335)) – 2017-01-25

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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 $x$?

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    The PIDs are artinian? – 2017-01-25
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    @MonsieurGalois A domain that is Artinian is a field. So **no**, PIDs are typically not Artinian. – 2017-01-25
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    Sorry, now I see why it has finite divisors. I didn't see that a PID is a UFD too. So the factorization has finite non-unit divisors, implying what you said. – 2017-01-25