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Let $R$ be a commutative integral domain, $I,J,K$ three ideals of $R$ with $I\neq (0)$ being finitely generated. Then does $IJ=IK$ imply $J=K$?

With Nakayama lemma, I can prove it if one of $J$ and $K$ equals to $R$. And I also know it holds when $R$ is a Prüfer domain or $I$ is singly generated.

Censi LI
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  • Maybe I did not fully understand your statement but $IJ=IK$ always (for any $J$ and $K$) holds whenever $I={0}$ ? – Clément Guérin Jun 03 '15 at 07:28
  • @ClémentGuérin Ah, I missed this case, now it has been fixed. Thank... – Censi LI Jun 03 '15 at 07:30
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    The example posted in the accepted answer is well known. Much more interesting is to find such an example in $K[X,Y]$ which is integrally closed but not Prufer (or in any other integrally closed domain which isn't Prufer). Unfortunately I don't have one. – user26857 Jun 03 '15 at 09:40
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    I forgot to mention the following result: an integral domain $R$ is Prufer iff $R$ is integrally closed and satisfies the cancellation property. (This shows why it's more interesting to impose $R$ integrally closed when look for an example as you wanted.) – user26857 Jun 03 '15 at 09:48
  • @user26857 Thank you very much, honestly I think what's in your comment is more interesting than what I posted. – Censi LI Jun 03 '15 at 09:54

2 Answers2

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I think taking $R:=F[X,Y]$, $I:=(X,Y)$, $J:=(X^2,Y^2)$ and $K:=(X^2,XY,Y^2)$ might be a more interesting counterexample, as user26857 noted in the comment.

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It doesn't always hold. For example, Let $R:=\mathbb Z[\sqrt{-3}]$, take $I:=(2,1+\sqrt{-3})$, $J:=(2)$ and $K:=(1+\sqrt{-3})$, then $IJ=IK$.