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Suppose $K$ is a finite extension of $\mathbb Q$ and $\mathcal O_K$ is the set of the elements of $K$ which are integral over $\mathbb Z$. Now let $I$ and $J$ be two ideals of $\mathcal O_K$ and $IR=JR$ where $R$ is a non-zero ideal of $\mathcal O_K$. Now I want to show that $I=J$.

But I really don't know how to do that.

Can anyone give me some hints?

Thank you very much.

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    @molan im incorrect, $SR\neq\mathcal{O}_K$ but it is principal fractional ideal. im just repeating the very first comment (i was trying to fast-track you through a little piece of alg number theory and messed up)2011-12-16

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Generally an ideal $\rm\:I\:$ of a commutative ring is cancellable, i.e. $\rm\: I\:J = I\:K\ \Rightarrow\ J = K\:,\:$ iff $\rm\:I\:$ is locally regular principal, i.e. in every localization at a maximal ideal the ideal $\rm\:I\:$ becomes a principal ideal generated by a non-zero-divisor. For a proof see Anderson and Roitman, A Characterization of Cancellation Ideals, PAMS, 1997. Domains where all nonzero ideals $\rm\:I\ne 0\:$ are cancellable are known as almost Dedekind domains. Below are some well-known equivalent characterizations from Larsen and McCarthy, Multiplicative Theory of Ideals, 1971, who employ the equivalent definition that a domain is almost Dedekind if it is locally Dedekind, i.e. every localization at a a maximal ideal is Dedekind. Below, recall that a Prüfer domain is a domain where every finitely generated ideal $\ne 0$ is cancellable. Here are $\approx 30$ equivalent characterizations of Prüfer domains.

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