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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.

molan
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  • How much do you know? If you know the definition of the class group and the fact that it is finite, you can choose an ideal in the inverse class of R and multiply both sides of IR=JR by that ideal. The result is $I(\alpha)=J(\alpha)$ for some principal ideal $(\alpha)$. You can finish up by showing the sets in $K$ obtained by multiplying both sides by $1/\alpha$ are just $I$ and $J$. – Barry Smith Dec 16 '11 at 13:00
  • I just know little about algebraic number theory. Do you know some proof which just needs definition and theorem from abstract algebra? Thank you. – molan Dec 16 '11 at 13:05
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    Doesn't this just follow (almost immedietly) from unique factorization of ideals? @BarrySmith – Fredrik Meyer Dec 16 '11 at 13:07
  • @FredrikMeyer Yes, I found the theorem about this in some books. But I didn't learn algebraic number theory, so I want to prove it by the definiton of ideal. – molan Dec 16 '11 at 13:10
  • @Fredrik: I didn't think I could assume (apparently correctly) that molan had much algebraic number theory background. And I generally think of the result I stated as a precursor to (i.e. can be used in the proof of) the Unique Factorization Theorem – Barry Smith Dec 16 '11 at 13:31
  • @BarrySmith Thank you very much. According to your hints, I find something of Dedekind domain may be help. Although I still don't know the detail of it, but I find some books about that, and I will try to understand that. Anyway, thank you very much. – molan Dec 16 '11 at 13:39
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    I was about to say: the most beautiful and profound way to understand why your question is true is to learn a little about Dedekind domains --- you won't regret it! I recommend Ireland and Rosen's book "A Classical Introduction to Modern Number Theory" for a friendly introduction. You'll see your statement proved as Proposition 12.2.6. – Barry Smith Dec 16 '11 at 13:54
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    @molan You cannot prove it simply "by the definition of ideal" since it is not true generally, even after adding the necessary hypothesis that $\ I,J \ne (0)$. – Bill Dubuque Dec 16 '11 at 14:11
  • @BarrySmith Thank you so much. And I will see it later. – molan Dec 16 '11 at 14:14
  • @BillDubuque Yes, you are right. I firstly want to show this by the hypothesis and don't use some properties from algebraic number theory. – molan Dec 16 '11 at 14:17
  • Let $S={a\in K\ | \ aR\subset\mathcal{O}_K}$. then $SR=\mathcal{O}_K$ and $I=IRS=JRS=J$. – yoyo Dec 16 '11 at 14:39
  • @yoyo How to show that $\mathcal O_K\subset SR$? – molan Dec 16 '11 at 14:58
  • @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) – yoyo Dec 16 '11 at 15:33

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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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