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I'm looking for a proof of the following well-known proposition. I checked some books on algebraic number theory but could not find it.

Proposition Let $L$ be a finite extension of an algebraic number field $K$. Let $A$ and $B$ be the rings of integers in $K$ and $L$ respectively. Let $I$ be an ideal of $A$. Then $I = IB \cap A$.

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    "Embedding an ideal to an extension of an alegebraic number field" $\rightarrow$ "Embedding an ideal in an extension of an alegebraic number field"2012-06-14
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    Have you checked the book on commutative algebras by Atiyah? It seems tp be an example of extension and contractin of ideals.2012-06-14
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    @awllower I just checked Atiyah & MacDonald and I don't think they have the proof. The proposition is about an extension of a Dedekind domain. They don't treat that.2012-06-14
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    Indeed the two subjects are different; however some similarities are still shared by them, right? I mean per chance one could find some similar proof to that one, for the localizations.2012-06-14
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    @awllower You are right. I found a proof using the localizations. Since B is faithfully flat over A by [1], the proposition follows(for example by Matsumura). [1]: http://math.stackexchange.com/questions/158406/is-the-ring-of-integers-in-a-relative-algebraic-number-field-faithfully-flat-ove2012-06-14
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    Related: https://math.stackexchange.com/questions/63828, http://math.stackexchange.com/questions/3853642017-02-10

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You can also prove the equality directly using properties of Dedekind domains.

Let $a\in IB\cap A$. For any maximal ideal $\mathfrak p$ of $A$ and for any maximal ideal $\mathfrak q$ of $B$ lying over $\mathfrak p$, we have $$ v_{\mathfrak p}(a)=v_{\mathfrak q}(a)/e_{\mathfrak q/\mathfrak p}\ge v_{\mathfrak q}(IB)/e_{\mathfrak q/\mathfrak p}=v_{\mathfrak p}(I).$$ So $a\in I$.

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    Thanks! This is very nice. Let me see if there are other proofs before I accept yours.2012-06-14
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    Sorry, I am somewhat unfamiliar with the valuations recently, but I think I got what you try to express. Per chance one could prove this by [Lemma3.18](http://books.google.com.tw/books?id=RT5R_29X69wC&pg=PA14&dq=let+R+be+any+integral+domain+and+two+ideals&hl=zh-TW&sa=X&ei=hOfaT5L4KauiiAeHkc3BCg&ved=0CD0Q6AEwAQ#v=onepage&q=let%20R%20be%20any%20integral%20domain%20and%20two%20ideals&f=false)?2012-06-15
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    @awllower: the statement that $IB\cap A=I$ in the general situation requires the faithful flatness of $A\to B$ which is true for Dedekind domains. So I don't think the Lemma you refer to is enough.2012-06-17
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    Ah, I see why I thought of that lemma as useful: I mixed the concepts between the intersection and the contraction in this case... Sorry for that. No wander the statement appeared a little immediate to me...2012-06-18