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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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    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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    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