# Embedding an ideal to an extension of an algebraic number field

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 $I = IB \cap A$.

• "Embedding an ideal to an extension of an alegebraic number field" $\rightarrow$ "Embedding an ideal in an extension of an alegebraic number field" – Makoto Kato Jun 14 '12 at 8:25
• Have you checked the book on commutative algebras by Atiyah? It seems tp be an example of extension and contractin of ideals. – awllower Jun 14 '12 at 11:07
• @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. – Makoto Kato Jun 14 '12 at 12:08
• 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. – awllower Jun 14 '12 at 12:13
• @awllower You are right. I found a proof using the localizations. Since B is faithfully flat over A by , the proposition follows(for example by Matsumura). : math.stackexchange.com/questions/158406/… – Makoto Kato Jun 14 '12 at 20:53

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 $a\in I$.
• @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. – user18119 Jun 17 '12 at 21:28