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 $I = IB \cap A$.

  • $\begingroup$ "Embedding an ideal to an extension of an alegebraic number field" $\rightarrow$ "Embedding an ideal in an extension of an alegebraic number field" $\endgroup$ – Makoto Kato Jun 14 '12 at 8:25
  • $\begingroup$ Have you checked the book on commutative algebras by Atiyah? It seems tp be an example of extension and contractin of ideals. $\endgroup$ – awllower Jun 14 '12 at 11:07
  • $\begingroup$ @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. $\endgroup$ – Makoto Kato Jun 14 '12 at 12:08
  • $\begingroup$ 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. $\endgroup$ – awllower Jun 14 '12 at 12:13
  • $\begingroup$ @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]: math.stackexchange.com/questions/158406/… $\endgroup$ – Makoto Kato Jun 14 '12 at 20:53

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 $a\in I$.

  • $\begingroup$ Thanks! This is very nice. Let me see if there are other proofs before I accept yours. $\endgroup$ – Makoto Kato Jun 14 '12 at 21:54
  • $\begingroup$ 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? $\endgroup$ – awllower Jun 15 '12 at 7:44
  • $\begingroup$ @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. $\endgroup$ – user18119 Jun 17 '12 at 21:28
  • $\begingroup$ 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... $\endgroup$ – awllower Jun 18 '12 at 18:01

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