Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $I = IB \cap A$.

share|improve this question
    
"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 [1], the proposition follows(for example by Matsumura). [1]: math.stackexchange.com/questions/158406/… –  Makoto Kato Jun 14 '12 at 20:53
add comment

1 Answer

up vote 4 down vote accepted

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 $a\in I$.

share|improve this answer
    
Thanks! This is very nice. Let me see if there are other proofs before I accept yours. –  Makoto Kato Jun 14 '12 at 21:54
    
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? –  awllower Jun 15 '12 at 7:44
    
@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
    
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... –  awllower Jun 18 '12 at 18:01
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.