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 $I = IB \cap A$.

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 $a\in I$. 
