Let $I\subseteq B$ be an ideal, we define the ideal norm of $I$ as the ideal in $A$ generated by the elements $N_{E/K}(\alpha)$ where $\alpha \in I.$ We denote it by $N_{E/K}(I).$
If $\mathfrak{p}$ is a prime ideal of $A$, then $A_{\mathfrak{p}}$ is a DVR. I want to see that the norm is multiplicative.
I know I have to use some kind of theorem that reduces to local cases, but I don't know how to manage this.
Of course if $I,J$ are principal ideals, then $N_{E/K}(IJ)=N_{E/K}(I)N_{E/K}(J)$ because of the multiplicity of the "number" norm, but, why this is happening when there are no principal ideals?
Thanks and sorry for my english.
By the Chinese remainder theorem it is enough to consider ideals of the form $I = P^n$ for prime ideals $P$. We show that $N(P^n)=N(P)^n$ by induction. The case $n=1$ is clear. Assume now that it holds for $n$. We want to show it for $n+1$. The map $A_P/P^{n+1}\rightarrow A_P/P^n$ is surjective with kernel $P^n/P^{n+1}$, which is a $1$-dimensional $A_P/P$-vector space having $N(P)=|A_P/P|$ elements. It follows that $$ N(P^{n+1})=N(P^n)N(P)=N(P)^{n+1}. $$