Why is the Ideal Norm Multiplicative? I had asked this question before and got a partial answer.  Let $A$ be a Dedekind domain with quotient field $K$, $L$ a finite separable extension of $K$ of degree $n$, and $B$ the integral closure of $A$ in $L$.  The ideal norm $N_{L/K}$ is usually defined multiplicatively by the formula $N_{L/K}(\mathfrak P) = P^f$, where $P = \mathfrak P \cap A$ and $f = f(\mathfrak P / P)$.  
Another way to define it is as follows: for fractional ideals $M$ and $N$ of $B$, the localizations $M_P$ and $N_P$ are free $A_P$-modules, so there is a $K$-module automorphism $\psi$ of $L$ for which $\psi(M_P) = N_P$.  One then defines the module index $[M_P : N_P]_{A_P}$ to be $Det(\psi)A$.  More generally one defines the module index $[M : N]_A$ to be $$ \prod\limits_P P^{\nu_P [M_P : N_P]_{A_P}}$$ If $M$ is a fractional ideal of $B$, one can define the norm $N_{L/K}(M)$ to be $[B : M]_A$.  (see Cassels and Frohlich, first section of Algebraic Number Theory)
I have seen a proof using Smith Canonical Form that these definitions agree when $M$ is a prime ideal, but I have yet to see a simple general proof that the norm (defined in the latter way) is multiplicative.  The "long" way of doing this is to look at the completions of $K$ with respect to the $\nu_P$, and then look at the ways in which $\nu_P$ may be extended to $L$ (which correspond to the prime ideals lying over $P$).  The special case $A = \mathbb{Z}$ is more clear, because the module index $[B : M]_{\mathbb{Z}}$ is just the ideal generated by the cardinality of $B/M$.
 A: Let me expand on my comment:
1) You use the CRT to reduce to the case of a prime power:  let
$$\mathfrak{a}=\prod_{i=1}^r\mathfrak{p_i}^{e_i}$$
with $\mathfrak{p}_i\ne\mathfrak{p}_j$ when $i\ne j$. Since each $\mathfrak{p_i}+\mathfrak{p_j}=\mathcal{O}$, we have
$$\mathcal{O}/\mathfrak{a}\cong\bigoplus_{i=1}^r\mathcal{O}/\mathfrak{p}_i^{e_i}\;.$$
2) You use Krull dimension 1 to break down $\mathfrak{p}^{n+1}/\mathfrak{p}^n\cong\mathbb{F}_{N(\mathfrak{p})}$ to reduce to the case of prime ideals since you can measure the index of $[\mathcal{O}:\mathfrak{p}^e]$ using towers
$$\mathfrak{p}\stackrel{i_0}{\longrightarrow}\mathfrak{p}^2\stackrel{i_1}{\longrightarrow}\ldots\stackrel{i_e}{\longrightarrow}\mathfrak{p}^e$$
where each $i_k$ is inclusion.
3) From there you just need to measure the size of relative field extensions. If $\lambda/\kappa$ is the residue field extension for $L/K$, then $\lambda$ is a vector space over $\kappa$ as well as over the base field, $\mathbb{F}_p$. But then $[\lambda:\mathbb{F}_p]=[\lambda:\kappa][\kappa:\mathbb{F}_p]$.
