Norm of ideals is multiplicative I have 2 questions:
1.) Why do we have, for a prime ideal $ 0 \not= \mathcal p \subset \mathcal O $, that $\mathcal p^n / p^{n+1} \cong \mathcal O/\mathcal p $ (as groups) ?
2.) And why does follow from this that the norm of ideals is multiplicative?
Thanks for any advice!
 A: Suppose that $R$ is a Dedekind ring. We define the norm of a nonzero ideal $\mathfrak{a}\subseteq R$ as follows 
$$N(\mathfrak{a}) = |R/\mathfrak{a}|$$
If $R$ is a number ring of some finite extension of $\mathbb{Q}$, then this norm is always finite. Let $\mathfrak{a},\mathfrak{b}\subseteq R$ be nonzero ideals of $R$. We have decompositions 
$$\mathfrak{a} = \prod_{\mathfrak{p}}\mathfrak{p}^{\alpha(\mathfrak{p})},\,\mathfrak{b} = \prod_{\mathfrak{p}}\mathfrak{p}^{\beta(\mathfrak{p})}$$
into products of prime ideals (the decompositions are essentially finite as usual). We have
$$\mathfrak{a}\cdot \mathfrak{b} = \prod_{\mathfrak{p}}\mathfrak{p}^{\alpha(\mathfrak{p}) + \beta(\mathfrak{p})}$$
Now by abstract form of Chinese remainder theorem 
$$R /(\mathfrak{a}\cdot \mathfrak{b}) = \prod_{\mathfrak{p}}R/\mathfrak{p}^{\alpha(\mathfrak{p}) + \beta(\mathfrak{p})}$$
and thus
$$N(\mathfrak{a}\cdot \mathfrak{b}) = \prod_{\mathfrak{p}}|R/\mathfrak{p}^{\alpha(\mathfrak{p}) + \beta(\mathfrak{p})}|$$
In order to prove that the norm is multiplicative it suffices to check that for a single prime ideal $\mathfrak{p}$ and nonnegative integer $k$ we have $|R/\mathfrak{p}^k| = |R/\mathfrak{p}|^k$. Indeed, if this is the case then
$$|R/\mathfrak{p}^{\alpha + \beta}| = |R/\mathfrak{p}|^{\alpha +\beta} = |R/\mathfrak{p}|^{\alpha}\cdot |R/\mathfrak{p}|^{\beta} = |R/\mathfrak{p}^{\alpha}|\cdot |R/\mathfrak{p}^{\beta}|$$
So let us prove this formula. For each $k$ we have composition series
$$\mathfrak{p}^{k} \subseteq \mathfrak{p}^{k-1} \subseteq ...\subseteq \mathfrak{p}^2 \subseteq \mathfrak{p} \subseteq R$$
Hence the cardinality of $R/\mathfrak{p}^k$ is the product of cardinalities of factors of this composition series. Thus 
$$|R/\mathfrak{p}^k|= |R/\mathfrak{p}|\cdot |\mathfrak{p}/\mathfrak{p}^2|\cdot ...\cdot |\mathfrak{p}^{k-1}/\mathfrak{p}^k|$$
Now if we know that $\mathfrak{p}^{n}/\mathfrak{p}^{n+1} \cong R/\mathfrak{p}$ for all $n$, then also
$$|\mathfrak{p}^{n}/\mathfrak{p}^{n+1}| = |R/\mathfrak{p}|$$
and hence $|R/\mathfrak{p}^k| = |R/\mathfrak{p}|^k$ for each $k$. Therefore, the multiplicativity of norms reduces to isomorphism
$$\mathfrak{p}^{n}/\mathfrak{p}^{n+1} \cong R/\mathfrak{p}$$
for positive integer $n$. Now this can be proven by picking $p\in \mathfrak{p}\setminus \mathfrak{p}^2$. Then we define
$$R/\mathfrak{p} \ni x\,\mathrm{mod}\,\mathfrak{p} \mapsto p^nx\,\mathrm{mod}\,\mathfrak{p}^{n+1} \in \mathfrak{p}^n/\mathfrak{p}^{n+1}$$
and this is an isomorphism. 


*

*For injectivity note that $p^nx \in \mathfrak{p}^{n+1}$ implies that 
$$(p^n)\cdot (x) \subseteq \mathfrak{p}^{n+1}$$ 
and now looking at prime factorization of these ideals and using the assumption that $p\not \in \mathfrak{p}^2$ you get that $x\in \mathfrak{p}$.

*For surjectivity we have $\mathfrak{p}^{n+1} \subsetneq(p^n)+ \mathfrak{p}^{n+1}\subseteq \mathfrak{p}^n$ which implies that 
$(p^n)+ \mathfrak{p}^{n+1}= \mathfrak{p}^n$ by uniqueness of decomposition into prime ideals in $R$. 
