Norm of ideals is multiplicative

Question

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!

Answer 1

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.

1. 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 $x\in \mathfrak{p}$.

2. 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 $R$.