# How to make sense of the ideal norm in a localization?

This is a follow-up to this question: Localization of finite modules, or: compatibility of ideal norms with localization at a prime number

Let $K$ be an algebraic number field and $N:\mathcal{I}(\mathcal{O}_K)\to \mathbb{Z}$ be $N(\mathfrak{a})=\# (\mathcal{O}_K/\mathfrak{a})$, where $\mathcal{I}(\mathcal{O}_K)$ denotes the set of non-zero ideals of $\mathcal{O}_K$.

Let $S\subset \mathcal{O}_K$ be a multiplicative subset. Is there a way to make sense of a "norm of ideals" in $S^{-1}\mathcal{O}_K$ that is "compatible" with the norm of $\mathcal{O}_K$?

I'm especially interested in the case $S=\mathbb{Z}\setminus p \mathbb{Z}$, where $p$ is a prime number.

In the previous question the answer by froggie immediately shows that we can't naively define it (in this special case) as the cardinality of the quotient.

I know the question is vaguely phrased, but I seem to recall reading about such a thing, and I can't remember where.

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For any non-zero ideal $\mathfrak a$ decomposed into $\prod_i \mathfrak m_i^{r_i}$ product of maximal ideals, define $\mathfrak a_S$ by removing those $\mathfrak m_i$ which meet $S$ and consider $$N_S(\mathfrak a):=N(\mathfrak a_S).$$

Edit Some formal properties:

• $N_S(\mathfrak a \mathfrak b)=N_S(\mathfrak a)N_S(\mathfrak b)$;

• $N_S(\mathfrak a)=1$ if and only if $S\cap \mathfrak a\ne\emptyset$ if and only of $\mathfrak a S^{-1}O_K=S^{-1}O_K$;

• Let $L/K$ be a finite extension. Then $$N_T(\mathfrak c)=N_S(N_{O_L/O_K}(\mathfrak c))$$
where $T=S$ is considered as a multiplicative subset of $O_L$;

• If $S=\mathbb Z\setminus p\mathbb Z$, then for any $n\in \mathbb Z$, write $n=mp^r$ with $m\in\mathbb Z$ prime to $p$, we have $$N_S(nO_K)=m^{[K:\mathbb Q]}$$ and for any $\alpha\in O_K$, $N_S(\alpha O_K)$ is the prime-to-$p$ part of $N_{L/K}(\alpha)\in \mathbb Z$.

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