Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 1 down vote accepted

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 $N_{L/K}(\alpha)\in \mathbb Z$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.