Let $K$ be a number field and let $R=\mathcal{O}_K$ be its ring of integers. Let $a_1, \ldots, a_h$ be ideals of $R$ generating the class group $Cl(K)$. Let $S$ denote the set of valuations corresponding to prime ideals dividing $a_1\cdots a_h$. Let $$R_S = \{a \in K \mid v(a)\geq 0,\ v \text{ non-Archimedean valuation on } K, \ v \notin S \}$$ Silverman (p.213 Arithmetic of Elliptic Curves) claims that $R_S$ has trivial class group. I’m sure this is a standard and simple fact, but I’m having trouble proving it.

We have an inclusion $R \to R_S$ and a map $Cl(R) \to Cl(R_S)$. Let $I$ be an ideal of $R_S$, we want to show it is principal. We have $I \cap R \equiv \prod a_i^{e_i}$ in the class group so there exists principal ideal $P,P’$ of $R$ such that $$P(I\cap R) = P’ \prod a_i^{e_i}$$ $$P(I\cap R)R_S = P’ \prod a_i^{e_i}R_S$$

Got stuck. Somehow the $a_iR_S$ become principal in $R_S$. I know $$R=\bigcap\limits_\mathfrak{p} R_\mathfrak{p}$$ $$R_S=\bigcap\limits_{\mathfrak{p} \notin S} R_\mathfrak{p}$$ Where $\mathfrak{p}$ ranges over all maximal ideals of $R$

  • $\begingroup$ @Watson thank you so much for the detailed answer! At first it was confusing the line J=xa_i (what is J, why only one a_i) but I figured out what you meant and basically the idea is the a_i become principal $\endgroup$ – usr0192 Nov 23 '18 at 22:39
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    $\begingroup$ Possibly related: math.stackexchange.com/questions/1144386 (recall that a Dedekind is a PID iff it is a UFD). $\endgroup$ – Watson Nov 27 '18 at 7:52

$ \newcommand{\O}{\mathcal O} \newcommand{\Cl}{\mathrm{Cl}} \newcommand{\p}{\mathfrak{p}} $ Let $K$ be a number field, and $a_1, ..., a_h$ be ideals of $\O_K$ which are representatives of the ideal classes of $\Cl(\O_K)$ (in particular, $h$ is the class number of $K$). Let $S$ be a finite set of places of $\O_K$, containing all the prime factors of all $a_i$, as well as all infinite places.

Claim : The ring $\O_{K, S}$ is a principal ideal domain.

Proof :

By proposition 23.2 b) and d), the ring $\O_{K, S}$ is a Dedekind domain, and every ideal of $\O_{K,S}$ is an extended ideal, i.e. of the form $J\O_{K,S}$ for some ideal $J \subset \O_K$.

By definition of the ideals $a_i$, there is an element $x \in K^{\times}$ and an index $i$ such that $$J = x a_i.$$ The ideal $a_i^h = y \O_K$ is principal (where $y \in \O_K$), since the class group is annihilated by $h$. One sees that $y \in a_i \cap \O_{K,S}^{\times}$ (because $a_i^h \subset a_i$ and the only primes dividing $y$ are in $S$), which implies that $a_i \O_{K,S} = (1)$ and finally $$J \O_{K,S} = x \O_{K,S} \subset \O_{K,S}$$ is a principal ideal, as claimed.


1) Instead of using proposition 23.2 above, one can first show that $\O_{K,S}$ is equal to a localization $\Sigma^{-1}\O_K$, namely for the multiplicative set $\Sigma$ of elements $a \in \O_K \setminus \{0\}$ such that the only prime factors of $a \O_K$ belong to $S$, that is $$\Sigma = \bigcap_{\p \not \in S} (\O_K \setminus \p) = \O_K \setminus \bigcup_{\p \not\in S} \p.$$ (There seems to be a typo on page 41 of this book. The "multiplicative" set they are considering doesn't even contain $1$).

Notice that we want to show the equality $$\left( \bigcap_{\p \not \in S} (\O_K \setminus \p) \right)^{-1} \O_K = \bigcap_{\p \not \in S} ((\O_K \setminus \p)^{-1} \O_K),$$ which is not obvious.

$$ $$ A) We clearly have $\Sigma^{-1} \O_K \subseteq \O_{K,S}$. The reverse inclusion is not obvious: it uses the fact that the class group of $\O_K$ is torsion (because finite) — see here. In fact the argument is similar as above: pick $x \in \O_{K,S}$. For every $\p \in S$ such that $v_{\p}(x) < 0$, write $\p^h = b_{\p} \O_K$ as an principal ideal. Then $b := \prod\limits_{v_{\p}(x) < 0} b_{\p}^{-v_{\p}(x)}$ is an element of $\O_K$ only divisible by primes in $S$, so that $b \in \Sigma$. Finally, $bx \in \O_K$, so we conclude that $x \in \Sigma^{-1} \O_K$ as desired.

[Here is a wrong proof that $\O_{K,S} \subset \Sigma^{-1} \O_K$ : if $x \in \O_{K,S}$ is non-zero, then fix $b_p \in p^{-v_p(x)}$ for every prime $p \in S$ such that $v_p(x)<0$. Then $b = \prod_p b_p \in \Sigma$ and $xb \in \O_K$, so that $x \in \Sigma^{-1} \O_K$. The problem is that $b$ may have prime factors outside $\{ \p \mid a_i \}$, so that $b \in \Sigma$ is not true].

See also here, or here about relation between overrings and localizations. Moreover, remark 2.1 here might be of interest (but there is no proof there). Finally, more general statements are given in theorem 2 here, mostly proven in Fossum, The Divisor Class Group of a Krull Domain.

$$ $$ B) Once you know that $\O_{K,S} \cong \Sigma^{-1}\O_K$, you can apply proposition 3.11.1) in Atiyah, MacDonald, Introduction to Commutative Algebra, to conclude that any ideal of $\O_{K,S}$ is indeed an extended ideal $J \O_{K,S}$, where $J$ is an ideal of $\O_K$. (Moreover, any localization of a Dedekind domain is still Dedekind, see here).

$$ $$ C) You can also look at prop. 4.19 here, but notice that they define $\O_{K,S}$ as a localization.

$$ $$ 2) Here we are considering the multiplicative set $\Sigma = \O_K \setminus \bigcup\limits_{\p \in \mathrm{Spec}(\O_K) \setminus S} \p.$ If instead you consider $\Sigma' = \O_K \setminus \p$ for a single fixed prime $\p$ of $\O_K$, you also get that $(\Sigma')^{-1}\O_K = (\O_K)_{\p}$ is a PID : it is even a DVR.

This is a result similar to the claim above, but which is in fact quite unrelated, since $\Sigma$ deals with the cofinite set $\mathrm{Spec}(\O_K) \setminus S$ of primes, while $\Sigma'$ deals with the finite set $\{\p\}$...

  • $\begingroup$ Further reference: proposition 5.3.6 in Heights in Diophantine Geometry. $\endgroup$ – Watson Dec 2 '18 at 9:59

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