I'm having a hard time understanding some notions of a paper I'm working on. Let $L/K$ be a finite normal extension of number fields and $S$ be a set of places of $K$ prime to $p$ where $p$ denotes an odd prime. The author defines a group

$$R_{K,S} = \{x \in \mathbb{Z}_p \otimes E_K, x \equiv 1 \mod \nu, \nu \in S\}$$ where $E_K$ denotes the groups of units of $K$. Several questions arise:

1) Is $R_{K,S}$ equal the groups of $S$-units of $K$, $\mathcal{O}_{K,S}^\times = \{a\in K, v_{\mathfrak{p}}(a) = 0 \ \forall \mathfrak{p} \notin S \}$ as defined on page 451 of Neukirch, Schmidt, Wingberg: Cohomology of Number Fields, tensored with $\mathbb{Z}_p$?

2) What "happens" when tensoring over $\mathbb{Z}_p$? I know that you can make any group into a $\mathbb{Z}_p$-module but is there any intuition why one would do that? Does the above group arise naturally as a kernel of some map?

3) Is there a more natural notion of the above group in terms of ideles?

4) A general question on Dirichlet's unit theorem for $S$-units: Is there a version without any constraints on $S$, i.e. without having to assume that the archimedian places are in $S$, i.e. $S_\infty \subset S$?

EDIT: Is $x \equiv 1 \mod \nu$ for $\nu \in S$ just an untypical notion of $S$ being a modulus $S = \prod_{\nu} \nu$?

Thank you for your help! :) Tom

EDIT (11/29/2013): For 1): My guess (above) is not right, but: Is $R_{K,S} = E_{K,S,1} \otimes \Bbb Z_p$, where $E_{K,S,1}$ denote the subgroup of $E_K$ that are principal units at every place $\nu \in S$?

For 2): I know now that tensoring over $\Bbb Z_p$ kills the prime-to-$p$-part and every module becomes a $\Bbb Z_p$-module, one can therefore speak of the $\Bbb Z_p$-rank of every module.

For 4): I now know that there can't be any version without the infinite places, because they are essential to the proof. But how can I actually use the Dirichlet unit theorem as stated in Neukirch, Schmidt, Wingberg: Cohomology of Number Fields 2nd edition 2008, proposition (8.7.2) to compute the $\Bbb Z_p$-rank of a $\Bbb Z_p$-submodule of $\mathcal{O}_{K,S}^\times$ of finite index?

  • $\begingroup$ Dear Tom: After your edits, I am no longer sure what you would like answered. Would you mind pointing it out exactly? $\endgroup$ Nov 30, 2013 at 20:32
  • $\begingroup$ @BrunoJoyal My main problem is to understand $R_{K,S}$ and the congruence notation above. $R_{K,S}$ seems to be a submodule of the units $\mathcal{O}_{K,S}^\times$ tensored with $\Bbb Z_p$ of $K$ and via Dirichlet unit theorem it is isomorphic or injects into something that I don't know. How would one write down $R_{K,S}$ as an idèle? $\endgroup$
    – BIS HD
    Dec 2, 2013 at 10:31
  • $\begingroup$ @BrunoJoyal Do you have an idea or a hint or a reference for me? That would help me very much and in addition the bounty won't get lost ;-) $\endgroup$
    – BIS HD
    Dec 6, 2013 at 16:03

1 Answer 1


Ok, let me give this a shot. First of all, I don't think the proposed definition of $R_{K, S}$ makes sense. On $\mathbf Z_p \otimes E_K$, we lose the residue class maps away from $p$. It should rather be

$$R_{K, S} = \{x \in E_K : x \equiv 1 \mod \nu, \forall \nu\in S\} \otimes_{\mathbf Z} {\mathbf Z_p}.$$

It is not true that $R_{K, S} = \mathcal O_{K, S}^\times \otimes \mathbf Z_p$. The rank of $R_{K, S}$ is bounded by the rank of $E_K$, and in general the rank of $\mathcal O_{K, S}^\times$ is much larger.

Tensoring with $\mathbf Z_p$, as you say, has the effect of killing the $p$-torsion. As for the question of why one would want to do that, I cannot really tell without seeing the context. In any case, I do not see a natural definition of $R_{K, S}$ in terms of idèles. I'm also not sure whether it appears naturally as the kernel of some map. (Sorry for all of these disappointing answers!)

As for Dirichlet's Unit Theorem, one does not in general assume that $S$ contains the archimedean places. For instance, the classical unit theorem (which states the finite-generatedness of $\mathcal O^\times_K$) is the case $S=\emptyset$, according to me. However, perhaps some authors use different definitions and consider the classical unit theorem to be the case where $S$ consists of all of the archimedean places. It really depends on the definitions at hand, but the statements should be the same in the end.


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