Let $L/\mathbb{Q}$ be a finite Galois extension with Galois group $G$. It is well known that the ring of integers $\mathcal{O}_L$ is free over its associated order $\mathfrak{A}_{L/\mathbb{Q}}=\{x\in \mathbb{Q}[G]\mid x\mathcal{O}_L\subseteq \mathcal{O}_L\}$ if

  1. $G$ is abelian (Leopoldt, 1959);
  2. $G$ is dihedral of order $2p$, where $p$ is a rational prime (Bergé, 1972);
  3. $G$ is the quaternion group of order $8$, and the extension is wild (Martinet, 1972).

In some other papers, I have found written that also the local counterparts are true ($L/\mathbb{Q}_p$ finite Galois extension with the same Galois group as before), and it seems that the authors suggest that these results are naturally implied by the global ones. But while they seem "folklore", these implication are not immediate to me. A couple of observations (I wish to thank Fabio Ferri for the precious discussion).

  1. It seems to me that only the number fields case is considered in the papers.
  2. Probably the key is to repeat the proof as it is for the $p$-adic case; for example, this works in the Martinet's case.
  3. There are other ways to get these results, like for example using Lettl's work on absolutely abelian extensions for Leopoldt's local case, or realise local extensions as completion of global ones with the same Galois group (here $p\ne 2$, we refer to Henniart, 2001). But I am interested in knowing if there is an immediate and "direct way".

Summarising, my question is: is it immediate that the global cases imply the local cases?

(More generally, a question could be: if every Galois extension of number fields with a certain type of Galois group admits freeness of the ring of integers over the associated order, then the same also holds for every Galois extension of local fields with the same Galois group?)

  • $\begingroup$ I have asked this question in MSE, but I had no replies, and I was suggested to ask the question here. The link is the following: math.stackexchange.com/questions/4131623/… $\endgroup$ May 17, 2021 at 17:47
  • 1
    $\begingroup$ precious = previous? (On US keyboards at least, c is next to v). $\endgroup$
    – KConrad
    May 17, 2021 at 21:43
  • $\begingroup$ No I meant precious, like, really useful! $\endgroup$ May 18, 2021 at 7:01
  • 1
    $\begingroup$ No, I do not. Consider writing to experts on Galois module structure, e.g., Boas Erez. $\endgroup$
    – KConrad
    May 22, 2021 at 2:33
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    $\begingroup$ When you say "In some other papers" can you give specific references? $\endgroup$ Jun 11, 2021 at 10:09

1 Answer 1


There are a number of things one can say about this.

First, some of the papers you mention do also cover the case of $p$-adic fields. For example, see Bergé's paper Sur l’arithmétique d’une extension diédrale Annales de l’institut Fourier, tome 22, no 2 (1972), p. 31-59. On page 32 you'll see that A can be a PID satisfying conditions 1), 2) & 3) listed there, so in particular one can take $A=\mathbb{Z}_{p}$.

Second, I'm not sure what's "wrong" with the method of realising a Galois extension of $p$-adic fields as the completion of a Galois extension of number fields with the same Galois group. Given the result you are aiming to prove, this seems entirely natural to me. The result of Henniart you mention is described in Chapter IX, $\S$5 (page 574) of Cohomology of number fields, (available here https://www.mathi.uni-heidelberg.de/~schmidt/NSW2e/) and gives the result you want for $p$ odd. If this doesn't satisfy your definition of immediate and "direct", then I'm not sure what does. Can you clarify?

Third, we know lots more in the $p$-adic case than we do for the number field case. For example, we know that for any (at most) tamely ramified finite Galois extension $L/K$ of $p$-adic fields, we have that $\mathcal{O}_L$ is free over the group ring $\mathcal{O}_K[\mathrm{Gal}(L/K)]$ (which is equal to the associated order in this case). There are lots more results, but I just give this as an example. But I think it's true that for (almost) any specific Galois group $G$ for which we know the freeness result for all $G$-extensions $K/\mathbb{Q}$, one can give a reference for a direct proof for the same result for $G$-extensions $L/\mathbb{Q}_p$ (assuming that these latter extensions even exist).

Actually, perhaps one exception to this last claim is $Q_8$-extensions $L/\mathbb{Q}_2$. As you say, one can easily adapt the number field proof to the $2$-adic setting or use the method mentioned two paragraphs above (we can't use Henniart's result here, but as there are only finitely many $Q_8$-extensions $L/\mathbb{Q}_2$ one can, for example, use databases of extensions of $\mathbb{Q}_2$ and $\mathbb{Q}$.)

I should say that Fabio Ferri (who is my PhD student), is aiming to post an preprint to the arXiv in the next couple of weeks, that in addition to its main results on the additive Galois module structure of $A_4$, $S_4$ and $A_5$-extensions of $\mathbb{Q}$, will also provide a nice summary / discussion of the relation between freeness in the number field and $p$-adic cases.

  • $\begingroup$ Thank you very much for your answer. Let me try to clarify my question. $\endgroup$ Jun 11, 2021 at 12:38
  • $\begingroup$ 1) That is absolutely my bad, in Bergé's paper we can take $A=\mathbb{Z}_p$, so there is no problem there. $\endgroup$ Jun 11, 2021 at 12:39
  • $\begingroup$ 2) There is no problem a priori in the use of Henniart's result (except when $p=2$ clearly), and of course, once you have this result, the proof in the local case is immediate and direct. But this result is quite complicated, it seems to need a big machinery behind, and finally, I am quite sure that this is not the result that other mathematicians had in mind when they used global to deduce local. $\endgroup$ Jun 11, 2021 at 12:43
  • $\begingroup$ For example, in the preface of Childs' book "Taming wild extensions: Hopf algebras and local Galois module theory", published in 2000, Childs says that the global case "immediately implies the corresponding result for abelian extensions of $\mathbb{Q}_p$. $\endgroup$ Jun 11, 2021 at 12:45
  • $\begingroup$ 3) Ok so with probably just one exception, where there is a proof in the global case, there is also a direct proof of a local case. Also, thank to Henniart's result, now we know that this is always true when $p\ne 2$, and there is no need to prove again directly the case also in the local setting. This is great. $\endgroup$ Jun 11, 2021 at 12:50

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