# von Staudt-Clausen over a totally real field

Before I ask the question, I need to recall what Bernoulli numbers $(B_k)_{k\in\mathbb{N}}$ are, and what von Staudt and Clausen discovered about them in 1840. The numbers $B_k\in\mathbb{Q}$ are the coefficients in the formal power series $${T\over e^T-1}=\sum_{k\in\mathbb{N}}B_k{T^k\over k!}$$ so that $B_0=1$, $B_1=-1/2$, and it is easily seen that $B_k=0$ for $k>1$ odd.

Theorem (von Staudt-Clausen, 1840) Let $k>0$ be an even integer, and let $p$ run through the primes. Then
$$B_k+\sum_{p-1|k}{1\over p}\in\mathbb{Z}.$$ John Coates remarked at a recent workshop that the analogue of this theorem for a totally real number field $F$ (other than $\mathbb{Q}$) is an open problem; even a weak analogue would imply Leopoldt's conjecture for $F$. I missed the opportunity of pressing him for details.

Question : What is the analogous statement over a totally real number field ?

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In your statement there is no dependence anywhere on an underlying field. What is the connection with fields? Is this the von Staudt-Clausen for some $p$-cyclotomic field? – Anweshi Jan 9 '10 at 16:41
There is a simple and standard relation between Bernoulli numbers and values of the Riemann zeta function at negative integers, and hence also between Bernoulli numbers and the p-adic L-function of Q. Totally real fields have classical and p-adic L-functions too (with "the same" special values at negative integers) and presumably this sort of thing is what Coates had in mind. I can't answer Dalawat's question though. – Kevin Buzzard Jan 9 '10 at 17:00
Since the relevance is for proving Leopoldt, may I ask the opinion of people about the following paper? arxiv.org/abs/0905.1274 – Anweshi Jan 9 '10 at 19:42
I want to stress that I have not even attempted to read Mihailescu's paper. But my understanding from talking to people who have attempted to read it is that the jury is still out. – Kevin Buzzard Jan 10 '10 at 0:19

I also can't answer the question, but I'll say some things that could help. One thing von Staudt-Clausen tells you is the denominator of the Bernoulli number $B_k$: it is precisely, the product of primes p for which $p-1\mid k$ (when $p-1\nmid k$, a result of Kummer says that $B_k/k$ is p-integral). As Buzzard commented, the Bernoulli numbers should be thought of (at least in this situation) as appearing in special values of p-adic L-functions, specifically, for k a positive integer $$\zeta_p(1-k)=(1-p^{k-1})(-B_k/k),$$ where $\zeta_p$ is the p-adic Riemann zeta function (see chapter II of Koblitz's "p-adic numbers, p-adic analysis, and zeta-functions", for example). For a totally real field F, a generalization of the p-adic Riemann zeta function exists, namely the p-adic Dedekind zeta function $\zeta_{F,p}$ (as proved independently by Deligne–Ribet (Inv Math 59), Cassou-Noguès (Inv Math 51), and Barsky (1978)). One link between these and the Leopoldt conjecture is through the p-adic analytic class number formula which is the main theorem of Colmez's "Résidue en s = 1 des fonctions zêta p-adiques" (Inv Math 91): $$\lim_{s\rightarrow1}(s-1)\zeta_{F,p}(s)=\frac{2^{[F:\mathbf{Q}]}R_phE_p}{w\sqrt{D}}$$ where h is the class number, $$E_p=\prod_{\mathfrak{p}\mid p}\left(1-\mathcal{N}(\mathfrak{p})^{-1}\right)$$ is a product of Euler-like factors, w = 2 is the number of roots of unity, D is the discriminant and $R_p$ is the interesting part here: the p-adic regulator (as Colmez notes, $\sqrt{D}$ and $R_p$ both depend on a choice of sign, but their ratio does not).

Theorem: The Leopoldt conjecture is equivalent to the non-vanishing of the p-adic regulator.

(For this, see, for example, chapter X of Neukirch-Schmidt-Wingberg's "Cohomology of number fields").

A clear consequence of this is that if $\zeta_{F,p}$ does not have a pole at s = 1, then the Leopoldt conjecture is false for (F, p). Perhaps an understanding of the denominators of values of $\zeta_{F,p}$ could lead to an understanding of the pole at s = 1 of $\zeta_{F,p}$.

Added (2010/04/09): So here's how you can use von Staudt–Clausen to see that the $p$-adic zeta function (of Q) has a pole at s = 1. It is clear from your statment of vS–C that it is saying that for $k\equiv0\text{ (mod }p-1)$, $B_k\equiv -1/p\text{ (mod }\mathbf{Z}_p)$ (i.e. it is not $p$-integral). Let $k_i=(p-1)p^i$, the $k_i$ is $p$-adically converging to 0, so $\zeta_p(1-k_i)$ is approaching $\zeta_p(1)$ (since $\zeta_p(s)$ is $p$-adically continuous, at least for $s\neq1$). By the aforementioned interpolation property of $\zeta_p(1-k)$, we have $$v_p(\zeta_p(1-k_i))=v_p(B_{k_i}/k_i)=-1-i\rightarrow -\infty$$ hence $1/\zeta_p(1-k_i)$ is approaching 0.

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I accept the answer as pointing in the right direction, but I would request a few edits (remove the first occurrence of "$p-1|k$" in the second sentence, insert a few dollar signs here and there, say that $w=2$, replace $E_p$ by $E_p(1)$, where $E_p(s)=\prod_{\mathfrak{p}| p}(1-(N\mathfrak{p})^{-s})$, etc.) – Chandan Singh Dalawat Jan 10 '10 at 7:01
done. hopefully a better answer will show up. – Rob Harron Jan 10 '10 at 7:19
Many thanks for your answer. – Chandan Singh Dalawat Jan 10 '10 at 9:27
Historical remark: Cassou-Nogues constructed the p-adic L-functions at about the same time as Deligne--Ribet and via a totally different method. – Kevin Buzzard Jan 10 '10 at 11:22
I've added an argument that allows you to see how von Staudt–Clausen let's you see the pole of the $p$-adic zeta function (thanks to Chan-Ho Kim and Rob Pollack for some helpful discussion). I also fully attributed the existence of the $p$-adic Dedekind zeta function. – Rob Harron Apr 9 '10 at 20:11