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 ?*



von Staudt-Clausen over a totally real field
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](http://www.ams.org/mathscinet-getitem?mr=579702)), Cassou-Noguès ([Inv Math 51](http://www.ams.org/mathscinet-getitem?mr=524276)), and Barsky ([1978](http://www.ams.org/mathscinet-getitem?mr=525346))). 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](http://www.ams.org/mathscinet-getitem?mr=922806)):
$$\lim\_{s\rightarrow1}(s-1)\zeta\_{F,p}(s)=\frac{2^{[F:\mathbf{Q}]}R\_phE\_p}{w\sqrt{D}}$$
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.


