MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
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? – 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
up vote 8 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.