p-adic Measure and Kummer’s Congruence

Abstract

In modern number theory, the p-adic method or p-adic way of thinking plays an important role. As an example, there are objects called p-adic L-functions which correspond to the Dirichlet L-functions, and in fact the natural setup to understand the Kummer congruence described in Sect. 3.2 is in the context of the p-adic L-functions.

These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

In modern number theory, the p-adic method or p-adic way of thinking plays an important role. As an example, there are objects called p-adic L-functions which correspond to the Dirichlet L-functions, and in fact the natural setup to understand the Kummer congruence described in Sect. 3.2 is in the context of the p-adic L-functions. To be precise, a modified version (by a suitable “Euler factor”) of Kummer’s congruence guarantees the existence of the p-adic L-function.

To discuss this aspect fully is beyond the scope of this book, but in this chapter we explain the p-adic integral expression of the Bernoulli number and prove Kummer’s congruence using it. Interested readers are advised to read books such as Iwasawa [51] , Washington [100] , Lang [66] .

We assume the basics of p-adic numbers. For this we refer readers to Serre [83, Ch. 1] or Gouvea [37]. The results in this chapter are not used in other chapters.

11.1 Measure on the Ring of p-adic Integers and the Ring of Formal Power Series

In this section we review the general correspondence between measures on the ring of p-adic integers Z _p and the ring of formal power series. We use this setup in the next section to define the Bernoulli measure on Z _p and to express Bernoulli numbers as integrals. This expression turns out to be very useful in proving Kummer’s congruence relation.

Let $\overline{\mathbf{Q}}_{p}$ be the algebraic closure of the field Q _p of p-adic numbers. The p-adic absolute value $\vert \vert$ of Q _p (normalized by $\vert p\vert = 1/p$) is extended uniquely to $\overline{\mathbf{Q}}_{p}$. We use the same notation $\vert \vert$ for this extension. Then $\overline{\mathbf{Q}}_{p}$ is not complete with respect to this absolute value, and the completion is denoted by C _p . The absolute value $\vert \vert$ also extends naturally to C _p . Let $\mathcal{O}_{p}$ be the ring of integers of C _p :

$$\displaystyle{\mathcal{O}_{p} =\{ x \in \mathbf{C}_{p}\mid \vert x\vert \leq 1\}.\qquad }$$

Remark 11.1.

Like the complex number field C, the field C _p is complete and algebraically closed. To do analysis in the p-adic setting, we need this big field.

First we review the general theory of measures on Z _p .

Denote the Z-module Z∕p ⁿ Z by X _n and the canonical map from X _n+1 to X _n by π _n+1, so $\pi _{n+1}: X_{n+1} \rightarrow X_{n}$ is defined by

$$\displaystyle{x\bmod p^{n+1}\mathbf{Z}\longmapsto x\bmod p^{n}\mathbf{Z}.}$$

The system of pairs (X _n , π _n ) gives a projective system and we have the projective limit $\displaystyle \lim_{\longleftarrow} \: X_{n}\:$:

$$\displaystyle{\lim _{\longleftarrow }\:X_{n} ={\Bigl \{ (x_{n}) \in \prod _{n\geq 1}X_{n}\mid \pi _{n+1}(x_{n+1}) = x_{n}\Bigr \}}.}$$

The ring of p-adic integers Z _p is identified with this projective limit $\displaystyle \lim_{\longleftarrow} \: X_{n}\:$.

Definition 11.2 (Measure on Z _p ).

A set of functions $\mu =\{\mu _{n}\}_{n=1}^{\infty }\,$ is called an $\mathcal{O}_{p}$-valued measure on Z _p if the following two conditions are satisfied:

• (i)  Each μ _n is an $\mathcal{O}_{p}$-valued function on X _n , $\mu _{n}: X_{n}\longrightarrow \mathcal{O}_{p}\,$.

• (ii) For any n ∈ N and x ∈ X _n , the distribution property

  $$\displaystyle{\mu _{n}(x) =\sum _{\begin{array}{c}y\in X_{n+1} \\ \pi _{n+1}(y)=x\end{array}}\,\mu _{n+1}(y)}$$

  holds.

The set of $\mathcal{O}_{p}$-valued measures on Z _p is denoted by $\,\mathcal{M}(\mathbf{Z}_{p},\mathcal{O}_{p})\,$. This has an $\mathcal{O}_{p}$-module structure. Further, the norm of $\mu =\{\mu _{n}\} \in \mathcal{M}(\mathbf{Z}_{p},\mathcal{O}_{p})\,$ is defined as

$$\displaystyle {\Vert \mu \Vert = \sup \limits_{n\in \mathbf{N},\;x\in X_{n}} \, \vert \mu_{n}(x) \vert.}$$

Also, the $\mathcal{O}_{p}$-module of continuous $\mathcal{O}_{p}$-valued functions on Z _p is denoted by $C(\mathbf{Z}_{p},\mathcal{O}_{p})$, and the norm $\Vert \varphi \Vert$ of an element $\varphi \in C(\mathbf{Z}_{p},\mathcal{O}_{p})$ is defined by

$$\displaystyle{\Vert \varphi \Vert = \sup \limits_{x\in \mathbf{Z}_{p}}\:\vert \varphi (x)\vert.}$$

For $\varphi \in C(\mathbf{Z}_{p},\mathcal{O}_{p})$ and $\mu =\{\mu _{n}\} \in \mathcal{M}(\mathbf{Z}_{p},\mathcal{O}_{p})\,$, the integral on Z _p is defined by

$$\displaystyle{\int _{\mathbf{Z}_{p}}\,\varphi (x)d\mu (x) =\lim _{n\rightarrow \infty }\,\sum _{r=0}^{p^{n}-1 }\varphi (r)\mu _{n}(r).}$$

(We use the abbreviated notation μ _n (r) for $\mu _{n}(r\bmod p^{n})$. A similar abbreviation will be used in the following.) The convergence of the limit on the right-hand side is guaranteed by the following estimate: when n < m, we have

$$\displaystyle\begin{array}{rcl} & & \Bigm \vert\sum _{r=0}^{p^{n}-1 }\varphi (r)\mu _{n}(r) -\sum _{l=0}^{p^{m}-1 }\varphi (l)\mu _{m}(l)\Bigm \vert {}\\ & & \quad =\;\biggm \vert\sum _{ r=0}^{p^{n}-1 }{\Bigl (\varphi (r)\mu _{n}(r) -\sum _{q=0}^{p^{m-n}-1 }\varphi (r + p^{n}q)\mu _{ m}(r + p^{n}q)\Bigr )}\Bigm \vert {}\\ & & \quad =\;\Bigm \vert\sum _{ r=0}^{p^{n}-1 }\left (\sum _{q=0}^{p^{m-n}-1 }\left (\varphi (r) -\varphi (r + p^{n}q)\right )\mu _{ m}(r + p^{n}q)\right )\Bigm \vert {}\\ & & \quad \leq \max \limits_{r,\;q}\left \vert \varphi (r) -\varphi (r + p^{n}q)\right \vert \Vert \mu \Vert. {}\\ \end{array}$$

For each natural number k, the binomial polynomial

$$\displaystyle{\binom{t}{k} = \frac{t(t - 1)\cdots (t - k + 1)} {k!} }$$

in t is a continuous function on Z _p .

To $\mu =\{\mu _{n}\} \in \mathcal{M}(\mathbf{Z}_{p},\mathcal{O}_{p})\,$ we associate $f \in \mathcal{O}_{p}[[X]]$ in the following manner. Set $\varLambda = \mathcal{O}_{p}[[X]]$, $\varLambda _{n} = ((1 + X)^{p^{n} } - 1)\varLambda$ and consider the projective system $\,\{(\varLambda /\varLambda _{n},\varpi _{n})\}\,$ by the natural map $\varpi _{n}:\varLambda /\varLambda _{n}\longrightarrow \varLambda /\varLambda _{n-1}$. Define f _n (X) ∈ Λ∕Λ _n by

$$\displaystyle{f_{n}(X) =\sum _{ r=0}^{p^{n}-1 }\mu _{n}(r)(1 + X)^{r} =\sum _{ r=0}^{p^{n}-1 }\sum _{k=0}^{r}\,\mu _{ n}(r)\binom{r}{k}X^{k} =\sum _{ k=0}^{p^{n}-1 }\,c_{n,k}X^{k}.}$$

Here we understand that the equalities are $\bmod \,\varLambda _{n}$ and put

$$\displaystyle{c_{n,k} =\sum _{ r=0}^{p^{n}-1 }\,\mu _{n}(r)\binom{r}{k}.}$$

Since we have

$$\displaystyle\begin{array}{rcl} (\varpi _{n}f_{n})(X)& =& \varpi _{n}\left (\sum _{r=0}^{p^{n}-1 }\,\mu _{n}(r)(1 + X)^{r}\right ) {}\\ & =& \varpi _{n}\left (\sum _{r^{{\prime}}=0}^{p^{n-1}-1 }\sum _{l=0}^{p-1}\,\mu _{ n}(r^{{\prime}} + p^{n-1}l)(1 + X)^{r^{{\prime}} }(1 + X)^{p^{n-1}l }\right ) {}\\ & =& \sum _{r^{{\prime}}=0}^{p^{n-1}-1 }\,\mu _{n-1}(r^{{\prime}})(1 + X)^{r^{{\prime}} } {}\\ & =& f_{n-1}(X), {}\\ \end{array}$$

the system (f _n ) is an element in the projective limit $\lim \limits_{\longleftarrow }\,\varLambda /\varLambda _{n}\,$. Now we have the isomorphism

$$\displaystyle{\varLambda \mathop{\cong}\lim _{\longleftarrow }\varLambda /\varLambda _{n},\quad \varLambda \ni g\longmapsto (g_{n}) \in \lim _{\longleftarrow }\varLambda /\varLambda _{n},}$$

where, for g ∈ Λ, the system (g _n ) is given by $g_{n} = g\bmod \varLambda _{n}$. Through this isomorphism, the above {f _n } corresponds to f ∈ Λ by

$$\displaystyle{f(X) =\sum _{ m=0}^{\infty }\:c_{ m}X^{m},}$$

where

$$\displaystyle\begin{array}{rcl} c_{m}& =& \lim _{n\rightarrow \infty }\:\sum _{r=0}^{p^{n}-1 }\,\mu _{n}(r)\binom{r}{m} {}\\ & =& \int _{\mathbf{Z}_{p}}\,\binom{x}{m}\,d\mu (x). {}\\ \end{array}$$

We therefore have obtained a map from $\mathcal{M}(\mathbf{Z}_{p},\mathcal{O}_{p})$ to $\mathcal{O}_{p}[[X]]$. An important fact is that this map gives a natural isomorphism between $\mathcal{M}(\mathbf{Z}_{p},\mathcal{O}_{p})$ and the ring of formal power series $\mathcal{O}_{p}[[X]]$, often referred to as the Iwasawa isomorphism. The way to associate a measure to an element in $\mathcal{O}_{p}[[X]]$ is described as follows.

For $f =\sum\nolimits_{ m=0}^{\infty }c_{m}X^{m} \in \mathcal{O}_{p}[[X]]$, define μ = {μ _n } by

$$\displaystyle{ \mu _{n}(r) = \frac{1} {p^{n}}\sum _{\zeta ^{p^{n}}=1}\,\zeta ^{-r}f(\zeta -1)\qquad (r \in X_{ n}), }$$

(11.1)

the sum running over all p ⁿ-th roots ζ of 1. Since $\vert \zeta -1\vert < 1$, f(ζ − 1) converges. For each m ≥ 0, we have

$$\displaystyle\begin{array}{rcl} \frac{1} {p^{n}}\sum _{\zeta ^{p^{n}}=1}\,\zeta ^{-r}(\zeta -1)^{m}& =& \frac{1} {p^{n}}\sum _{\zeta ^{p^{n}}=1}\,\sum _{j=0}^{m}\:\zeta ^{-r}\binom{m}{j}(-1)^{m-j}\zeta ^{j} {}\\ & =& \sum _{\stackrel{0\leq j\leq m}{j\equiv r\bmod p^{n}}}\,\binom{m}{j}(-1)^{m-j}. {}\\ \end{array}$$

So this is contained in $\mathcal{O}_{p}$. In particular, if p ⁿ > r > m, then this is zero. When ζ is a primitive p ^ν-th root of 1 (ν ≥ 1), the equality

$$\displaystyle{\vert \zeta -1\vert ^{\varphi (p^{\nu })} =\vert p\vert \qquad (\varphi \;\mbox{ is the Euler function})}$$

holds and hence

$$\displaystyle{\vert (\zeta -1)^{m}\vert =\vert p^{m/\varphi (p^{\nu })}\vert.}$$

From this, we conclude that p ^e divides the quantity

$$\displaystyle{\sum _{\begin{array}{c}0\leq j\leq m \\ j\equiv r\bmod p^{n} \end{array}}\,\binom{m}{j}(-1)^{m-j}}$$

for e = m∕ϕ(p ⁿ) − n. Therefore,

$$\displaystyle{\mu _{n}(r) =\sum _{ m=0}^{\infty }\,c_{ m}\left ( \frac{1} {p^{n}}\sum _{\zeta ^{p^{n}}=1}\,\zeta ^{-r}(\zeta -1)^{m}\right )}$$

is convergent and the value is in $\mathcal{O}_{p}$. To check the distribution property (ii) of the measure, we need to calculate the following value:

$$\displaystyle\begin{array}{rcl} \sum \limits_{y\in X_{n+1},\;\pi _{n+1}(y)=x}\,\mu _{n+1}(y)& =& \sum \limits_{a\bmod p}\,\mu _{n+1}(x + p^{n}a) {}\\ & =& \frac{1} {p^{n+1}}\sum \limits_{\zeta ^{p^{n+1}}=1}\left (\sum \limits_{a\bmod p}\,\zeta ^{-(x+p^{n}a) }\right )f(\zeta -1). {}\\ \end{array}$$

Using the identity

$$\displaystyle{\sum _{a\bmod p}\,\zeta ^{-p^{n}a } = \left \{\begin{array}{@{}l@{\quad }l@{}} 0\quad &\mbox{ if $\zeta ^{p^{n} }\neq 1$}, \\ p\quad &\mbox{ if $\zeta ^{p^{n} } = 1$} \end{array} \right.}$$

for a p ⁿ⁺¹-th root ζ of 1, we have

$$\displaystyle{\sum _{\begin{array}{c}a\bmod p \\ \zeta ^{p^{n+1}}=1\end{array}}\,\zeta ^{-x-p^{n}a } = p\sum _{\zeta ^{p^{n}}=1}\zeta ^{-x},}$$

so we have

$$\displaystyle{\sum _{\begin{array}{c}y\in X_{n+1} \\ \pi _{n+1}(y)=x\end{array}}\,\mu _{n+1}(y) =\mu _{n}(x)\;}$$

which is to be proved. If we define the formal power series $\tilde{f} \in \mathcal{O}_{p}[[X]]$ corresponding to this measure defined as before, then the coefficients c _k ^′ of X ^k of this series are given by

$$\displaystyle\begin{array}{rcl} c_{k}^{{\prime}}& =& \lim _{ n\rightarrow \infty }\sum _{r=0}^{p^{n}-1 }\mu _{n}(r)\binom{r}{k} {}\\ & =& \lim _{n\rightarrow \infty }\sum _{m=0}^{\infty }c_{ m}\sum _{r=0}^{p^{n}-1 }\binom{r}{k}\sum _{{ 0\leq j\leq m \atop j\equiv r\bmod p^{n}} }\binom{m}{j}(-1)^{m-j}. {}\\ \end{array}$$

We fix k. To calculate the coefficient of c _m in the expression of c _k ^′ in the right-hand side above, we fix m. We have $\binom{r}{k} = 0$ for k > r so we may assume that k ≤ r. Taking n big enough, we assume that m < p ⁿ. Then, if $j \equiv r\bmod p^{n}$ for some j with 0 ≤ j ≤ m, we have j = r since we also have 0 ≤ r ≤ p ⁿ − 1 by definition. So we may assume that k ≤ r = j ≤ m. So the coefficient of c _m is given by

$$\displaystyle{\sum _{r=k}^{m}\binom{r}{k}\binom{m}{r}(-1)^{m-r} =\sum _{ i=0}^{m-k}\binom{m - k}{i}\binom{m}{k}(-1)^{m-k-i} = \left \{\begin{array}{cl} 1&\text{ if }m = k, \\ 0&\text{ if }m\neq k. \end{array} \right.}$$

Hence we have $c_{k}^{{\prime}} = c_{k}$. So we have $\tilde{f} = f$ and two mappings are inverse with each other and we see that the set $\mathcal{M}(\mathbf{Z}_{p},\mathcal{O}_{p})$ of $\mathcal{O}_{p}$-valued measures and the space of formal power series $\mathcal{O}_{p}[[X]]$ are bijective.

More precisely, we can introduce a product for both spaces and show that these are isomorphic as $\mathcal{O}_{p}$ algebras, as given in the following theorem whose complete proof is omitted (see e.g. Lang [66, Ch.4] ).

For two measures $\mu,\;\nu \in \mathcal{M}(\mathbf{Z}_{p},\mathcal{O}_{p})$, we define an $\mathcal{O}_{p}$-valued function (μ ∗ ν) _n on X _n by

$$\displaystyle{ (\mu {\ast}\nu )_{n}(x) =\sum _{ y=0}^{p^{n}-1 }\:\mu _{n}(y)\nu _{n}(x - y)\qquad (x \in X_{n}). }$$

(11.2)

Then μ ∗ ν = { (μ ∗ ν) _n } becomes an element of $\mathcal{M}(\mathbf{Z}_{p},\mathcal{O}_{p})$. We call this a convolution product of μ and ν. The set $\mathcal{M}(\mathbf{Z}_{p},\mathcal{O}_{p})\,$ becomes an $\mathcal{O}_{p}$ algebra by this product μ ∗ ν.

Theorem 11.3 (Iwasawa isomorphism).

Between the space $\,\mathcal{M}(\mathbf{Z}_{p},\mathcal{O}_{p})\,$ of $\mathcal{O}_{p}$ -valued measures and the ring of formal power series $\mathcal{O}_{p}[[X]]$ , there is an $\mathcal{O}_{p}$ algebra isomorphism $P: \mathcal{M}(\mathbf{Z}_{p},\mathcal{O}_{p})\longrightarrow \mathcal{O}_{p}[[X]]$ given by

$$\displaystyle{\mathcal{M}(\mathbf{Z}_{p},\mathcal{O}_{p}) \ni \mu =\{\mu _{n}\}\mathop{ \longmapsto } ^{P}f(X) =\sum _{ m=0}^{\infty }\,c_{ m}X^{m} \in \mathcal{O}_{ p}[[X]].}$$

Here, c _m is determined by μ:

$$\displaystyle{c_{m} =\int _{\mathbf{Z}_{p}}\,\binom{x}{m}\,d\mu (x),}$$

and conversely μ _n is determined by f:

$$\displaystyle{\mu _{n}(x) = \frac{1} {p^{n}}\sum _{\zeta ^{p^{n}}=1}\,\zeta ^{-x}f(\zeta -1).}$$

For convenience of the description below, we recall Mahler’s¹ theorem giving the necessary and sufficient condition for an $\mathcal{O}_{p}$-valued function on Z _p to be continuous.

Theorem 11.4.

The function $\varphi: \mathbf{Z}_{p}\longrightarrow \mathcal{O}_{p}$ is continuous if and only if it can be written as

$$\displaystyle{\varphi (x) =\sum _{ n=0}^{\infty }\:a_{ n}\binom{x}{n},\qquad a_{n} \in \mathcal{O}_{p},\quad \vert a_{n}\vert \longrightarrow 0.}$$

If this is the case, the coefficients a _n are uniquely determined by $\varphi$ and given by

$$\displaystyle{a_{n} =\sum _{ k=0}^{n}(-1)^{n-k}\binom{n}{k}\varphi (k).}$$

We omit the proof (cf. Lang [66, §4.1] ).

If we use Theorem 11.4, we can understand a part of Theorem 11.3 more intuitively as follows. Fix x ₀ ∈ Z. Denote by $\varphi$ the characteristic polynomial of $x_{0} + p^{n}\mathbf{Z}_{p}$. Then by the definition of the p-adic measure, we see easily that

$$\displaystyle{\int _{\mathbf{Z}_{p}}\varphi (x)d\mu (x) =\mu _{n}(x_{0}).}$$

So if we replace $\varphi (x)$ by the expansion $\varphi (x) =\sum\nolimits_{ m=0}^{\infty }a_{m}\binom{x}{m}$ in Theorem 11.4, we have

$$\displaystyle{\mu _{n}(x_{0}) =\sum _{ m=0}^{\infty }a_{ m}c_{m} =\sum _{ m=0}^{\infty }c_{ m}\sum _{k=0}^{m}(-1)^{m-k}\binom{m}{k}\varphi (k).}$$

Now, for any ζ with $\zeta ^{p^{n} } = 1$, we have

$$\displaystyle{\sum _{m=0}^{\infty }c_{ m}(\zeta -1)^{m} =\sum _{ m=0}^{\infty }c_{ m}\sum _{k=0}^{m}\binom{m}{k}(-1)^{m-k}\zeta ^{k}.}$$

Since $\varphi (k) = 1$ if $k \equiv x_{0}\bmod p^{n}$ and $\varphi (k) = 0$ otherwise, we have

$$\displaystyle{ \frac{1} {p^{n}}\sum _{\zeta ^{p^{n}}=1}\zeta ^{-x_{0} }\sum _{m=0}^{\infty }c_{ m}(\zeta -1)^{m} =\sum _{ m=0}^{\infty }c_{ m}\sum _{k=0}^{m}\binom{m}{k}(-1)^{m-k}\varphi (k).}$$

So we get the expression of μ(x) by f in Theorem 11.3.

We describe here several useful properties of the correspondence P in Theorem 11.3 between measures and formal power series. Let the maximal ideal of $\mathcal{O}_{p}$ be

$$\displaystyle{\mathcal{P} =\{ z \in \mathcal{O}_{p}\mid \vert z\vert < 1\}.}$$

For $z \in \mathcal{P}$, define the function (1 + z)^x in x by

$$\displaystyle{(1 + z)^{x}:=\sum _{ n=0}^{\infty }\:\binom{x}{n}z^{n}.}$$

By Mahler’s theorem, (1 + z)^x is a continuous function of x ∈ Z _p . When x is a non-negative integer, this definition of (1 + z)^x coincides with the usual binomial expansion . We have the relation

$$\displaystyle{ (1 + z)^{x}(1 + z)^{x^{{\prime}} } = (1 + z)^{x+x^{{\prime}} }\qquad (x,\;x^{{\prime}}\in \mathbf{Z}_{ p}). }$$

(11.3)

This is obvious for x,  x ^′ ∈ N, and the general case for x,  x ^′ ∈ Z _p follows from the fact that the set N of natural numbers is dense in Z _p .

In the following, we list several properties of measures and corresponding power series, which will be used later.

Property (1).

Let $z \in \mathcal{P}$. If μ corresponds to f (i.e. Pμ = f), then

$$\displaystyle{f(z) =\int _{\mathbf{Z}_{p}}\,(1 + z)^{x}\,d\mu (x).}$$

In particular, by putting z = 0,

$$\displaystyle{f(0) =\int _{\mathbf{Z}_{p}}\,d\mu (x).}$$

Proof.

Writing $f(X) = \sum\limits_{n=0}^{\infty }\:c_{ n}X^{n}\:$, we have by Theorem 11.3

$$\displaystyle\begin{array}{rcl} \int _{\mathbf{Z}_{p}}\,(1 + z)^{x}\,d\mu (x)& =& \int _{\mathbf{ Z}_{p}}\:\sum _{n=0}^{\infty }\binom{x}{n}z^{n}\,d\mu (x) {}\\ & =& \sum _{n=0}^{\infty }\,z^{n}\int _{ \mathbf{Z}_{p}}\:\binom{x}{n}\,d\mu (x) {}\\ & =& \sum _{n=0}^{\infty }\,c_{ n}z^{n} = f(z). {}\\ \end{array}$$

 □ 

We call the map λ from $C(\mathbf{Z}_{p},\mathcal{O}_{p})$ to $\mathcal{O}_{p}$ a bounded linear functional on $C(\mathbf{Z}_{p},\mathcal{O}_{p})$ if the following conditions (i), (ii) are satisfied:

• (i) For any $\varphi,\;\varphi ^{{\prime}}\in C(\mathbf{Z}_{p},\mathcal{O}_{p})$ and any $a,\;b \in \mathcal{O}_{p}$,

  $$\displaystyle{\lambda (a\varphi + b\varphi ^{{\prime}}) = a\lambda (\varphi ) + b\lambda (\varphi ^{{\prime}}).}$$

• (ii) There exists a positive constant M > 0 such that for any $\varphi \in C(\mathbf{Z}_{p},\mathcal{O}_{p})$,

  $$\displaystyle{\vert \lambda (\varphi )\vert \leq M\Vert \varphi \Vert.}$$

The norm of λ is defined by

$$\displaystyle{\Vert \lambda \Vert =\mathop{ \mathrm{sup}}_{\stackrel{\varphi \in C(\mathbf{Z}_{p},\mathcal{O}_{p})}{\varphi \neq 0}}\:\frac{\vert \lambda (\varphi )\vert } {\Vert \varphi \Vert }.}$$

Let λ be a bounded linear functional on $C(\mathbf{Z}_{p},\mathcal{O}_{p})$. For x ∈ X _n  = Z∕p ⁿ Z, write the characteristic function of x + p ⁿ Z _p as $\varphi _{x,\,n}$. If we put

$$\displaystyle{\mu _{n}(x) =\lambda (\varphi _{x,\,n}),}$$

then μ = {μ _n } is an $\mathcal{O}_{p}$-valued measure on Z _p (i.e. $\mu \in \mathcal{M}(\mathbf{Z}_{p},\mathcal{O}_{p})$). Conversely, given $\mu =\{\mu _{n}\} \in \mathcal{M}(\mathbf{Z}_{p},\mathcal{O}_{p})$, if we put

$$\displaystyle{\lambda (\varphi ) =\int _{\mathbf{Z}_{p}}\,\varphi (x)\,d\mu (x),}$$

then λ is a bounded linear functional on $C(\mathbf{Z}_{p},\mathcal{O}_{p})$. This correspondence between λ and μ is easily seen to be one to one.

Moreover, for $h \in C(\mathbf{Z}_{p},\mathcal{O}_{p})$ and $\mu \in \mathcal{M}(\mathbf{Z}_{p},\mathcal{O}_{p})\,$, the map

$$\displaystyle{\varphi \longmapsto \int _{\mathbf{Z}_{p}}\,\varphi (x)h(x)\,d\mu (x),\qquad (\varphi \in C(\mathbf{Z}_{p},\mathcal{O}_{p}))}$$

is a bounded linear functional on $C(\mathbf{Z}_{p},\mathcal{O}_{p})$. Let hμ be the corresponding measure. It is an interesting problem to compute the formal power series corresponding to the measure hμ when μ corresponds to $f = P\mu \in \mathcal{O}_{p}[[X]]$. Properties (2) and (3) below give examples of this correspondence.

For $f \in \mathcal{O}_{p}[[X]]$, put

$$\displaystyle{ (\mathbb{U}f)(X) = f(X) -\frac{1} {p}\sum _{\zeta ^{p}=1}\,f(\zeta (1 + X) - 1). }$$

(11.4)

Since

$$\displaystyle{\frac{1} {p}\sum _{\zeta ^{p}=1}\,\left (\zeta (1 + X) - 1)\right )^{l} \in \mathbf{Z}_{ p}[X]}$$

for non-negative integers l, we have $\mathbb{U}f \in \mathcal{O}_{p}[[X]]$.

Property (2).

Let $f \in \mathcal{O}_{p}[[X]]$ and μ _f be the corresponding measure. Also, let ψ be the characteristic function of Z _p ^×. Then the formal power series corresponding to the measure ψ μ _f is $\mathbb{U}f$, i.e., $\psi \mu _{f} =\mu _{\mathbb{U}f}$. More precisely, we have for any $\varphi \in C(\mathbf{Z}_{p},\mathcal{O}_{p})\,$

$$\displaystyle{\int _{\mathbf{Z}_{p}}\,\varphi (x)\psi (x)\,d\mu _{f}(x) =\int _{\mathbf{Z}_{p}}\,\varphi (x)\,d\mu _{\mathbb{U}f}(x).}$$

This can also be written as

$$\displaystyle{\int _{\mathbf{Z}_{p}^{\times }}\,\varphi (x)\,d\mu _{f}(x) =\int _{\mathbf{Z}_{p}}\,\varphi (x)\,d\mu _{\mathbb{U}f}(x).}$$

Proof.

Write the power series corresponding to the measure ψ μ _f as g. When $z \in \mathcal{P}$, by Property (1) we have

$$\displaystyle{g(z) =\int _{\mathbf{Z}_{p}}\,(1 + z)^{x}\psi (x)\,d\mu _{ f}(x).}$$

Let ζ be a pth root of 1. Regarding ψ also as a function on Z∕p Z via $\psi (a\bmod p) =\psi (a + p\mathbf{Z}_{p})$, and putting

$$\displaystyle{\hat{\psi }(\zeta ) = \frac{1} {p}\sum _{a\in \mathbf{Z}/p\mathbf{Z}}\,\psi (a)\zeta ^{-a},}$$

(Fourier transform on Z∕p Z) we have

$$\displaystyle{\psi (a) =\sum _{\zeta ^{p}=1}\,\hat{\psi }(\zeta )\zeta ^{a}}$$

by a simple calculation (inverse Fourier transform). Since

$$\displaystyle{\hat{\psi }(\zeta ) = \left \{\begin{array}{ll} -\frac{1} {p}\quad &\qquad \text{if }\zeta \neq 1, \\ \frac{p-1} {p} \quad &\qquad \text{if }\zeta = 1, \end{array} \right.}$$

by the definition of ψ, we obtain

$$\displaystyle\begin{array}{rcl} g(z)& =& \int _{\mathbf{Z}_{p}}\,(1 + z)^{x}\psi (x)\,d\mu _{ f}(x) {}\\ & =& \int _{\mathbf{Z}_{p}}\,(1 + z)^{x}\sum _{ \zeta ^{p}=1}\,\hat{\psi }(\zeta )\zeta ^{x}\,d\mu _{ f}(x) {}\\ & =& \sum _{\zeta ^{p}=1}\,\hat{\psi }(\zeta )\int _{\mathbf{Z}_{p}}\,(1 + z)^{x}\zeta ^{x}\,d\mu _{ f}(x) {}\\ & =& \sum _{\zeta ^{p}=1}\,\hat{\psi }(\zeta )\int _{\mathbf{Z}_{p}}\,{\Bigl (1 + (\zeta (1 + z) - 1)\Bigr )}^{x}\,d\mu _{ f}(x) {}\\ & =& \sum _{\zeta ^{p}=1}\,\hat{\psi }(\zeta )f(\zeta (1 + z) - 1) = f(z) -\frac{1} {p}\sum _{\zeta ^{p}=1}\,f(\zeta (1 + z) - 1). {}\\ \end{array}$$

This shows $g = \mathbb{U}f$. (Here we define the power ζ ^x for x ∈ Z _p by

$$\displaystyle{\zeta ^{x} = (1 +\zeta -1)^{x} =\sum _{ n=0}^{\infty }\,\binom{x}{n}(\zeta -1)^{n}.}$$

If we choose a ∈ Z so that x − a ∈ p Z _p , we have ζ ^x = ζ ^a.) □ 

Define the differential operator D on the ring of formal power series $\mathcal{O}_{p}[[X]]$ by

$$\displaystyle{D = (1 + X)D_{X},\qquad \mbox{ where}\quad D_{X} = \frac{d} {dX}.}$$

Property (3).

For $f \in \mathcal{O}_{p}[[X]]$, the power series corresponding to the measure xμ _f is Df. Hence the power series corresponding to the measure x ^k μ _f   (k natural number) is D ^k f and the equalities

$$\displaystyle{\int _{\mathbf{Z}_{p}}\,x^{k}\,d\mu _{ f}(x) =\int _{\mathbf{Z}_{p}}\,d\mu _{D^{k}f}(x) = (D^{k}f)(0)}$$

hold.

Proof.

It is enough to show this when k = 1. Let $g \in \mathcal{O}_{p}[[X]]$ be the power series corresponding to the measure xμ _f . By Property (1), we have for $z \in \mathcal{P}$

$$\displaystyle{g(z) =\int _{\mathbf{Z}_{p}}\,x(1 + z)^{x}\,d\mu _{ f}(x).}$$

Put $f(X) =\sum\nolimits_{ n=0}^{\infty }\,a_{n}X^{n},\ g(X) =\sum\nolimits_{ n=0}^{\infty }\,b_{n}X^{n}$. Using

$$\displaystyle{X\binom{X}{n} = (n + 1)\binom{X}{n + 1} + n\binom{X}{n}}$$

and Theorem 11.3, we have

$$\displaystyle\begin{array}{rcl} b_{n}& =& \int _{\mathbf{Z}_{p}}\,\binom{x}{n}\,d\mu _{g}(x) =\int _{\mathbf{Z}_{p}}\,\binom{x}{n}x\,d\mu _{f}(x) {}\\ & =& (n + 1)\int _{\mathbf{Z}_{p}}\binom{x}{n + 1}d\mu _{f}(x) + n\int _{\mathbf{Z}_{p}}\binom{x}{n}d\mu _{f}(x) {}\\ & =& (n + 1)a_{n+1} + na_{n}. {}\\ \end{array}$$

On the other hand, Df is computed as

$$\displaystyle\begin{array}{rcl} (Df)(X)& =& ((1 + X)D_{X}f)(X) {}\\ & =& (1 + X)(a_{1} + 2a_{2}X +\, \cdots \, + na_{n}X^{n-1} +\, \cdots \,) {}\\ & =& \sum _{n=0}^{\infty }\,((n + 1)a_{ n+1} + na_{n})X^{n}. {}\\ \end{array}$$

This gives g = Df. □ 

In general, for a power series f(X), we define a new power series f ^∗(Z) in Z by setting X = e ^Z − 1:

$$\displaystyle{ f^{{\ast}}(Z) = f(e^{Z} - 1). }$$

(11.5)

For example, when

$$\displaystyle{f(X) = (1 + X)^{a} =\sum _{ n=0}^{\infty }\,\binom{a}{n}X^{n},}$$

we have

$$\displaystyle{f^{{\ast}}(Z) = e^{aZ} =\sum _{ n=0}^{\infty }\,\frac{a^{n}Z^{n}} {n!}.}$$

Note the identity

$$\displaystyle{ (D_{Z}^{k}f^{{\ast}})(0) = (D^{k}f)(0) }$$

(11.6)

since

$$\displaystyle{D_{Z}f^{{\ast}}(Z) = (1 + X)D_{ X}f(X) = Df(X).}$$

The next property is the basis of the fact that the isomorphism P in Theorem 11.3 is an $\mathcal{O}_{p}$ algebra isomorphism.

Property (4).

Let the measures μ,  ν correspond respectively to the power series $f,\;g \in \mathcal{O}_{p}[[X]]$ (i.e., μ = μ _f ,  ν = μ _g ). Then the power series corresponding to the convolution μ ∗ ν is fg:

$$\displaystyle{\mu _{f} {\ast}\mu _{g} =\mu _{fg}.}$$

Proof.

By Eq. (11.1), we have

$$\displaystyle\begin{array}{rcl} \mu _{n}(r)& =& \frac{1} {p^{n}}\sum _{\zeta ^{p^{n}}=1}\,\zeta ^{-r}f(\zeta -1), {}\\ \nu _{n}(k - r)& =& \frac{1} {p^{n}}\sum _{\zeta ^{p^{n}}=1}\,\zeta ^{-k+r}g(\zeta -1). {}\\ \end{array}$$

Substituting this into the right-hand side of (11.2), we obtain

$$\displaystyle\begin{array}{rcl} (\mu {\ast}\nu )_{n}(k)& =& \sum _{r=0}^{p^{n}-1 }\, \frac{1} {p^{n}}\sum _{\zeta ^{p^{n}}=1}\,\zeta ^{-r}f(\zeta -1) \frac{1} {p^{n}}\sum _{\xi ^{p^{n}}=1}\,\xi ^{-k+r}g(\xi -1) {}\\ & =& \frac{1} {p^{n}}\sum _{\zeta }\sum _{\xi }\,f(\zeta -1)g(\xi -1)\xi ^{-k} \cdot \frac{1} {p^{n}}\sum _{r=0}^{p^{n}-1 }\,(\xi /\zeta )^{r} {}\\ & =& \frac{1} {p^{n}}\sum _{\zeta }\,f(\zeta -1)g(\zeta -1)\zeta ^{-k} {}\\ & =& \mu _{fg,\:n}(k). {}\\ \end{array}$$

Here ζ and ξ run through all p ⁿ-th roots of 1. From this, Property (4) follows. □ 

11.2 Bernoulli Measure

We define a specific measure called the Bernoulli measure. Recall that the first Bernoulli polynomial is by definition equal to

$$\displaystyle{B_{1}(x) = x -\frac{1} {2}.}$$

In the following, p denotes an odd prime. For each natural number n and $x \in X_{n} = \mathbf{Z}/p^{n}\mathbf{Z}$, set

$$\displaystyle{E_{n}(x) = B_{1}\left (\left \{ \frac{x} {p^{n}}\right \}\right ),}$$

where in the right-hand side, we regard x as an integer representing $x\bmod p^{n}$, and for w ∈ R, {w} is the real number satisfying 0 ≤ { w} < 1 and w −{ w} ∈ Z (the fractional part of w). Then E = { E _n } is a measure on Z _p but is not $\mathcal{O}_{p}$-valued. We modify this as follows in order to have an $\mathcal{O}_{p}$-valued measure. Take an invertible element c in Z _p (i.e. c ∈ Z _p ^×), and for x ∈ X _n  = Z∕p ⁿ Z, let

$$\displaystyle{E_{c,n}(x) = E_{n}(x) - cE_{n}(c^{-1}x).}$$

We understand c ⁻¹ x as an element in X _n  = Z∕p ⁿ Z. It is easy to see that E _c  = { E _c, n } is an $\mathcal{O}_{p}$-valued measure. We call this the Bernoulli measure.

Proposition 11.5.

• (1) The formal power series corresponding to the Bernoulli measure E _c is given by

  $$\displaystyle{f_{c}(X) = \frac{1} {X} - \frac{c} {(1 + X)^{c} - 1}.}$$

• (2) Let k be a natural number. For c ∈ Z _p ^× with c ^k ≠ 1, we have

  $$\displaystyle{\frac{B_{k}} {k} = \frac{(-1)^{k}} {1 - c^{k}}\int _{\mathbf{Z}_{p}}\,x^{k-1}\,dE_{ c}.}$$

  In particular, if $p - 1 \nmid k$ , then B _k ∕k ∈ Z _(p) .

Proof.

• (1) Since c ∈ Z _p ^×, we see f _c  ∈ Z _p [[X]], the first two terms of f _c (X) being

  $$\displaystyle{f_{c}(X) = \frac{c - 1} {2} + \frac{1 - c^{2}} {12} X +\, \cdots \,.}$$

  Let μ = {μ _n } be the measure on Z _p corresponding to f _c by Theorem 11.3. For $r \in X_{n} = \mathbf{Z}/p^{n}\mathbf{Z}$ we have

  $$\displaystyle\begin{array}{rcl} \mu _{n}(r)& =& \frac{1} {p^{n}}\sum _{\zeta ^{p^{n}}=1}\,\zeta ^{-r}f_{ c}(\zeta -1) {}\\ & =& \frac{1} {p^{n}}f_{c}(0) + \frac{1} {p^{n}}\sum _{\zeta ^{p^{n}}=1,\:\zeta \neq 1}\,\zeta ^{-r}\left ( \frac{1} {\zeta -1} - \frac{c} {\zeta ^{c} - 1}\right ). {}\\ \end{array}$$

  Now we use Lemma 8.5 on p. 110. For $\zeta ^{p^{n} } = 1,\;\zeta \neq 1\,$ and f = p ⁿ, the lemma gives

  $$\displaystyle{ \frac{1} {\zeta ^{c} - 1} = \frac{1} {f}\sum _{j=1}^{f-1}\,j\zeta ^{cj}}$$

  since (c, p) = 1. By this, if we choose l so that $cl \equiv k\bmod p^{n},\;0 \leq l < p^{n}$, we obtain

  $$\displaystyle\begin{array}{rcl} \frac{1} {f}\sum _{\zeta ^{p^{n}}=1,\:\zeta \neq 1}\,\zeta ^{-k} \frac{c} {\zeta ^{c} - 1}& =& \frac{c} {f^{2}}\sum _{\zeta ^{p^{n}}=1}\,\zeta ^{-k}\sum _{ j=1}^{f-1}\,j\zeta ^{cj} -\frac{c(f - 1)} {2f} {}\\ & =& \frac{cl} {f} -\frac{c(f - 1)} {2f} {}\\ & =& c\left \{\frac{c^{-1}k} {p^{n}} \right \} -\frac{c} {2} + \frac{c} {2f} {}\\ \end{array}$$

  and by substituting this into the formula for μ _n (r) above and noting that f _c (0) = (c − 1)∕2, we have

  $$\displaystyle\begin{array}{rcl} \mu _{n}(r)& =& \frac{c - 1} {2f} + \left (\left \{ \frac{r} {p^{n}}\right \} -\frac{1} {2} + \frac{1} {2f} - c\left \{\frac{c^{-1}r} {p^{n}} \right \} + \frac{c} {2} - \frac{c} {2f}\right ) {}\\ & =& \left (\left \{ \frac{r} {p^{n}}\right \} -\frac{1} {2}\right ) - c\left (\left \{\frac{c^{-1}r} {p^{n}} \right \} -\frac{1} {2}\right ). {}\\ \end{array}$$

  By the definition of the Bernoulli measure, we conclude μ _n (r) = E _c, n (r), i.e., μ = E _c and the power series corresponding to E _c is f _c .

  The proof of (2) goes as follows. By Property (3) and Eq. (11.6) we have

  $$\displaystyle{\int _{\mathbf{Z}_{p}}\,x^{k-1}\,dE_{ c} = (D^{k-1}f_{ c})(0) = (D_{Z}^{k-1}f_{ c}^{{\ast}})(0).}$$

  Here by definition (11.5), we have

  $$\displaystyle\begin{array}{rcl} f_{c}^{{\ast}}(Z)& =& f_{ c}(e^{Z} - 1) = \frac{1} {e^{Z} - 1} - \frac{c} {e^{cZ} - 1} {}\\ & =& \sum _{n=1}^{\infty }\:(1 - c^{n})(-1)^{n}B_{ n}\frac{Z^{n-1}} {n!}, {}\\ \end{array}$$

  so we have

  $$\displaystyle{(D_{Z}^{k-1}f_{ c}^{{\ast}})(0) = (1 - c^{k})(-1)^{k}\frac{B_{k}} {k} }$$

  and thus

  $$\displaystyle{\int _{\mathbf{Z}_{p}}\,x^{k-1}\,dE_{ c} = (1 - c^{k})(-1)^{k}\frac{B_{k}} {k}.}$$

  This gives (2). □ 

11.3 Kummer’s Congruence Revisited

The “right” formulation of Kummer’s congruence is the following.

Theorem 11.6.

Suppose p is an odd prime.

• (1) Assume that m is a positive even integer such that $p - 1 \nmid m$ . Then B _m ∕m ∈ Z _(p) .

• (2) Let a be a positive integer, and m and n positive even integers satisfying $m \equiv n\bmod (p - 1)p^{a-1}$ and $m\not\equiv 0\bmod (p - 1)$ . Then we have

  $$\displaystyle{(1 - p^{m-1})\frac{B_{m}} {m} \equiv (1 - p^{n-1})\frac{B_{n}} {n} \ \bmod p^{a}.}$$

To prove this, we need the following integral expression of the Bernoulli number, a refined version of Proposition 11.5 (2).

Proposition 11.7.

Let k be a positive even integer and take c ∈ Z _p ^× . Then we have

$$\displaystyle{(1 - c^{k})(1 - p^{k-1})\frac{B_{k}} {k} =\int _{\mathbf{Z}_{p}^{\times }}\,x^{k-1}\,dE_{ c}.}$$

Proof.

The power series that corresponds to the Bernoulli measure E _c is f _c in Proposition 11.5. As in (11.4), define from f _c a new power series g by

$$\displaystyle{g(X) = \mathbb{U}f_{c}(X) = f_{c}(X) -\frac{1} {p}\sum _{\zeta ^{p}=1}\,f_{c}(\zeta (1 + X) - 1).}$$

We have $g \in \mathcal{O}_{p}[[X]]$ and so we let μ = μ _g be the measure on $\mathcal{O}_{p}$ obtained from g. By Property (2) on p. 192 we have

$$\displaystyle{\int _{\mathbf{Z}_{p}^{\times }}\,x^{k-1}\,dE_{ c} =\int _{\mathbf{Z}_{p}}\,x^{k-1}\,d\mu.}$$

Further, using Property (3) on p. 194 and (11.6) one sees

$$\displaystyle{\int _{\mathbf{Z}_{p}}\,x^{k-1}\,d\mu = (D^{k-1}g)(0) = (D_{ Z}^{k-1}g^{{\ast}})(0).}$$

We compute the value $\,(D_{Z}^{k-1}g^{{\ast}})(0)\,$. First,

$$\displaystyle{g^{{\ast}}(Z) = \frac{1} {e^{Z} - 1} - \frac{c} {e^{cZ} - 1} -\frac{1} {p}\sum _{\zeta ^{p}=1}\left ( \frac{1} {\zeta e^{Z} - 1} - \frac{c} {\zeta ^{c}e^{cZ} - 1}\right ).}$$

Here, since

$$\displaystyle{\frac{1} {p}\sum _{\zeta ^{p}=1}\, \frac{1} {\zeta X - 1} = \frac{1} {X^{p} - 1},}$$

we get

$$\displaystyle\begin{array}{rcl} g^{{\ast}}(Z)& =& \frac{1} {e^{Z} - 1} - \frac{c} {e^{cZ} - 1} -\left ( \frac{1} {e^{pZ} - 1} - \frac{c} {e^{cpZ} - 1}\right ) {}\\ & =& \sum _{k=0}^{\infty }\,(1 - c^{k})(-1)^{k}\frac{B_{k}} {k!} Z^{k-1} -\sum _{ k=0}^{\infty }\,(1 - c^{k})(-1)^{k}\frac{B_{k}} {k!} (pZ)^{k-1} {}\\ & =& \sum _{k=1}^{\infty }\,(1 - c^{k})(1 - p^{k-1})(-1)^{k}\frac{B_{k}} {k!} Z^{k-1}. {}\\ \end{array}$$

Hence if k is even we have

$$\displaystyle{(D_{Z}^{k-1}g^{{\ast}})(0) = (1 - c^{k})(1 - p^{k-1})\frac{B_{k}} {k},}$$

and the proposition is established. □ 

Proof of Theorem 11.6.

The first assertion is already given in Theorem 3.2, but we give here an alternative proof for that too. Since we assumed $m\not\equiv 0\bmod p - 1$, we can take c ∈ Z such that (c, p) = 1 and $c^{m}\not\equiv 1\bmod p$. For instance one may take a primitive root $\bmod \,p$. From the proposition above, we have

$$\displaystyle{(1 - c^{n})(1 - p^{n-1})\frac{B_{n}} {n} =\int _{\mathbf{Z}_{p}^{\times }}\,x^{n-1}\,dE_{ c}}$$

and

$$\displaystyle{(1 - c^{m})(1 - p^{m-1})\frac{B_{m}} {m} =\int _{\mathbf{Z}_{p}^{\times }}\,x^{m-1}\,dE_{ c}.}$$

The assumption $m \equiv n\bmod (p - 1)p^{a-1}$ gives $c^{n-m} \equiv 1\bmod p^{a}\,$, and since we assumed (1 − c ^m, p) = 1, we have also (1 − c ⁿ, p) = 1. Since E _c is an $\mathcal{O}_{p}$ measure, the above integral values are in $\mathcal{O}_{p}$ and we see that B _n ∕n and B _m ∕m ∈ Z _(p). Since $\,x^{m-1} \equiv x^{n-1}\bmod p^{a}\,$ if x ∈ Z _p ^×, and since E _c is an $\mathcal{O}_{p}$-valued measure, we have

$$\displaystyle{(1 - c^{n})\left ((1 - p^{n-1})\frac{B_{n}} {n} - (1 - p^{m-1})\frac{B_{m}} {m} \right ) \in p^{a}\mathcal{O}_{ p}.}$$

The left-hand side being contained in Z _p , we conclude

$$\displaystyle{(1 - p^{n-1})\frac{B_{n}} {n} - (1 - p^{m-1})\frac{B_{m}} {m} \in p^{a}\mathbf{Z}_{ p}.}$$

This proves the theorem. □ 

Theorem 3.2 is a corollary of Theorem 11.6. Indeed, if a < m ≤ n, then by Theorem 11.6, we have

$$\displaystyle\begin{array}{rcl} & & (1 - p^{m-1})\frac{B_{m}} {m} - (1 - p^{n-1})\frac{B_{n}} {n} {}\\ & & \quad = (1 - p^{m-1})\left (\frac{B_{m}} {m} -\frac{B_{n}} {n} \right ) + \frac{B_{n}} {n} (p^{n-m} - 1)p^{m-1} {}\\ & & \quad \equiv 0\bmod p^{a}. {}\\ \end{array}$$

Since p − 1 ∤ n, we have B _n ∕n ∈ Z _(p) by Theorem 11.6. Since a ≤ m − 1, we have $p^{m-1}B_{n}/n \in p^{a}\mathbf{Z}_{(p)}$. Hence we have

$$\displaystyle{\frac{B_{m}} {m} -\frac{B_{n}} {n} \equiv 0\bmod p^{a}.}$$

Exercise 11.8.

Give an example of an odd prime p and integers 2 ≤ a = m < n such that the congruence in Theorem 3.2 does not hold. Check that for the same choice of a, n, m and p, the congruence of Theorem 11.6 surely holds.

Hint: For example, put p = 5, a = m = 2 and n = 22 and use the following values:

$$\displaystyle{B_{2} = \frac{1} {6},\qquad B_{22} = \frac{854513} {138}.}$$

Exercise 11.9.

Show that the Bernoulli number B _n is given by the limit (p-adic limit in Q _p )

$$\displaystyle{\lim _{m\rightarrow \infty } \frac{1} {p^{m}}\sum _{i=0}^{p^{m}-1 }i^{n}.}$$

(For a function $f\,:\, \mathbf{Z}_{p} \rightarrow \mathbf{Q}_{p}$ with a suitable condition, the limit

$$\displaystyle{\lim _{m\rightarrow \infty } \frac{1} {p^{m}}\sum _{i=0}^{p^{m}-1 }f(i)}$$

is sometimes referred to as the Volkenborn integral of f over Z _p . See [94, 95] for details.)