p-adic exponential and p-adic logarithm

Let $p$ be a prime number and let ${ℂ}_p$ be the field of complex $p$-adic numbers (http://planetmath.org/ComplexPAdicNumbers5).

Definition 1.

The $p$-adic exponential is a function ${\exp }_p\mathrm{:}R\mathrm{\to }{\mathrm{C}}_p$ defined by

$${\exp }_p(s)=\sum _{n=0}^{\mathrm{\infty }}\frac{s^n}{n!}$$

where

The domain of ${\exp }_p$ is restricted because the radius of convergence of the series ${\sum }_{n=0}^{\mathrm{\infty }}{z}^n/n!$ over ${ℂ}_p$ is precisely $r=p^{-1/(p-1)}$. Recall that, for $z\in {\mathbb{Q}}_p$, we define

$${|z|}_p=\frac{1}{p^{{\nu }_p(z)}}$$

where ${\nu }_p(z)$ is the largest exponent $\nu$ such that $p^{\nu }$ divides $z$. For example, if $p\ge 3$, then ${\exp }_p$ is defined over $p{\mathbb{Z}}_p$. However, $e={\exp }_p(1)$ is never defined, but ${\exp }_p(p)$ is well-defined over ${ℂ}_p$ (when $p=2$, the number $e^4\in {ℂ}_2$ because ).

Definition 2.

The $p$-adic logarithm is a function ${\log }_p\mathrm{:}S\mathrm{\to }{\mathrm{C}}_p$ defined by

$${\log }_p(1+s)=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n+1}\frac{s^n}{n}$$

where

We extend the $p$-adic logarithm to the entire $p$-adic complex field ${\mathrm{C}}_p$ as follows. One can show that:

$${ℂ}_p=\{p^t\cdot w\cdot u:t\in \mathbb{Q},w\in W,u\in U\}=p^{\mathbb{Q}}\times W\times U$$

where $W$ is the group of all roots of unity of order prime to $p$ in ${\mathrm{C}}_p^{\mathrm{\times }}$ and $U$ is the open circle of radius centered at $z\mathrm{=}\mathrm{1}$:

We define ${\log }_p\mathrm{:}{\mathrm{C}}_p\mathrm{\to }{\mathrm{C}}_p$ by:

$${\log }_p(s)=lo{g}_p(u)$$

where $s\mathrm{=}{p}^r\mathrm{\cdot }w\mathrm{\cdot }u$, with $w\mathrm{\in }W$ and $u\mathrm{\in }U$.

Proposition (Properties of ${\exp }_p$ and ${\log }_p$).

With ${\exp }_p$ and ${\log }_p$ defined as above:

1. If ${\exp }_p(s)$ and ${\exp }_p(t)$ are defined then ${\exp }_p(s+t)={\exp }_p(s){\exp }_p(t)$.

2. ${\log }_p(s)=0$ if and only if $s$ is a rational power of $p$ times a root of unity.

3. ${\log }_p(xy)={\log }_p(x)+{\log }_p(y)$, for all $x$ and $y$.

4. If then

   $${\exp }_p({\log }_p(1+s))=1+s,{\log }_p({\exp }_p(s))=s.$$

In a similar way one defines the general $p$-adic power by:

$$s^z={\exp }_p(z{\log }_p(s))$$

where it makes sense.