$p$-adic logarithm

Question

For $q\in\mathbb{C}_p$ such that $|q|_p < 1$ show that there exists a unique logarithm $\log_q:\mathbb{C}_p^{*}\to\mathbb{C}_p$ with
(i) $\log_q(q)=0$
(ii) $\forall x\in\mathbb{C}_p$, $|x|_p<1:$ $\log_p(1+x)=\log_q(1+x)$

The definitions:
$\log_p(1+x)=\sum\limits_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}$; $\log_p(p)=0$; $\log_p : \mathbb{C}_p^{*}\to\mathbb{C}_p$; $|\cdot|_p : \mathbb{C}_p\to\mathbb{Q}$; $|p|_p:=\frac{1}{p}$.

Further I know from different lemmata tha: $\log_p(1+x)=0\Leftrightarrow x = 0 \Leftrightarrow u = 0$, since we can write $y = x+1 = p^r*w*u$.

My idea is to simply define $$\log_q(x)=\left\{\begin{array}{ll} \log_p(x), & x\not=q \\ 0, & x = q\end{array}\right. $$ But it seems to be to easy. Could someone maybe give me a hint for this task?

Best, Luca

Answer 1

First, let me point out that your claim that $\log(1+x)=0$ if and only if $x=0$ is incorrect, because if $\zeta$ is a $p^m$-th root of unity, then $v(\zeta-1)>0$, or in your notation, $|\zeta-1|<1$. And of course $\log(\zeta)=0$.

So your definition seems to want to run as follows: Choose $q$ with $|q|<1$, i.e. $v(q)>0$, fixed from now on. Now, you probably know that the residue field is the algebraic closure of $\mathbb F_p$, and that every element of this field can be lifted uniquely to a root of unity of order prime to $p$. Thus there is a well-defined multiplicative map that takes a unit $z$ (with $|z|=1$ to the unique root of unity that’s congruent to $z$ modulo the maximal ideal. Let’s call this map $z\mapsto\zeta_z$. So when $|z|=1$, you have $z=\zeta_zz'$, where $z'$ is a principal unit, i.e. $|z'-1|<1$, so you can apply the analytic logarithm to $z'$. Finally, for a general $z\in\mathbb C_p$, I think you’ve recognized that $z=q^\lambda z_0$ for $\lambda=m/n\in\mathbb Q$, up to an uncertainty of a factor of an $n$th root of unity, and for $|z_0|=1$. Now, any logarithm has to vanish on such an $n$-th root of unity, as you know, so your logarithm is well-defined: $\log_q(z)=\log(\zeta_{z_0})$.