On the $p$-adic logarithm

Question

It is well known that $p$-adic logarithm define an isomorphism from $B(1,p^{-1/p-1})$ to $B(0,p^{-1/p-1})$. I would like to know if is it possible to extend somehow this logarithm to the set of $x$ such that $|x|_p=1$. My idea would be to find a minmal $n$ such that $x^n \in B(1,p^{-1/p-1})$ and define $\log_p(x)=\frac{\log_p(x^n)}{n}$. But I am not able to find such a $n$. So I would like to know if it is indeed possible to extend $\log_p$ to this set.

Answer 1

For $p>2,$ it is a standard fact that

$\{x\in \Bbb Q_p: |x|_p=1\} = \Bbb Z_p^\times \simeq \mu_{p-1}(\Bbb Q_p) \times \{x\in \Bbb Q_p: |x-1|_p <1\}$

as topological groups. The second factor on the right hand side are the principal units, often denoted $U_1$, and since the $p$-adic absolute value on $\Bbb Q_p$ takes only integer powers of $p$ as value, it is identical to your $B(1, p^{-\frac{1}{p-1}})$.

To extend the logarithm $log:U_1 \rightarrow \Bbb Q_p$ to the full set $\Bbb Z_p^\times$, you therefore "only" have to define it on $\mu_{p-1}(\Bbb Q_p)$, the $(p-1)$-th roots of unity. Indeed, the only $n$ you have to choose for your proposed extension formula is $n=p-1$. However, if you want the extension to be a a group homomorphism (from the multiplicative group on the left to the additive on the right), you'll have to define it as $log(\zeta) = 0$ for all $\zeta \in \mu_{p-1}(\Bbb Q_p)$, because those $\zeta$'s are torsion, whereas the only torsion element on the right hand side is $0$. (Or, using your extension formula, $log(\zeta) = \frac{log(\zeta^{p-1})}{p-1} = \frac{log(1)}{p-1} = 0$.)

In particular, an extension of the logarithm to $\{x\in \Bbb Q_p: |x|_p=1\}$ cannot be at the same time a group homomorphism and injective.