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.

  • $\begingroup$ There is a standard extension of the $p$-adic logarithm to $\Bbb Q_p^*$. $\endgroup$ – Angina Seng Jun 20 '18 at 2:19
  • $\begingroup$ Yes, but I would like somehow that the $log_p$ stay an isomorphism on my set. $\endgroup$ – Pierre21 Jun 20 '18 at 2:21
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    $\begingroup$ This is one of my previous questions, which looks related: math.stackexchange.com/questions/2353389/… $\endgroup$ – Tob Ernack Jun 20 '18 at 2:26
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    $\begingroup$ So you can define $\log_p$ as a power series on the set $\{x: |x - 1|_p \lt 1\}$, and you can extend its domain to $\mathbb{Q}_p^*$ by defining $\log_p(p) = 0$, and in general for $x = up^n$ you would have $\log_p(x) = \log_p(u)$, where $u$ is a unit in $\mathbb{Z}_p^*$. Each unit can be written as $u = \zeta u'$ where $\zeta$ is a root of unity and $u'$ is a principal unit. Then $\log_p(u) = \log_p(\zeta) + \log_p(u') = \log_p(u')$ since the log is zero on the roots of unity. $\endgroup$ – Tob Ernack Jun 20 '18 at 2:35
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    $\begingroup$ I don't think the logarithm usually gives you an isomorphism because it is not injective (any root of unity in $\mathbb{Z}_p$ will be mapped to $0$). It is injective on the principal units though. $\endgroup$ – Tob Ernack Jun 20 '18 at 2:58

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.

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  • $\begingroup$ Thak you very much that's very clear !! $\endgroup$ – Pierre21 Jun 20 '18 at 5:43
  • $\begingroup$ Do you have any references that talk about all this results ? I would be very interested to read them. $\endgroup$ – Pierre21 Jun 20 '18 at 5:51
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    $\begingroup$ Any book about $p$-adics (F. Gouvea, A. Robert) and/or local fields (Serre, or corresponding chapters in Neukirch's Algebraic Number Theory or Class Field Theory). There's probably even dozens of decent lecture notes out there. Also, math.stackexchange.com/q/73565/96384 and math.stackexchange.com/q/1298669/96384. $\endgroup$ – Torsten Schoeneberg Jun 20 '18 at 5:59
  • $\begingroup$ Super ! Thanks ! $\endgroup$ – Pierre21 Jun 20 '18 at 6:03

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