Kernel of $p$-adic logarithm. I'm completely clueless as to how to answer the following question:
Let $K$ be a field of characteristic zero which is complete with respect to a non-Archimedean aboslute value $|\cdot|$. Let $U_1=\{x\in K:\ |x-1|<1\}$. Show that the kernel of $\log:\ U_1\ \longrightarrow\ K$ is equal to the set of $p$-power roots of unity in $K$ if $|\cdot|$ is equivalent to the $p$-adic absolute value on $\mathbb{Q}$.
Any suggestions?
 A: The $p$-adic logarithm is defined by the power series
$$\log(1+x) = \sum_{n\geq 1} \frac{(-1)^{n+1}}{n} x^n = x - \frac{x^2}{2} + \frac{x^3}{3} \pm \ldots$$
which converges for $|x| < 1$. If, however, $|x| < p^{-1/(p-1)}$, then the summands $\frac{(-1)^{n+1}}{n} x^n$ for $n \geq 2$ have strictly smaller absolute value than $x$, and therefore $|\log(1+x)| = |x|$ by the properties of non-archimedean absolute values.
Now let $1+x$ be in the kernel of $\log: U_1 \to K$. For $n$ sufficiently large, we have $|x|^{p^n} < p^{-1/(p-1)}$. Then
$$(1+x)^{p^n}-1 = \sum_{j=1}^{p^n} \binom{p^n}{j} x^j$$
has absolute value $< p^{-1/(p-1)}$ (since for $0< j < p^n$ we have $|\binom{p^n}{j}x^j| \leq |px| < |p| \leq p^{-1/(p-1)}$), and therefore
$$|(1+x)^{p^n} - 1| = |\log((1+x)^{p^n})| = |p^n \log(1+x)| = 0.$$
Thus, $(1+x)$ is a $p^n$-th root of unity.
Conversely, every $p$-power root of unity in $K$ is a torsion element of $U_1$. But $K$ has characteristic zero, so $(K,+)$ is torsion-free. Therefore all $p$-power roots of unity are in the kernel of the homomorphism $\log: U_1 \to K$.
A: There are many ways of approaching this question, beyond @marlu’s excellent response. My favorite goes this way:
First, show that
\begin{align}
\log(1+x)&=\lim_{n\to\infty}\frac{(1+x)^{p^n}-1}{p^n}\,,
\end{align}
where the convergence is $p$-adic coefficientwise. This is a nice easy exercise.(*)
Next, expand this limit as an infinite product:
\begin{align}
\log(1+x)&=x\prod_{m=1}^\infty\frac{P_m(x)}{pP_{m-1}(x)}\,,
\end{align}
where $P_m(x)=(1+x)^{p^m}-1$. Note that $P_m/P_{m-1}$ is a monic $\mathbb Z$-polynomial of degree $p^m-p^{m-1}$, whose roots are the primitive $p^m$-th roots of unity. (Indeed, it’s a translate of the $p^m$-cyclotomic polynomial.) And the infinite product is also convergent, in the coefficientwise $p$-adic topology. If $k$ is any finite extension of $\mathbb Q_p$, with ring of (local) integers $\mathfrak o$ and the maximal ideal $\mathfrak m$ of $\mathfrak o$, that topology is the right one for substituting elements of $\mathfrak m$ into series. So, for elements $\xi$ of $\mathfrak m$, $\log(1+\xi)$ is expressible via that infinite product, and is zero if and only if $1+\xi$ is a $p$-power root of unity.
(*) This definition of the $p$-adic logarithm corresponds to the nonstandard definition of the real logarithm:
\begin{align}
\log(x)&=\lim_{n\to\infty} n\left( x^{1/n}-1\right)\,,
\end{align}
which Kenneth Ireland used at least for introducing the logarithm and exponential functions when he was teaching at Brown.
A: Doesn't the introduction of infinite products hint at an exponential function ? Actually, in the domain $v(x) > v(p)/(p-1)$ (see Marlu's answer) the exponential power series converges, hence in this neighborhood of $0$, $(K^*,\cdot)$ is locally isomorphic to $(K, +)$ via exp-log . But $L$ has no torsion since $K$ has characteristic $0$.
