$p$-adic logarithm, $|\log_p(1 + x)|_p = |x|_p$?

Question

Define the $p$-adic logarithm$$\log_p(1 + x) = \sum_{i =1}^\infty (-1)^{i-1}x^i/i.$$How do I see that if $p > 2$ and $|x|_p < 1$, then $|\log_p(1 + x)|_p = |x|_p$?

Answer 1

Here is an outline of a proof. First recall that the disc of convergence of $\log_p(1+x)$ is $D(0,1) = p\mathbb{Z}_p$. Let $v_p$ denote the $p$-adic valuation. Given $x \in p\mathbb{Z}_p$, then $v_p(x) \geq 1$.

($1$) Show that $v_p\left(\frac{x^n}{n}\right) > v_p(x)$ for all $n \geq 2$. Intuitively, this should hold because $x$ has at least $1$ factor of $p$, so $x^n$ has at least $n$ factors of $p$, while the number of factors of $p$ in $n$ grows at most logarithmically. More formally, write $n = up^k$ for some $k \in \mathbb{Z}_{\geq 0}$ and unit $u \in \mathbb{Z}_p^\times$. Then \begin{align*} v_p\left(\frac{x^n}{n}\right) = n v_p(x) - v_p(n) = up^k v_p(x) - k \, . \end{align*}

($2$) Recall that $v_p(a \pm b) \geq \min\{v_p(a),v_p(b)\}$. Show if $a,b \in \mathbb{Q}_p$ and $v_p(a) \neq v_p(b)$, then $v_p(a \pm b) = \min\{v_p(a),v_p(b)\}$.

$(3)$ Consider a partial sum $$ s_N = \sum_{n = 1}^N (-1)^{n-1} \frac{x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \pm \frac{x^N}{N} \, . $$ Since $v_p\left(\frac{x^n}{n}\right) > v_p(x)$ for all $n \geq 2$ by $(1)$, then $$ v_p(s_N) = \min\left\{v_p(x), v_p\left(\frac{x^2}{2}\right), \ldots, v_p\left(\frac{x^N}{N}\right)\right\} = v_p(x) $$ by $(2)$, so $|s_N|_p = |x|_p$ for all $N$.

$(4)$ Since the absolute value $|\cdot|_p$ is continuous, then $$ |x|_p = \lim_{N \to \infty} |s_N|_p = |\lim_{N \to \infty} s_N|_p = |\log_p(1+x)|_p $$ as desired.