A question about the $p$-adic product formula for $\log(1+X)$ and $p$-adic geometry

As you recognize in a comment, the right ring is the subring of $\Bbb Q[[x]]$ consisting of all $f=\sum_0^\infty a_ix^i$ for which $\liminf_iv(a_i)/i=0$. These are precisely the $\Bbb Q$-power series that give you a convergent series when you substitute $x\to\xi$, no matter the $\xi\in\Bbb C_p$ with $v(\xi)>0$. Here, of course, I’m using the additive valuation $v$ on $\Bbb C_p$, and we might as well normalize so that $v(p)=1$. I think that for the purpose of this answer, I’ll call our ring $\Bbb Q_p\{\{x\}\}$.

What I want to do, because that’s more comfortable for me, is move the focus of attention from $1$ to $0$, so that instead of multiplying $\xi$ and $\eta$ elements close to $1$, to get $\xi\eta$, I’ll call them $1+x=\xi$ and $1+y$, and “multiply” them as $(1+x)(1+y)-1=x+y+xy$. This is the formal group of multiplication. I’ll write $M(x,y)=x+y+xy$, and define an endomorphism of $M$ as any $\Bbb C_p$-series, $\psi$, with $p$-integral coefficients, such that $M\bigl(\psi(x),\psi(y)\bigr)=\psi\bigl(M(x,y)\bigr)$. It’s a wonderful fact that all $M$-endomorphisms are in $\Bbb Z_p[[x]]$, and that for every $\gamma\in\Bbb Z_p$, there is an endomorphism $\psi$ with $\psi'(0)=\gamma$. One calls this endomorphism $[\gamma]_M$. You see immediately that for any positive integer $n$, $[n]_M(x)=(1+x)^n-1$, in particular for $n=p^m$. These are the polynomials that appear in @EinarRødland’s answer.

I’ll use the same shift of coordinates to call $\text{Log}(x)=\sum_{n=1}^\infty(-1)^{n+1}x^n/n$. This counts as a homomorphism from $M$ to the additive formal group $A(x,y)=x+y$, that is, $\text{Log}\bigl(M(x,y)\bigr)=\text{Log}(x)+\text{Log}(y)$.

You do see that for $z\in\Bbb Z_p$, $(1+x)^z-1$ makes sense all right, ’cause for each $i\ge1$, the binomial coefficient $\binom zi$ is in $\Bbb Z_p$; here’s one way of showing this. Since $z\in\Bbb Z_p$, there’s a sequence $\{n_j\}$ of positive integers $n_j$ with $z$ as limit. The associated binomial functions $\binom xi=x(x-1)(x-2)\cdots(x-i+1)/i!$ are polynomials and thus continuous as functions $\Bbb Q_p\to\Bbb Q_p$, and take $\Bbb Z$-values when $x$ is evaluated to any positive integer. Thus $\binom zi\in\Bbb Z_p$.

Perhaps without noticing it, we used the topology of coefficientwise convergence in $\Bbb Q\{\{x\}\}$ in the preceding argument. (If you look closely enough, this is also the topology of uniform convergence on proper subdiscs of the unit disk in $\Bbb C_p$.) We can use the same notion of convergence to show that $$ \text{Log}(x)=\lim_m\frac{[p^m]_M(x)}{p^m}\,. $$ I was hoping to show this by looking at the binomial coefficients, but easiest seems to be to differentiate each $[p^m](x)/p^m$ to get $(1+x)^{p^m-1}$, which have limit $(1+x)^{-1}$, $p$-adically, coefficientwise.

You look at $[p^m](x)/p^m$, and realize that it’s a finite product $$ \frac{[p^m](x)}{p^m} = x\prod_{k=1}^m\frac{[p^k](x)}{p[p^{k-1}](x)}\,, $$ in which the factors are polynomials that have limit $1$, $p$-adically, coefficientwise. Of course in the first factor, $[p^0](x)=x$, the identity endomorphism. This is exactly what @EinarRødland has said.

I’ve left out a lot of details, I know. But it all works out fine, and gives you, I think, a clear idea of what the roots of the logarithm Log are. This should answer most of your questions. But I should point out that the roots of the Log, i.e. the numbers $\zeta_{p^n}-1$ do not “converge to the boundary”, although their valuations $v(\zeta-1)$ do converge to zero. It’s an entirely different thing. The roots of unity in $\Bbb C_p$ are a discrete group, in strong contrast to the complex situation.

For $x \in p\Bbb{Z}_p$ $$\log(1+x) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}x^k, \qquad \exp(p^nx) =\sum_{k=0}^\infty \frac{p^{nk}x^k}{k!}= \exp'(p^nx)$$ $$ (1+x)^{p^n} = \exp(p^n \log(1+x)), \qquad g_x(y) = \exp(y \log(1+x))$$ $$ \log(1+x) =g_x'(0)=\lim_{n \to \infty} \frac{\exp(p^n \log(1+x))-1}{p^n}=\lim_{n \to \infty} \frac{(1+x)^{p^n}-1}{p^n}$$ $$= \lim_{n \to \infty} x\prod_{m=1}^n \frac{(1+x)^{p^m}-1}{p((1+x)^{p^{m-1}}-1)}$$

With $\zeta_{p^m}\in \overline{\Bbb{Q}}_p$ a primitive $p^m$-th root of unity, the roots of $(1+x)^{p^m}-1$ are the $\zeta_{p^m}^a-1, a \in 0 \ldots p^m-1$ so that $$\frac{(1+x)^{p^m}-1}{p((1+x)^{p^{m-1}}-1)}= \prod_{a=1,\ p\ \nmid\ a}^{p^m} (1-\frac{x}{\zeta_{p^m}^a-1})\in \Bbb{Q}_p[x], \qquad v_p(\zeta_{p^m}^a-1) = \frac1{(p-1)p^{m-1}}$$ which means $1-\frac{x}{\zeta_{p^m}^a-1}$ looks like $1-xp^{-\epsilon}$ and it is only the product of all its conjugates that $\to 1$.

Because $\frac{(1+x)^{p^n} -1}{p^n } \in x+x^2 \Bbb{Q}_p[[x])$ the ideal it generates is $(x)$.

Not sure if this really answers your question, but here's at least a derivation of the formula (assuming the product should be for all $n\ge0$ rather than $n\ge1$).

First, using $\Phi_0(X)=X$, note that the partial products $$ \Psi_n(X) = X\cdot\prod_{k=0}^{n-1} Q_k(X) = \Phi_0(X)\cdot\prod_{k=0}^{n-1}\frac{\Phi_{k+1}(X)}{p\Phi_k(X)} = \frac{\Phi_n(X)}{p^n} = \frac{(1+X)^{p^n}-1}{p^n}. $$ So, the claim is that $\Psi_n(X)\rightarrow\log(1+X)$ as $n$ increases.

Let's expand $$ \Psi_n(X) = \frac{(1+X)^{p^n}-1}{p^n} = \sum_{i=1}^{p^n} \frac{\binom{p^n}{i}}{p^n} X^i = \sum_{i=1}^{p^n} \frac{(p^n-1)\cdots(p^n-i+1)}{1\cdots(i-1)}\cdot\frac{X^i}{i}. $$ It is tempting to reduce this modulo $p^n$ and expect the terms to become $(-1)^{i-1}X^i/i\mod p^n$, but for $i\ge p$ there will be multipla of $p$ in the denominator. However, if we look at terms of degree at most $k$ modulo $p^m$, the terms will remain constant for sufficiently large $n$: I think $n\ge m+k/(p-1)$ should suffice, where the extra margin is to overcome the number of times $p$ divides the denominator.