Reordering proof of $p$-adic logarithm fundamental property

Question

I am reading a book on $p$-adic analysis and am currently stuck at the proof of the following theorem:

Theorem The $p$-adic logarithm satisfies the fundamental property $$\ln_p(xy) = \ln_p(x) + \ln_p(y)$$

As a reminder: the $p$-adic logarithm is defined as $$\ln_p(x) = \sum_{n = 1}^\infty (-1)^{n + 1} \frac{(x-1)^n}n$$

The proof they give is the following:

The following identity $$\log(1 + X) + \log(1 + Y) - \log(1 + X + Y + XY) = 0$$ holds for formal power series. One can check directly (by expanding and reordering terms) that all coefficients of the resulting series reduce to zero.

Where $\log(1 + X)$ is defined as $$\log(1 + X) = \sum_{n = 1}^\infty (-1)^{n + 1} \frac{X^n}n$$

I understand that you must be able to rearrange the above equation/series to get zero, as the identity can be proven through other ways. However, I'm a bit confused as to what specific reordering they are referring to here. I've tried developing the partial sums till $n = 4$ but wasn't able to find anything interesting.

Answer 1

I agree that kind of "formal to numerical" reasoning is dubious. What book are you reading that in?

A different proof, based on differentiation and recentering of $p$-adic power series, is Theorem 8.5 here.

Answer 2

I disagree that the “formal to numerical” reasoning is dubious. But it is tricky to pin down precisely what happens.

The simple story is the following: the formal power series $S(x,y)=\log(1+x)+\log(1+y)-\log(1+x+y+xy)$ is such that, when $z,w \in \mathbb{C}$ have small enough absolute value, $S(z,w)=0$ (by the property of the logarithm). By the theory of holomorphic functions, this means that the power series $S$ itself is zero. But when $z,w \in \mathfrak{m}_{\mathbb{C}_p}$, $S(z,w)$ (which is zero since $S=0$) is equal to $\ln_p(1+z)+\ln_p(1+w)-\ln_p(1+z+w+zw)$. QED.

Now, how to formalize that? The issue is that it’s not clear that the formal series $\log(1+P(x,y))$ (where $P$ is a polynomial with $P(0,0)=0$) evaluates at $(z,w)$ (when defined) to $\log_{an}(1+P(z,w))$ (where $\log_{an}$ is an “analytic logarithm”, ie an actual function instead of a power series).

To prove it, we need to introduce a bit of notation.

Let $K$ be a field with a nontrivial absolute value $|\cdot|$ and completion $\overline{K}$.

Call a power series $f(x_1,\ldots,x_n) =\sum_{\alpha \in \mathbb{N}^n}{a_{\alpha}x^{\alpha}} \in K[[x_1,\ldots,x_n]]$ tame if there exists $r_1,\ldots,r_n >0$ such that $|a_{\alpha}| = O(r_1^{-\alpha_1} \ldots r_n^{-\alpha_n})$.

We can then consider the function $E(f): (z_1,\ldots,z_n) \in D(f):=\cup_r{B(0,r_1) \times B(0,r_n)} \longmapsto f(z_1,\ldots,z_n)$, where the balls are open balls in $\overline{K}$, and we take the reunion over all $r$ as above.

It’s not hard to see that if $f,g$ are tame, then $f+g$ (resp. $fg$) is tame, $D(f+g) \supset D(f) \cap D(g)$ (resp. $D(fg) \supset D(f) \cap D(g)$), and the functions $E(f+g)$ and $E(f)+E(g)$ (resp. $E(fg)$ and $E(f)E(g)$) are (well defined and) equal on $D(f) \cap D(g)$.

The key lemma is the following one.

Lemma: let $f \in K[[T]]$ be tame. Let $P(x_1,\ldots,x_n)$ be a polynomial with entries in $K$ and $P(0)=0$. Then

1. $f \circ P$ is tame.
2. The functions $E(f \circ P)$ and $E(f) \circ P$ coincide on $D(0,r_1) \times \ldots \times D(0,r_n)$ for some $r_1, \ldots, r_n >0$.
3. if moreover $K$ is non-Archimedean and the coefficients of $P$ are in $O_K$, then $D(f \circ P) \supset [D(f)\cap D(0,1)]^n$ and the functions $E(f \circ P)$ and $E(f) \circ P$ coincide on $[D(f) \cap D(0,1)]^n$.

[Remark: when we apply it, $n=2$, $K$ is $\mathbb{C}$ or $\mathbb{C}_p$, and $f(T)=\log(1+T)$, $P(x,y) \in \{x,y,x+y+xy\}$.]

Proof: I only prove 1) and 2), the proof of 3) is similar but we can replace most of the sums with sups, which makes everything easier (the binomial coefficients lie in the unit ball, $M=1$ and we can drop $r$ from all the estimates).

Let $r >0$ be the number of nonzero coefficients of $P$ and $M \geq 1$ be at least the maximum of their absolute values.

So we can write $P(X)=\sum_{i=1}^r{p_{i}X^{\beta_i}}$.

Write $f=\sum_{n \geq 0}{a_nT^n}$, and let $1 \geq s>0$ be such that $|a_n| = O(s^{-n})$.

If $|z_1|, \ldots, |z_n| < s/Mr$, then $|P(z)| < s$ and thus $E(f) \circ P$ is defined at $z$.

Now, $f \circ P = \sum_{k \geq 0}{a_kP^k}=\sum_{k \geq 0}{\sum_{k=m_1+\ldots+m_r}{a_k\binom{k}{m_1,\ldots,m_r}p_1^{m_1} \ldots p_r^{m_r}X^{m_1\beta_1+\ldots m_r\beta_r}}}.$

In this sum, in any term of degree $k$, the coefficient has absolute value at most $Ck^r(rM/s)^k$ (where $|a_k| \leq Cs^{-k}$). Moreover, in this infinite sum, there are only at most $(|\alpha|+1)^{r}$ instances of a term of given (multi-degree) $\alpha$.

In particular, in the formal series $f \circ P$, a coefficient of degree $k$ has absolute value at most $C(k+1)^{2r}(rM/s)^k$, hence $f \circ P$ is tame and $D(f \circ P) \supset D(0,s/rM)^n$.

Let $(f \circ P)_k$ be the truncation of the formal series $(f \circ P)$ to its terms of degree $\leq k$.

Then by definition, $E(f \circ P)(z_1,\ldots,z_n)$ is the limit of the $(f\circ P)_k(z_1,\ldots,z_n)$.

Now, since $\sum_{d>k}{a_dP^d}$ contains no term of degree $\leq k$, it follows that $(f \circ P)_k-\sum_{d=0}^k{a_kP^k}$ is sum of a subset of the $a_d\binom{d}{m_1,\ldots,m_r}p_1^{m_1} \ldots p_r^{m_r}X^{m_1 \beta_1+\ldots m_r\beta_r}$ where $d > k$ and $m_1+\ldots+m_r=d$.

In particular, if $|z_i| \leq \sigma/Mr \leq 1$ with $\sigma < s$, then \begin{align*}|(f \circ P)_k(z_1,\ldots,z_n)-\sum_{d=0}^k{a_dP^d(z_1,\ldots,z_n)}|& \leq \sum_{d > k}{Cs^{-d}\sum_{d=m_1+\ldots+m_r}{\binom{d}{m_1,\ldots,m_r} M^{m_1\beta_1+\ldots+m_r\beta_r}(\sigma/Mr)^{m_1\beta_1+\ldots+m_r\beta_r}}}\\ & \leq C\sum_{m_1+\ldots m_r > k}{\binom{m_1+\ldots+m_r}{m_1,\ldots,m_r}(\sigma/s)^{m_1|\beta_1|+\ldots+m_r|\beta_r|}}\\ & \leq C\sum_{d > k}{d^r(\sigma/s)^d}. \end{align*}

When $k \rightarrow \infty$, we find $E(f \circ P)(z_1,\ldots,z_n)=(E(f) \circ P) (z_1,\ldots,z_n)$.

This proves 1) and 2).