I'm wondering if my proof (attempt) of the identity $\log((1+X)(1+Y))=\log(1+X)+\log(1+Y)$ (on $\mathbb{Q}[\![X,Y]\!]$) is correct.
First, I'm thinking $\mathbb{Q}[\![X,Y]\!]$ as the proyective limit of $\mathbb{Q}[X,Y]/\langle X,Y\rangle^{N}$. So, in order to show $\log((1+X)(1+Y))=\log(1+X)+\log(1+Y)$, it is enough to show the identity as polynomials, this is,
\begin{equation}\label{log}
\sum_{n=1}^{N}\frac{(-1)^{n-1}}{n}(X+Y+XY)\equiv \sum_{n=1}^{N}\frac{(-1)^{n-1}}{n}X^{N}+\sum_{n=1}^{N}\frac{(-1)^{n-1}}{n}Y^{N}\mod\langle X,Y\rangle^{N+1} \qquad[I]
\end{equation}
for every $N\geq 1$. In other words, we have to show that the terms of order $\,\leq N$ coincide for the two power series.
First, before showing my attempt, let me explain why I considered it.
Lets start with $N=1$. The RHS of (I) is $X+Y+XY$ which agrees with the LHS up to the order $1$. For $N=2$, the RHS is $X+Y+XY \textbf{$-\frac{1}{2}X^{2}-\frac{1}{2}Y^{2}-XY-\frac{1}{2}X^{2}Y^{2}-X^{2}Y-XY^{2}$}$
which agrees with the LHS up to the order $2$. Here, we see that dealing with the case $N+1$, we can rewrite the LHS in the following way: one part has terms of order $\leq N$, another part has terms of order $N+1$, and there is a part of order $\geq N+1$, but we are only interested in the first two parts, this motivates us to do induction on $N$. So, lets do this. I already showed the case $N=1$, so assume (I) for $N$, and lets check it for $N+1$. The RHS is:
\begin{eqnarray}
\sum_{n=1}^{N+1}\frac{(-1)^{n-1}}{n}(X+Y+XY)^{n}&=&\sum_{n=1}^{N}\frac{(-1)^{n-1}}{n}(X+Y+XY)^{n}+\frac{(-1)^{N}}{N+1}(X+Y+XY)^{N+1}\\
&=&\sum_{n=1}^{N}\frac{(-1)^{n-1}}{n}X^{N}+\sum_{n=1}^{N}\frac{(-1)^{n-1}}{n}Y^{N}+\frac{(-1)^{N}}{N+1}(X+Y+XY)^{N+1}
\end{eqnarray}
where the last line is due to the inductive hypothesis. Finally, the only terms in $\frac{(-1)^{N}}{N+1}(X+Y+XY)^{N+1}$ of order $N+1$ are $\frac{(-1)^{N}}{N+1}(X^{N+1}+Y^{N+1})$, and the rest have order $\geq N+2$; in conclusion,
$$\sum_{n=1}^{N+1}\frac{(-1)^{n-1}}{n}(X+Y+XY)^{n}=\sum_{n=1}^{N}\frac{(-1)^{n-1}}{n}X^{N}+\sum_{n=1}^{N}\frac{(-1)^{n-1}}{n}Y^{N}$$
up to the order $N+1$, so the result follows.
I'm dealing with the $p$-adic logarithm, I need to show that the identity $\log(1+x)(1+y)=\log(1+x)+\log(1+y)$ holds for every $x,y$ with $|x|_{p},|y|_{p}<1$, where $|-|_{p}$ is the $p$-adic absolute value. This is not so important since the proof is purely algebraic, however, if the identity holds for formal power series, it holds $p$-adically. A different approach I was thinking is first looking at the identity over $\mathbb{C}$, then use the series representation of each side, and invoke the identity theorem to conclude that the coefficients are equal for every $n$, thus the identity holds in the ring of formal power series.
Is my proof correct? Also, is this second approach correct? I've seen a proof of the identity (in the $p$-adics) using derivation, however I'm not allowed to use it, since is not part of the book.
Any comment, suggestion, correction is appreciated; thanks
Best Answer
It's a nice idea but your induction doesn't work. The inductive hypothesis gives that the sum of the first $N$ terms of the LHS is equal to the sum of the first $N$ terms of the RHS $\bmod (X, Y)^{N+1}$ but in the inductive step you need to know what is happening $\bmod (X, Y)^{N+2}$. Your proof "proved too much": note that if it worked it would've worked with $XY$ replaced by any sum of higher-order terms whatsoever, since you could've started the induction at $N = 1$.
Using the identity theorem over $\mathbb{C}$ will work fine. For a purely algebraic proof I would take the formal derivative of both sides in $X$; it's not hard to show from the power series definition that
$$\frac{\partial}{\partial X} \log \left( (1 + X)(1 + Y) \right) = \frac{1 + Y}{(1 + X)(1 + Y)} = \frac{1}{1 + X}$$
and it's very easy to show that the same holds for $\frac{\partial}{\partial x} \left( \log (1 + X) + \log (1 + Y) \right)$, so from here it only remains to show that the "constant terms" match, which means plugging in $X = 0$ and checking that $\log (1 + Y) = \log (1 + Y)$. I guess this is the same proof you said you weren't allowed to use but this is by far the simplest way to prove a lot of formal power series identities, much simpler than the identity theorem.