# $p$-adic logarithm is a homomorphism, formal power series proof

Consider the $p$-adic logarithm defined by the series $$\log (1+x) = \sum_{n\ge 1} (-1)^{n+1} \frac{x^n}{n}.$$ It converges for $|x|_p < 1$, and if $|x|_p < 1$ and $|y|_p < 1$, then we have $$\log ((1+x)\cdot (1+y)) = \log (1+x) + \log (1+y).$$ One way to show it is to note that in the ring of formal power series $\mathbb{Q} [[X,Y]]$ (where $\log (1+X)$ is defined by the same formula) we have $$\log ((1+X)\cdot (1+Y)) = \log (1+X) + \log (1+Y).$$

How does one see that this formal identity indeed implies the identity above?

We have to see that $$\sum_{n\ge 1} (-1)^{n+1}\,\frac{(x+y+xy)^n}{n} = \sum_{n\ge 1} (-1)^{n+1}\,\left(\frac{x^n}{n} + \frac{y^n}{n}\right).$$ Let us expand the term $(x+y+xy)^n$: $$(x+y+xy)^n = \sum_{i_1 + i_2 + i_3 = n} {n \choose i_1, i_2, i_3} \, x^{i_1}\,y^{i_2}\,(xy)^{i_3} = \sum_{i_1 + i_2 + i_3 = n} {n \choose i_1, i_2, i_3}\,x^{i_1+i_3}\,y^{i_2+i_3} = \sum_{i\ge 0} \sum_{j\ge 0} {n \choose n-j, n-i, i+j-n}\,x^i\,y^j.$$ We have then $$\sum_{n\ge 1} (-1)^{n+1}\,\frac{(x+y+xy)^n}{n} = \sum_{n\ge 1} \sum_{i\ge 0} \sum_{j\ge 0} \frac{(-1)^{n+1}}{n}\,{n \choose n-j, n-i, i+j-n}\,x^i\,y^j.$$ Now the order of sums $\sum_{n\ge 1} \sum_{i\ge 0} \sum_{j\ge 0}$ may be changed (I will go back to this point below) to obtain $$\sum_{i\ge 0} \sum_{j\ge 0} \sum_{n\ge 1} \frac{(-1)^{n+1}}{n}\,{n \choose n-j, n-i, i+j-n}\,x^i\,y^j,$$ and we have to see that the numbers $$c_{ij} = \sum_{n\ge 1} \frac{(-1)^{n+1}}{n}\,{n \choose n-j, n-i, i+j-n}$$ satisfy $$c_{ij} = \begin{cases} (-1)^{m+1}/m, & \text{if }i = m, j = 0 \text{ or } i = 0, j = m,\\ 0, & \text{otherwise}. \end{cases}$$ But we already know that it's true thanks to the formal identity in $\mathbb{Q} [[X,Y]]$, so we are done.

The only non-formal step in the above is changing the order of sums. Recall that in the non-archimedian case, we have $$\sum_{i\ge 0} \sum_{j\ge 0} x_{ij} = \sum_{j\ge 0} \sum_{i\ge 0} x_{ij}$$ if $|x_{ij}| \to 0$ as $\max (i,j) \to \infty$.

In the above case, we may note that $$\left|\sum_{j\ge 0} \frac{(-1)^{n+1}}{n}\,{n \choose n-j, n-i, i+j-n}\,x^i\,y^j\right|_p \xrightarrow{\max (n,i) \to \infty} 0$$ (by the way, is it completely obvious?) so that $$\sum_{n\ge 1} \sum_{i\ge 0} \sum_{j\ge 0} \frac{(-1)^{n+1}}{n}\,{n \choose n-j, n-i, i+j-n}\,x^i\,y^j = \sum_{i\ge 0} \sum_{n\ge 1} \sum_{j\ge 0} \frac{(-1)^{n+1}}{n}\,{n \choose n-j, n-i, i+j-n}\,x^i\,y^j = \sum_{i\ge 0} \sum_{j\ge 0} \sum_{n\ge 1} \frac{(-1)^{n+1}}{n}\,{n \choose n-j, n-i, i+j-n}\,x^i\,y^j$$ (we swap the two inner sums in the second equality since they are finite).

My question is the following: all these details look a bit messy. Is there a shorter justification of the transition from the formal identity to the corresponding identity with $p$-adic series?

Koblitz in his GTM 58 book says that since in the non-archimedian situation, any convergent series converges after an arbitrary reordering, we can automatically assume that we may write $$\sum_{n\ge 1} (-1)^{n+1}\,\frac{(x+y+xy)^n}{n} = \sum_{i\ge 0}\sum_{j\ge 0} c_{ij}\,x^i\,y^j,$$ for some $c_{ij}$. Maybe I am missing something obvious, and the above change of summation order indeed doesn't require any explicit justifications?

Thank you.

• I would try and use that $\exp(x)=\sum_{n=0}^\infty \frac{x^n}{n!}$ maps $p\Bbb{Z}_p$ into $1+p\Bbb{Z}_p$, and gives a 2-sided inverse to the logarithm. Archimedean considerations alone would prove the desired identities in the ring of formal power series, no? Proving that $\exp(x)$ is a homomorphism is easier (just a reordering and binomial theorem), so the inverse has to be a homomorphism, too. May be some key link is broken here? – Jyrki Lahtonen Aug 14 '17 at 19:48
• @Jyrki Lahtonen There is a problem with this argument. exp and log are inverse, but only as functions $B (0, p^{-1/(p-1)}) \leftrightarrow B (1, p^{-1/(p-1)})$. And log is a homomorphism on the whole ball $B (1,1)$. – AAA Aug 15 '17 at 9:34
• @Jyrki Lahtonen Here is a dumb example: for the $2$-adic logarithm we have $\log (1-2) = 0$, so that $\exp (\log (1-2)) = 1 \ne -1$. – AAA Aug 15 '17 at 9:37
• Ok. Thanks for the reminder! But, wouldn't it be enough for an identity of power series to hold in a smaller ball to imply that the same identity holds in the ring of formal power series? – Jyrki Lahtonen Aug 15 '17 at 12:12
• @Jyrki Lahtonen That is true, but then justifying this using properties of analytic functions is almost the same as justifying the argument of Prof. Lubin below :) – AAA Aug 15 '17 at 20:18

It’s always possible that I have misunderstood the thrust of your question, but perhaps this argument will satisfy the preconditions you have set:

Set $G(x,y)=\log\bigl[(1+x)(1+y)\bigr]$ and $H(x,y)=\log(1+x)+\log(1+y)$. Take the derivative of each with respect to $x$. From $G$, you get $$\frac1{(1+x)(1+y)}\frac\partial{\partial x}\bigl[(1+x)(1+y)\bigr]=\frac1{1+x}\,,$$ while from $H$ you get, of course, $\frac1{1+x}$. So $G$ and $H$ differ by a $y$-series: $$\log\bigl[(1+x)(1+y)\bigr]=K(y)+\log(1+x)+\log(1+y)\,.$$ Now substitute $x=0$ and get $K=0$.

• Thank you very much! Fist of all, I must say that I am new to the $p$-adic stuff, and it is great to have you on this website, I have learned a lot from reading your answers. Your proof is very good. I think some authors give the formal power series argument to avoid discussing derivatives of $p$-adic functions. I wanted to understand if the transition from the formal identity is indeed automatic, or requires some thought... – AAA Aug 14 '17 at 0:13
• @AAA: Derivatives are tricky to reason with -- there are lots of locally constant functions, so some function having derivative zero does not imply it is a constant. In particular, I don't believe Lubin's argument is valid at all -- the fact that $G$ and $H$ have the same derivative in the first argument is nowhere near enough to imply that you can write $G(x,y)-H(x,y) = K(y)$. – Hurkyl Aug 14 '17 at 0:55
• @Hurkyl This argument works with some care. See Theorem 8.5 in math.uconn.edu/~kconrad/blurbs/gradnumthy/infseriespadic.pdf – AAA Aug 14 '17 at 1:04
• @AAA: It's the omission of the care that I'm complaining about. – Hurkyl Aug 14 '17 at 1:06
• The aspect of the question that I was addressing, @Hurkyl, was the formal identity expressing the homomorphism. I think that this is enough to guarantee the truth of the nonformal statement on any set $I_r\times I_r$, where $0<r<1$ and $I_r=\{z:|z|\le r\}$. For this, you need to make use of knowledge of how fast the coefficients of the logarithm grow. – Lubin Aug 14 '17 at 13:06

For a shorter proof, the main idea is simple: you evaluate the formal identity at $(X,Y) = (x,y)$ to get the special identity. Everything else is technical detail.

To address a concern in the comments, the point is that evaluation is continuous. If $\sum a_k x^k$ is a convergent power series in a topological ring $R$, then any continuous homomorphism $\varphi : R \to S$ will satisfy

$$\varphi\left( \sum_{k=0}^{\infty} a_k x^k \right) = \varphi\left( \lim_{n \to \infty} \sum_{k=0}^{n} a_k x^k \right) = \lim_{n \to \infty} \varphi\left( \sum_{k=0}^{n} a_k x^k \right) \\= \lim_{n \to \infty} \sum_{k=0}^n \varphi(a_k) \varphi(x)^k = \sum_{k=0}^\infty \varphi(a_k) \varphi(x)^k$$

In particular, if $\log(1+z)$ is defined in $R$, then $\log(1 + \varphi(z))$ is defined in $S$ and $\varphi(\log(1+z)) = \log(1+\varphi(z))$.

The main technical obstacle is the fact that the power series $\log(1+T)$ does not have $\mathbb{Z}_p$-integral coefficients, so we can't directly invoke many of the usual facts about power series rings.

So, one must instead develop enough of the theory of convergent formal power series to show the homomorphisms involved are defined and continuous. Unfortunately, I don't recall how straightforward this is; but maybe your source already has the relevant theorems.

• In order to see that $\log ((1+x)\cdot (1+y))$ may be written as some series of the form $\sum_{i\ge 0}\sum_{j\ge 0} c_{ij}\,x^i\,y^j$, one still needs to justify why the necessary rearrangement of terms is valid in the analytic context. I guess my question boils down to that. – AAA Aug 14 '17 at 1:46
• @AAA: I see what you mean; I would be inclined to avoid that issue entirely by arguing $\log((1+X)(1+Y)$ is the image of $\log(1+T)$ under an an evaluation homomorphism $\mathbb{Q}[[T]]' \to \mathbb{Q}[[X,Y]]'$, and showing that composing two evaluations gives an evaluation. (where by the prime I mean convergent series) – Hurkyl Aug 14 '17 at 1:51
• @AAA: No wait, I'm making things too complicated. Even in the ring of power series, you still have $\log((1+X)(1+Y)) = -\sum (-1)^n ((1+X)(1+Y)-1)^n/n$. You don't need to rewrite $\log((1+x)(1+y))$ at all. – Hurkyl Aug 14 '17 at 3:02
• @AAA: I've rewritten by post a fair bit. – Hurkyl Aug 14 '17 at 3:21