It seems like there are two ways to make sense out of powers in the $p$-adic context, so I wonder if there's any difference. Here is some background (I hope the experts will excuse me for writing out certain obvious things, and I also hope there are no mistakes).

  • First of all, we have the following formal identity in the ring of power series $\mathbb{Q} [[X,Y]]$: $$\tag{0} \sum_{n\ge 0} {Y\choose n}\,X^n = \exp (Y\cdot \log (1+X)),$$ where the "binomial coefficients" are defined as polynomials $${Y \choose n} = \frac{Y\,(Y-1)\cdots (Y-n+1)}{n!} \in \mathbb{Q} [Y].$$

  • Let $\alpha\in\mathbb{Z}_p$. Then also ${\alpha \choose n} \in \mathbb{Z}_p$ for all $n = 0,1,2,3,\ldots$, and we see that for $|x|_p < 1$ the binomial series $$\tag{1} (1+x)^\alpha = \sum_{n\ge 0} {\alpha\choose n} \, x^n$$ converges.

  • Under the same assumptions, we may consider the $p$-adic series $$\tag{2} \exp (\alpha\cdot \log (1+x)).$$ However, even though $\log (1+x)$ again converges for $|x|_p < 1$, the $p$-adic exponential series converges only for $|x|_p < p^{-1/(p-1)}$, and $\log$ and $\exp$ are inverse to each other only as functions $1 + B (0,p^{-1/(p-1)}) \leftrightarrow B (0,p^{-1/(p-1)})$, so I suppose we have to restrict our attention to $|x|_p < p^{-1/(p-1)}$ to make sure that everything works.

Here's a couple of questions:

  • What is the easiest way to show that (1) and (2) coincide for $|x|_p < p^{-1/(p-1)}$? My suggestion is calculate the Taylor coefficients, which works both formally and non-formally. For $f (X) = \exp (Y\cdot \log (1+X))$ we have $$f^{(n)} (X) = \exp (Y\cdot \log (1+X))\cdot\frac{Y\,(Y-1)\cdots (Y-n+1)}{(1+X)^n},$$ so we see that the coefficients $f^{(n)} (0)/n!$ are exactly what we need... I guess it is easy enough?

  • Which of the two definitions is better: (1) or (2)? Well, if I understand well (?), (1) has a bigger domain of convergence. Is that correct?

Note that if we are interested in $\mathbb{Q}_2$ and $\alpha = 1/2$ (the typical application to finding square roots), then it seems like both approaches give the same result. In the denominators of ${1/2 \choose n}$ we have powers of $2$; namely, $$\left|{1/2 \choose n}\right|_2 = |1/2|^n_2\cdot |n!|_2 = 2^{2n-s_2 (n)},$$ meaning that the series converges for $x\in 8\mathbb{Z}_2$. Similarly, if we consider $$\exp (1/2\cdot \log (1+x))$$ then it also converges precisely for $x\in 8\mathbb{Z}_2$ (the logarithm converges on $4\mathbb{Z}_2$, but then there is the $1/2$ factor which leads to some troubles with convergency of the exponential, unless we take $x \in 8\mathbb{Z}_2$)... Well, it's not like we should expect something different, knowing what the squares in $\mathbb{Z}_2$ are.

Thank you for your answers and comments.

  • $\begingroup$ For $\alpha \in \mathbb{Q}$ you can use $\mathbb{C}$ to say the formal series coincide. Otherwise you can ask whether the p-adic functions are p-adically continuous in $\alpha$. Another question : if $\alpha \in 1/2$ then what is the structure of things like $\mathbb{Z}_p^\times/(\mathbb{Z}_p^\times)^2$ ? And what are the branches of $\log$ in $\mathbb{Q}_p, \mathbb{C}_p$, is there an equivalent of $2ik\pi$ ? $\endgroup$ – reuns Aug 17 '17 at 14:29
  • $\begingroup$ @reuns As for the questions, (1) $\mathbb{Z}_p / (\mathbb{Z}_p^\times)^2$, it is either the cyclic group $C_2$ for odd $p$, or $C_2\times C_2$ for $p = 2$. It is an application of Hensel's lemma. (2) There is no (lowbrow) $p$-adic analogue of $2i k \pi$. $\endgroup$ – AAA Aug 17 '17 at 14:43
  • $\begingroup$ @reuns It feels morally wrong to show that some formal identity holds simply because it holds in $\mathbb{C}$ :-) After all, the proof for $\mathbb{C}$ itself that we assume to know most likely works formally, and it is the formal identity that should explain all its non-formal incarnations. $\endgroup$ – AAA Aug 17 '17 at 14:47
  • $\begingroup$ Actually, it is an elementary exercise in (formal) undergraduate calculus to show that for $f (X) = \exp (Y\cdot \log (1+X))$ we have $f^{(n)} (X) = \exp (Y\cdot \log (1+X))\cdot\frac{Y\,(Y-1)\cdots (Y-n+1)}{(1+X)^n}$, so setting $X = 0$ we see that the Taylor coefficients are exactly what we need... The same argument works non-formally. Thus the two series coincide. I guess it is easy enough. $\endgroup$ – AAA Aug 17 '17 at 15:23
  • 2
    $\begingroup$ @reuns I believe it follows from Strassman's Theorem that there are no "multiple zeroes" for the exponential function in $\mathbb{Q}_p$. Indeed, for odd $p$, if $f(x) = e^{x/p} - 1 = \sum_{k=1}^\infty \frac{x^k}{p^k k!}$, then we have $a_N = \max |a_k| = a_1$, implying $f$ has at most one zero, namely $0$, in $\mathbb{Z}_p$, where it converges. For $\mathbb{C}_p$, the Weierstrass Preparation Theorem would have to be used probably, and, for $p=2$, $f(x)=e^{x/4}-1$ would have to be used for convergence purposes. The same result should hold. $\endgroup$ – Paul LeVan Aug 17 '17 at 19:50

We’re dealing with a principal unit $z=1+x$ where $|x|<1$, or equivalently $v_p(x)>0$; and with a $p$-adic integer $\alpha$. You ask about two methods of calculating $z^\alpha$, but you don’t say anything about the definition of this construct.

For my money, $z^\alpha$ is to be defined this way: Let $\{n_i\}$ be a sequence of positive integers with $p$-adic limit $\alpha$. Then $z^\alpha$ is defined to be $\lim_iz^{n_i}$. Of course you have to verify that the sequence $\{z^{n_i}\}$ is $p$-adically convergent, and that the result does not depend on the choice of the sequence $\{n_i\}$. From a computational standpoint, you might find that using this definition directly is a faster way of calculating $z^\alpha$ than the Binomial. For your sequence of positive integers, just use what you get by cutting off the $p$-ary expansion of $\alpha$.

I guess that your task is to show that the Binomial series approximation to $z^\alpha$ agrees with the definition for all $z$ in the open unit disk. As you recognize, any method involving the exponential series will be of much lesser applicability.

One other remark: if you’re only calculating $z^{m/n}$, surely the fastest method of computation is to use Newton-Raphson on $X^n=z^m$.

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    $\begingroup$ Thank you very much! You guessed the source of my confusion! I've seen several times the series $(1+x)^\alpha$ or $\exp (\alpha\,\log (1+x))$ being used to calculate powers, but it felt like a trick. The actual point is that these things are continuous in $\alpha$. E.g. it turns out (after writing down some identities with binomial coefficients) that $|(1+x)^\alpha-(1+x)^\beta|_p \le |\alpha-\beta|_p\cdot C$, where $C$ is some real constant depending on $x$ (actually, $C = |x|_p = 1/p^{-n}$ for $n\ge 1$ when $x\in p\mathbb{Z}_p$, but it's not necessary to restrict our attention to this case). $\endgroup$ – AAA Aug 18 '17 at 19:48
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    $\begingroup$ Right. You don’t even need to know the binomial coefficients explicitly, just that in $(s+t)^p$, all the intermediate coefficients are divisible by $p$. Same for the exponent $q=p^m$, in case you need to look at $(s+t)^q$: still all middle coeffs are divisible by $p$. $\endgroup$ – Lubin Aug 18 '17 at 20:59

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