Definition of $a^b$ for $a\in C_p$ and $b\in\mathbb{Z}_p^*$

Question

In a paper I found the notion $(1+T)^b\in \mathbb{Q}_p[\![T]\!]$, where $b\in\mathbb{Z}_p^*$. It is implied that this powerseries converges for elements with norm $<1$. I am now trying to define $a^b$ for $a\in C_p$, the completion of the algebraic closure of $\mathbb{Q}_p$, and $a\in \mathbb{Q}_p[\![T]\!]$. I tried this by defining $b^a:= \exp(b \log(a))$ but the $p$-adic exponential function is only defined on elements with norm $<p^{-1/(p-1)}$. I would like to define $a^b$ for elements $a\in C_p$ with $|a|<1$. Is this possible and if yes how?

Answer 1

This is something I know a little about.

(1) If you’re taking about series over a $p$-adic field, $(1+x)^z$ is well defined as a series in $\Bbb Z_p[[x]]$ when $z\in\Bbb Z_p$. Note that the exponent need not be in $\Bbb Z_p^*$. Note also that the coefficients of the series are in $\Bbb Z_p$, not merely in $\Bbb Q_p$. As Jyrki suggested, one way to construct this series is to find a sequence of positive integers $\{n_i\}$ whose limit is the $p$-adic integer $z$, and form the limit of the series (they are actually polynomials!) $(1+x)^{n_i}$, that is $$ \text{when }\lim_in_i=z\,,\qquad(1+x)^z=\lim_i(1+x)^{n_i}\,. $$ The limit of series is to be taken coefficientwise, not uniformly in $i$.Of course there’s a lot to be proved: that the limit of the polynomials does exist, and that the final result doesn’t depend on the sequence of integers going to $z$. None of this is hard.

(2) Another way of defining $(1+x)^z$ is to use the good old Binomial expansion: $$ (1+x)^z=\sum_{n=0}^\infty\binom znx^n\,,\quad\text{where }\binom zn=\frac{z(z-1)\cdots(z-n+1)}{n!}\,. $$ Here, you must note that if $z\in\Bbb Z_p$, the polynomial $\binom zn$ takes values in $\Bbb Z_p$. Why? Because if $z\in\Bbb Z^{\ge0}$, then $\binom zn\in\Bbb Z$, and you use continuity of the polynomial function $\binom zn$.

(3) If you’re worrying about the domain of convergence of the series $(1+x)^z$, where $z\in\Bbb Z_p$, then since the coefficients are all in $\Bbb Z_p$, you have a series convergent for all $x$ with $|x|_p<1$. If $z\notin\Bbb Z^{\ge0}$, then I believe the domain of convergence will be no larger than what I just specified.

(4) If you want to define $(1+a)^b$ for $a\in\Bbb C_p$ and $|a|_p<1$, you may start by taking the logarithm of $1+a$: this is well defined, using the familiar series for $\log(1+x)$. But you must be warned that the range (image) of the log is all of $\Bbb C_p$ (!!) This is easily seen, using the Newton polygon of $\log(1+x)-\zeta$ for your arbitrary given $\Bbb C_p$-number $\zeta$. And of course the logarithm is many-to-one (in fact $\infty$-to-one) in this context, so you’re not going to be able to make a choice of an antilogarithmic value. I don’t think that you’ll get far in defining $(1+a)^b$ when $b$ is not a $p$-adic integer—there’s some formal-group magic going on there.

(5) Returning to the definability of the logarithm, you see that $\log(\xi)$ must be zero for any root of unity $\xi$. You may be disappointed at this, but the fact makes it possible to define the logarithm as $\log:\Bbb C_p^*\to\Bbb C_p$ by the simple expedient of proclaiming that $\log(p)=1$. Arbitrary, yes, but also natural. With this, you define $\log(p^\mu)=\mu$ for any rational number $\mu$; the multivaluedness of $z^\mu$ doesn’t matter ’cause two different valued are related by a root of unity, and these have trivial logarithm. Then every $z\in\Bbb C_p^*$ may be written $z=p^\mu\xi(1+a)$ with $\mu$ rational, $\xi$ a root of unity, and $|a|<1$. for this, you need that the residue field of $\Bbb C_p$ is the union (direct limit) of finite fields. And so the log of $z$ is well defined, as a $\Bbb C_p$-number.