finding "exp(1)" in the p-adic numbers

Question

Can anyone give me some kind of description when, in $\mathbb{C}_p$(the completion of the algebraic closure of the p-adic numbers), there is an element $x$ which satisfies $ Log_p(x) = 1, $ and in case there is, how can one find him?(Obviously, if exists, this element is not unique.)

Answer 1

The following is essentially a quote of A Course in $p$-adic Analysis, section 5.4.1.

The problem is that the radius of convergence of the $\exp_p$ series verifies $$r_2=\frac12,\qquad\qquad \frac1p<r_p<1,\qquad (p>2).$$ A possible solution is taking $$\exp_p(1)=\exp_p(n/n)=\root n\of{\exp_p(n)}$$ for $n$ ($p$-adically)small enough. Namely, for $p>3$ we have $|p|_p=1/p<r_p$, the series of $\exp_p(p)$ converges and we can define $$\exp_p(1)=\root p\of{\exp_p(p)}.$$ (some root, no canonical choice)

Similarly, $\exp_2(1)=\root 4\of{\exp_2(4)}$ works.

If you are interested in concrete cases, PARI/GP can do calculations with $p$-adic numbers: https://www.math.lsu.edu/~verrill/teaching/math7280/padics_in_pari.pdf.

Answer 2

There’s another way of finding numbers $e$ such that $\log_p(e)=1$, more direct but I guess depending on things you may not have seen. Just take the $p$-adic logarithm series, $$ L(1+x)=\log(x)=x-x^2/2+x^3/3-x^4/4+\cdots\,, $$ which of course is just the series you know from Calculus. Now there are direct ways of finding roots of the equation $G(x)=L(x)-1=0$. You draw the Newton Polygon of $G$, this amounts to plotting the origin plus all the points $(n,v_p(1/n))$, where $v_p$ is the additive $p$-adic valuation, measuring how divisible by powers of $p$ a number is. The points that count are $(0,0)$ and all $(p^k,-k)$ for $k\ge0$. Draw a segment from each of these points to the next, and that’s the Newton Polygon. The segments are of length $p$, $p^2-p$, $p^3-p^2$, etc., with slopes the negative reciprocals of these numbers. There’s a $\mathbb Q_p$-factor, monic (in fact irreducible in this case), for each segment, of degree equal to the width of the segment, and with all roots of $p$-adic valuation the negative of the slope. In particular, you get an irreducible monic polynomial factor of degree $p$ with roots $\sigma$ of valuation $1/p$. These roots $\sigma$ have the property that $\log_p(1+\sigma)=1$.

With a computation package, I found that for $p=3$, this first monic irreducible polynomial is very close to $X^3+3X^2+3X+15$, so if you start with one of the roots of this polynomial, you can apply Newton-Raphson (the reciprocal of the derivative of your series is $1+x$, so the computation is easy), to get a root very quickly.