I'm stuck in some propertie about the $p$-adic logarithm. The propertie comes from a proposition in a Book by Dwork which I'm studying. The proposition says:
If $v_{p}(x)>\frac{1}{p-1}$, then $v_{p}(\log(1+x))=v_{p}(x)$. And if
$$\frac{1}{p^{s}(p-1)}<v_{p}(x)<\frac{1}{p^{s-1}(p-1)}$$
for some $s\geq 1$, then
$$v_{p}(\log(1+x))=p^{s}v_{p}(x)-s\text{.}$$
Here $v_{p}$ denote the $p-$adic valuation and we define
$$\log(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}x^{n}$$
which converges for $|x|_{p}<1$, or, equivalently, for $v_{p}(x)>0$.
I have the first part, and the reasoning suggest that we must verify that for every $n<p^{s}$ we have $p^{s}v_{p}(x)-s<p^{n}v_{p}(x)-v_{p}(n)$. But I don't have succeed.
I was trying to play with the inequalities, but the most closely to my objective is that $nv_{p}(x)<p^{s}v_{p}(x)$, and since $n<p^{s}$ we have $v_{p}(n)<s$ and so $nv_{p}(x)-s<p^{s}v_{p}(x)-v_{p}(x)$.
Can anyone give me some hint? I will appreciate.
Thanks
Best Answer
I’ll give but a hint. Look at the actual series for $\log(1+x)$. Now, given a number $x$ for which $v_p(x)$ is in one of the specified open intervals, see what the value of each monomial in the series turns out to be. You’ll see that only one monomial takes on a minimum value. Now use the fact that even in an infinite sum, if one term has a $v_p$-value (say $\upsilon$) less than all the others, then the $v_p$-value of the sum is $\upsilon$.