# Valuation of the p adic logarithm

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 $$\frac{1}{p^{s}(p-1)} for some $$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 $$n we have $$p^{s}v_{p}(x)-s. But I don't have succeed. I was trying to play with the inequalities, but the most closely to my objective is that $$nv_{p}(x), and since $$n we have $$v_{p}(n) and so $$nv_{p}(x)-s.

Can anyone give me some hint? I will appreciate. Thanks

• What is the relation between the $\,\log(1+x)\,$ and $\,\log(1+x^p)\,$ series? – Somos Dec 8 '18 at 19:50

## 1 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 $$\upsilon$$.