Valuation of a particular element

Question

I am tying to compute the valuation of a particular element of $\mathbb{Q}_p$.

I am trying to compute $\operatorname{val}_p(P)$ where $P=\frac{\log(1+p^2)}{\log(1+p)}$ and $\log$ is the $p$-adic $\log$ map.

So $\operatorname{val}_p(P)=\operatorname{val}_p(\log(1+p^2))-\operatorname{val}_p(\log(1+p))$. But I think

$\operatorname{val}_p(\log(1+p^2))=\operatorname{val}_p(1+p^2)=0$ and $\operatorname{val}_p(\log(1+p))=0$, so $\operatorname{val}_p(P)=0$.

Is my solution right?

Answer 1

Recall that the logarithm's definition

$$\log(1+x)=\sum_{n=1}^\infty {(-1)^{n+1}x^n\over n}$$

gives the first term in the series for $\log(1+p^k)$ as having (additive) valuation $k$, since

$$\log(1+p^k)=-p^k+O(p^{k+1}), p>2$$

and the strong triangle inequality says this is all you get in this case. If $p=2$ you get $\log(1+p)$ to have valuation $2$ since the first two terms cancel, and the fourth reaches the minimum valuation at ${2^4\over 4}=4$ (valuation $2$) so that $\log\left({1+p^2\over 1+p}\right)$ has (additive) valuation $2-1=1$ in the odd prime case, and (additive) valuation $2-2=0$.