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Let $ K $ be a $ \mathfrak{p} $-adic extension of $ \mathbb{Q}_p $ with ramification index $ e $. Denote the extension of the $ p $-adic valuation $ v_p $ to $ K $ again by $ v_p $, then $ v_\mathfrak{p} = ev_p $ is the normalized valuation on $ K $. For integers $ j $ > 1 and $ v_p(z) > \frac{e}{p-1} $ we have (after some calculation, not so difficult) $$ v_p \big( \frac{z^j}{j} \big) - v_p(z) > 0, $$ hence $ v_p(\log(1+z)) = v_p(z) $. I'm not sure how this follows? And also, the conclusion is that for large $ n $, $ \log $ maps $ U^{(n)} $ into $ \mathfrak{p}^n $, is this because $ v_p $ is injective?

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I'll use "absolute value" notation $|\ |_p$ instead of valuation notation $v_p$.

$\log(1+z)=z-z^2/2+z^3/3-\cdots$. Then $|z^j/j|_p<|z|_p$ for $j>1$. As the valuation is ultrametric then $|z|_p>|-z^2/2+z^3/3-\cdots|$, In an ultrametric valuation $|a|>|b|$ implies $|a+b|=|a|$, so applying this so $a=z$ and $b=-z^2/2+z^3/3-\cdots$ we get $|\log(1+z)|_p=|z|_p$.

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  • $\begingroup$ Thank you, I just realized this as well, but it's good to have confirmation. What about the second part? $\endgroup$ – user228960 May 17 '18 at 16:07
  • $\begingroup$ I'm not sure "into" here connotes injectivity. As far as I can tell, your second question follows from $v_p(a)\ge n/e$ implies $v_p(\log(1+a))\ge n/e$ for $n\gg0$. $\endgroup$ – Angina Seng May 17 '18 at 16:32
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Unlike @LordSharktheUnknown, I’ll hold to the additive notation $v=v_p$, under which $v_p(p)=1$. For the purpose of talking about the logarithm, the normalized valuation is not of interest. After all, $v$ extends very nicely and without ambiguity throughout the algebraic closure, even unto the completion thereof, $\Bbb C_p$.

The burden of your question seems to be just why $v(\log(1+z))=v(z)$ when $v(z)>\frac1{p-1}$. You see this by looking closely at the series expansion of $\log(1+z)$: $$ \log(1+z) = z-\frac{z^2}2+\cdots+\frac{z^p}p+\cdots\,, $$ where, first, the coefficients of monomials of degree $d$ with $1\le d<p$ all are $p$-adic units, so with valuation zero, so that $v(z^n/n)=nv(z)$ for these values of $d$, and $v\bigl(\sum_1^{p-1}z^n/n\bigr)$ is the smallest of these, namely $v(z)$. I say that for $n>p$, we also have $v(z^n/n)>v(z)$: this is the same as $(n-1)v(z)>v(n)$, easily enough seen from the inequalities $n>p$ and $v(z)>\frac1{p-1}$. Thus in the infinite sum, the monomial $z$ dominates, in the sense that $v(z)$ is less than $v(z^n/n)$ for all $n>1$. But the point of a nonarchimedean valuation is that in any sum, if one term has smaller $v$-value than any of the others, the valuation of this term is the valuation of the sum.


Let me point out here that there’s a more advanced expression for $\log(1+z)$ that lets you know at a glance what $v(\log(z))$ is: it’s a convergent infinite product in $\Bbb Q_p[[z]]$, namely $$ \log(1+z) = z\prod_{n=1}^\infty\frac{(1+z)^{p^n}-1}{p\bigl((1+z)^{p^{n-1}}-1\bigr)}\,. $$ Here each fraction is actually a $\Bbb Q_p$-polynomial related to the cyclotomic polynomial $\Phi_{p^n}$, and the convergence is not in the usual power-series topology, but rather $p$-adically coefficientwise. This corresponds to uniform convergence on proper open subdisks of the $p$-adic unit disc of $\Bbb C_p$. The first factor, for instance, is $$ \frac{(1+z)^p-1}{pz}=1+\frac{p-1}2z+\frac{(p-1)(p-2)}6z^2+\cdots+z^{p-2}+\frac{z^{p-1}}p\,, $$ which you easily see is a $p$-adic unit if $v(z)>\frac1{p-1}$. The other factors also have constant term $1$ and are $p$-integral except for the last monomial, which is $$\frac{z^{p^n-p^{n-1}}}p\,.$$ All this tells you what the roots of the logarithm are, and more. But it certainly says that $v(\log(1+z))=v(z)$ under the conditions you are interested in.

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