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Nov 22, 2021 at 2:18 comment added Lubin I have not checked your formula, @jjcale , but I have to remind you that when $v(z)$ is equal to the valuation of a root of the function (log in this case), the Newton copolygon does not give much information, because isn that case, there are two (or more) monomials whose evealuations give the same $v$-value. Here, I think the information from the N-copoly is misleading for $v(x)=1/2$. Check it out!
Nov 21, 2021 at 15:01 comment added jjcale @Lubin : Is the following formular true for your example ? : $$ \log_2(1+\mathfrak{m}) = \coprod_{a \in \{0,1\}^5} ((\sum_{k=1}^5 a_k \log_2(1+2^{k/6})) + 2^{7/6} R)$$
Aug 11, 2017 at 14:31 comment added Lubin Thanks, @KConrad: In cyclotomic fields, call $\pi_n=\zeta_{p^n}-1$, so that $v(\pi_n)=1/((p-1)p^{n-1})$. Then for $p>2$, $v(\log(1+\pi_n^2))=1-n+2/(p-1)$; for $p=2$, $v(\log(1+\pi_n^3))=7/2-n$, formula valid for $n\ge5$. Perhaps I should have avoided all the copolygon talk by pointing out that for the logarithm, the crucial monomials are those in powers of $p$, and you need only see which of these monomials dominates when you plug in your $\lambda$.
Aug 11, 2017 at 5:37 comment added KConrad Since the question expressed specific interest in $p$-power cyclotomic extensions of $\mathbf Q_p$, it would be good if the specific example at the start were taken from such a field.
Aug 11, 2017 at 4:23 history answered Lubin CC BY-SA 3.0