TITLE: Is it possible to describe the image of the $p$-adic logarithm on $1+\mathfrak{m}$, where $\mathfrak{m}$ is the maximal ideal of a $p$-adic field?
QUESTION [10 upvotes]: Let $R$ be the ring of integers of some finite extension of $\mathbb{Q}_p$. In particular, I'm interested in the case $R = \mathbb{Z}_p[\zeta_{p^k}]$ (the totally ramified extension of $\mathbb{Z}_p$ of degree $(p-1)p^{k-1}$)
Let $\pi$ be a uniformizer of $R$, and $\mathfrak{m} = (\pi)$ the maximal ideal.
The $p$-adic logarithm on $1+\mathfrak{m}$ is the function:
$$\log_p : 1+\mathfrak{m}\rightarrow \mathfrak{m}$$
given by
$$\log_p(1+x) = \sum_{n\ge 1}\frac{(-1)^{n-1}}{n}x^n$$
which clearly converges on $1+\mathfrak{m}$, and whose image lies in $\mathfrak{m}$.
If the ramification index of $R/\mathbb{Z}_p$ is $e$, then 
$$\log_p|_{1+\mathfrak{m}^r} : 1+\mathfrak{m}^r\longrightarrow\mathfrak{m}^r$$
is an isomorphism for any $r > e/(p-1)$, but is in general neither injective nor surjective on $\mathfrak{m}$.
Is it possible to describe in general the image $\log_p(1+\mathfrak{m})\subset\mathfrak{m}$ of $\log_p$? (or at least in the case of $R = \mathbb{Z}_p[\zeta_{p^k}]$?

REPLY [4 votes]: It is certainly not true that the logarithm maps into the maximal ideal of the integers, when the ramification index is large. For instance, if $p=2$ and $\lambda^6=2$, you get $v_2(\log(1+\lambda))=-5/3$, where $v_2$ is
the $2$-adic valuation normalized so that $v_2(2)=1$.
Indeed, for all $z$ with $v_p(z)$ unequal to any number $\frac1{(p-1)p^m}$, $v_p(\log(1+z))$ is given by the (function whose graph is the boundary of the) Newton copolygon.
The copolygon encodes the same information as the polygon, but in a way that may be more useful for some purposes. Starting with a convergent power series $f=\sum_nc_nx^n$, you take, in the right-hand Cartesian half-plane, the intersection of all the sets given by $\eta\le n\xi+v(c_n)$. The result is a closed convex set with polygonal boundary. The vertices have for their $\xi$-coordinates the negative slopes of the Newton polygon’s segments, and the segments have for their slopes the first coordinates of the Newton polygon’s vertices. Call $\psi$ the function whose graph is the boundary of the copolygon.
Now, if $v(\lambda)$ is unequal to the $\xi$-coordinate of any vertex of the copolygon of $f$, then $v(f(\lambda))=\psi(v(\lambda)$, as is easily seen.
The fact that the copolygon of the logarithmic series $x-x^2/2+x^3/3-\cdots$ descends infinitely far in the open right-hand half-plane shows
that the possible valuations of $\log(1+\lambda)$ are all real numbers. But if you don’t like copolygon talk, you can prove directly by looking at the polygon of $\log(1+x)-\mu$, for a $\mu$ in an algebraic closure of $\Bbb Q_p$, or even $\mu\in\Bbb C_p$, that the logarithm maps onto the specified algebraically closed field.