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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 $R = \mathbb{Z}_p[\zeta_{p^k}]$?

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  • $\begingroup$ Let $p^k (p^f-1)$ be the order of the torsion subgroup of $R^\times$ and $r > e/(p-1)$ . Then $\log_p(1+\mathfrak{m})$ is the disjoint union of $p^{f(r-1)-k}$ translations of $\mathfrak{m}^r$ . $\endgroup$
    – jjcale
    Nov 22, 2021 at 18:56

2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$
    – KConrad
    Aug 11, 2017 at 5:37
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    $\begingroup$ 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 $\lambda$. $\endgroup$
    – Lubin
    Aug 11, 2017 at 14:31
  • $\begingroup$ @Lubin : Is the following formular true for your example ? : $$ \log_2(1+\mathfrak{m}) = \coprod_{a \in \{0,1\}^5} ((\sum_{k=1}^5 a_k \log_2(1+2^{k/6})) + 2^{7/6} R)$$ $\endgroup$
    – jjcale
    Nov 21, 2021 at 15:01
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    $\begingroup$ 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! $\endgroup$
    – Lubin
    Nov 22, 2021 at 2:18
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Let $R$ be the ring $\mathbb{Z}_p[\zeta]$, where $\zeta = \zeta_{p^k}$. Let $\pi$ be the uniformizer $\zeta - 1$ of $R$, so that $|\pi|^{(p-1)p^{k-1}} = |p|$.

Write $D_r$ for the disk $\{x \in R: |x - 1| < |\pi|^r\}$. Then the morphism $\log$ is an isomorphism on $D_{p^{k - 1}}$, which has index $p^{p^{k-1}}$ in $D_0$.

On the other hand, the roots of unity in $D_0$ are exactly all the $p^k$-th roots of unity. Consequently, in the case $k = 1$, one concludes that $D_0$ is isomorphic to $\mathbb{Z}/p\mathbb{Z} \times D_1$, and the image of the $\log$ map is the disk $\{x \in R: |x| < |\pi|\}$.


To get a feeling of what happens in the general case, let us look at the case $p = 3$ and $k = 2$. The group $\mu$ of roots of unity in $D_0$ has $9$ elements, but the quotient $D_0 / D_3$ has $27$ elements. So the quotient $D_0 / \mu D_3$ has three elements, generated by $1 - \pi$. One calculates: \begin{eqnarray*} \log(1 - \pi) & = & \log(1 - \pi)(1 + \pi) \\ & = & \frac{1}{3}\log(1-\pi^2)^3 \\ & = & \frac{1}{3}\log(1 - \pi^6 + \cdots) \\ & = & \frac{1}{3}(-\pi^6 + \cdots) \end{eqnarray*} which has absolute value $1$.

So in this case, the image of $\log$ consists of three disks, contained in $\mathfrak{m}$, $1 + \mathfrak{m}$, $2 + \mathfrak{m}$, respectively, and all of the same size as $D_3$.

The general case should be similar. For given $p$ and $k$, the image can be calculated with an algorithm. It then depends on what other information one needs.

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