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


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 '17 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 '17 at 14:31

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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