Frobenius-corrected logarithm (${\mathcal{L}}_f(t)=\frac{1}{p}\log \frac{f(t{)}^p}{f([\pi ]t)}$)

Assume $K_{\mathfrak{p}}={\mathbb{Q}}_p$, $p>2$, $f(t)\in {\mathbb{Z}}_p[[t]{]}^{\times }$, and $[\pi ]t\equiv t^p(\mathrm{mod}p)$. Then the displayed ratio lies in $1+p{\mathbb{Z}}_p[[t]]$, so its logarithm divided by $p$ is integral. A formal-group isomorphism over the completed unramified coefficient ring transfers it to an integral Amice transform. A norm relation makes the corresponding p-adic measure supported on units. Its $k$th moment has the Euler correction $1-{\pi }^k/p$.

Let $R={\mathbb{Z}}_p[[T]]$, $p$ odd, and $\varphi (T)=(1+T{)}^p-1$. An element $g\in R$ is the Frobenius-corrected logarithm of a unit $f$ with $f(0)\in 1+p{\mathbb{Z}}_p$ if and only if $g^{\prime }(0)=0$. Necessity follows by differentiating at zero, using ${\varphi }^{\prime }(0)=p$. For sufficiency put $H=\exp (pg)\in 1+pR$ and solve $f^p=H\varphi (f)$. Choose $f_0=\exp (p{g}_0/(p-1))$. The linear equation reduces to $H_1=0$, so $f_1$ is free. In degree $n\ge 2$ the unknown $f_n$ has coefficient $p{f}_0^{p-1}(1-p^{n-1})$. The remaining terms are divisible by $p$ because reduction of any integral truncated $f$ satisfies $f^p\equiv \varphi (f)(\mathrm{mod}p)$. Division gives a unique integral $f_n$. Two lifts differ by $(1+T{)}^c$, $c\in {\mathbb{Z}}_p$, as follows from the logarithmic equation in the coordinate $z=\log (1+T)$. If also $Sg=0$, the Coleman trace operator gives $\log (Nf/f)=pSg=0$. As $Nf/f\in 1+pR$, injectivity of the p-adic logarithm implies $Nf=f$.