Coleman norm operator ($\phi (\mathcal{N}f)={\prod }_{{\xi }^p=1}f(\xi (1+T)-1)$)

The determinant norm for the finite free extension $R/\phi (R)$ defines $\mathcal{N}$ on units. After adjoining the p-th roots of unity, the norm is the displayed product of conjugates. A Coleman power series of a norm-compatible unit sequence satisfies $\mathcal{N}f=f$. For symmetric cyclotomic units, the product relation follows directly from ${\prod }_{{\xi }^p=1}(1-{\xi }^a{X}^a)=1-X^{ap}$ when $p\nmid a$.

Every unit $\overline{f}\in {\mathbb{F}}_p[[T]{]}^{\times }$ has a unique Coleman norm operator-fixed lift to ${\mathbb{Z}}_p[[T]{]}^{\times }$. Choose any lift $f$ and use $Nf\equiv f(\mathrm{mod}p)$. The consecutive quotients of the iterates are $N^n(Nf/f)\in 1+p^{n+1}R$ by Coleman norm contraction, so they converge coefficientwise uniformly in the p-adic integers. Their limit is norm-fixed by continuity. Two such lifts have norm-fixed quotient in $1+pR$ and therefore coincide. Uniqueness also makes the lifting map multiplicative and independent of the initial lift.

Let $W$ be the subgroup fixed by the Coleman norm operator. The Coleman norm-fixed lift gives the unique factorization $f=w(\overline{f})v$, with $v\in 1+pR$. Multiplication consequently gives the displayed direct product of groups, while reduction identifies $W$ with ${\mathbb{F}}_p[[T]{]}^{\times }$. This is a decomposition of unit groups, not an identification of a power-series ring with a product of groups.

For odd $p$, the Coleman norm operator satisfies the displayed inclusion for every $k\ge 1$. Write $M_h$ for multiplication by $h$ on the finite free module $R$ over $\varphi (R)$. Its trace is $p\varphi (S(h))$, where $S$ is the Coleman trace operator. In $\det (I+p^k{M}_h)$ the linear term is therefore divisible by $p^{k+1}$, and all higher terms are divisible by $p^{2k}$. Congruence reflection for Frobenius substitution gives the assertion. A norm-fixed element of $1+pR$ lies in $1+p^{k}R$ for every $k$ and hence equals $1$.