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Put $X=1+T$ and let $\phi$ be the Frobenius substitution on cyclotomic power series, $\phi (g)(T)=g(X^p-1)$. It is injective: modulo $p$ it is the injective map $\overline{g}(T)\mapsto \overline{g}(T^p)$. If $\phi (g)=0$, this makes every coefficient of $g$ divisible by $p$; dividing by $p$ and repeating makes every coefficient zero.

We prove that $R$ is a finite free module over $S=\phi (R)$ of rank $p$, with basis $1,X,\dots ,X^{p-1}$. Modulo $p$, every series has a unique decomposition over ${\mathbb{F}}_p[[T^p]]$ in the basis $1,T,\dots ,T^{p-1}$. Changing to the powers of $X=1+T$ is an invertible triangular change of basis. Thus, modulo $p$, every $h\in R$ can be expressed as ${\sum }_{i=0}^{p-1}{X}^i\phi (h_i)$. Lift the coefficients to $R$, subtract this expression, and divide the remainder by $p$. Repeating and summing the p-adically convergent coefficient series givesUniqueness follows by reduction modulo $p$ and successive division by $p$. This proves finite freeness, not just a formal invariance of the right side in the question.

For $f\in R^{\times }$, multiplication by $f$ is an invertible $S$-linear map $m_f:R\to R$. Its determinant $b={\det }_S(m_f)$ belongs to $S^{\times }$. DefineThis is the norm for a finite free ring extension, pulled back along the injective substitution, and it is multiplicative and unit-valued.

To identify its determinant with the desired product, extend coefficients to $\mathcal{O}={\mathbb{Z}}_p[{\mu }_p]$. The substitutions $X\mapsto \xi X$ for ${\xi }^p=1$ are continuous automorphisms of $\mathcal{O}[[T]]$ fixing $\phi (\mathcal{O}[[T]])$. Evaluation of a series at $\xi X-1$ is legitimate in the $(\xi -1,T)$-adic topology, because the constant term $\xi -1$ is topologically nilpotent. On fraction fields this is a degree-$p$ extension with these $p$ distinct automorphisms: finite freeness bounds the degree by $p$, while the automorphisms force degree at least $p$. The determinant norm therefore equals the product of its conjugates. Norms commute with this scalar extension, soas required. Finally, any other unit-valued map satisfying this identity has the same image under $\phi$ for every $f$, and injectivity of $\phi$ makes it identical to $N$. The Coleman norm operator exists and is unique.

Put $X=1+T$ and $Y=X^p-1$, so that $\varphi (f)=f(Y)$. The essential integrality statement for the Frobenius substitution on cyclotomic power series is the decomposition into a finite free module:To prove it, reduce modulo $p$. Since $Y\equiv T^p$, grouping the coefficients of a formal power series according to their exponents modulo $p$ gives the unique expression ${\sum }_{j=0}^{p-1}{T}^j{h}_j(T^p)$. The change from $1,T,\dots ,T^{p-1}$ to $1,X,\dots ,X^{p-1}$ is a triangular matrix with diagonal entries $1$, so this is also a basis over ${\mathbb{F}}_p[[T^p]]$. Lift these $p$ coefficient series to $R$, subtract their contribution from $f$, divide the remaining error by $p$, and repeat. The sums of the successive lifts converge in the finer of the topologies on integral formal power series, because $R$ is complete in that topology. This proves existence of the decomposition over $R$. A relation among its summands reduces to the zero relation modulo $p$, forcing each coefficient series to be divisible by $p$. Repetition forces divisibility by every power of $p$, hence every coefficient series is zero. This proves uniqueness and, in particular, injectivity of $\varphi$.

For $f={\sum }_{i=0}^{p-1}{X}^i\varphi (f_i)$, defineThis is a ${\mathbb{Z}}_p$-linear map. To calculate the average over roots of unity, work over $\mathcal{O}={\mathbb{Z}}_p[{\zeta }_p]$. The substitutions $T\mapsto \zeta X-1$ are legitimate: their constant terms are topologically nilpotent, so the defining coefficient sums converge in $\mathcal{O}$. Since $(\zeta X{)}^p=X^p$, each $\varphi (f_i)$ is unchanged by the substitution. The sums of the powers of the roots of unity are $p$ for $i=0$ and zero for $1\le i<p$. ConsequentlyThus the average is integral and belongs to $\varphi (R)$, and the constructed Coleman trace operator has the required property. Injectivity of $\varphi$ forces uniqueness of any map satisfying that property. Finally, the decomposition of $\varphi (g)$ has $f_0=g$ and all other coefficients zero, giving $S\circ \varphi ={\mathrm{id}}_R$.

We will also use the trace of multiplication by $h$ on this finite free module. After extending scalars to split the conjugates, that trace is the sum just computed. Hence

Set $X=1+T$ and let $\phi :R\to R$ be the Frobenius substitution on cyclotomic power series, $\phi (g)(T)=g(X^p-1)$. It is injective: modulo $p$ the substitution is $g(T)\mapsto g(T^p)$, which is injective; if $\phi (g)=0$, this first gives $g\in pR$, and iterating gives $g\in {\bigcap }_r{p}^{r}R=0$.

We first prove that every formal power series has a unique expansionModulo $p$, the ring ${\mathbb{F}}_p[[T]]$ is free over ${\mathbb{F}}_p[[T^p]]$ with basis $1,T,\dots ,T^{p-1}$: group the coefficients by their exponents modulo $p$. Replacing this basis by $1,X,\dots ,X^{p-1}$ gives an invertible triangular change of basis with diagonal entries one. Lift the resulting coefficients to $R$, subtract their displayed expression from $f$, divide the remainder by $p$ and repeat. The p-adic limits of these corrections give the required $f_i$ because $R$ is p-adically complete. If an expression is zero, reduction modulo $p$ forces all $f_i$ to be divisible by $p$; repetition forces every $f_i$ to vanish. This proves both existence and uniqueness without dividing coefficients by $p$.

Define the Coleman trace operator by $\psi (f)=f_0$. To verify its averaging formula, work temporarily over ${\mathbb{Z}}_p[{\mu }_p][[T]]$, interpreting the substitutions in the $(\xi -1,T)$-adic topology. The argument $\xi X-1$ is topologically nilpotent there, so substituting into an arbitrary formal power series is legitimate. MoreoverThe sum of ${\xi }^i$ over the p-th roots of unity is $p$ at $i=0$ and zero at $1\le i<p$. ThereforeThe right side belongs to $R$, so the apparent denominator causes no integrality problem. Injectivity of $\phi$ proves that this is the unique map $\psi$ with the required identity. The construction also proves $\psi (\phi (g)f)=g\psi (f)$ and $\psi (\phi (g))=g$.

Write a bar for reduction modulo $p$. For $n\ge 1$,The binomial theorem in characteristic $p$ givessince $\left(\genfrac{}{}{0ex}{}{p-1}{i}\right)\equiv (-1{)}^i(\mathrm{mod}p)$ and $p$ is odd. Taking the zeroth basis coefficient yieldsThe endpoint case $n=1$ includes the constant term $T^0=1$.