Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 26 / 1

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 $T^0=1$.