Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 27 / 1

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.

The shifted cyclotomic polynomial ${\Phi }_{p^{n+1}}(1+T)$ is Eisenstein at $p$: its constant coefficient is $p$, and modulo $p$ it is $T^{p^n(p-1)}$. Thus $[K_n:{\mathbb{Q}}_p]=p^n(p-1)$, ${\zeta }_n-1$ is a uniformizer, and $[K_n:K_{n-1}]=p$ for $n\ge 1$. The $p$ automorphisms over $K_{n-1}$ send ${\zeta }_n$ to $\xi {\zeta }_n$ for $\xi \in {\mu }_p$.

Since ${\zeta }_n-1$ has positive valuation, evaluating an integral formal power series there converges. A unit series has unit constant coefficient and evaluates to a unit. Evaluate the Coleman norm operator identity at $T={\zeta }_n-1$:The final equality uses norm fixation and the compatible choice of roots. HenceIn particular, norm fixation of a single Coleman power series produces a norm-compatible sequence of local units.