Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 29 / 1

The structural facts used about a non-Archimedean local field are that it is complete for a normalized discrete valuation $v:K^{\times }\to \mathbb{Z}$, its valuation ring ${\mathcal{O}}_K=\{x:v(x)\ge 0\}$ is a discrete valuation ring, its maximal ideal is ${\mathfrak{m}}_K=\pi {\mathcal{O}}_K$ for a uniformizer $\pi$, and its residue field is a finite field $k={\mathbb{F}}_q$. Polynomial operations are continuous for this valuation topology.

A simple-root version of Hensel lemma is: if $f\in {\mathcal{O}}_K[X]$, $a_1\in {\mathcal{O}}_K$, $f(a_1)\in {\mathfrak{m}}_K$, and $f^{\prime }(a_1)\notin {\mathfrak{m}}_K$, then there is a unique $a\in {\mathcal{O}}_K$ with $f(a)=0$ and $a\equiv a_1(\mathrm{mod}{\mathfrak{m}}_K)$.

Construct $a_n$ with $f(a_n)\in {\pi }^n{\mathcal{O}}_K$. Given $a_n$, put $a_{n+1}=a_n+{\pi }^n{b}_n$. Taylor expansion over the valuation ring givesThe omitted terms contain ${\pi }^{2n}$, and $2n\ge n+1$ for $n\ge 1$. The residue of $f^{\prime }(a_n)$ remains the nonzero residue of $f^{\prime }(a_1)$, so choose $b_n$ withThis makes $f(a_{n+1})\in {\pi }^{n+1}{\mathcal{O}}_K$. The congruences $a_{n+1}\equiv a_n(\mathrm{mod}{\pi }^n)$ make $(a_n)$ a Cauchy sequence; completeness gives its limit $a\in {\mathcal{O}}_K$. Continuity and $v(f(a_n))\ge n$ give $f(a)=0$, and the first congruence gives its specified residue class.

For uniqueness, if $a,b$ are two roots in that class, polynomial division of their difference givesThe second factor is a unit, since $f^{\prime }(b)$ is a unit and $a-b\in {\mathfrak{m}}_K$. Thus $a=b$. This proves both existence and uniqueness in the stated Hensel lemma.

Write the finite residue field as $k={\mathbb{F}}_q$, with $q=p^f$ and residue characteristic $p$. Every $\bar{a}\in k$ satisfies ${\bar{a}}^q=\bar{a}$. The polynomial $F(X)=X^q-X$ has derivative $q{X}^{q-1}-1$, whose reduction is $-1$. Hence Hensel lemma gives a unique lift $[\bar{a}]\in {\mathcal{O}}_K$ such thatThis includes $[0]=0$ and $[1]=1$. If $\bar{a}\ne 0$, its lift is a unit and satisfies $[\bar{a}{]}^{q-1}=1$.

The product $[\bar{a}][\bar{b}]$ also satisfies $X^q=X$ and has residue $\bar{a}\bar{b}$. Uniqueness therefore gives $[\bar{a}\bar{b}]=[\bar{a}][\bar{b}]$. Reduction is a left inverse of this map, so the map is injective. Restricting to nonzero residues proves the requested multiplicative Teichmuller lift:Its image is the group of $(q-1)$st roots of unity in $K$. Indeed a root of unity of that order is a unit satisfying $X^q=X$, hence is the unique lift of its residue. The only finite-field fact used here is $a^q=a$; for $a\ne 0$ it follows from Lagrange's theorem in the multiplicative group of order $q-1$.

Now suppose $\mathrm{char}K=p>0$. The Teichmuller lifts constructed in (b) are additive as well as multiplicative: if $a^q=a$ and $b^q=b$, then the characteristic-$p$ Frobenius identity gives $(a+b{)}^q=a^q+b^q=a+b$. Uniqueness of the lift with prescribed residue yields $[\bar{a}+\bar{b}]=[\bar{a}]+[\bar{b}]$. Their set is therefore a finite coefficient field in positive-characteristic local fields, and reduction identifies it with $k$.

Choose a uniformizer $\pi$. For $x\in {\mathcal{O}}_K$, let $a_0$ be the lift of its residue and write $x=a_0+\pi x_1$ with $x_1\in {\mathcal{O}}_K$. Repeat with $x_1$, then $x_2$, and so on. This givesThe remainder tends to zero, and completeness gives $x={\sum }_{j\ge 0}{a}_j{\pi }^j$. The expansion is unique: in a nonzero difference of two series the first different coefficient is a nonzero element of the coefficient field, hence a unit, and so gives exactly the valuation of that difference.

Consequently substitution $t\mapsto \pi$ defines a bijection $k[[t]]\to {\mathcal{O}}_K$. It is a ring homomorphism because the coefficient lift is a field homomorphism, polynomial operations agree before passage to limits, and multiplication is continuous. Every element of $K$ becomes integral after multiplication by a sufficiently large power of $\pi$. Allowing finitely many negative powers therefore extends substitution to the field isomorphismIt preserves the normalized discrete valuation, since the first nonzero coefficient determines the valuation on each side. The results used are the structural facts stated in (a), its proved Hensel lemma, the proved multiplicative lifting in (b), and the characteristic-$p$ identity $(a+b{)}^q=a^q+b^q$. No classification theorem for local fields is needed.