Local Fields

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 27 November 2014

Local Fields

A discrete valuation on a field $F$ is a surjective map $v:F\to \mathbb{Z}\cup \left(\infty \right),$ such that:

(i)	$v\left(0\right)=\infty ,$
(ii)	$v:F^{\times }\to \mathbb{Z}$ is a surjective homomorphism,
(iii)	$v(x+y)\ge \text{inf}\{v\left(x\right),v\left(y\right)\}$ for all $x,y\in F\text{.}$

Examples

(1)

$\text{(}t\text{-adic}$ valuation) Consider rational functions in $t$ over a field $F\text{.}$ Every non-zero rational function $f\left(t\right)\in F\left(t\right)$ can be written in the form $f\left(t\right)=t^n\frac{a\left(t\right)}{b\left(t\right)}$ where $a\left(t\right)$ and $b\left(t\right)$ are polynomials with non-zero constant term and $n\in \mathbb{Z}\text{.}$ Putting $v\left(f\left(t\right)\right)=n$ gives a discrete valuation on $F\left(t\right)\text{.}$

(2)

$\text{(}p\text{-adic}$ valuation) Fix a prime $p\text{.}$ Every non-zero rational number $f\in \mathbb{Q}$ can be written in the form $f=p^n\frac{a}{b}$ where $a$ and $b$ are integers relatively prime to $p$ and $n\in \mathbb{Z}\text{.}$ Putting $v\left(f\right)=n$ gives a discrete valuation on $\mathbb{Q}\text{.}$

Let $R$ be an integral domain. A function $v:R\to \mathbb{Z}\cup \left(\infty \right),$ satisfying (i)-(iii) above is called a discrete valuation on $R\text{.}$ A discrete valuation on an integral domain extends uniquely to a discrete valuation on its quotient field.

Examples

(1)	The $t\text{-adic}$ valuation $v$ on $F\left[t\right],$ where $v\left(a\left(t\right)\right)$ is the highest power of $t$ dividing $a\left(t\right)\in F\left[t\right]$ extends to the $t\text{-adic}$ valuation on $F\left(t\right)\text{.}$
(2)	The $p\text{-adic}$ valuation $v$ on $\mathbb{Z},$ where $v\left(a\right)$ is the highest power of $p$ dividing $a\in \mathbb{Z}$ extends to the $p\text{-adic}$ valuation on $\mathbb{Q}\text{.}$

Given a field $F$ and a discrete valuation $v$ on $F$ we set $$\begin{array}{rcl}{\displaystyle 𝔇}& =& {\displaystyle \{x\in F:v\left(x\right)\ge 0\},}\\ {\displaystyle 𝔅}& =& {\displaystyle \{x\in F:v\left(x\right)\ge 1\},}\\ {\displaystyle U}& =& {\displaystyle \{x\in F:v\left(x\right)=0\}\text{.}}\end{array}$$ Then $𝔇$ is a ring, called the valuation ring. It is a local ring with maximal ideal $𝔅\text{.}$ The set $U$ is the group of units of $𝔇\text{.}$ The field $k=𝔇/𝔅$ is called the residue class field associated to the valuation. An element $\pi$ in $F$ with valuation $v\left(\pi \right)=1$ is called a uniformizer. The ring $𝔇$ is principal ideal domain and the non-zero ideals of $𝔇$ are ${𝔅}^n=\left({\pi }^n\right)=\{x\in F:v\left(x\right)\ge n\}\text{.}$

Example

(1)	In the rational function case the valuation ring $𝔇$ is the localization of $F\left[t\right]$ at the ideal $\left(t\right)$ generated by $t\text{.}$ The residue class field is isomorphic to $F\text{.}$ The canonical map is given by evaluation at $t=0\text{.}$
(2)	In the $p\text{-adic}$ case $𝔇$ is the localization of $\mathbb{Z}$ at the ideal $\left(p\right)\text{.}$ The residue class field is the finite field with $p$ elements.

Exercise. An element $x$ of $F$ can be expressed uniquely in the form $x=u{\pi }^n,$ $u\in U$ a unit and $n\in \mathbb{Z}\text{.}$

Fix $\rho >1\text{.}$ We define a function $\mid \hspace{0.17em}\mid :F\to [0,\infty ),$ from $F$ to the non-negative real numbers by $$\mid x\mid =\frac{1}{{\rho }^{v\left(x\right)}}\text{for}\hspace{0.17em}x\in F\text{.}$$ This satisfies

(1)	$\mid x\mid =0$ if and only if $x=0\text{.}$
(2)	$\mid xy\mid =\mid x\mid \mid y\mid$ for all $x,y\in F\text{.}$
(3)	$\mid x+y\mid \le \text{max}\{\mid x\mid ,\mid y\mid \}$ for all $x,y\in F\text{.}$

The last inequality implies

(3)'	$\mid x+y\mid \le \mid x\mid +\mid y\mid \text{.}$

A function satisfying (1), (2) and (3) is called a non-archimedean absolute value. Function which satisfy (1), (2) and (3)' but not (3) are called archimedean absolute values. An absolute value gives a metric $d(x,y)=\mid x-y\mid$ on $F\text{.}$ If it is non-archimedean we have $d(x,z)\le \text{max}\{d(x,y),d(y,z)\}\text{.}$

The non-archimedean metric defined by $v$ makes $F$ into a topological field. Note the topology does not depend on the choice of $\rho \text{.}$ In this topology the open balls about $a\in F$ are the sets $\{x\in F:v(x-a)>n\}$ and the closed balls are the sets $\{x\in F:v(x-a)\ge n\}\text{.}$ But $\{x\in F:v(x-a)>n\}=\{x\in F:v(x-a)\ge n+1\}\text{.}$ Thus each open ball is closed. Hence each open set is closed, i.e the topology is totally disconnected.

If $F$ is complete with respect to the metric $\mid \hspace{0.17em}\mid$ we call $v$ a complete discrete valuation. Since different choices of $\rho >1$ give equivalent metrics this notion is well defined. We can always complete of $F$ with respect to the metric $\mid \hspace{0.17em}\mid \text{.}$ The Cauchy sequences in $F$ form a ring under pointwise operations. The null sequences form a maximal ideal. The completion $\stackrel{\sim }{F}$ of $F$ is the quotient field. The field $F$ is naturally embedded in the quotient field via the constant sequences. The the valuation $F$ extends uniquely to the completion. In fact if a is determined by Cauchy sequence $\left\{a_n\right\},$ the sequence $\left\{v\left(a_n\right)\right\}$ will be eventually constant and $v\left(\alpha \right)=\underset{n\to \infty }{\text{lim}}v\left(a_n\right)$ is well defined.

The valuation ring of the completion is the closure of the valuation ring $𝔇$ of $F$ in $\stackrel{\sim }{F}\text{.}$ The residue class rings of $F$ and $\stackrel{\sim }{F}$ completion are naturally isomorphic.

Let $\pi$ be a uniformiser of $F$ and $A$ a set of residue class representatives in $𝔇$ of $k\text{.}$ Then all sums ${\displaystyle \sum _{n\gg -\infty }^{\infty }}{a}_n{\pi }^n,$ with each $a_n\in A,$ converge in $\stackrel{\sim }{F}\text{.}$ Every element of $\stackrel{\sim }{F}$ is uniquely represented by such a sum.

The valuation ring of the completion is all the sums ${\displaystyle \sum _{n=0}^{\infty }{a}_n{\pi }^n}\text{.}$

Examples

(1)

The completion of $F\left[t\right]$ with respect to the $t\text{-adic}$ valuation is the ring of formal power series in $t$ with coefficients in $F,$ $$F\left[\left[t\right]\right]=\{\sum _{n=0}^{\infty }{a}_n{t}^n:a_n\in F\}\text{.}$$ The completion of $F\left(t\right)$ is $F\left(\left(t\right)\right),$ the field of formal Laurent series with a pole at zero, $$F\left(\left(t\right)\right)=\{\sum _{n\gg -\infty }^{\infty }{a}_n{t}^n:a_n\in F\}\text{.}$$ The valuation is given by $$v(\sum _{n\gg -\infty }^{\infty }{a}_n{t}^n=\text{min}\{n:a_n\ne 0\}\text{.})$$

(2)

The completion of $\mathbb{Z}$ with respect to the $p\text{-adic}$ topology is the ring of $p\text{-adic}$ integers, $${\mathbb{Z}}_p=\{\sum _{n=0}^{\infty }{a}_n{p}^n:a_n\in \{0,1,\dots ,p-1\}\}\text{.}$$ The completion of $\mathbb{Q},$ is the field of $p\text{-adic}$ numbers, $${\mathbb{Q}}_p=\{\sum _{n\gg -\infty }^{\infty }{a}_n{p}^n:a_n\in \{0,1,\dots ,p-1\}\}\text{.}$$ The valuation is given by $$v\left(\sum _{n\gg -\infty }^{\infty }{a}_n{p}^n\right)=\text{min}\{n:a_n\ne 0\}\text{.}$$

Local Field. A field which is complete with respect to a complete valuation and has finite residue class field is called is called a local field.

(1)	A local field of characteristic $p>0$ is of the form $F\left(\left(t\right)\right),$ with $F$ finite of characteristic $p\text{.}$
(2)	The local fields of characteristic zero are the finite algebraic extensions of ${\mathbb{Q}}_p\text{.}$

We can make a canonical choice of absolute value in a local field $F\text{.}$ Let $q$ denote the order of its residue class field. Then define $$\mid x\mid =\frac{1}{q^{v\left(x\right)}}\text{for}\hspace{0.17em}x\in F\text{.}$$

The valuation ring of a local field $𝔇,$ and its ideals ${𝔅}^n,$ are compact. They form a compact system of neighbourhoods of zero in $F\text{.}$ Hence $F$ is a locally compact Hausdorff topological group under addition. We can fix a Haar measure $\mu$ on $F$ by setting $\mu \left(𝔇\right)=1\text{.}$

For all $x\in F,$ and measurable $A,$ we have $\mu \left(xA\right)=\mid x\mid \mu \left(A\right)\text{.}$

		Proof. (Sketch)
	This trivial for $x=0\text{.}$ For $x\ne 0,$ $\mu \prime \left(A\right)=\mu \left(xA\right)$ is also a Haar measure. Thus $\mu \left(xA\right)=c\left(x\right)\mu \left(A\right)$ for some positive real $c\left(x\right)$ Now deduce $c\left(xy\right)=c\left(x\right)c\left(y\right)$ for all $x,y\in F\text{.}$ Then by taking $A=𝔇$ show that $c\left(u\right)=1$ for $u\in U,$ and (count cosets), $c\left(\pi \right)=\frac{1}{q}\text{.}$ The result now follows. $\square$

The sets $1+{𝔅}^n,$ $n>0$ form a compact system of neighbourhoods of the identity in $F^{\times }\text{.}$ Thus $F^{\times }$ is a locally compact Hausdorff topological group under multiplication.

Haar measure on $F^{\times }$ is given by $\frac{dx}{\mid x\mid }\text{.}$

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