Extending the p-adic norm to the field of real numbers

One of the most beautiful problems that I've encountered was the famous Monsky theorem that asserts that one can not tile a rectangle by an odd number of equal area triangles.

This problem is wildly discussed in the forums, yet it is highly sophisticated and requires a wide algebra knowledge.

Among thing required is the fact that one can extend the p-adic absolute from the rational field to the real field, I would like to present this extension here.

I will assume that the reader is familiar with some basic algebra materials that one ought to learn in a second year of undergraduate studies, such as:

Linear algebra,group theory , determinant map, field extension, set theory.

The p-adic norm on the rationals:

As an application of the fundamental theorem of arithmatics every rational number $q\in Q$ can be written:

where

the p-adic absolute value of

is define to be:

$|q|_p=\frac{1}{p^a}$

basic facts:

2. $|q_1+q_2|_p\leq max(|q_1|_p ,|q_2|_p)\leq |q_1|_p+|q_2|_p$ (ultrametric inequality)

Proof of (2) : assume $|q_1|_p\leq |q_2|_p$

then $q_1=p^k\frac{a}{b}\, \, \, \, q_2=p^l \frac{c}{d}\, \, \, l\leq k , \, \, a,b,c,d \, \, \, prime \, \, to\, \, p$

we have :
$q_1+q_2=p^l(p^{k-l}\frac{a}{b}+\frac{c}{d})=p^l(\frac{p^{k-l}ad+cb}{bd})$ . but $\ p^{k-l}(ad+cb)$ is an integer, so
altogether $\ |p^{k-l}(ad+cb)|_p\leq 1, |bd|_p=1, |p^l|_p=|q_2|_p$ , and the proof is finished.
(because of property 1)

actually one can prove $|q_1|_p<|q_2|_p,\, \, \, |q_1+q_2|_p=|q_2|$

Our goal now is to construct an new absolute value on $\mathbb{R}$ , ||r||_p

such that it will still have properties 1 and 2 (on the real numbers )and Will be an extension of our absolute value on the rationals for all

$q\in Q$ .

first step: extending to finite extensions. The Norm

let us assume we have two fields $F\subseteq E$ where is a finite extension of

given an element $x\in E$ , considering

as a vector space over

multiplying by

is a linear transformation! so we can note it as T_x

. being a finite dimensional linear transformation

its have a well define determinant. the norm of

is defined to be N(x)=det(T_x)

practically one can calculate this norm by considering what multiplying by

does to basis elements.

Example: let us consider the quadratic extension $Q(\sqrt{a}):Q, \sqrt{a}\notin Q$ every element can be written as $x+\sqrt{a}y\, \, \, \, \, \, \, \, x,y\in Q$ . let see what happens when multiplying with the basis element

$T_{x+\sqrt{a}y}[1]=x+\sqrt{a}y\,\, \, \, \, \, \, \, \, \, T_{x+\sqrt{a}y}[\sqrt{a}]=[x+\sqrt{a}y]\sqrt{a}=ay+\sqrt{a}x$ .

writing it as a matrix we have:

$T_{x+\sqrt{a}y}=\left \begin{pmatrix} x&ay \\ y& x \end{pmatrix}$

which means : $N(x+\sqrt{a}y)=det(T_{x+\sqrt{a}y})=x^2-ay^2$

consider the case $a=-1\,\,\, F=\mathbb{R}$ !

because of the multiplicative nature of the determinant it is not hard to show that :

$N(wv)=N(w)N(v)\, \, for \, \, all\, \, w,v\in E$ .

also note $N(d)=d^n \, \, \, \, n=[E:F], d\in F$ .

now we can advance one little step towards our destination.

suppose we have $Q\subseteq F\subseteq E$ where

is a finite extension of

, and that somehow

we managed to extend the p-adic valuation on

, and we would like to further extend it to

the field

, we define it to be:

$||x||_p =\sqrt[n]{|N(x)|_p}\, \, \,\, \, \, \, \, \, \, \, \, \, n=E:F$

it is easy to see that this absolute value is still the same if $x\in F$ , and still have the multiplicative nature that the absolute value should have.

the hard part is to show that the new absolute value still satisfy the ultrametric inequality, but
i will not give it here, i refer you to (*)

second step: extending the norm to transcendental extension

assume $Q\subseteq F\subseteq E$ where we managed to define the p-adic absolute value on

and we want to extend it to

, but this time $E=F(\alpha)$ and $\alpha$ is a transcendental element over

what can we do?

this time it turns out that we have a lot of freedom. first pick some c>0

, it is known that

every element $x\in E$ can be uniquely written as

$x=\frac{p(\alpha)}{h(\alpha)}\, \, \, \, \, \, p(\alpha)=\sum_{i=0}^{n}a_i \alpha ^i\: \: \:h(\alpha)=\sum_{i=0}^{k}b_j \alpha ^i\: \: \: a_i,b_i \in F$ .

we define

$||x||_p=\frac{max_{i}(c^i|a_i|_p)}{max_{j}(c^j|b_j|_p)}$ .

this is a little complicated but one can check this to be an extension of the p-adic valuation.

so far we have managed to extended the p-adic valuation a 'little further' than we had,

how can we be sure that we can do this procedure for until we reach the whole real numbers

the magic words are Zorn's lemma!

we define a partial order on the set of all p-adic absolute values that are defined on some subfield of the real numbers much similar to how we use Zorn's lemma in the Hahn-Banach Theorem

$||_1\leq ||_2\Leftrightarrow domain||_1\subseteq domain||_2\: \: \: |x|_2=|x|_1 \, \, \, for\, \, \, \, all \, \, \, \, x\in domain||_1$

it is easy to see that every totally ordered subset have an upper bound and that we have a maximal element.

Claim: this maximal element is defined over all the real numbers.

Proof: assume not , we have ||_p

and it is defined on $F\subset \mathbb{R}$ then one can extend by adding to

an element $\alpha\in \mathbb{R} \, \, \, \, \, \, \alpha \notin F$ , $F(\alpha)/F$ can be algebraic or transcendental extension

but we still know how to extend the absolute value on it! contradiction to the maximality.

(*) for further information I advise you to read Cassel's local Fields book chapter 6-7

Remark 1: There is of course lots of ways to extend the norm and the absolute value!

Remark 2: The use of the axiom of choice is seem to be very needed, for example:no one can prove the existence of nontrivial automorphisms of the complex field with out it.

Nice problem:

Try proving that you can not tile a rectangle with odd number of $1,2,\sqrt{5}$ - right triangles

without Monsky Theorem

Of Ravens and Writing Desks - על עורבים ושולחנות כתיבה

Wednesday, September 28, 2011

Extending the p-adic norm to the field of real numbers

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