User:Marty/UCSC Math 203 Fall 2008/Week 8

In the exercises of Chapter 5 of Atiyah-Macdonald, the theory of valuation rings is developed "from scratch". In order to motivate this development, and to place it in context, we discuss a few facts about absolute values, and their relation to valuation rings.

Absolute values
Suppose that $K$ is a field. An absolute value (or multiplicative norm) on $K$ is a function $N \colon K \rightarrow \RR$ which satisfies the following:
 * $N$ is a norm on the set $K$:
 * $N(x) \geq 0$ for all $x \in K$.
 * $N(x) = 0$ if and only if $x = 0$.
 * $N(x+y) \leq N(x) + N(y)$ for all $x \in K$. (The triangle inequality).
 * $N$ is multiplicative: $N(xy) = N(x) N(y)$, for all $x,y \in K$.

There is a stark dichotomy among the absolute values, which we describe here. Note that there is a unique ring homomorphism $\iota \colon \ZZ \rightarrow K$, called the characteristic homomorphism. Its image is either a finite field $\FF_p$ (for some prime number $p$) or $\iota$ is injective.

Definition: An absolute value $N$ on a field $K$, is called Archimedean if $\iota(\ZZ)$ is unbounded. In other words, there are integers of arbitrarily large absolute value (when they are "interpreted" as elements of $K$).

If $K$ is a field of characteristic $p$ (prime), then every absolute value $N$ is non-Archimedean since $\iota(\ZZ)$ is finite. The usual absolute value on $\QQ$ is Archimedean, since $\vert n \vert$ can be arbitrarily large as for various integers $n \in \ZZ$.

The differences between Archimedean and non-Archimedean absolute values are dramatic. Perhaps the most important fact about non-Archimedean absolute values is that they satisfy a huge strengthening of the trinagle inequality, known sometimes as the ultra-metric triangle inequality:

Proposition: If $N$ is a non-Archimedean absolute value on a field $K$, then for all $x,y \in K$, $N(x + y) \leq max(N(x), N(y))$.

Proof: First, one can prove that if $N$ is non-Archimedean, then $N(n) \leq 1$ for every nonzero integer $n$ (consider $N(n^k)$ for natural numbers $k$). Consider any two elements $x,y \in K$, and let $k$ be a natural number. Then, we find that: $$N(x+y)^k = N \left( \sum_i \left( {k \atop i} x^i y^{k-i} \right) \right) \leq \sum_i N(x^i) N(y^{k-i}) \leq (k+1) max \{ N(x), N(y) \}^k.$$ Taking $k^{th}$ roots of both sides demonstrates that: $$N(x+y) \leq \sqrt[k]{k+1} \cdot max \{N(x), N(y) \}.$$ Taking the limit as $k$ approaches infinity yields the ultrametric inequality.

While Archimedean absolute values are interesting in their own right, and they are familiar from classical analysis, non-Archimedean absolute values also play an important role and are connected closely to valuations (as defined in Atiyah-Macdonald).

Proposition: If $N$ is a non-Archimedean absolute value on a field $K$, then consider the function $v \colon K^\times \rightarrow \RR$ given by $v(x) = - log(N(x))$. Then $v$ is a valuation.

Proof: This is a routine verification, given the previous proposition that $N(x + y) \leq max(N(x), N(y))$.

Fix a non-Archimedean absolute value $N$ on a field $K$; let $v(x) = - log(N(x))$ be the associated valuation. We find the following:
 * The valuation ring $A \subset K$, defined as those elements with nonnegative valuation, is also given by: $$A = \{ x \in K \mbox{ such that } N(x) \leq 1 \}.$$ Thus the valuation ring is the "closed unit disc" in $K$.
 * The maximal ideal $M \subset A$ is the open unit disc in $K$, consisting of elements of positive valuation.

Now, it is a theorem of Ostrowski (Sound has a nice short treatment at ) that every absolute value on $\QQ$ is equivalent to one of the following:
 * A power $\vert \bullet \vert^\alpha$ of the usual absolute value, where $0 \leq \alpha \leq 1$. (When $\alpha = 0$, this is the "trivial" absolute value, given by $N(x) = 0$ for $x = 0$ and $N(x) = 1$ for $x \neq 0$.
 * A power $\vert \bullet \vert_p^\alpha$ of the "$p$-adic" absolute value, where $0 \leq \alpha < \infty$. Here, the $p$-adic absolute value is uniquely characterized by its multiplicativity, the fact that $\vert q \vert = 1$ for all primes $q \neq p$, and the fact that $\vert p \vert = p^{-1}$.  ($p$ is "small").

Recall that the real numbers may be constructed as "equivalence classes of Cauchy sequences" of rational numbers, using the usual absolute value to define "Cauchy sequence". However, in the construction, any absolute value may be used instead. Using the $p$-adic absolute value instead yields the fields of $p$-adic numbers, denoted $\QQ_p$, one for every prime number $p$. (Nonzero powers of absolute values yield the same topology, and hence the same notion of "Cauchy sequence" and the same resulting field). The $p$-adic fields are often analytically simpler than the real field, since the triangle inequality is so much stronger on $\QQ_p$ than it is on $\RR$.

The following is a major theme in number theory: problems about rational numbers are often hard. (For example, do there exist rational numbers $x,y,z$ such that $x^2 + 12 y^2 + 7 z^2 = 34$?) Problems about real numbers or $p$-adic numbers are often easy (and provably algorithmically solvable). Some problems (like the one above) can be answered with a local-global principle:
 * If one can find $p$-adic numbers $x,y,z$, such that $x^2 + y^2 + 7 z^2 = 34$, for all prime numbers $p$, and one can find real numbers $x,y,z$ such that $x^2 + 12 y^2 + 7 z^2 = 34$, then one can find rational numbers $x,y,z$ such that $x^2 + 12 y^2 + 7 z^2 = 34$.