Examples of Valuations

The archetypal example of an archimedean valuation is the absolute value on the complex numbers. It is essentially the only one:

Theorem 15.3.1 (Gelfand-Tornheim)   Any field $K$ with an archimedean valuation is isomorphic to a subfield of  $\mathbf{C}$, the valuation being equivalent to that induced by the usual absolute value on  $\mathbf{C}$.

We do not prove this here as we do not need it. For a proof, see [Art59, pg. 45, 67].

There are many non-archimedean valuations. On the rationals $\mathbf{Q}$ there is one for every prime $p>0$, the $p$-adic valuation, as in Example 15.2.9.

Theorem 15.3.2 (Ostrowski)   The nontrivial valuations on $\mathbf{Q}$ are those equivalent to $\left\vert \cdot \right\vert _p$, for some prime $p$, and the usual absolute value $\left\vert \cdot \right\vert _\infty$.

Remark 15.3.3   Before giving the proof, we pause with a brief remark about Ostrowski. According to

   http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Ostrowski.html

Ostrowski was a Ukrainian mathematician who lived 1893-1986. Gautschi writes about Ostrowski as follows: ``... you are able, on the one hand, to emphasise the abstract and axiomatic side of mathematics, as for example in your theory of general norms, or, on the other hand, to concentrate on the concrete and constructive aspects of mathematics, as in your study of numerical methods, and to do both with equal ease. You delight in finding short and succinct proofs, of which you have given many examples ...'' [italics mine]

We will now give an example of one of these short and succinct proofs.

Proof. Suppose $\left\vert \cdot \right\vert$ is a nontrivial valuation on $\mathbf{Q}$.

Nonarchimedean case: Suppose $\left\vert c\right\vert\leq 1$ for all $c\in\mathbf{Z}$, so by Lemma 15.2.10, $\left\vert \cdot \right\vert$ is nonarchimedean. Since $\left\vert \cdot \right\vert$ is nontrivial, the set

$$\mathfrak{p}=\{a\in\mathbf{Z}: \left\vert a\right\vert<1\}$$

is nonzero. Also $\mathfrak{p}$ is an ideal and if $\left\vert ab\right\vert<1$, then $\left\vert a\right\vert\left\vert b\right\vert=\left\vert ab\right\vert<1$, so $\left\vert a\right\vert<1$ or $\left\vert b\right\vert<1$, so $\mathfrak{p}$ is a prime ideal of  $\mathbf{Z}$. Thus $\mathfrak{p}=p\mathbf{Z}$, for some prime number $p$. Since every element of $\mathbf{Z}$ has valuation at most $1$, if $u\in\mathbf{Z}$ with $\gcd(u,p)=1$, then $u\not\in\mathfrak{p}$, so $\left\vert u\right\vert=1$. Let $\alpha=\log_{\left\vert p\right\vert}\frac{1}{p}$, so $\left\vert p\right\vert^\alpha = \frac{1}{p}$. Then for any $r$ and any $u\in\mathbf{Z}$ with $\gcd(u,p)=1$, we have

$$\left\vert up^r\right\vert^{\alpha} = \left\vert u\right\vert^{\a... ...= \left\vert p\right\vert^{\alpha r} = p^{-r} = \left\vert up^r\right\vert _p.$$

Thus $\left\vert \cdot \right\vert^{\alpha} = \left\vert \cdot \right\vert _p$ on $\mathbf{Z}$, hence on $\mathbf{Q}$ by multiplicativity, so $\left\vert \cdot \right\vert$ is equivalent to $\left\vert \cdot \right\vert _p$, as claimed.

Archimedean case: By replacing $\left\vert \cdot \right\vert$ by a power of $\left\vert \cdot \right\vert$, we may assume without loss that $\left\vert \cdot \right\vert$ satisfies the triangle inequality. We first make some general remarks about any valuation that satisfies the triangle inequality. Suppose $a\in\mathbf{Z}$ is greater than $1$. Consider, for any $b\in\mathbf{Z}$ the base-$a$ expansion of $b$:

$$b = b_m a^m + b_{m-1} a^{m-1} + \cdots + b_0,$$

where

$$0 \leq b_j < a \qquad (0\leq j \leq m),$$

and $b_m\neq 0$. Since $a^m\leq b$, taking logs we see that $m\log(a)\leq \log(b)$, so

$$m \leq \frac{\log(b)}{\log(a)}.$$

Let $$M=\max_{1\leq d<a}\left\vert d\right\vert$$. Then by the triangle inequality for $\left\vert \cdot \right\vert$, we have

$$\left\vert b\right\vert \leq \left\vert b_m\right\vert{a}^m + \cdots + \left\vert b_1\right\vert\left\vert a\right\vert + \left\vert b_0\right\vert$$

$$\leq M\cdot (\left\vert a\right\vert^m + \cdots + \left\vert a\right\vert + 1)$$

$$\leq M\cdot (m+1)\cdot \max(1,\left\vert a\right\vert^m)$$

$$\leq M\cdot\left(\frac{\log(b)}{\log(a)} + 1\right) \cdot \max\left(1,\left\vert a\right\vert^{\log(b)/\log(a)}\right),$$

where in the last step we use that $m\leq \frac{\log(b)}{\log(a)}$. Setting $b=c^n$, for $c\in\mathbf{Z}$, in the above inequality and taking $n$th roots, we have

$$\left\vert c\right\vert \leq \left( M\cdot \left(\frac{\log(c^n)}{\log(a)}+1\right)\cdot \max(1,\left\vert a\right\vert^{log(c^n)/\log(a)})\right)^{1/n}$$

$$= M^{1/n}\cdot\left( \frac{\log(c^n)}{\log(a)}+1\right)^{1/n}\cdot \max\left(1,\left\vert a\right\vert^{\log(c^n)/\log(a)}\right)^{1/n}.$$

The first factor $M^{1/n}$ converges to $1$ as $n\to \infty$, since $M\geq 1$ (because $\left\vert 1\right\vert = 1$). The second factor is

$$\left(\frac{\log(c^n)}{\log(a)}+1\right)^{1/n} = \left(n \cdot \frac{\log(c)}{\log(a)}+1\right)^{1/n}$$

which also converges to $1$, for the same reason that $n^{1/n}\to 1$ (because $\log(n^{1/n})=\frac{1}{n}\log(n)\to 0$ as $n\to \infty$). The third factor is

$$\max\left(1,\left\vert a\right\vert^{\log(c^n)/\log(a)}\right)^{1... ...t\vert^{\log(c)/\log(a)} & \text{if }\left\vert a\right\vert\geq 1. \end{cases}$$

Putting this all together, we see that

$$\left\vert c\right\vert \leq \max\left(1,\left\vert a\right\vert^{\frac{\log(c)}{\log(a)}}\right).$$

Our assumption that $\left\vert \cdot \right\vert$ is nonarchimedean implies that there is $c\in\mathbf{Z}$ with $c>1$ and $\left\vert c\right\vert>1$. Then for all $a\in\mathbf{Z}$ with $a>1$ we have

$$1 < \left\vert c\right\vert \leq \max\left(1,\left\vert a\right\vert^{\frac{\log(c)}{\log(a)}}\right),$$ (15.5)

so $1<\left\vert a\right\vert^{\log(c)/\log(a)}$, so $1<\left\vert a\right\vert$ as well (i.e., any $a\in\mathbf{Z}$ with $a>1$ automatically satisfies $\left\vert a\right\vert>1$). Also, taking the $1/\log(c)$ power on both sides of (15.3.1) we see that

$$\left\vert c\right\vert^{\frac{1}{\log(c)}} \leq \left\vert a\right\vert^{\frac{1}{\log(a)}}.$$ (15.6)

Because, as mentioned above, $\left\vert a\right\vert>1$, we can interchange the roll of $a$ and $c$ to obtain the reverse inequality of (15.3.2). We thus have

$$\left\vert c\right\vert = \left\vert a\right\vert^{\frac{\log(c)}{\log(a)}}.$$

Letting $\alpha=\log(2)\cdot \log_{\left\vert 2\right\vert}(e)$ and setting $a=2$, we have

$$\left\vert c\right\vert^{\alpha} = \left\vert 2\right\vert^{\frac... ...ert}(e)}\right)^{\log(c)} = e^{\log(c)} = c = \left\vert c\right\vert _\infty.$$

Thus for all integers $c\in\mathbf{Z}$ with $c>1$ we have $\left\vert c\right\vert^{\alpha} = \left\vert c\right\vert _{\infty}$, which implies that $\left\vert \cdot \right\vert$ is equivalent to $\left\vert \cdot \right\vert _\infty$. $\qedsymbol$

Let $k$ be any field and let $K=k(t)$, where $t$ is transcendental. Fix a real number $c>1$. If $p=p(t)$ is an irreducible polynomial in the ring $k[t]$, we define a valuation by

$$\left\vert p^a \cdot \frac{u}{v}\right\vert _p = c^{-\deg(p)\cdot a},$$ (15.7)

where $a\in\mathbf{Z}$ and $u,v\in k[t]$ with $p\nmid u$ and $p\nmid v$.

Remark 15.3.4   This definition differs from the one page 46 of [Cassels-Frohlich, Ch. 2] in two ways. First, we assume that $c>1$ instead of $c<1$, since otherwise $\left\vert \cdot \right\vert _p$ does not satisfy Axiom 3 of a valuation. Also, we write $c^{-\deg(p)\cdot a}$ instead of $c^{-a}$, so that the product formula will hold. (For more about the product formula, see Section 20.1.)

In addition there is a a non-archimedean valuation $\left\vert \cdot \right\vert _\infty$ defined by

$$\left\vert\frac{u}{v}\right\vert _\infty = c^{\deg(u)-\deg(v)}.$$ (15.8)

This definition differs from the one in [Cas67, pg. 46] in two ways. First, we assume that $c>1$ instead of $c<1$, since otherwise $\left\vert \cdot \right\vert _p$ does not satisfy Axiom 3 of a valuation. Here's why: Recall that Axiom 3 for a non-archimedean valuation on $K$ asserts that whenever $a\in K$ and $\left\vert a\right\vert\leq 1$, then $\left\vert a+1\right\vert\leq 1$. Set $a=p-1$, where $p=p(t)\in K[t]$ is an irreducible polynomial. Then $\left\vert a\right\vert=c^0 = 1$, since $\ord _p(p-1) = 0$. However, $\left\vert a+1\right\vert = \left\vert p-1+1\right\vert = \left\vert p\right\vert=c^1<1$, since $\ord _p(p) = 1$. If we take $c>1$ instead of $c<1$, as I propose, then $\left\vert p\right\vert=c^1>1$, as required.

Note the (albeit imperfect) analogy between $K=k(t)$ and $\mathbf{Q}$. If $s=t^{-1}$, so $k(t)=k(s)$, the valuation $\left\vert \cdot \right\vert _{\infty}$ is of the type (15.3.3) belonging to the irreducible polynomial $p(s)=s$.

The reader is urged to prove the following lemma as a homework problem.

Lemma 15.3.5   The only nontrivial valuations on $k(t)$ which are trivial on $k$ are equivalent to the valuation (15.3.3) or (15.3.4).

For example, if $k$ is a finite field, there are no nontrivial valuations on $k$, so the only nontrivial valuations on $k(t)$ are equivalent to (15.3.3) or (15.3.4).