Valuations

Definition 15.1.1 (Valuation)   A $\left\vert \cdot \right\vert{}$ on a field $K$ is a function defined on $K$ with values in $\mathbf{R}_{\geq 0}$ satisfying the following axioms:

(1)

$\left\vert a\right\vert = 0$ if and only if $a = 0$,

(2)

$\left\vert ab\right\vert=\left\vert a\right\vert\left\vert b\right\vert$, and

(3)

there is a constant $C\geq 1$ such that $\left\vert 1+a\right\vert\leq C$ whenever $\left\vert a\right\vert\leq 1$.

The is the valuation for which $\left\vert a\right\vert=1$ for all $a\neq 0$. We will often tacitly exclude the trivial valuation from consideration.

From (2) we have

$$\left\vert 1\right\vert = \left\vert 1\right\vert\cdot \left\vert 1\right\vert,$$

so $\left\vert 1\right\vert = 1$ by (1). If $w\in K$ and $w^n=1$, then $\vert w\vert=1$ by (2). In particular, the only valuation of a finite field is the trivial one. The same argument shows that $\vert-1\vert=\vert 1\vert$, so

$$\vert-a\vert = \vert a\vert$$   all $$a \in K.$$

Definition 15.1.2 (Equivalent)   Two valuations $\left\vert \cdot \right\vert{}_1$ and $\left\vert \cdot \right\vert{}_2$ on the same field are if there exists $c>0$ such that

$$\left\vert a\right\vert _2 = \left\vert a\right\vert _1^c$$

for all $a\in K$.

Note that if $\left\vert \cdot \right\vert{}_1$ is a valuation, then $\left\vert \cdot \right\vert{}_2=\left\vert \cdot \right\vert{}_1^c$ is also a valuation. Also, equivalence of valuations is an equivalence relation.

If $\left\vert \cdot \right\vert{}$ is a valuation and $C$ is the constant from Axiom (3), then there is a $c>0$ such that $C^c=2$ (i.e., $c=\log(C)/\log(2)$). Then we can take $2$ as constant for the equivalent valuation $\left\vert \cdot \right\vert{}^c$. Thus every valuation is equivalent to a valuation with $C=2$. Note that if $C=1$, e.g., if $\left\vert \cdot \right\vert{}$ is the trivial valuation, then we could simply take $C=2$ in Axiom (3).

Proposition 15.1.3   Suppose $\left\vert \cdot \right\vert{}$ is a valuation with $C=2$. Then for all $a, b\in K$ we have

$$\vert a + b\vert \leq \vert a\vert + \vert b\vert .$$ (15.1)

Proof. Suppose $a_1, a_2\in K$ with $\vert a_1\vert\geq\vert a_2\vert$. Then $a=a_2/a_1$ satisfies $\vert a\vert\leq 1$. By Axiom (3) we have $\vert 1+a\vert\leq 2$, so multiplying by $a_1$ we see that

$$\vert a_1+ a_2\vert\leq 2\vert a_1\vert = 2\cdot\max\{\vert a_1\vert,\vert a_2\vert\}.$$

Also we have

$$\vert a_1+ a_2 + a_3 + a_4\vert\leq 2\cdot\max\{\vert a_1+a_2\ver... ...eq 4\cdot \max\{\vert a_1\vert,\vert a_2\vert,\vert a_3\vert,\vert a_4\vert\},$$

and inductively we have for any $r>0$ that

$$\vert a_1 + a_2 + \cdots + a_{2^r}\vert \leq 2^r\cdot\max{\vert a_j\vert}.$$

If $n$ is any positive integer, let $r$ be such that $2^{r-1}\leq n\leq 2^r$. Thenn

$$\vert a_1 + a_2 + \cdots + a_{n}\vert \leq 2^r\cdot \max\{\vert a_j\vert\} \leq 2n\cdot \max\{\vert a_j\vert\},$$

since $2^r\leq 2n$. In particular,

$$\vert n\vert \leq 2n\cdot \vert 1\vert = 2n n>0 .$$ (15.2)

Applying (15.1.2) to $$\left\vert\binom{n}{j}\right\vert$$ and using the binomial expansion, we have for any $a, b\in K$ that

$$\vert a+b\vert^n = \left\vert\sum_{j=0}^n \binom{n}{j} a^j b^{n-j}\right\vert$$

$$\leq 2(n+1)\max_j\left\{ \left\vert\binom{n}{j}\right\vert \left\vert a\right\vert^j\left\vert b\right\vert^{n-j}\right\}$$

$$\leq 2(n+1)\max_j\left\{ 2 \binom{n}{j} \left\vert a\right\vert^j\left\vert b\right\vert^{n-j}\right\}$$

$$\leq 4(n+1)\max_j\left\{ \binom{n}{j} \left\vert a\right\vert^j\left\vert b\right\vert^{n-j}\right\}$$

$$\leq 4(n+1)(\left\vert a\right\vert+\left\vert b\right\vert)^n.$$

Now take $n$th roots of both sides to obtain

$$\vert a+b\vert \leq \sqrt[n]{4(n+1)}\cdot (\vert a\vert + \vert b\vert).$$

We have by elementary calculus that

$$\lim_{n\to \infty} \sqrt[n]{4(n+1)} = 1,$$

so $\vert a+b\vert\leq \vert a\vert+\vert b\vert$. (The ``elementary calculus'': We instead prove that $\sqrt[n]{n}\to 1$, since the argument is the same and the notation is simpler. First, for any $n\geq 1$ we have $\sqrt[n]{n}\geq 1$, since upon taking $n$th powers this is equivalent to $n\geq 1^n$, which is true by hypothesis. Second, suppose there is an $\varepsilon >0$ such that $\sqrt[n]{n}\geq 1+\varepsilon$ for all $n\geq 1$. Then taking logs of boths sides we see that $\frac{1}{n}\log(n)\geq \log(1+\varepsilon ) > 0$. But $\log(n)/n\to 0$, so there is no such $\varepsilon$. Thus $\sqrt[n]{n}\to 1$ as $n\to \infty$.) $\qedsymbol$

Note that Axioms (1), (2) and Equation (15.1.1) imply Axiom (3) with $C=2$. We take Axiom (3) instead of Equation (15.1.1) for the technical reason that we will want to call the square of the absolute value of the complex numbers a valuation.

Lemma 15.1.4   Suppose $a, b\in K$, and $\left\vert \cdot \right\vert{}$ is a valuation on $K$ with $C\leq 2$. Then

$$\Bigl\vert\vert a\vert - \vert b\vert\Bigr\vert \leq \left\vert a-b\right\vert.$$

(Here the big absolute value on the outside of the left-hand side of the inequality is the usual absolute value on real numbers, but the other absolute values are a valuation on an arbitrary field $K$.)

Proof. We have

$$\vert a\vert = \vert b + (a-b)\vert \leq \vert b\vert + \vert a-b\vert,$$

so $\vert a\vert-\vert b\vert\leq \left\vert a-b\right\vert$. The same argument with $a$ and $b$ swapped implies that $\vert b\vert-\vert a\vert\leq \left\vert a-b\right\vert$, which proves the lemma. $\qedsymbol$