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

$\displaystyle \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

$\displaystyle \vert-a\vert = \vert a\vert$   all $\displaystyle 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

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

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

$\displaystyle \vert a + b\vert \leq \vert a\vert + \vert b\vert$   (triangle inequality)$\displaystyle .$ (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

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

Also we have

$\displaystyle \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

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

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

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

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

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

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

$\displaystyle \vert a+b\vert^n$ $\displaystyle = \left\vert\sum_{j=0}^n \binom{n}{j} a^j b^{n-j}\right\vert$    
  $\displaystyle \leq 2(n+1)\max_j\left\{ \left\vert\binom{n}{j}\right\vert \left\vert a\right\vert^j\left\vert b\right\vert^{n-j}\right\}$    
  $\displaystyle \leq 2(n+1)\max_j\left\{ 2 \binom{n}{j} \left\vert a\right\vert^j\left\vert b\right\vert^{n-j}\right\}$    
  $\displaystyle \leq 4(n+1)\max_j\left\{ \binom{n}{j} \left\vert a\right\vert^j\left\vert b\right\vert^{n-j}\right\}$    
  $\displaystyle \leq 4(n+1)(\left\vert a\right\vert+\left\vert b\right\vert)^n.$    

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

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

We have by elementary calculus that

$\displaystyle \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

$\displaystyle \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

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

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$

William Stein 2004-05-06