Types of Valuations

We define two important properties of valuations, both of which apply to equivalence classes of valuations (i.e., the property holds for $\left\vert \cdot \right\vert{}$ if and only if it holds for a valuation equivalent to $\left\vert \cdot \right\vert{}$).

Definition 15.2.1 (Discrete)   A valuation $\left\vert \cdot \right\vert{}$ is if there is a $\delta>0$ such that for any $a\in K$

$$1-\delta < \left\vert a\right\vert < 1+\delta \implies \vert a\vert=1.$$

Thus the absolute values are bounded away from $1$.

To say that $\left\vert \cdot \right\vert{}$ is discrete is the same as saying that the set

$$G=\bigl\{ \log\left\vert a\right\vert : a \in K, a\neq 0 \bigr\} \subset \mathbf{R}$$

forms a discrete subgroup of the reals under addition (because the elements of the group $G$ are bounded away from 0).

Proposition 15.2.2   A nonzero discrete subgroup $G$ of $\mathbf{R}$ is free on one generator.

Proof. Since $G$ is discrete there is a positive $m\in G$ such that for any positive $x\in G$ we have $m\leq x$. Suppose $x\in G$ is an arbitrary positive element. By subtracting off integer multiples of $m$, we find that there is a unique $n$ such that

$$0\leq x-nm <m.$$

Since $x-nm\in G$ and $0<x-nm<m$, it follows that $x-nm=0$, so $x$ is a multiple of $m$. $\qedsymbol$

By Proposition 15.2.2, the set of $\log\left\vert a\right\vert$ for nonzero $a\in K$ is free on one generator, so there is a $c<1$ such that $\left\vert a\right\vert$, for $a\neq 0$, runs precisely through the set

$$c^\mathbf{Z}= \{c^m : m\in \mathbf{Z}\}$$

(Note: we can replace $c$ by $c^{-1}$ to see that we can assume that $c<1$).

Definition 15.2.3 (Order)   If $\left\vert a\right\vert = c^m$, we call $m=\ord (a)$ the of $a$.

Axiom (2) of valuations translates into

$$\ord (ab) = \ord (a) + \ord (b).$$

Definition 15.2.4 (Non-archimedean)   A valuation $\left\vert \cdot \right\vert{}$ is if we can take $C=1$ in Axiom (3), i.e., if

$$\vert a + b\vert \leq \max\bigl\{\vert a\vert,\vert b\vert\bigr\}.$$ (15.3)

If $\left\vert \cdot \right\vert{}$ is not non-archimedean then it is .

Note that if we can take $C=1$ for $\left\vert \cdot \right\vert{}$ then we can take $C=1$ for any valuation equivalent to $\left\vert \cdot \right\vert{}$. To see that (15.2.1) is equivalent to Axiom (3) with $C=1$, suppose $\vert b\vert\leq \vert a\vert$. Then $\vert b/a\vert\leq 1$, so Axiom (3) asserts that $\vert 1+b/a\vert\leq 1$, which implies that $\vert a+b\vert \leq \vert a\vert = \max\{\vert a\vert,\vert b\vert\}$, and conversely.

We note at once the following consequence:

Lemma 15.2.5   Suppose $\left\vert \cdot \right\vert{}$ is a non-archimedean valuation. If $a, b\in K$ with $\vert b\vert<\vert a\vert$, then $\vert a+b\vert=\vert a\vert.$

Proof. Note that $\vert a+b\vert\leq \max\{\vert a\vert,\vert b\vert\} = \vert a\vert$, which is true even if $\vert b\vert=\vert a\vert$. Also,

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

where for the last equality we have used that $\vert b\vert<\vert a\vert$ (if $\max\{\vert a+b\vert,\vert b\vert\} = \vert b\vert$, then $\vert a\vert\leq \vert b\vert$, a contradiction).

$\qedsymbol$

Definition 15.2.6 (Ring of Integers)   Suppose $\left\vert \cdot \right\vert$ is a non-archimedean absolute value on a field $K$. Then

$$\O = \{a\in K : \vert a\vert\leq 1\}$$

is a ring called the of $K$ with respect to $\left\vert \cdot \right\vert{}$.

Lemma 15.2.7   Two non-archimedean valuations $\left\vert \cdot \right\vert{}_1$ and $\left\vert \cdot \right\vert{}_2$ are equivalent if and only if they give the same $\O$.

We will prove this modulo the claim (to be proved later in Section 16.1) that valuations are equivalent if (and only if) they induce the same topology.

Proof. Suppose suppose $\left\vert \cdot \right\vert{}_1$ is equivalent to $\left\vert \cdot \right\vert{}_2$, so $\left\vert \cdot \right\vert{}_1 = \left\vert \cdot \right\vert{}_2^c$, for some $c>0$. Then $\left\vert c\right\vert _1 \leq 1$ if and only if $\left\vert c\right\vert _2^c \leq 1$, i.e., if $\left\vert c\right\vert _2 \leq 1^{1/c}=1$. Thus $\O _1 = \O _2$.

Conversely, suppose $\O _1 = \O _2$. Then $\vert a\vert _1<\vert b\vert _1$ if and only if $a/b\in \O _1$ and $b/a\not\in \O _1$, so

$$\vert a\vert _1<\vert b\vert _1 \iff \vert a\vert _2 < \vert b\vert _2.$$ (15.4)

The topology induced by $\vert$  $\vert _1$ has as basis of open neighborhoods the set of open balls

$$B_1(z,r) = \{x \in K : \vert x-z\vert _1<r \},$$

for $r>0$, and likewise for $\vert$  $\vert _2$. Since the absolute values $\vert b\vert _1$ get arbitrarily close to 0, the set $\mathcal{U}$ of open balls $B_1(z,\vert b\vert _1)$ also forms a basis of the topology induced by $\vert$  $\vert _1$ (and similarly for $\vert$  $\vert _2$). By (15.2.2) we have

$$B_1(z,\vert b\vert _1) = B_2(z,\vert b\vert _2),$$

so the two topologies both have $\mathcal{U}$ as a basis, hence are equal. That equal topologies imply equivalence of the corresponding valuations will be proved in Section 16.1. $\qedsymbol$

The set of $a\in \O$ with $\vert a\vert<1$ forms an ideal $\mathfrak{p}$ in $\O$. The ideal $\mathfrak{p}$ is maximal, since if $a\in \O$ and $a\not\in\mathfrak{p}$ then $\vert a\vert=1$, so $\vert 1/a\vert = 1/\vert a\vert = 1$, hence $1/a\in \O$, so $a$ is a unit.

Lemma 15.2.8   A non-archimedean valuation $\left\vert \cdot \right\vert{}$ is discrete if and only if $\mathfrak{p}$ is a principal ideal.

Proof. First suppose that $\left\vert \cdot \right\vert{}$ is discrete. Choose $\pi \in \mathfrak{p}$ with $\vert\pi\vert$ maximal, which we can do since

$$S=\{\log\vert a\vert : a \in \mathfrak{p}\} \subset (-\infty,1],$$

so the discrete set $S$ is bounded above. Suppose $a\in\mathfrak{p}$. Then

$$\left\vert\frac{a}{\pi}\right\vert = \frac{\left\vert a\right\vert}{\left\vert\pi \right\vert} \leq 1,$$

so $a/\pi\in \O$. Thus

$$a = \pi \cdot \frac{a}{\pi} \in \pi \O .$$

Conversely, suppose $\mathfrak{p}=(\pi)$ is principal. For any $a\in\mathfrak{p}$ we have $a=\pi b$ with $b\in\O$. Thus

$$\vert a\vert = \vert\pi\vert\cdot \vert b\vert \leq \vert\pi\vert < 1.$$

Thus $\{\vert a\vert : \vert a\vert<1\}$ is bounded away from $1$, which is exactly the definition of discrete. $\qedsymbol$

Example 15.2.9   For any prime $p$, define the $p$-adic valuation $\left\vert \cdot \right\vert{}_p:\mathbf{Q}\to\mathbf{R}$ as follows. Write a nonzero $\alpha\in K$ as $p^n\cdot \frac{a}{b}$, where $\gcd(a,p)=\gcd(b,p)=1$. Then

$$\left\vert p^n\cdot \frac{a}{b}\right\vert _p := p^{-n} = \left(\frac{1}{p}\right)^{n}.$$

This valuation is both discrete and non-archimedean. The ring $\O$ is the local ring

$$\mathbf{Z}_{(p)} = \left\{\frac{a}{b}\in\mathbf{Q}: p\nmid b\right\},$$

which has maximal ideal generated by $p$. Note that $\ord (p^n\cdot \frac{a}{b}) = p^n.$

We will using the following lemma later (e.g., in the proof of Corollary 16.2.4 and Theorem 15.3.2).

Lemma 15.2.10   A valuation $\left\vert \cdot \right\vert{}$ is non-archimedean if and only if $\vert n\vert\leq 1$ for all $n$ in the ring generated by $1$ in $K$.

Note that we cannot identify the ring generated by $1$ with  $\mathbf{Z}$ in general, because $K$ might have characteristic $p>0$.

Proof. If $\left\vert \cdot \right\vert{}$ is non-archimedean, then $\vert 1\vert\leq 1$, so by Axiom (3) with $a=1$, we have $\vert 1+1\vert\leq 1$. By induction it follows that $\vert n\vert\leq 1$.

Conversely, suppose $\vert n\vert\leq 1$ for all integer multiples $n$ of $1$. This condition is also true if we replace $\left\vert \cdot \right\vert{}$ by any equivalent valuation, so replace $\left\vert \cdot \right\vert{}$ by one with $C\leq 2$, so that the triangle inequality holds. Suppose $a\in K$ with $\vert a\vert\leq 1$. Then by the triangle inequality,

$$\left\vert 1+a\right\vert^n = \left\vert(1+a)^n\right\vert$$

$$\leq \sum_{j=0}^n \left\vert\binom{n}{j}\right\vert \left\vert a\right\vert$$

$$\leq 1 + 1 + \cdots + 1 = n.$$

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

$$\left\vert 1+a\right\vert \leq \sqrt[n]{n},$$

and take the limit as $n\to \infty$ to see that $\left\vert 1+a\right\vert \leq 1$. This proves that one can take $C=1$ in Axiom (3), hence that $\left\vert \cdot \right\vert{}$ is non-archimedean. $\qedsymbol$