Topology

A valuation $\left\vert \cdot \right\vert$ on a field $K$ induces a topology in which a basis for the neighborhoods of $a$ are the

$$B(a,d) = \{x\in K : \left\vert x-a\right\vert < d\}$$

for $d>0$.

Lemma 16.1.1   Equivalent valuations induce the same topology.

Proof. If $\left\vert \cdot \right\vert _1=\left\vert \cdot \right\vert _2^r$, then $\left\vert x-a\right\vert _1 < d$ if and only if $\left\vert x-a\right\vert _2^r<d$ if and only if $\left\vert x-a\right\vert _2<d^{1/r}$ so $B_1(a,d) = B_2(a,d^{1/r})$. Thus the basis of open neighborhoods of $a$ for $\left\vert \cdot \right\vert _1$ and $\left\vert \cdot \right\vert _2$ are identical. $\qedsymbol$

A valuation satisfying the triangle inequality gives a metric for the topology on defining the distance from $a$ to $b$ to be $\left\vert a-b\right\vert$. Assume for the rest of this section that we only consider valuations that satisfy the triangle inequality.

Lemma 16.1.2   A field with the topology induced by a valuation is a , i.e., the operations sum, product, and reciprocal are continuous.

Proof. For example (product) the triangle inequality implies that

$$\left\vert(a+\varepsilon )(b+\delta) - ab\right\vert \leq \left\v... ...rt\delta\right\vert + \left\vert b\right\vert\left\vert\varepsilon \right\vert$$

is small when $\left\vert\varepsilon \right\vert$ and $\left\vert\delta\right\vert$ are small (for fixed $a, b$). $\qedsymbol$

Lemma 16.1.3   Suppose two valuations $\left\vert \cdot \right\vert _1$ and $\left\vert \cdot \right\vert _2$ on the same field $K$ induce the same topology. Then for any sequence $\{x_n\}$ in $K$ we have

$$\left\vert x_n\right\vert _1 \to 0 \iff \left\vert x_n\right\vert _2 \to 0.$$

Proof. It suffices to prove that if $\left\vert x_n\right\vert _1\to 0$ then $\left\vert x_n\right\vert _2\to 0$, since the proof of the other implication is the same. Let $\varepsilon >0$. The topologies induced by the two absolute values are the same, so $B_2(0,\varepsilon )$ can be covered by open balls $B_1(a_i,r_i)$. One of these open balls $B_1(a,r)$ contains 0. There is $\varepsilon '>0$ such that

$$B_1(0,\varepsilon ') \subset B_1(a,r)\subset B_2(0,\varepsilon ).$$

Since $\left\vert x_n\right\vert _1\to 0$, there exists $N$ such that for $n\geq N$ we have $\left\vert x_n\right\vert _1 <\varepsilon '$. For such $n$, we have $x_n\in B_1(0,\varepsilon ')$, so $x_n\in B_2(0,\varepsilon )$, so $\left\vert x_n\right\vert _2<\varepsilon$. Thus $\left\vert x_n\right\vert _2\to 0$. $\qedsymbol$

Proposition 16.1.4   If two valuations $\left\vert \cdot \right\vert _1$ and $\left\vert \cdot \right\vert _2$ on the same field induce the same topology, then they are equivalent in the sense that there is a positive real $\alpha$ such that $\left\vert \cdot \right\vert _1 = \left\vert \cdot \right\vert _2^{\alpha}$.

Proof. If $x\in K$ and $i=1,2$, then $\left\vert x^n\right\vert _i \to 0$ if and only if $\left\vert x\right\vert _i^n\to 0$, which is the case if and only if $\left\vert x\right\vert _i<1$. Thus Lemma 16.1.3 implies that $\left\vert x\right\vert _1<1$ if and only if $\left\vert x\right\vert _2<1$. On taking reciprocals we see that $\left\vert x\right\vert _1>1$ if and only if $\left\vert x\right\vert _2>1$, so finally $\left\vert x\right\vert _1 = 1$ if and only if $\left\vert x\right\vert _2=1$.

Let now $w,z\in K$ be nonzero elements with $\left\vert w\right\vert _i \neq 1$ and $\left\vert z\right\vert _i\neq 1$. On applying the foregoing to

$$x = w^m z^n \qquad (m,n\in\mathbf{Z})$$

we see that

$$m\log\left\vert w\right\vert _1 + n\log\left\vert z\right\vert _1 \geq 0$$

if and only if

$$m\log\left\vert w\right\vert _2 + n\log\left\vert z\right\vert _2 \geq 0.$$

Dividing through by $\log\left\vert z\right\vert _i$, and rearranging, we see that for every rational number $\alpha=-n/m$,

$$\frac{\log\left\vert w\right\vert _1}{\log \left\vert z\right\ver... ...{\log\left\vert w\right\vert _2}{\log \left\vert z\right\vert _2} \geq \alpha.$$

Thus

$$\frac{\log\left\vert w\right\vert _1}{\log \left\vert z\right\vert _1} = \frac{\log\left\vert w\right\vert _2}{\log \left\vert z\right\vert _2},$$

so

$$\frac{\log\left\vert w\right\vert _1}{\log \left\vert w\right\vert _2} = \frac{\log\left\vert z\right\vert _1}{\log \left\vert z\right\vert _2}.$$

Since this equality does not depend on the choice of $z$, we see that there is a constant $c$ ( $=\log\left\vert z\right\vert _1/\log \left\vert z\right\vert _2$) such that $\log\left\vert w\right\vert _1/\log \left\vert w\right\vert _2 = c$ for all $w$. Thus $\log\left\vert w\right\vert _1 = c\cdot \log\left\vert w\right\vert _2$, so $\left\vert w\right\vert _1 = \left\vert w\right\vert _2^c$, which implies that $\left\vert \cdot \right\vert _1$ is equivalent to $\left\vert \cdot \right\vert _2$. $\qedsymbol$