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

$\displaystyle 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

$\displaystyle \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

$\displaystyle \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

$\displaystyle 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

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

we see that

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

if and only if

$\displaystyle 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$,

$\displaystyle \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

$\displaystyle \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

$\displaystyle \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$

William Stein 2004-05-06