# properties of non-archimedean valuations

If $K$ is a field, and $\left\lvert\cdot\right\rvert$ a nontrivial non-archimedean valuation (or absolute value) on $K$, then $\left\lvert\cdot\right\rvert$ has some properties that are counterintuitive (and that are false for archimedean valuations).

###### Theorem 1.

Let $K$ be a field with a non-archimedean absolute value $\left\lvert\cdot\right\rvert$. For $r>0$ a real number, $x\in K$, define

 $\displaystyle B(x,r)=\{y\in K\ \mid\ \left\lvert x-y\right\rvert $\displaystyle\bar{B}(x,r)=\{y\in K\ \mid\ \left\lvert x-y\right\rvert\leq r\},% \ \text{the closed ball of radius r at x}$

Then

1. 1.

$B(x,r)$ is both open and closed;

2. 2.

$\bar{B}(x,r)$ is both open and closed;

3. 3.

If $y\in B(x,r)$ (resp. $\bar{B}(x,r)$) then $B(x,r)=B(y,r)$ (resp. $\bar{B}(x,r)=\bar{B}(y,r)$);

4. 4.

$B(x,r)$ and $B(y,r)$ (resp. $\bar{B}(x,r)$ and $\bar{B}(y,r)$) are either identical or disjoint;

5. 5.

If $B_{1}=B(x,r)$ and $B_{2}=B(y,s)$ are not disjoint, then either $B_{1}\subset B_{2}$ or $B_{2}\subset B_{1}$;

6. 6.

If $(x_{n})$ is a sequence of elements of $K$ with $\lim_{n\to\infty}x_{n}=0$, then $\sum_{n=1}^{\infty}x_{n}$ is Cauchy (and thus if $K$ is complete, a sufficient condition for convergence of a series is that the terms tend to zero)

Proof.  We start by proving (3). Suppose $y\in B(x,r)$. If $z\in B(x,r)$, then since the absolute value is non-archimedean, we have

 $\left\lvert z-y\right\rvert=\left\lvert(z-x)+(x-y)\right\rvert\leq\max(\left% \lvert z-x\right\rvert,\left\lvert x-y\right\rvert)

so that $z\in B(y,r)$. Clearly $x\in B(y,r)$, so reversing the roles of $x$ and $y$, we see that $B(x,r)=B(y,r)$. Finally, replacing $B$ by $\bar{B}$ and $<$ by $\leq$, we get equality of closed balls as well.

(4) is now trivial: If $B(x,r)\cap B(y,r)\neq\emptyset$, choose $z\in B(x,r)\cap B(y,r)$; then by (3), $B(x,r)=B(z,r)=B(y,r)$. An identical argument proves the result for closed balls.

To prove (5), choose $z\in B_{1}\cap B_{2}$. Assume first that $r\leq s$; then $B(z,r)=B_{1}$, and $B(z,r)\subset B(z,s)=B_{2}$, so that $B_{1}\subset B_{2}$. If $s\leq r$, then we have identically that $B_{2}\subset B_{1}$. (Note that (4) is a special case when $r=s$).

(1) and (2) now follow: for (1), note that $B(x,r)$ is obviously open; its complement consists of a union of open balls of radius $r$ disjoint with $B(x,r)$ and its complement is therefore open. Thus $B(x,r)$ is closed. For (2), $\bar{B}(x,r)$ is obviously closed; to see that it is open, take any $y\in\bar{B}(x,r)$; then $\bar{B}(x,r)=\bar{B}(y,r)$ and thus $B(y,s)\subset\bar{B}(y,r)$ for $s is an open neighborhood of $y$ contained in $\bar{B}(x,r)$, which is therefore open.

Finally, to prove (6), we must show that given $\epsilon$, we can find $N>0$ sufficiently large such that $\left\lvert\sum_{i=m}^{n}x_{i}\right\rvert<\epsilon$ whenever $m,n>N$. Simply choose $N$ such that $\left\lvert x_{i}\right\rvert<\epsilon$ for $i>N$; then

 $\left\lvert\sum_{i=m}^{n}x_{i}\right\rvert\leq\max(\left\lvert x_{m}\right% \rvert,\ldots,\left\lvert x_{n}\right\rvert)<\epsilon$
Title properties of non-archimedean valuations PropertiesOfNonarchimedeanValuations 2013-03-22 18:01:11 2013-03-22 18:01:11 rm50 (10146) rm50 (10146) 6 rm50 (10146) Theorem msc 13F30 msc 13A18 msc 12J20 msc 11R99 CompleteUltrametricField