# complete ultrametric field

A field $K$ equipped with a non-archimedean valuation$|\cdot|$  is called a non-archimedean field or also an ultrametric field, since the valuation the ultrametric$d(x,\,y):=|x\!-\!y|$  of $K$.

###### Theorem.

Let $(K,\,d)$ be a complete (http://planetmath.org/Complete) ultrametric field.  A necessary and sufficient condition for the convergence of the series

 $\displaystyle a_{1}\!+\!a_{2}\!+\!a_{3}\!+\ldots$ (1)

in $K$ is that

 $\displaystyle\lim_{n\to\infty}a_{n}\;=\;0.$ (2)

Proof.  Let $\varepsilon$ be any positive number.  When (1) converges, it satisfies the Cauchy condition and therefore exists a number $m_{\varepsilon}$ such that surely

 $|a_{m+1}|\;=\;\left|\sum_{j=1}^{m+1}a_{j}-\sum_{j=1}^{m}a_{j}\right|\;<\;\varepsilon$

for all  $m\geqq m_{\varepsilon}$;  thus (2) is necessary.  On the contrary, suppose the validity of (2).  Now one may determine such a great number $n_{\varepsilon}$ that

 $|a_{m}|\;<\;\varepsilon\qquad\forall m\,\geqq\,n_{\varepsilon}.$

No matter how great is the natural number $n$, the ultrametric then guarantees the inequality

 $|a_{m}\!+\!a_{m+1}\!+\ldots+\!a_{m+n}|\;\leqq\;\max\{|a_{m}|,\,|a_{m+1}|,\,% \ldots,\,|a_{m+n}|\}\;<\;\varepsilon$

always when  $m\geqq n_{\varepsilon}$.  Thus the partial sums of (1) form a Cauchy sequence, which converges in the complete field.  Hence the series (1) converges, and (2) is sufficient.

 Title complete ultrametric field Canonical name CompleteUltrametricField Date of creation 2013-03-22 14:55:37 Last modified on 2013-03-22 14:55:37 Owner pahio (2872) Last modified by pahio (2872) Numerical id 15 Author pahio (2872) Entry type Theorem Classification msc 12J10 Classification msc 54E35 Related topic Series Related topic NecessaryConditionOfConvergence Related topic ExtensionOfValuationFromCompleteBaseField Related topic PropertiesOfNonArchimedeanValuations Defines ultrametric field Defines non-archimedean field