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