integral element

An element $a$ of a field $K$ is an integral element of the field $K$, iff

$$|a|\le 1$$

for every non-archimedean valuation  $|\cdot |$  of this field.

The set $𝒪$ of all integral elements of $K$ is a subring (in fact, an integral domain) of $K$, because it is the intersection of all valuation rings in $K$.

Examples

1. $K=\mathbb{Q}$.  The only non-archimedean valuations of $\mathbb{Q}$ are the $p$-adic valuations  $|\cdot {|}_p$  (where $p$ is a rational prime) and the trivial valuation (all values are 1 except the value of 0).  The valuation ring ${𝒪}_p$ of  $|\cdot {|}_p$  consists of all so-called p-integral rational numbers whose denominators are not divisible by $p$.  The valuation ring of the trivial valuation is, generally, the whole field.  Thus, $𝒪$ is, by definition, the intersection of the ${𝒪}_p$’s for all $p$;  this is the set of rationals whose denominators are not divisible by any prime, which is exactly the set $\mathbb{Z}$ of ordinary integers.

2. If $K$ is a finite field, it has only the trivial valuation.  In fact, if $|\cdot |$ is a valuation and $a$ any non-zero element of $K$, then there is a positive integer $m$ such that  $a^m=1$,  and we have  ${|a|}^m=|a^m|=|1|=1$,  and therefore  $|a|=1$.  Thus, $|\cdot |$ is trivial and  $𝒪=K$.  This means that all elements of the field are integral elements.

3. If $K$ is the field ${\mathbb{Q}}_p$ of the $p$-adic numbers (http://planetmath.org/NonIsomorphicCompletionsOfMathbbQ), it has only one non-trivial valuation, the $p$-adic valuation, and now the ring $𝒪$ is its valuation ring, which is the ring of $p$-adic integers (http://planetmath.org/PAdicIntegers);  this is visualized in the 2-adic (dyadic) case in the article “$p$-adic canonical form”.