An element $a$ of a field $K$ is an integral element of the field $K$ , iff $$|a| \leq 1$$ for every non-archimedean valuation $|\cdot|$ of this field.

The set $\mathcal{O}$ 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 $\mathcal{O}_p$ 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, $\mathcal{O}$ is, by definition, the intersection of the $\mathcal{O}_p$ '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 $\mathcal{O} = K$ . 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, it has only one non-trivial valuation, the $p$ -adic valuation, and now the ring $\mathcal{O}$ is its valuation ring, which is the ring of $p$ -adic integers; this is visualized in the 2-adic (dyadic) case in the article ``$p$ -adic canonical form''.