# integral element

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. 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. 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. 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 $\mathcal{O}$ 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”.

Title integral element IntegralElement 2013-03-22 14:15:56 2013-03-22 14:15:56 pahio (2872) pahio (2872) 31 pahio (2872) Definition msc 12E99 PAdicCanonicalForm PAdicValuation KummersCongruence