ring of exponent

Definition.  Let $\nu$ be an exponent valuation of the field $K$. The subring

$${𝒪}_{\nu }:=\{\alpha \in K\mathrm{⋮}\nu (\alpha )\geqq 0\}$$

of $K$ is called the $\nu$.  It is, naturally, an integral domain.  Its elements are called $\nu$.

Theorem 1.  The ring of the exponent $\nu$ of the field $K$ is integrally closed in $K$.

Theorem 2.  The ring ${𝒪}_{\nu }$ only one prime element $\pi$, when one does not regard associated elements as different.  Any non-zero element $\alpha$ can be represented uniquely with a $\pi$ in the form

$$\alpha =\epsilon {\pi }^m,$$

where $\epsilon$ is a unit of ${𝒪}_{\nu }$ and  $m=\nu (\alpha )\geqq 0$.  This means that $𝒪$ is a UFD.

Remark 1.  The prime elements $\pi$ of the ring ${𝒪}_{\nu }$ are characterised by the equation  $\nu (\pi )=1$  and the units  $\epsilon$ the equation  $\nu (\epsilon )=0$.

Remark 2.  In an algebraically closed field $\mathrm{\Omega }$, there are no exponents (http://planetmath.org/ExponentValuation).  In fact, if there were an exponent $\nu$ of $\mathrm{\Omega }$ and if $\pi$ were a prime element of the ring of the exponent, then, since the equation  $x^2-\pi =0$  would have a root (http://planetmath.org/Equation) $\varrho$ in $\mathrm{\Omega }$, we would obtain  $2\nu (\varrho )=\nu ({\varrho }^2)=\nu (\pi )=1$;  this is however impossible, because an exponent attains only integer values.

Theorem 3.  Let  ${𝔒}_1,\mathrm{\dots },{𝔒}_r$ be the rings of the different exponent valuations ${\nu }_1,\mathrm{\dots },{\nu }_r$ of the field $K$.  Then also the intersection

$$𝔒:=\bigcap _{i=1}^r{𝔒}_i$$

is a subring of $K$ with unique factorisation (http://planetmath.org/UFD).  To be precise, any non-zero element $\alpha$ of $𝔒$ may be uniquely represented in the form

$$\alpha =\epsilon {\pi }_1^{n_1}\mathrm{\cdots }{\pi }_r^{n_r},$$

in which $\epsilon$ is a unit of $𝔒$,  the integers $n_1,\mathrm{\dots },n_r$ are nonnegative and${\pi }_1,\mathrm{\dots },{\pi }_r$ are coprime prime elements of $𝔒$ satisfying