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

of $K$ is called the ring of the exponent $\nu$.  It is, naturally, an integral domain.  Its elements are called integral with respect to $\nu$.

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

Theorem 2.  The ring $\mathcal{O}_{\nu}$ contains 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 fixed $\pi$ in the form

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

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

Remark 2.  In an algebraically closed field $\Omega$, there are no exponents.  In fact, if there were an exponent $\nu$ of $\Omega$ 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 $\varrho$ in $\Omega$, 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  $\mathfrak{O}_{1},\,\ldots,\,\mathfrak{O}_{r}$ be the rings of the different exponent valuations $\nu_{1},\,\ldots,\,\nu_{r}$ of the field $K$.  Then also the intersection

is a subring of $K$ with unique factorisation.  To be precise, any non-zero element $\alpha$ of $\mathfrak{O}$ may be uniquely represented in the form

in which $\varepsilon$ is a unit of $\mathfrak{O}$,  the integers $n_{1},\,\ldots,\,n_{r}$ are nonnegative and $\pi_{1},\,\ldots,\,\pi_{r}$ are fixed coprime prime elements of $\mathfrak{O}$ satisfying