Definition. Let $\nu$ be an exponent valuation of the field $K$ . The subring $$\mathcal{O}_\nu \;:=\; \{\alpha \in K\,\vdots\;\; \nu(\alpha) \geqq 0\}$$ 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 $$\alpha \;=\; \varepsilon\pi^m,$$ 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 $$\mathfrak{O} \;:=\; \bigcap_{i=1}^r\mathfrak{O}_i$$ 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 $$\alpha \;=\; \varepsilon\pi_1^{n_1}\cdots\pi_r^{n_r},$$ 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