exponent valuation

Definition.  A function $\nu$ defined in a field $K$ is called an exponent valuation or shortly an exponent of the field, if it satisfies the following conditions:

1. $\nu (0)=\mathrm{\infty }$  and $\nu (\alpha )$ runs all rational integers when $\alpha$ runs the nonzero elements of $K$.

2. $\nu (\alpha \beta )=\nu (\alpha )+\nu (\beta )$.

3. $\nu (\alpha +\beta )\geqq \min \{\nu (\alpha ),\nu (\beta )\}$.
   

Note that because of the discrete value set $\mathbb{Z}$, an exponent valuation belongs to the discrete valuations, andbecause of notational causes, to the order valuations.

Properties.
$\nu \mathbf{}\mathrm{(}\mathrm{1}\mathrm{)}\mathrm{=}\mathrm{0}$
$\nu \mathbf{}\mathrm{(}\mathrm{-}\alpha \mathrm{)}\mathrm{=}\nu \mathbf{}\mathrm{(}\alpha \mathrm{)}$
$\nu \mathbf{}\mathrm{\left(}{\displaystyle \frac{\alpha }{\beta }}\mathrm{\right)}\mathrm{=}\nu \mathbf{}\mathrm{(}\alpha \mathrm{)}\mathrm{-}\nu \mathbf{}\mathrm{(}\beta \mathrm{)}$
$\nu \mathbf{}\mathrm{(}{\alpha }^n\mathrm{)}\mathrm{=}n\mathbf{}\nu \mathbf{}\mathrm{(}\alpha \mathrm{)}$
$\nu \mathbf{}\mathrm{(}{\alpha }_{\mathrm{1}}\mathrm{+}\mathrm{\dots }\mathrm{+}{\alpha }_n\mathrm{)}\mathrm{\geqq }\min \mathbf{}\mathrm{\{}\nu \mathbf{}\mathrm{(}\alpha \mathrm{)}\mathrm{,}\mathrm{\dots }\mathrm{,}\nu \mathbf{}\mathrm{(}{\alpha }_n\mathrm{)}\mathrm{\}}$
$\nu \mathbf{}\mathrm{(}\alpha \mathrm{+}\beta \mathrm{)}\mathrm{=}\min \mathbf{}\mathrm{\{}\nu \mathbf{}\mathrm{(}\alpha \mathrm{)}\mathrm{,}\nu \mathbf{}\mathrm{(}\beta \mathrm{)}\mathrm{\}}\mathit{\hspace{1em}}\mathrm{\text{if}}\mathbf{}\nu \mathbf{}\mathrm{(}\alpha \mathrm{)}\mathrm{\ne }\nu \mathbf{}\mathrm{(}\beta \mathrm{)}$

Example.  If an integral domain $𝒪$ has a divisor theory  ${𝒪}^{*}\to 𝔇$, then for each prime divisor $𝔭$ there is an exponent valuation ${\nu }_{𝔭}$ of the quotient field $K$ of $𝒪$.  It is given by

$${\nu }_{𝔭}(\xi )=:{\nu }_{𝔭}(\alpha )-{\nu }_{𝔭}(\beta )\text{ when }\xi =\frac{\alpha }{\beta }\text{ with }\alpha ,\beta \in {𝒪}^{*}.$$

Hence, ${𝔭}^{{\nu }_{𝔭}(\alpha )}$ exactly divides $\alpha$.  Apparently, ${\nu }_{𝔭}(\xi )$ does not depend on the quotient form $\frac{\alpha }{\beta }$ for $\xi$.  It is not hard to show that ${\nu }_{𝔭}$ defined above is an exponent of the field $K$.

Different prime divisors $𝔭$ and $𝔮$ determine different exponents ${\nu }_{𝔭}$ and ${\nu }_{𝔮}$, since the condition 3 of the definition of divisor theory (http://planetmath.org/DivisorTheory) guarantees such an element $\gamma$ of $𝒪$ which in divisible by $𝔭$ but not by $𝔮$; then  ${\nu }_{𝔭}(\gamma )\geqq 1$,  ${\nu }_{𝔮}(\gamma )=0$.

Theorem.  Let  ${\nu }_1,\mathrm{\dots },{\nu }_r$  be different exponents of a field $K$.  Then for arbitrary set  $n_1,\mathrm{\dots },n_r$  of integers, there exists in $K$ an element $\xi$ such that

$${\nu }_1(\xi )=n_1,\mathrm{\dots },{\nu }_r(\xi )=n_r.$$

The proof of this theorem is found in [1].