# 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. 1.

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

2. 2.

$\nu(\alpha\beta)\;=\;\nu(\alpha)\!+\!\nu(\beta)$.

3. 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  , and because of notational causes, to the order valuations.

Properties.
$\nu(1)=0$
$\nu(-\alpha)=\nu(\alpha)$
$\displaystyle\nu\left(\frac{\alpha}{\beta}\right)=\nu(\alpha)\!-\!\nu(\beta)$
$\nu(\alpha^{n})=n\,\nu(\alpha)$
$\nu(\alpha_{1}+\ldots+\alpha_{n})\geqq\min\{\nu(\alpha),\,\ldots,\,\nu(\alpha_% {n})\}$
$\nu(\alpha\!+\!\beta)=\min\{\nu(\alpha),\,\nu(\beta)\}\quad\mbox{if}\;\;\;\nu(% \alpha)\neq\nu(\beta)$

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

 $\nu_{\mathfrak{p}}(\alpha)\;=:\;\begin{cases}&\infty\quad\mbox{when }\alpha=0,% \\ &\max\;\{k\in\mathbb{Z}\,\vdots\;\;\mathfrak{p}^{k}\mid(\alpha)\}\;\;\mbox{ % when }\,\alpha\neq 0;\end{cases}$
 $\nu_{\mathfrak{p}}(\xi)\;=:\;\nu_{\mathfrak{p}}(\alpha)-\nu_{\mathfrak{p}}(% \beta)\;\mbox{ when }\;\xi=\frac{\alpha}{\beta}\mbox{ with }\,\alpha,\,\beta% \in\mathcal{O}^{*}.$

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

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

Let  $\nu_{1},\,\ldots,\,\nu_{r}$  be different exponents of a field $K$.  Then for arbitrary set  $n_{1},\,\ldots,\,n_{r}$  of integers, there exists in $K$ an element $\xi$ such that

 $\nu_{1}(\xi)\;=\;n_{1},\;\;\ldots,\;\;\nu_{r}(\xi)\;=\;n_{r}.$

The proof of this theorem is found in .

## References

• 1 S. Borewicz & I. Safarevic: Zahlentheorie.  Birkhäuser Verlag. Basel und Stuttgart (1966).
 Title exponent valuation Canonical name ExponentValuation Date of creation 2013-03-22 17:59:31 Last modified on 2013-03-22 17:59:31 Owner pahio (2872) Last modified by pahio (2872) Numerical id 12 Author pahio (2872) Entry type Definition Classification msc 13F30 Classification msc 13A18 Classification msc 12J20 Classification msc 11R99 Synonym exponent of field Related topic DiscreteValuation Related topic OrderValuation Related topic UltrametricTriangleInequality Related topic DivisorTheoryAndExponentValuations Related topic DivisorTheory Defines exponent of a field Defines exponent of the field