PlanetMath: exponent valuation

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exponent valuation

(Definition)

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

$\nu(0) = \infty$ and $\nu(\alpha)$ runs all rational integers when $\alpha$ runs the non-zero elements of .
$\nu(\alpha\beta) = \nu(\alpha)+\nu(\beta)$ .
$\nu(\alpha+\beta) \geqq \min\{\nu(\alpha),\,\nu(\beta)\}$ .

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.
$\nu(1) = 0$
$\nu(-\alpha) = \nu(\alpha)$
$\displaystyle\nu\left(\frac{\alpha}{\beta}\right) = \nu(\alpha)\!-\!\nu(\beta)$
$\nu(\alpha^n) = n\,\nu(\alpha)$
$\nu(\alpha_1+\ldots+\alpha_n) \geqq \min\{\nu(\alpha),\,\ldots,\,\nu(\alpha_n)\}$
$\nu(\alpha+\beta) = \min\{\nu(\alpha),\,\nu(\beta)\}$ 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 of $\mathcal{O}$ . It is given by

$\begin{displaymath} \nu_{\mathfrak{p}}(\alpha)\, := \, \begin{cases} & \infty \q... ...\mid (\alpha)\} \;\; \mbox{ when } \,\alpha \neq 0; \end{cases}\end{displaymath}$

$\displaystyle \nu_{\mathfrak{p}}(\xi) \,:=\, \nu_{\mathfrak{p}}(\alpha)-\nu_{\mathfrak{p}}(\beta) \;$ when $\displaystyle \; \xi = \frac{\alpha}{\beta}$ with $\displaystyle \,\alpha,\,\beta \in \mathcal{O}^*.$

Hence, $\mathfrak{p}^{\nu_{\mathfrak{p}}(\alpha)}$ strictly 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

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 guarantees such an element $\gamma$ of $\mathcal{O}$ which in divisible by $\mathfrak{p}$ but not by $\mathfrak{q}$ ; then $\nu_{\mathfrak{p}}(\gamma) \geqq 1$ , $\nu_{\mathfrak{q}}(\gamma) = 0$ .

Theorem. Let $\nu_1,\,\ldots,\,\nu_r$ be different exponents of a field . Then for arbitrary set $n_1,\,\ldots,\,n_r$ of integers, there exists in an element $\xi$ such that

$\displaystyle \nu_1(\xi) = n_1,\;\;\ldots,\;\;\nu_r(\xi) = n_r.$

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

Bibliography

1: S. BOREWICZ & I. SAFAREVIC: Zahlentheorie. Birkhäuser Verlag. Basel und Stuttgart (1966).

"exponent valuation" is owned by pahio.

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Other names:

exponent of field

Also defines:

exponent of a field, exponent of the field

Keywords:

exponent

This object's parent.

Attachments:: ring of exponent (Definition) by pahio
continuation of exponent (Definition) by pahio

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Cross-references: proof, divisible, quotient, strictly divides, quotient field, prime divisor, divisor theory, integral domain, properties, order valuations, discrete valuations, discrete, integers, rational, exponent, field, function
There are 9 references to this entry.

This is version 8 of exponent valuation, born on 2008-04-15, modified 2008-05-16.
Object id is 10503, canonical name is ExponentValuation2.
Accessed 686 times total.

Classification:

AMS MSC:	11R99 (Number theory :: Algebraic number theory: global fields :: Miscellaneous)
	12J20 (Field theory and polynomials :: Topological fields :: General valuation theory)
	13A18 (Commutative rings and algebras :: General commutative ring theory :: Valuations and their generalizations)
	13F30 (Commutative rings and algebras :: Arithmetic rings and other special rings :: Valuation rings)

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