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[parent] exponent valuation (Definition)

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) = \infty$ and $ \nu(\alpha)$ runs all rational integers when $ \alpha$ runs the non-zero 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, 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 $ K$ 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 $ 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 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 $ K$. Then for arbitrary set $ n_1,\,\ldots,\,n_r$ of integers, there exists in $ K$ 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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See Also: discrete valuation, order valuation, ultrametric triangle inequality, divisor theory and exponent valuations, divisor theory

Other names:  exponent of field
Also defines:  exponent of a field, exponent of the field
Keywords:  exponent

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ring of exponent (Definition) by pahio
continuation of exponent (Definition) by pahio
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Cross-references: 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 477 times total.

Classification:
AMS MSC11R99 (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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