PlanetMath: ring of exponent

(more info)

Math for the people, by the people.

Main Menu

sections Encyclopædia
Papers
Books
Expositions

meta Requests (205)
Orphanage
Unclass'd
Unproven (490)
Corrections (8)
Classification

talkback Polls
Forums
Feedback
Bug Reports

downloads Snapshots
PM Book

information News
Docs
Wiki
ChangeLog
TODO List
Legalese
About

Entry average rating: No information on entry rating

ring of exponent

(Definition)

Definition. Let $\nu$ be an exponent valuation of the field . The subring

$\displaystyle \mathcal{O}_\nu := \{\alpha \in K\,\vdots\;\; \nu(\alpha) \geqq 0\}$

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 is integrally closed in .

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

$\displaystyle \alpha = \varepsilon\pi^m,$

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 . Then also the intersection

$\displaystyle \mathfrak{O} := \bigcap_{i=1}^r\mathfrak{O}_i$

is a subring of

with unique factorisation. To be precise, any non-zero element $\alpha$ of $\mathfrak{O}$ may be uniquely represented in the form

$\displaystyle \alpha = \varepsilon\pi_1^{n_1}\cdots\pi_r^{n_r},$

in which $\varepsilon$ is a unit of $\mathfrak{O}$ , $n_1,\,\ldots,\,n_r$ non-negative integers and $\pi_1,\,\ldots,\,\pi_r$ fixed coprime prime elements of $\mathfrak{O}$ satisfying

$\begin{displaymath} \nu_i(\pi_j)\, = \, \delta_{ij} = \begin{cases} & 1 \;\;\mbox{for }\, i = j,\ & 0 \;\;\mbox{for }\, i \neq j. \end{cases}\end{displaymath}$

"ring of exponent" is owned by pahio.

(view preamble)

See Also: discrete valuation ring, local ring

Also defines:

ring of an exponent, ring of the exponent, integral with respect to an exponent

This object's parent.

Log in to rate this entry.
(view current ratings)

Cross-references: coprime, intersection, integer, algebraically closed, equation, UFD, unit, associated elements, prime element, ring, integrally closed, integral domain, subring, field, exponent valuation
There are 3 references to this entry.

This is version 8 of ring of exponent, born on 2008-04-16, modified 2008-05-08.
Object id is 10507, canonical name is RingOfExponent.
Accessed 420 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)

Pending Errata and Addenda

None.[ View all 1 ]

Discussion

forum policy

No messages.

Interact