PlanetMath (more info)
 Math for the people, by the people.
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
sections
Encyclopædia
Papers
Books
Expositions

meta
Requests (211)
Orphanage
Unclass'd (1)
Unproven (472)
Corrections (36)
Classification

talkback
Polls
Forums
Feedback
Bug Reports

downloads
Snapshots
PM Book

information
News
Docs
Wiki
ChangeLog
TODO List
Legalese
About
Owner confidence rating: Very high Entry average rating: No information on entry rating
[parent] valuation domain is local (Theorem)
Theorem 1   Every valuation domain is a local ring.

Proof. Let $ R$ be a valuation domain and $ K$ its field of fractions. We shall show that the set of all non-units of $ R$ is the only maximal ideal of $ R$.

Let $ a$ and $ b$ first be such elements of $ R$ that $ a-b$ is a unit of $ R$; we may suppose that $ ab \neq 0$ since otherwise one of $ a$ and $ b$ is instantly stated to be a unit. Because $ R$ is a valuation domain in $ K$, therefore e.g. $ \frac{a}{b}\in R$. Because now $ \frac{a-b}{b} = 1-\frac{a}{b}$ and $ (a-b)^{-1}$ belong to $ R$, so does also the product $ \frac{a-b}{b}\cdot(a-b)^{-1} = \frac{1}{b}$, i.e. $ b$ is a unit of $ R$. We can conclude that the difference $ a-b$ must be a non-unit whenever $ a$ and $ b$ are non-units.

Let $ a$ and $ b$ then be such elements of $ R$ that $ ab$ is its unit, i.e. $ a^{-1}b^{-1}\in R$. Now we see that

$\displaystyle a^{-1} = b\cdot a^{-1}b^{-1}\in R,\,\,\,b^{-1} = a\cdot a^{-1}b^{-1}\in R ,$
and consequently $ a$ and $ b$ both are units. So we conclude that the product $ ab$ must be a non-unit whenever $ a$ is an element of $ R$ and $ b$ is a non-unit.

Thus the non-units form an ideal $ \mathfrak{m}$. Suppose now that there is another ideal $ \mathfrak{n}$ of $ R$ such that $ \mathfrak{m}\subset\mathfrak{n}\subseteq R$. Since $ \mathfrak{m}$ contains all non-units, we can take a unit $ \varepsilon$ in $ \mathfrak{n}$. Thus also the product $ \varepsilon^{-1}\varepsilon$, i.e. 1, belongs to $ \mathfrak{n}$, or $ R\subseteq\mathfrak{n}$. So we see that $ \mathfrak{m}$ is a maximal ideal. On the other hand, any maximal ideal of $ R$ contains no units and hence is contained in $ \mathfrak{m}$; therefore $ \mathfrak{m}$ is the only maximal ideal.



"valuation domain is local" is owned by pahio.
(view preamble)

View style:

See Also: valuation ring, valuation determined by valuation domain, henselian field


This object's parent.
Log in to rate this entry.
(view current ratings)

Cross-references: contained, contains, ideal, difference, product, unit, maximal ideal, field of fractions, local ring, valuation domain
There is 1 reference to this entry.

This is version 7 of valuation domain is local, born on 2004-12-27, modified 2005-04-26.
Object id is 6599, canonical name is ValuationDomainIsLocal.
Accessed 1361 times total.

Classification:
AMS MSC: 16U10 (Associative rings and algebras :: Conditions on elements :: Integral domains)
 13G05 (Commutative rings and algebras :: Integral domains)
 13F30 (Commutative rings and algebras :: Arithmetic rings and other special rings :: Valuation rings)
Pending Errata and Addenda
None.
Discussion
Style: Expand: Order:
forum policy

No messages.

Interact
post | correct | update request | prove | add result | add corollary | add example | add (any)