valuation domain is local


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

Let a and b first be such elements of R that a-b is a unit of R; we may suppose that  ab0  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.  abR.  Because now  a-bb=1-ab  and  (a-b)-1  belong to R, so does also the producta-bb(a-b)-1=1b,  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-1b-1R.  Now we see that


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 𝔪.  Suppose now that there is another ideal 𝔫 of R such that  𝔪𝔫R.  Since 𝔪 contains all non-units, we can take a unit ε in 𝔫.  Thus also the product ε-1ε, i.e. 1, belongs to 𝔫, or  R𝔫.  So we see that 𝔪 is a maximal ideal.  On the other hand, any maximal ideal of R contains no units and hence is contained in 𝔪; therefore 𝔪 is the only maximal ideal.

Title valuation domain is local
Canonical name ValuationDomainIsLocal
Date of creation 2013-03-22 14:54:49
Last modified on 2013-03-22 14:54:49
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 10
Author pahio (2872)
Entry type Theorem
Classification msc 13F30
Classification msc 13G05
Classification msc 16U10
Related topic ValuationRing
Related topic ValuationDeterminedByValuationDomain
Related topic HenselianField