valuation domain is local

Theorem.

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  $ab\ne 0$  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

$$a^{-1}=b\cdot a^{-1}{b}^{-1}\in R,b^{-1}=a\cdot a^{-1}{b}^{-1}\in R,$$

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 $𝔪$.  Suppose now that there is another ideal $𝔫$ of $R$ such that  $𝔪\subset 𝔫\subseteq R$.  Since $𝔪$ contains all non-units, we can take a unit $\epsilon$ in $𝔫$.  Thus also the product ${\epsilon }^{-1}\epsilon$, i.e. 1, belongs to $𝔫$, or  $R\subseteq 𝔫$.  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