valuation domain is local

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

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

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.

Title valuation domain is local ValuationDomainIsLocal 2013-03-22 14:54:49 2013-03-22 14:54:49 pahio (2872) pahio (2872) 10 pahio (2872) Theorem msc 13F30 msc 13G05 msc 16U10 ValuationRing ValuationDeterminedByValuationDomain HenselianField