Editing Valuation ring (section)

== Dominance and integral closure ==
The [[unit (ring theory)|units]], or invertible elements, of a valuation ring are the elements ''x'' in ''D'' such that ''x''<sup>&nbsp;−1</sup> is also a member of ''D''. The other elements of ''D'' – called nonunits – do not have an inverse in ''D'', and they form an [[ideal (ring theory)|ideal]] ''M''. This ideal is maximal among the (totally ordered) ideals of D. Since ''M'' is a [[maximal ideal]], the [[quotient ring]] ''D''/''M'' is a field, called the '''residue field''' of ''D''.

In general, we say a [[local ring]] <math>(S,\mathfrak{m}_S)</math> dominates a local ring <math>(R,\mathfrak{m}_R)</math> if <math>S \supseteq R</math> and <math>\mathfrak{m}_S \cap R = \mathfrak{m}_R</math>; in other words, the inclusion <math>R \subseteq S</math> is a [[local ring homomorphism]]. Every local ring <math>(A, \mathfrak{p})</math> in a field ''K'' is dominated by some valuation ring of ''K''. Indeed, the set consisting of all subrings ''R'' of ''K'' containing ''A'' and <math>1 \not\in \mathfrak{p}R</math> is [[empty set|nonempty]] and is inductive; thus, has a maximal element <math>R</math> by [[Zorn's lemma]]. We claim ''R'' is a valuation ring. ''R'' is a local ring with maximal ideal containing <math>\mathfrak{p}R</math> by maximality. Again by maximality it is also integrally closed. Now, if <math>x \not\in R</math>, then, by maximality, <math>\mathfrak{p}R[x] = R[x]</math> and thus we can write:
:<math>1 = r_0 + r_1 x + \cdots + r_n x^n, \quad r_i \in \mathfrak{p}R</math>.

Since <math>1 - r_0</math> is a unit element, this implies that <math>x^{-1}</math> is integral over ''R''; thus is in ''R''. This proves ''R'' is a valuation ring. (''R'' dominates ''A'' since its maximal ideal contains <math>\mathfrak{p}</math> by construction.)

A local ring ''R'' in a field ''K'' is a valuation ring if and only if it is a maximal element of the set of all local rings contained in ''K'' partially ordered by dominance. This easily follows from the above.{{efn|Proof: if ''R'' is a maximal element, then it is dominated by a valuation ring; thus, it itself must be a valuation ring. Conversely, let ''R'' be a valuation ring and ''S'' a local ring that dominates ''R'' but not ''R''. There is ''x'' that is in ''S'' but not in ''R''. Then <math>x^{-1}</math> is in ''R'' and in fact in the maximal ideal of ''R''. But then <math>x^{-1} \in \mathfrak{m}_S</math>, which is absurd. Hence, there cannot be such ''S''.}}

Let ''A'' be a subring of a field ''K'' and <math>f: A \to k</math> a [[ring homomorphism]] into an [[algebraically closed field]] ''k''. Then ''f'' extends to a ring homomorphism <math>g: D \to k</math>, ''D'' some valuation ring of ''K'' containing ''A''. (Proof: Let <math> g: R \to k </math> be a maximal extension, which clearly exists by Zorn's lemma. By maximality, ''R'' is a local ring with maximal ideal containing the [[kernel (algebra)|kernel]] of ''f''. If ''S'' is a local ring dominating ''R'', then ''S'' is algebraic over ''R''; if not, <math>S</math> contains a polynomial ring <math>R[x]</math> to which ''g'' extends, a contradiction to maximality. It follows <math>S/\mathfrak{m}_S</math> is an algebraic field extension of <math>R/\mathfrak{m}_R</math>. Thus, <math>S \to S/\mathfrak{m}_S \hookrightarrow k</math> extends ''g''; hence, ''S'' = ''R''.)

If a subring ''R'' of a field ''K'' contains a valuation ring ''D'' of ''K'', then, by checking Definition 1, ''R'' is also a valuation ring of ''K''. In particular, ''R'' is local and its maximal ideal contracts to some prime ideal of ''D'', say, <math>\mathfrak{p}</math>. Then <math>R = D_\mathfrak{p}</math> since <math>R</math> dominates <math>D_\mathfrak{p}</math>, which is a valuation ring since the ideals are totally ordered. This observation is subsumed to the following:{{sfn|Zariski|Samuel|1975|loc=Ch. VI, Theorem 3}} there is a [[bijection|bijective]] correspondence <math>\mathfrak{p} \mapsto D_\mathfrak{p}, \operatorname{Spec}(D) \to</math> the set of all subrings of ''K'' containing ''D''. In particular, ''D'' is integrally closed,{{sfn|Efrat|2006|p=38}}{{efn|To see more directly that valuation rings are integrally closed, suppose that ''x''<sup>''n''</sup>&nbsp;+&nbsp;''a''<sub>1</sub>''x''<sup>''n''−1</sup>&nbsp;+ ... +&nbsp;''a''<sub>0</sub>&nbsp;{{=}}&nbsp;0.  Then dividing by
''x''<sup>''n''−1</sup> gives us ''x'' {{=}} −''a''<sub>1</sub>&nbsp;−&nbsp;...&nbsp;−&nbsp;{{nowrap|''a''<sub>0</sub>''x''<sup>−''n''+1</sup>}}. If ''x'' were not in ''D'', then ''x''<sup>−1</sup> would be in ''D'' and this would express ''x'' as a finite sum of elements in ''D'', so that ''x'' would be in ''D'', a contradiction.}} and the [[Krull dimension]] of ''D'' is the number of proper subrings of ''K'' containing ''D''.

In fact, the [[integral closure]] of an integral domain ''A'' in the field of fractions ''K'' of ''A'' is the [[intersection (set theory)|intersection]] of all valuation rings of ''K'' containing ''A''.{{sfn|Matsumura|1989|loc=Theorem 10.4}} Indeed, the integral closure is contained in the intersection since the valuation rings are integrally closed. Conversely, let ''x'' be in ''K'' but not integral over ''A''. Since the ideal <math>x^{-1} A[x^{-1}]</math> is not <math>A[x^{-1}]</math>,{{efn|In general, <math>x^{-1}</math> is integral over ''A'' if and only if <math>xA[x] = A[x].</math>}} it is contained in a maximal ideal <math>\mathfrak{p}</math>. Then there is a valuation ring ''R'' that dominates the localization of <math>A[x^{-1}]</math> at <math>\mathfrak{p}</math>. Since <math>x^{-1} \in \mathfrak{m}_R</math>, <math>x \not\in R</math>.

The dominance is used in [[algebraic geometry]]. Let ''X'' be an algebraic variety over a field ''k''. Then we say a valuation ring ''R'' in <math>k(X)</math> has "center ''x'' on ''X''" if <math>R</math> dominates the local ring <math>\mathcal{O}_{x, X}</math> of the structure sheaf at ''x''.{{sfn|Hartshorne|1977|loc=Ch II. Exercise 4.5}}