Valuations
LetKbe a field We say that an integral domainO⊆Kis avaluation ringofKif
A valuation ring O is defined such that for every element x in its field K, either x or x - 1 belongs to O This property ensures that K serves as the field of fractions of O Therefore, an integral domain O qualifies as a valuation ring if it meets these criteria within its field of fractions.
In the context of a valuation ring O of a field K, we define V as K × /O ×, where R × represents the group of units of a ring R The valuation induced by O is represented by the natural map ν: K × → V While V could be expressed in a multiplicative form, we adhere to the conventional additive notation We refer to V as the group of values of O, and by convention, we extend the valuation ν to include all of K by setting ν(0) = ∞.
For elementsaO × ,bO × ofV, defineaO × ≤bO × ifa − 1 b∈O, and putv0} ThenPis the set of nonunits ofO From (1.1.1), it follows thatPis an ideal, and hence the unique maximal ideal ofO Ifν(a)> ν(b), thenab − 1 ∈P, whenceν(1+ab − 1 ) =0 and thereforeν(a+b) =ν(b) To summarize:
Lemma 1.1.2 IfO is a valuation ring with valuation ν, thenO has a unique maximal ideal P={x∈O|ν(x)>0}and(1.1.1)is an equality unless, perhaps, ν(a) =ν(b).
A valuation ring \( O \) of a field \( K \) is established through the natural map \( K^\times \rightarrow K^\times / O^\times \), which defines a valuation Conversely, a nontrivial homomorphism \( \nu \) from \( K^\times \) into a totally ordered additive group \( G \) that satisfies \( \nu(a+b) \geq \min\{\nu(a), \nu(b)\} \) allows us to define \( O_\nu \) as the set \( \{x \in K^\times \mid \nu(x) \geq 0\} \cup \{0\} \) It can be verified that \( O_\nu \) forms a valuation ring of \( K \), and \( \nu \) induces an order-preserving isomorphism from \( K^\times / O^\times \) to its image It is important to note that if \( \nu \) is replaced by \( \nu^n \) for any positive integer \( n \), the same valuation of \( K \) is obtained.
The maximal ideal of O ν is denoted as P ν, comprising elements x in K where ν(x) > 0, with the residue field of ν defined as F ν := O ν / P ν A valuation ν of K is considered a k-valuation if it contains a subfield k and ν(x) = 0 for all x in k ×, resulting in F ν being an extension of k, which in our case is a finite extension.
Our first main result on valuations is the extension theorem, but first we need a few preliminaries.
Lemma 1.1.3 Let R be a subring of a ring S and let x∈S Then the following conditions are equivalent:
1 x satisfies a monic polynomial with coefficients in R,
3 x lies in a subring that is a finitely generated R-submodule.
Proof The implications (1)⇒(2) ⇒ (3) are clear To prove (3)⇒(1), let
{x 1 , ,x n }be a set ofR-module generators for a subringS 0 containingx, then there are elementsa i j ∈Rsuch that xx i ∑ n j=1 a i j x j for 1≤i≤n.
Multiplying the matrix (δijx - aij) by its transposed matrix of cofactors results in f(x)xj = 0 for all j, where f(X) is the monic polynomial det(δijX - aij) and δij represents the Kronecker symbol Consequently, we deduce that f(x)S0 = 0, and since 1 ∈ S0, it follows that f(x) = 0.
In the context of rings R and S, an element x in S is considered integral over R if it meets specific criteria Furthermore, the set S is deemed integral over R if every element within S satisfies this integrality condition with respect to R Additionally, if R[x] and R[y] are finitely generated R-modules, they possess generators that play a crucial role in their structure.
The ring R[x,y] is generated by the elements {x_i y_j}, demonstrating that the sum and product of integral elements remain integral, thus forming a subring ˆR of all elements in S that are integral over R If an element x in S satisfies the polynomial equation x^n + ∑(from i=0 to n-1) a_i x^i = 0 with coefficients a_i in ˆR, then x is integral over the finitely generated R-module ˆR_0 := R[a_0, , a_{n-1}] Furthermore, if {b_1, , b_m} are generators for the R-module ˆR_0, then the set {b_i x_j | 1 ≤ i ≤ m, 0 ≤ j < n} generates the R-module ˆR_0[x].
Corollary 1.1.4 The set of all elements of S integral over R forms a subringR,ˆ and any element of S integral overR is already inˆ R.ˆ
The ring ˆRis called theintegral closureofRinS If ˆR=R, we say thatRis integrally closedinS IfS is otherwise unspecified, we take it to be the field of fractions ofR.
A ring \( R \) is defined as a local ring if it possesses a unique maximal ideal \( M \), where every element outside \( M \) is a unit In this context, if \( R \) is an integral domain with a prime ideal \( P \), the localization \( R_P \) at \( P \) forms a local subring within the field of fractions, comprising all elements of the form \( x/y \) where \( y \) is an element of \( P \).
Lemma 1.1.5 (Nakayama’s Lemma) Let R be a local ring with maximal ideal
P and let M be a nonzero finitely generated R-module Then PMM.
Proof LetM=Rm 1 +ããã+Rm n , wherenis minimal, and putM 0 :=Rm 2 +ããã+
Rm n ThenM 0 is a proper submodule IfM=PM, we can write m 1 ∑ n i=1 a i m i with a i ∈P, but 1−a 1 is a unit since R is a local ring, and we obtain the contradiction m 1 = (1−a 1 ) −1
The Valuation Extension Theorem states that if R is a subring of a field K and P is a nonzero prime ideal of R, then there exists a valuation ring O within K This valuation ring O has a maximal ideal M, satisfying the inclusion R⊆O⊆K, and the intersection of M with R equals the prime ideal P.
Proof Consider the set of pairs (R ,P ) where R is a subring of K and
P is a prime ideal of R We say that (R ,P ) extends (R ,P ) and write (R ,P )≥(R ,P )ifR ⊇R andP ∩R =P This relation is a partial order.
By Zorn’s lemma, there is a maximal extension(O,M)of(R,P).
We begin by noting that when M=0, it follows that O=K Upon confirming that M is contained in the intersection of MO and O, we establish that the pair (O M, MO M) is greater than or equal to (O, M) Given our choice of (O, M) as maximal, we deduce that O is a local ring with maximal ideal M Considering an element x in K, if M generates a proper ideal M1 of O[x^−1], then (O[x^−1], M1) is greater than or equal to (O, M), as M is a maximal ideal of O The maximality of (O, M) further implies that x^−1 is an element of O Otherwise, there exists an integer n and elements a_i in M such that
SinceOis a local ring, 1−a 0 is a unit Dividing (∗) by(1−a 0 )x −n , we find that xis integral overO In particular,O[x]is a finitely generatedO-module Now the maximality of(O,M)and (1.1.5) imply thatx∈O.
Corollary 1.1.7 Suppose that k⊆K are fields and x∈K If x is transcendental over k, there exists a k-valuationν of K withν(x)>0 If x is algebraic over k, ν(x) =0for all k-valuationsν.
Proof Ifxis transcendental overk, apply (1.1.6) withO:=k[x]andP:= (x)to obtain ak-valuationνwithν(x)>0 Conversely, if
∑ n i= 0 a i x i =0 witha i ∈kanda n =0, and ifνis ak-valuation, then we have ν(a n x n ) =nν(x) =ν i