Generating sequences and key polynomials
aa r X i v : . [ m a t h . A C ] J u l GENERATING SEQUENCES AND KEY POLYNOMIALS
M. S. BARNAB´E AND J. NOVACOSKI
Abstract.
The main goal of this paper is to study the different definitions ofgenerating sequences appearing in the literature. We present these definitionsand show that under certain situations they are equivalent. We also present anexample that shows that they are not, in general, equivalent. We also presentthe relation of generating sequences and key polynomials. Introduction
Given a valuation ν on a ring R , a generating sequence is a subset Q of R thatcompletely determines the valuation ν . The formalization of this idea appears indifferent works in slightly different ways. One of the main goals of this paper is tostablish the relation between these different definitions.The concept of graded algebra is closely related to generating sequences. Es-sentially, a generating sequence is a set whose images generate the graded algebra.Graded algebras play an important role to understand extensions of valuations (seefor instance, [3], [4], [11] and [12]). Graded algebras are also central objects in theapproach of Teissier for local uniformization (see [9] and [10]).Let R be a ring and ν a valuation on R . We denote by N the set of non-negativeintegers. For Q ⊆ R we will denote by N Q = { λ : Q −→ N | λ ( Q ) = 0 for only finitely many Q ∈ Q } and Q λ = Y λ ( Q ) =0 Q λ ( Q ) ∈ R. For each γ ∈ ν ( R ), we consider the sets P γ = { f ∈ R | ν ( f ) ≥ γ } and P + γ = { f ∈ R | ν ( f ) > γ } . The graded ring of R associated to ν is defined asgr ν ( R ) = M γ ∈ ν ( R ) P γ / P + γ . Mathematics Subject Classification.
Primary 13A18.
Key words and phrases.
Valuations, associated graded ring, generating sequences, keypolynomials.During the realization of this project Novacoski was supported by a research grant fromFunda¸c˜ao de Amparo `a Pesquisa do Estado de S˜ao Paulo (process number 2017/17835-9).
The addition on gr ν ( R ) is given by the abelian group structure and the multiplica-tion is given explicitly by (cid:16) f + P + ν ( f ) (cid:17) · (cid:16) g + P + ν ( g ) (cid:17) := (cid:16) f g + P + ν ( f )+ ν ( g ) (cid:17) and extending it to gr ν ( R ) in the obvious way.In [4], the definition of generating sequence is a slight variation of the following(see discussion in Section 5). A set Q ⊆ R satisfies (GS1) if for every γ ∈ ν ( R )the abelian group P γ is generated by n a Q λ | ν ( a Q λ ) ≥ γ where λ ∈ N Q and a ∈ R × o . If ν is centered on R (i.e., ν ( f ) ≥ f ∈ R ), then P γ and P + γ are idealsof R . Moreover, if we set m := P +0 = { f ∈ R | ν ( f ) > } , then m is a prime ideal and for each γ ∈ ν ( R ) we have that P γ / P + γ is an R/ m -module. In this case, gr ν ( R ) is an R/ m -algebra, that will be called the gradedalgebra of R associated to ν . The definition of generating sequence in [2] and[8] is the following. If ν is centered at R , then Q ⊆ R satisfies (GS2) if for every γ ∈ ν ( R ) the ideal P γ is generated by n Q λ | ν ( Q λ ) ≥ γ where λ ∈ N Q o . We observe that if ν is centered, then (GS1) implies (GS2) and that (GS2) onlymakes sense for centered valuations. Moreover, when dealing with key polynomialsfor a valuation ν on K [ x ], the case when ν is centered is not interesting (see Lemma5.1).The definition of a generating sequence in [1], [5] and [13] is the following. If ν is centered at R , then Q ⊆ R satisfies (GS3) if the setin ν ( Q ) := { in ν ( Q ) | Q ∈ Q } generates gr ν ( R ) as an R/ m -algebra.One of the main goals of this paper is to stablish the relation between thesedifferent definitions. More specifically, we prove the following. Theorem 1.1.
Let R be a ring, ν a valuation on R and Q a subset of R . (i): If ν is centered and Q satisfies (GS2) , then for every γ ∈ ν ( R ) the group P γ / P + γ is generated by n a Q λ + P + γ | λ ∈ N Q , ν ( Q λ ) = γ and ν ( a ) = 0 o . In particular, Q satisfies (GS3) . (ii): If Q satisfies (GS3) , then the semigroup ν ( R ) is generated by ν ( Q ) := { ν ( Q ) | Q ∈ Q } . Moreover, if R is a domain with K = Quot( R ) , Kν = R/ m and ν ( Q ) generates ν ( R ) , then (GS3) is satisfied. ENERATING SEQUENCE 3
In Section 3 we present an example that shows that the converse of Theorem 1.1 (i) does not hold in general.We also study the relation of generating sequences and key polynomials. Theinteresting case of study for key polynomials is for valuations ν on K [ x ] which arenot trivial on K . Hence, we want to compare sequences of key polynomials whichare complete and sequences satisfying (GS1) . We make a change on (GS1) when R = K [ x ]. We will say that Q satisfies (GS1 ∗ ) if for every f ∈ K [ x ] there exist a , . . . , a r ∈ K and λ , . . . , λ r ∈ N Q , such that f = r X i =1 a i Q λ i with ν (cid:16) a i Q λ i (cid:17) ≥ ν ( f ) , for every i, ≤ i ≤ r, and for every i , 1 ≤ i ≤ r , if λ i ( Q ) = 0, then deg( Q ) ≤ deg( f ). Since K [ x ] × = K × ,the only difference between (GS1) and (GS1 ∗ ) is the condition on the degrees.Another important result of this paper is the following. Theorem 1.2.
Let Q be a set of key polynomials for K [ x ] . Then Q is complete ifand only if Q satisfies (GS1 ∗ ) . This paper is divided as follows. In Section 2 we present a few basic results aboutgraded algebras. In Section 3 we present the proof of Theorem 1.1 as well as theexample that shows that its converse is not satisfied. In Section 4 we present thedefinition and the main results about key polynomials (as in [7]). We also presentsome results that do not appear in [7] but will be needed here. Finally, in Section5 we present the relation between key polynomials and generating sequences.2.
Preliminaries
Let R be a ring and ν a valuation on R . For f ∈ R we will denote by in ν ( f ) theimage of f in P ν ( f ) / P + ν ( f ) ⊆ gr ν ( R ) . We have the following properties in gr ν ( R ). Lemma 2.1.
Take f, g ∈ R . (i): in ν ( f ) · in ν ( g ) = in ν ( f g ) . (ii): in ν ( f ) = in ν ( g ) if and only if ν ( f ) = ν ( g ) and ν ( f − g ) > ν ( f ) . (iii): If ν ( f ) < ν ( g ) , then in ν ( f ) = in ν ( f + g ) . (iv): in ν ( f + g ) = in ν ( f ) + in ν ( g ) if and only if ν ( f ) = ν ( g ) = ν ( f + g ) .Proof. Item (i) follows directly from the definition. For (ii) we observe that if ν ( f ) = ν ( g ), then in ν ( f ) = in ν ( g ) because they are in different components ofgr ν ( R ). Moreover, in the case ν ( f ) = ν ( g ), we have that in ν ( f ) = in ν ( g ) meansthat ( f − g ) ∈ P + ν ( f ) , which means that ν ( f − g ) > ν ( f ).Assume that ν ( f ) < ν ( g ). Then ν ( f + g ) = ν ( f ) < ν ( g ), hence by item (ii) wehave in ν ( f ) = in ν ( f + g ). In order to prove (iv) , we observe that if ν ( f ) = ν ( g ) = M. S. BARNAB´E AND J. NOVACOSKI ν ( f + g ), thenin ν ( f + g ) = ( f + g ) + P + ν ( f ) = ( f + P + ν ( f ) ) + ( g + P + ν ( g ) )= in ν ( f ) + in ν ( g ) . The converse follows directly from (iii) . (cid:3) Lemma 2.2.
Assume that in ν ( f ) = in ν r X i =1 f i ! for some f , . . . , f r ∈ R, with ν ( f ) = ν ( f i ) for every i , ≤ i ≤ r . Then there exists I ⊆ { , . . . , r } such that in ν ( f ) = X i ∈ I in ν ( f i ) . Proof.
We will prove by induction on r . If r = 1, there is nothing to prove. Assumethat r > r −
1. If ν ( f ) < ν r X i =2 f i ! , then by Lemma2.1 (iii) we have in ν ( f ) = in ν f + r X i =2 f i ! = in ν ( f ) . On the other hand, if ν ( f ) = ν r X i =2 f i ! , then by Lemma 2.1 (iv) we havein ν ( f ) = in ν f + r X i =2 f i ! = in ν ( f ) + in ν r X i =2 f i ! and the result follows by the induction hypothesis. (cid:3) Proof of Theorem 1.1
Proof of Theorem 1.1.
In order to prove (i) , assume that Q satisfies (GS2) . Take γ ∈ ν ( R ) and choose f ∈ R such that ν ( f ) = γ . Since P γ is generated as an idealby n Q λ | λ ∈ N Q and ν (cid:16) Q λ (cid:17) ≥ γ o there exist a , . . . , a r ∈ R and λ , . . . , λ r ∈ N Q such that(1) f = r X i =1 a i Q λ i and ν (cid:16) Q λ i (cid:17) ≥ γ = ν ( f ) for every i, ≤ i ≤ r. Let I = { i ∈ { , . . . , r } | ν ( a i ) = 0 and ν (cid:16) Q λ i (cid:17) = γ } . It follows from (1) that I = ∅ . Then ν f − X i ∈ I a i Q λ i ! = ν X i/ ∈ I a i Q λ i ! > γ. ENERATING SEQUENCE 5
Therefore, f − X i ∈ I a i Q λ i ∈ P + γ . Moreover, in this case (using Lemma 2.2)in ν ( f ) = in ν X i ∈ I a i Q λ i ! = X i ∈ I ′ in ν (cid:16) a i Q λ i (cid:17) = X i ∈ I ′ a i (in ν ( Q )) λ i for some I ′ ⊆ I . Therefore, Q satisfies (GS3) .In order to prove (ii) assume that Q ⊆ R satisfies (GS3) . Take γ ∈ ν ( R ) and f ∈ R such that ν ( f ) = γ . By our assumption, there exist a , . . . , a r ∈ R \ m and λ , . . . , λ r ∈ N Q such that(2) in ν ( f ) = r X i =1 a i in ν (cid:16) Q λ i (cid:17) . Since in ν ( f ) is homogeneous, we can assume that the elements on the right of (2)are homogeneous and of the same degree of in ν ( f ). This means that for each ofsuch i we have γ = ν ( f ) = ν (cid:16) Q λ i (cid:17) = X λ i ( Q ) =0 λ i ( Q ) ν ( Q ) , which is what we wanted to prove.Now assume that Kν = R/ m and that ν ( R ) is generated by ν ( Q ). For f ∈ R there exist n , . . . , n r ∈ N and Q , . . . , Q r ∈ Q such that ν ( f ) = r X i =1 n i ν ( Q i ) = ν r Y i =1 Q n i i ! Since Kν = R/ m , there exists z ∈ R \ m such that zν = f r Y i =1 Q n i i ν. This means that ν f − z r Y i =1 Q n i i ! > ν ( f ) and consequentlyin ν ( f ) = in ν z r Y i =1 Q n i i ! = z r Y i =1 (in ν ( Q i )) n i . (cid:3) Example 3.1.
This is to show that the converse of Theorem 1.1 (i) does not holdin general. Consider a field k and the valuation on k [ x, y ] induced by the embeddingof k [ x, y ] in k (( t Q )) defined by x t and y ∞ X i =1 t i = t + t + t + t + . . . For every p ( x, y ) ∈ k [ x, y ] there exists n ∈ N and a n ∈ k such that ν ( p ( x, y ) − a n x n ) > n . M. S. BARNAB´E AND J. NOVACOSKI
Indeed, we can write p t, ∞ X i =1 t i ! = ∞ X k = n a k t k . Hence, ν ( p ( x, y ) − a n x n ) = ν t p t, ∞ X i =1 t i ! − a n t n ! > n . Since ν ( a n x n ) = n we have thatin ν ( p ) = in ν ( a n x n ) = a n in ν ( x ) n . Therefore, Q = { x } satisfies (GS3) .However, y ∈ P and since x e y are algebraically independent over k , for every p , . . . , p r ∈ k [ x, y ] we have y = r X i =1 p i x i . Therefore, Q = { x } does not satisfy (GS2) .4. Key polynomials
In order to define a key polynomial, we will need to define the number ǫ ( f ) for f ∈ K [ x ]. Let Γ ′ = Γ ⊗ Q be the divisible hull of Γ. For a polynomial f ∈ K [ x ]and k ∈ N , we consider ∂ k ( f ) := 1 k ! d k fdx k , the so called Hasse-derivative of f of order k . Let ǫ ( f ) = max k ∈ N (cid:26) ν ( f ) − ν ( ∂ k f ) k (cid:27) ∈ Γ ′ . Definition 4.1.
A monic polynomial Q ∈ K [ x ] is said to be a key polynomial (of level ǫ ( Q )) if for every f ∈ K [ x ] if ǫ ( f ) ≥ ǫ ( Q ), then deg( f ) ≥ deg( Q ).The next result is a characterization for ǫ ( f ). Consider an extension µ of ν to K [ x ] (here K denotes an algebraic closure of K ). For a polynomial f ∈ K [ x ], wedefine δ ( f ) = max { µ ( x − a ) | a is a root of f } . Proposition 4.2 (Proposition 3.1 of [6]) . If f ∈ K [ x ] is a monic polynomial, then ǫ ( f ) = δ ( f ) . Remark 4.3.
The condition of being monic can be dropped in the propositionabove. This follows from the fact that for every c ∈ K × we have that ǫ ( f ) = max k ∈ N (cid:26) ν ( f ) − ν ( ∂ k f ) k (cid:27) = max k ∈ N (cid:26) ν ( cf ) − ν ( ∂ k ( cf )) k (cid:27) = ǫ ( cf )and since f and cf have the same roots, we have δ ( f ) = δ ( cf ). Corollary 4.4.
For f, g ∈ K [ x ] we have ǫ ( f g ) = max { ǫ ( f ) , ǫ ( g ) } . ENERATING SEQUENCE 7
Proof.
We have that a is a root of f g if and only if it is a root of f or g . Hence ǫ ( f g ) = δ ( f g ) = max { µ ( x − a ) | a is a root of f g } = max { µ ( x − a ) | a is a root of f or g } = max { δ ( f ) , δ ( g ) } = max { ǫ ( f ) , ǫ ( g ) } . (cid:3) Corollary 4.5.
Every key polynomial is irreducible.Proof.
Assume that Q is a key polynomial and suppose it is not irreducible. Write Q = f g where deg( f ) < deg( Q ) and deg( g ) < deg( Q ). Then by the previous result ǫ ( Q ) = max { ǫ ( f ) , ǫ ( g ) } . This implies that ǫ ( f ) = ǫ ( Q ) or ǫ ( g ) = ǫ ( Q ). Sincedeg( f ) < deg( Q ) and deg( g ) < deg( Q ) this is a contradiction to the fact that Q isa key polynomial. (cid:3) Remark 4.6.
Corollary 4.5 was proved in [7] (Proposition 2.4 (ii) ). However, theproof above is simpler.
Lemma 4.7 (Lemma 2.3 of [7]) . Let Q be a key polynomial and take f, g ∈ K [ x ] such that deg( f ) < deg( Q ) and deg( g ) < deg( Q ) . Then for ǫ := ǫ ( Q ) and any k ∈ N we have the following: (i): ν ( ∂ k ( f g )) > ν ( f g ) − kǫ (ii): If ν Q ( f Q + g ) < ν ( f Q + g ) and k ∈ I ( Q ) := n i | ǫ ( f ) = ν ( f ) − ν ( ∂ i f ) i o , then ν ( ∂ k ( f Q + g )) = ν ( f Q ) − kǫ ; (iii): If h , . . . , h s are polynomials such that deg( h i ) < deg( Q ) for every i = 1 , . . . , s and s Y i =1 h i = qQ + r with deg( r ) < deg( Q ) and r = 0 , then ν ( r ) = ν s Y i =1 h i ! < ν ( qQ ) . Proposition 4.8 (Proposition 2.6 of [7]) . If Q is a key polynomial, then ν Q is avaluation of K [ x ] . Proposition 4.9 (Proposition 2.10 of [7]) . For two key polynomials
Q, Q ′ ∈ K [ x ] we have the following: (i): If deg( Q ) < deg( Q ′ ) , then ǫ ( Q ) < ǫ ( Q ′ ) ; (ii): If ǫ ( Q ) < ǫ ( Q ′ ) , then ν Q ( Q ′ ) < ν ( Q ′ ) ; (iii): If deg( Q ) = deg( Q ′ ) , then (3) ν ( Q ) < ν ( Q ′ ) ⇐⇒ ν Q ( Q ′ ) < ν ( Q ′ ) ⇐⇒ ǫ ( Q ) < ǫ ( Q ′ ) . Corollary 4.10.
Let Q and Q ′ be key polynomials such that ǫ ( Q ) ≤ ǫ ( Q ′ ) . Forevery f ∈ K [ x ] , if ν Q ( f ) = ν ( f ) , then ν Q ′ ( f ) = ν ( f ) . M. S. BARNAB´E AND J. NOVACOSKI
Proof.
It follows from Proposition 4.9 that if ǫ ( Q ) ≤ ǫ ( Q ′ ), then ν Q ′ ( Q ) = ν ( Q ).Since deg( Q ) ≤ deg( Q ′ ), for every f i ∈ K [ x ] with deg( f i ) < deg( Q ) we have ν Q ′ ( f i ) = ν ( f i ). Hence ν Q ′ ( f i Q i ) = ν ( f i Q i ).Take f ∈ K [ x ] such that ν Q ( f ) = ν ( f ) and let f = f + f Q + . . . + f n Q n be the Q -expansion of f . Then ν Q ′ ( f ) ≥ min ≤ i ≤ n { ν Q ′ ( f i Q i ) } = min ≤ i ≤ n { ν ( f i Q i ) } = ν Q ( f ) = ν ( f ) . Since ν Q ′ ( f ) ≤ ν ( f ) for every f ∈ K [ x ] we have our result. (cid:3) Definition 4.11.
A set Q ⊆ K [ x ] is called a complete set for ν if for every f ∈ K [ x ] there exists Q ∈ Q with deg( Q ) ≤ deg( f ) such that ν Q ( f ) = ν ( f ). If theset Q admits an order under which it is well-ordered, then it is called a completesequence. Theorem 4.12 (Theorem 1.1 of [7]) . Every valuation ν on K [ x ] admits a completeset Q of key polynomials. Moreover, Q can be chosen to be well-ordered with respectto the order given by Q < Q ′ if ǫ ( Q ) < ǫ ( Q ′ ) . Remark 4.13.
In [7], the definition of complete sequence does not require thatdeg( Q ) ≤ deg( f ) as in Definition 4.11 above. This property is important and theproof of Theorem 1.1 in [7] guarantees that the obtained sequence satisfies theadditional property.5. Generating sequences vs key polynomials
In this section we will discuss the relation between key polynomials and gener-ating sequences. We start with the following easy lemma.
Lemma 5.1.
A valuation ν on K [ x ] is centered if and only if ν ( a ) = 0 for every a ∈ K \ { } and ν ( x ) ≥ .Proof. Assume that ν is centered. In particular, ν ( x ) ≥
0. Moreover, since ν iscentered, ν ( a ) ≥ a ∈ K \ { } . If ν ( a ) >
0, then ν ( a − ) = − ν ( a ) < , which is a contradiction. Hence, ν ( a ) = 0.For the converse, assume that ν ( x ) ≥ ν ( a ) = 0 for every a ∈ K \ { } . Forevery p ( x ) = a + . . . + a n x n ∈ K [ x ] we have ν ( p ( x )) ≥ min { ν ( a i x i ) } ≥ . Hence, ν is centered. (cid:3) In order to prove Theorem 1.2, we will need a few results. Our first result showsthat any complete set (independently of being formed by key polynomials) satisfies (GS1 ∗ ) . ENERATING SEQUENCE 9
Proposition 5.2. If Q ⊆ K [ x ] is a complete set for ν , then Q satisfies (GS1 ∗ ) .Proof. We will prove by induction on the degree of f . If deg( f ) = 1, then f = x − a for some a ∈ K . By our assumption, there exists x − b ∈ Q such that β := ν ( x − a ) = ν x − b ( x − a ) = min { ν ( x − b ) , ν ( b − a ) } . This implies that ν ( x − b ) ≥ β , ν ( b − a ) ≥ β and that p = ( x − b ) + ( b − a ), whichis what we wanted to prove.Assume now that for k ∈ N , for every f ∈ K [ x ] of deg( f ) < k our result issatisfied. Let f be a polynomial of degree k . Since Q is a complete set for ν , thereexists q ∈ Q such that deg( q ) ≤ deg( f ) and ν q ( f ) = ν ( f ). Let f = f + f q + . . . + f s q s be the q -expansion of f . Since deg( q ) ≤ deg( f ), we have deg( f i ) < deg( f ) = k forevery i , 1 ≤ i ≤ s . By the induction hypothesis, there exist a , . . . , a r , . . . , a s , . . . , a sr s ∈ K and λ , . . . , λ r , . . . , λ s , . . . , λ sr s ∈ N Q , such that for every i , 0 ≤ i ≤ s , f i = r i X j =1 a ij Q λ ij with ν (cid:16) a ij Q λ ij (cid:17) ≥ ν ( f i ) for every j, ≤ j ≤ r i , and deg( Q ) ≤ deg( f i ) ≤ deg( f ) for every polynomial Q for which λ ij ( Q ) = 0 forsome i, j . This implies that f = s X i =0 r i X j =1 a ij Q λ ij q i = X ≤ i ≤ s, ≤ j ≤ r i a ij Q λ ′ ij , where λ ′ ij ( q ′ ) = ( λ ij ( q ′ ) + i if q ′ = qλ ij ( q ′ ) if q ′ = q . Moreover, since ν q ( f ) = min ≤ i ≤ s { ν ( f i q i ) } = ν ( f ) and ν (cid:16) a ij Q λ ij (cid:17) ≥ ν ( f i ) , for every i, ≤ i ≤ n and j, ≤ j ≤ r i , we have ν ( f ) ≤ ν ( f i ) + iν ( q ) ≤ ν (cid:16) a ij Q λ ij (cid:17) + iν ( q ) = ν (cid:16) a ij Q λ ′ ij (cid:17) , for every i , 0 ≤ i ≤ s and j , 1 ≤ j ≤ r i , which is what we wanted to prove. (cid:3) The next result gives a converse for Proposition 5.2.
Proposition 5.3.
Assume that Q is a subset of K [ x ] with the following properties: • ν Q is a valuation for every Q ∈ Q ; • for every finite subset F ⊆ Q , there exists Q ′ ∈ F such that ν Q ′ ( Q ) = ν ( Q ) for every Q ∈ F ; • (GS1 ∗ ) is satisfied. Then Q is a complete set for ν .Proof. Take any polynomial f ∈ K [ x ] and let β := ν ( f ). Then, there exist a , . . . , a r ∈ K and λ , . . . , λ r ∈ N Q such that f = r X i =1 a i Q λ i with ν (cid:16) a i Q λ i (cid:17) ≥ β, for every i, ≤ i ≤ r, and deg( Q ) ≤ deg( f ) for every Q ∈ Q for which λ i ( Q ) = 0 for some i , 1 ≤ i ≤ r .Let F := { Q ∈ Q | λ i ( Q ) = 0 for some i, ≤ i ≤ n } . Since F is finite, there exists Q ′ ∈ F such that ν Q ′ ( Q ) = ν ( Q ) for every Q ∈ F . Inparticular, ν (cid:16) a i Q λ i (cid:17) = ν Q ′ (cid:16) a i Q λ i (cid:17) for every i , 1 ≤ i ≤ r . Then β ≤ min ≤ i ≤ n n ν (cid:16) a i Q λ i (cid:17)o = min ≤ i ≤ n n ν Q ′ (cid:16) a i Q λ i (cid:17)o ≤ ν Q ′ ( f ) ≤ ν ( f ) = β. Therefore, ν Q ( f ) = ν ( f ) and this concludes the proof. (cid:3) Proof of Theorem 1.2. If Q is a complete set for ν , then by Proposition 5.2 Q satisfies (GS1 ∗ ) .To prove the converse, we observe that since every element Q in Q is a keypolynomial, by Proposition 4.8, we have that ν Q is a valuation. Moreover, since Q is ordered by Q < Q ′ if and only if ǫ ( Q ) < ǫ ( Q ′ ), for every finite set F we canchoose Q ′ ∈ F such that ǫ ( Q ) ≤ ǫ ( Q ′ ) for every Q ∈ F . Applying Corollary 4.10we obtain that ν Q ′ ( Q ) = ν ( Q ) for every Q ∈ F . The result now follows from Proposition 5.3. (cid:3)
An interesting consequence of Theorem 1.2 is the following.
Corollary 5.4.
For every valuation ν on K [ x ] , there exists a set of key polynomials Q ⊆ K [ x ] such that Q satisfies (GS1*) . Moreover, this set can be chosen to bewell-ordered with respect to the order Q < Q ′ if ǫ ( Q ) < ǫ ( Q ′ ) .Proof. It follows directly from Theorems 4.12 and 1.2. (cid:3)
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