Limit key polynomials as p -polynomials
aa r X i v : . [ m a t h . A C ] S e p LIMIT KEY POLYNOMIALS ARE p -POLYNOMIALS MICHAEL DE MORAES AND JOSNEI NOVACOSKI
Abstract.
The main goal of this paper is to characterize limit key polyno-mials for a valuation ν on K [ x ]. We consider the set Ψ α of key polynomialsfor ν of degree α . We set p be the exponent characteristic of ν . Our firstmain result (Theorem 1.1) is that if Q α is a limit key polynomial for Ψ α , thenthe degree of Q α is p r α for some r ∈ N . Moreover, in Theorem 1.2, we showthat there exist Q ∈ Ψ α and Q α a limit key polynomial for Ψ α , such that the Q -expansion of Q α only has terms which are powers of p . Introduction
The concept of key polynomials was introduced in [8] and [9] in order to un-derstand extensions of a valuation ν on a field K to the field K ( x ). The idea isthat for a given valuation ν on K ( x ), a key polynomial Q ∈ K [ x ] for ν allows usto augment ν , i.e., build a new valuation ν ′ with ν ( f ) ≤ ν ′ ( f ) for every f ∈ K [ x ](which we denote by ν ≤ ν ′ ) and ν ( Q ) < ν ′ ( Q ). MacLane proved that if ν isdiscrete, then every valuation ν on K ( x ), extending ν , can be built by startingwith a monomial valuation and using a sequence (of order type at most ω ) builtiteratively to obtain ν .A major development was presented by Vaqui´e in [12] and [13]. He introducedthe concept of limit key polynomial, and proved that if we allow these objects inthe sequence, then we can drop the assumption of ν being discrete in MacLane’smain result (in this case, the order type of the sequence can be larger than ω ).An alternative definition of key polynomials was introduced in [4] and [11] (in[4] they are called abstract key polynomials). The main difference between thesetwo objects is that key polynomial ν (as in Maclane and Vaqui´e’s work) allowsus to augment ν , while a key polynomial for ν (as in [4] and [11]) allows us to truncate ν . In particular, if we consider the valuation ν ′ on K [ x ] obtained from ν by the MacLane-Vaqui´e’s method, then ν ≤ ν ′ and by the method in [4] and[11], we obtain ν ′ ≤ ν . Because of this, key polynomials (as in [4] and [11]) arebetter to stablish the relation to other similar objects in the literature, such as pseudo-convergent sequences as defined in [6] and minimal pairs as defined in [1].The relations between these objects were explored in [10] and [11]. Mathematics Subject Classification.
Primary 13A18.
Key words and phrases.
Valuations, Key polynomials, Limit key polynomials, p -polynomials.During the realization of this project the second author was supported by a grant fromFunda¸c˜ao de Amparo `a Pesquisa do Estado de S˜ao Paulo (process number 2017/17835-9). The main pourpose of the work presented here is to understand the structureof limit key polynomials. These objects are main obstacles in some problems con-cerning valuations. One example of this is the local uniformization problem, whichis open in positive characteristic. One of the main problems in handling valuationsis the existence of defect. For instance, in [2] it is proved that if an extension doesnot have defect, then one can extend local uniformization. Hence, one is led to askthe relation between defect and the degrees of limit key polynomials.For a valuation ν on a field K we denote the residue field of ν by Kν . Then the characteristic exponent of ( K, ν ) is defined as char( Kν ) if char( Kν ) > Kν ) = 0. It is well-known that the defect of an extension is a power of p , thecharacteristic exponent of ν . If one wants to relate the defect to the degrees of limitkey polynomials it is natural to ask whether the degree of a limit key polynomialfor Ψ α is of the form p r α for some r ∈ N . This is our first main result.Throughout this paper we will consider a rank one valuation ν on K [ x ]. For f ∈ K [ x ] \ { } we define the level ǫ ( f ) of f by(1) ǫ ( f ) = max b ∈ N (cid:26) ν ( f ) − ν ( ∂ b f ) b (cid:27) , where ∂ b f is the Hasse derivative of f of order b . Take Q ∈ K [ x ] \ K a monicpolynomial. We say that Q is a key polynomial for ν if for every f ∈ K [ x ] \ { } of degree smaller than deg( Q ) we have ǫ ( f ) < ǫ ( Q ).For any f, g ∈ K [ x ], with q = 0, since K [ x ] is a euclidian domain, there existuniquely determined f , . . . , f n ∈ K [ x ] with f i = 0 or deg( f i ) < deg( q ) for every i ,0 ≤ i ≤ n , such that f = f n q n + . . . + f q + f . This expression is called the q -expansion of f . In this case, we set deg q ( f ) := n .For each q ∈ K [ x ] the map ν q ( f n q n + . . . + f q + f ) := min ≤ i ≤ n (cid:8) ν ( f i q i ) (cid:9) is well-defined. This map is called the truncation of ν on q . For polynomials f, q ∈ K [ x ] \ { } we define S q ( f ) := { i | ν q ( f ) = ν ( f i Q i ) } and δ q ( f ) := max S q ( f ) . For α ∈ N we consider Ψ α the set of key polynomials for ν of degree α . Assumethat Ψ α does not have a largest element (with respect to the order Q ≤ Q ′ if andonly if ǫ ( Q ) ≤ ǫ ( Q ′ )). Consider the set S α := { f ∈ K [ x ] | ν Q ( f ) < ν ( f ) for every Q ∈ Ψ α } . A monic polynomial Q α ∈ K [ x ] is a limit key polynomial for Ψ α if it belongsto S α and has the least degree among polynomials in S α . One of the main goals ofthis paper is to prove the following. IMIT KEY POLYNOMIALS ARE p -POLYNOMIALS 3 Theorem 1.1.
Assume that Ψ α admits a limit key polynomial. Then there exists r ∈ N such that for every Q ∈ Ψ α and every limit key polynomial Q α for Ψ α the Q -expansion of Q α is of the form Q α = Q p r + a p r − Q p r − + . . . + a Q + a . Moreover, for every Q ∈ Ψ α , we have ν Q ( Q α ) = p r ν ( Q ) . The ideia of the proof of Theorem 1.1 (which is presented in Section 3) is thefollowing. We let Q α = a d Q d + a d − Q d − + . . . + a be the Q -expansion of Q α . We set δ := δ Q ( Q α ). Then we show that d = δ (Propostion 3.7). This means that Q α = a δ Q δ + a δ − Q δ − + . . . + a . Then we show that a δ = 1 (Proposition 3.10) and that δ = p r for some r ∈ N (Proposition 3.11).The relation of key polynomials and pseudo-convergent sequences is studied in[11]. There, it is shown that any polynomial of smallest degree not fixed by apseudo-convergent sequence is a limit key polynomial. Moreover, in [6] it is shownthat such polynomials can be chosen to be p -polynomials , i.e., of the form a p r x p r + a p r − x p r − . . . + a x + a . Our next result generalizes this for every limit key polynomial.
Theorem 1.2.
Assume that Ψ α admits a limit key polynomial. Then there exist r ∈ N , Q ∈ Ψ α and a limit key polynomial Q α for Ψ α such that the Q -expansion of Q α is of the form Q α = Q p r + a p r − Q p r − + . . . + a Q + a . The main idea for the proof of Theorem 1.2 (which is presented in Section 4)is the following. Since rk( ν ) = 1 and the set ν (Ψ α ) is bounded (Lemma 3.5) ithas a supremum. We set B = sup( ν (Ψ α )). Then, for Q ∈ Ψ α and Q α a limit keypolynomial for Ψ α , if a monomial a i Q i in the Q -expansion of Q α has value greaterthan p r B , then Q α − a i Q i is also a limit key polynomial for Ψ α (Lemma 3.6). Wethen study the behaviour of the values of the coefficients in the expansions of Q α in elements of Ψ α . Using this we show that, for sufficiently large Q ∈ Ψ α the terms a i Q i in the Q -expansion of Q α for which i is not a power of p (and i = 0) havevalue greater than p r B . Hence, we can eliminate them to obtain the desired limitkey polynomial for Ψ α .The reason to present a form of limit key polynomials as in Theorem 1.2 isbecause the roots of such polynomials are simpler. Another reason for this is toclassify the defect of an extension. In [7], Kuhlmann presents a classification ofArtin-Schreier defect extensions as dependent or independent. We hope that the MICHAEL DE MORAES AND JOSNEI NOVACOSKI characterization of limit key polynomials as in Theorem 1.2 will allow us to presentsimilar classifications for defect extensions which are not necessarily Artin-Schreier.We observe that similar results as the ones presented here appear in [5]. The maindifference is that there they consider a specific type of sequence of key polynomialswhile here we consider the set of all key polynomials. Also, our proofs are simplerand more fundamental. 2.
Preliminaires
Definition 2.1.
Take a commutative ring R with unity. A valuation on R isa mapping ν : R −→ Γ ∞ := Γ ∪ {∞} where Γ is an ordered abelian group (andthe extension of addition and order to ∞ in the obvious way), with the followingproperties: (V1): ν ( f g ) = ν ( f ) + ν ( g ) for every f, g ∈ R . (V2): ν ( f + g ) ≥ min { ν ( f ) , ν ( g ) } for every f, g ∈ R . (V3): ν (1) = 0 and ν (0) = ∞ .2.1. Key polynomials.
In this section we discuss the basics about key polynmialsas presented in [11].
Remark 2.2.
The equality (1) implies that(2) ν ( ∂ b f ) ≥ ν ( f ) − bǫ ( f ) , for every b ∈ N . If ǫ ( Q ) > ǫ ( f ), then for every b ∈ N we have ν ( ∂ b f ) > ν ( f ) − bǫ ( Q ) . Notation 2.3.
For f ∈ K [ x ] \ { } we denote by I ( f ) := { b ∈ N | the equality holds in (2) } . Lemma 2.4.
Let f, g ∈ K [ x ] \ { } . We have (3) ǫ ( f g ) ≤ max { ǫ ( f ) , ǫ ( g ) } . Moreover, if ǫ ( f ) > ǫ ( g ) , then ǫ ( f g ) = ǫ ( f ) .Proof. Let ǫ = max { ǫ ( f ) , ǫ ( g ) } . For b ∈ N and i ∈ N with i ≤ b , we have(4) ν ( ∂ b − i f ∂ i g ) = ν ( ∂ b − i f ) + ν ( ∂ i g ) ≥ ν ( f ) − ( b − i ) ǫ + ν ( g ) − iǫ = ν ( f g ) − bǫ. Hence ν ( ∂ b ( f g )) = ν b X i =0 ∂ b − i f ∂ i g ! ≥ min ≤ i ≤ b { ν ( ∂ b − i f ∂ i g ) } ≥ ν ( f g ) − bǫ, and consequently ǫ ( f g ) ≤ max { ǫ ( f ) , ǫ ( g ) } .If ǫ ( f ) > ǫ ( g ), for b ∈ I ( f ) the equality (4) holds if and only if i = 0. Hence ν ( ∂ b ( f g )) = min ≤ i ≤ b { ν ( ∂ b − i f ∂ i g ) } = ν ( g∂ b f ) = ν ( f g ) − bǫ ( f ) , and therefore ǫ ( f g ) = ǫ ( f ) . (cid:3) IMIT KEY POLYNOMIALS ARE p -POLYNOMIALS 5 Remark 2.5.
The equality in (3) is always satisfied. The proof of it can be foundin [3] (Corollary 4.4). However, in this paper we will only need the inequality above.
Corollary 2.6. If Q ∈ K [ x ] is a key polynomial for ν , then Q is irreducible.Proof. Suppose, aiming for a contradiction, that Q is not irreducible. Write Q = f g with f, g ∈ K [ x ] of degree smaller that deg( Q ). By Lemma 2.4 we have ǫ ( Q ) ≤ ǫ ( f )or ǫ ( Q ) ≤ ǫ ( g ). This is a contradiction to the definition of key polynomial. (cid:3) For the remaining of this section, let Q be a key polynomial for ν and set ǫ := ǫ ( Q ). Remark 2.7.
The following properties are satisfied. (i): If S Q ( f ) consiste of a unique element, then ν Q ( f ) = ν ( f ). (ii): If g ∈ K [ x ] is such ν Q ( g ) < ν Q ( f ), then ν Q ( f + g ) = ν Q ( g ) and S Q ( f + g ) = S Q ( g ). (iii): If g ∈ K [ x ] is such ν Q ( g ) ≤ ν Q ( f ) and δ Q ( g ) / ∈ S Q ( f ), then ν Q ( f + g ) = ν Q ( g ) and δ Q ( g ) ∈ S Q ( f + g ). (iv): If g ∈ K [ x ] is such ν Q ( f + g ) = ν Q ( g ) and δ Q ( g ) ≥ δ Q ( f ), then δ Q ( f + g ) ≤ δ Q ( g ). Lemma 2.8.
Let f, g ∈ K [ x ] be polynomials of degree smaller than deg( Q ) and f g = qQ + r the Q -expansion of f g . We have ν ( f g ) = ν ( r ) < ν ( qQ ) , e hence δ Q ( f g ) = 0 .Proof. Let γ = min { ν ( f g ) , ν ( r ) } . It is enough to show that ν ( qQ ) > γ . Since thedegrees of f , g , q and r are smaller than deg( Q ) and Q is a key polynomial, Lemma2.4 gives us ǫ = ǫ ( qQ ) > ǫ ′ := max { ǫ ( f g ) , ǫ ( r ) } . Let b ∈ I ( qQ ). Then ν ( qQ ) − bǫ = ν ( ∂ b ( qQ )) ≥ min { ν ( ∂ b ( f g )) , ν ( ∂ b r ) } ≥ γ − bǫ ′ > γ − bǫ, and hence ν ( qQ ) > γ . (cid:3) Proposition 2.9.
The map ν Q is a valuation of K [ x ] .Proof. For f, g ∈ K [ x ] write f = f + . . . + f n Q n and g = g + . . . + , g m Q m the Q -expansions of f and g , respectively. We can assume, completing with zeroterms if necessary, that m = n . By definition we have ν Q (1) = 0 and ν Q (0) = ∞ .Since ν Q ( f + g ) = min ≤ i ≤ m { ν (cid:0) ( f i + g i ) Q i (cid:1) } ≥ min ≤ i,j ≤ n { ν ( f i Q i ) , ν ( g j Q j ) }≥ min { ν Q ( f ) , ν Q ( g ) } , MICHAEL DE MORAES AND JOSNEI NOVACOSKI we obtain the property (V1) .Let r = δ Q ( f ) and s = δ Q ( g ). We shall show that ν Q ( f g ) = ν Q ( f r Q r g s Q s ) = ν Q ( f ) + ν Q ( g ) . By Lemma 2.8, we have ν Q ( f r Q r g s Q s ) = ν Q ( f ) + ν Q ( g ) and δ Q ( f r Q r g s Q s ) = r + s. For i, j , 0 ≤ i, j ≤ n if i / ∈ S Q ( f ) or j / ∈ S Q ( g ), then ν Q ( f i Q i g j Q j ) > ν Q ( f r Q r g s Q s ) . If i ∈ S Q ( f ), j ∈ S Q ( f ) and ( i, j ) = ( r, s ), then ν Q ( f i Q i g j Q j ) = ν Q ( f r Q r g s Q s ) and δ Q ( f i Q i g j Q j ) = i + j < r + s. Hence, by Remark 2.7 (ii) and (iii) , we have the property (V2) . Therefore, ν Q isa valuation. (cid:3) Key polynomials of the same degree.
For this section, let Q , Q ∈ K [ x ]be key polynomials for ν with same degree. Assume that ν ( Q ) ≥ ν ( Q ) and let Q = Q + h be the Q -expansion of Q . Lemma 2.10.
We have δ Q ( Q ) = 1 and ν ( h ) ≥ ν Q ( Q ) = ν ( Q ) = ν Q ( Q ) .Proof. If ν ( h ) < ν ( Q ), then ν ( Q ) = min { ν ( Q ) , ν ( h ) } = ν ( h ) < ν ( Q ) whichwould contradict ν ( Q ) ≥ ν ( Q ). Hence, ν ( h ) ≥ ν ( Q ). Consequently ν Q ( Q ) =min { ν ( Q ) , ν ( h ) } = ν ( Q ) and δ Q ( Q ) = 1.If ν ( Q ) = ν ( Q ) the analogous argument would give ν Q ( Q ) = ν ( Q ) = ν ( Q ).If ν ( Q ) < ν ( Q ), the Remark 2.7 (i) gives us ν ( h ) = ν ( Q ) < ν ( Q ). Since Q = Q − h is the Q -expansion of Q , we have ν Q ( Q ) = ν ( h ) = ν ( Q ) and thisconcludes the proof. (cid:3) Lemma 2.11. If b ∈ I ( Q ) , then ν ( ∂ b Q ) = ν ( ∂ b Q ) .Proof. Since deg( h ) < deg( Q ), ν ( h ) ≥ ν ( Q ) and b ∈ I ( Q ), we have ν ( ∂ b h ) > ν ( h ) − bǫ ( Q ) ≥ ν ( Q ) − bǫ ( Q ) = ν ( ∂ b Q ) . Therefore, ν ( ∂ b Q ) = ν ( ∂ b Q + ∂ b h ) = ν ( ∂ b Q ). (cid:3) Lemma 2.12.
Let Q , Q ∈ K [ x ] be key polynomials for ν of the same degree.Then (i): if ν ( Q ) = ν ( Q ) , then ǫ ( Q ) = ǫ ( Q ) and I ( Q ) = I ( Q ) ; and (ii): if ν ( Q ) < ν ( Q ) , then ǫ ( Q ) < ǫ ( Q ) .Proof. Take b ∈ I ( Q ). By Lemma 2.11 we have ν ( ∂ b Q ) = ν ( ∂ b Q ). Since ν ( Q ) = ν ( Q ), we have ν ( ∂ b Q ) = ν ( ∂ b Q ) = ν ( Q ) − bǫ ( Q ) = ν ( Q ) − bǫ ( Q ) , and hence ǫ ( Q ) ≥ ǫ ( Q ). The same argument gives us ǫ ( Q ) ≥ ǫ ( Q ). Hence ǫ ( Q ) = ǫ ( Q ). Since ν ( ∂ b Q ) = ν ( ∂ b Q ), ν ( Q ) = ν ( Q ) and ǫ ( Q ) = ǫ ( Q ), we IMIT KEY POLYNOMIALS ARE p -POLYNOMIALS 7 have b ∈ I ( Q ) and hence I ( Q ) ⊆ I ( Q ). Since we can do the same reasoningexchanging Q and Q , we conclude that I ( Q ) = I ( Q ). This shows (i) .In order to prove (ii) take b ∈ I ( Q ). By Lemma 2.11 we have ν ( ∂ b Q ) = ν ( ∂ b Q ). Since ν ( Q ) > ν ( Q ) we have ν ( ∂ b Q ) = ν ( ∂ b Q ) = ν ( Q ) − bǫ ( Q ) < ν ( Q ) − bǫ ( Q ) , and consequently ǫ ( Q ) > ǫ ( Q ). (cid:3) Corollary 2.13. If ν ( Q ) < ν ( Q ) , then I ( Q ) ∩ I ( Q ) has at most one element.Proof. If I ( Q ) ∩ I ( Q ) = ∅ , then the result is trivially satisfied. Suppose that I ( Q ) ∩ I ( Q ) = ∅ and let b, b M ∈ I ( Q ) ∩ I ( Q ) where b M = max I ( Q ) ∩ I ( Q ).We will show that b = b M . By Lemma 2.11 we have ν ( ∂ b Q ) = ν ( ∂ b Q ) and ν ( ∂ b M Q ) = ν ( ∂ b M Q ). Hence, ν ( Q ) − bǫ ( Q ) = ν ( Q ) − bǫ ( Q ) , and consequently(5) ν ( Q ) − ν ( Q ) = b ( ǫ ( Q ) − ǫ ( Q )) . Substituting ν ( Q ) = ν ( ∂ b M Q ) + b M ǫ ( Q ) and ν ( Q ) = ν ( ∂ b M Q ) + b M ǫ ( Q ) in(5) and taking into account that ν ( ∂ b M Q ) = ν ( ∂ b M Q ) we obtain b M ( ǫ ( Q ) − ǫ ( Q )) = b ( ǫ ( Q ) − ǫ ( Q )) . By Lemma 2.12 (ii) we have ǫ ( Q ) < ǫ ( Q ) and hence b = b M . (cid:3) Lemma 2.14.
For every f ∈ K [ x ] we have ν Q ( f ) ≤ ν Q ( f ) ≤ ν ( f ) .Proof. Let f = f n Q n + . . . + f 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 ) . because ν Q ( Q ) = ν ( Q ). (cid:3) Lemma 2.15.
Suppose that ν ( Q ) < ν ( Q ) and take f ∈ K [ x ] . Then (i): if S Q ( f ) = { } , then ν Q ( f ) < ν Q ( f ) ; and (ii): if ν Q ( f ) < ν ( f ) , then ν Q ( f ) < ν Q ( f ) .Proof. Let f = n X i =0 f i Q i be the Q -expansion of f . Set a := n X i =1 f i Q i = Q n X i =1 f i Q i − ! =: Q b. MICHAEL DE MORAES AND JOSNEI NOVACOSKI
Since S Q ( f ) = { } , Lemmas 2.10 and 2.14 together with ν Q ( b ) ≥ ν Q ( b ) and ν Q ( Q ) = ν ( Q ) < ν ( Q ) imply that ν Q ( f ) = ν Q ( a ) = ν Q ( Q ) + ν Q ( b ) > ν Q ( Q ) + ν Q ( b ) = ν Q ( a ) . Since deg( f ) < deg( Q ) = deg( Q ) we have ν Q ( f ) = ν Q ( f ) ≥ ν Q ( a ) andhence ν Q ( f ) > ν Q ( a ). Consequently ν Q ( f ) = ν Q ( f + a ) = ν Q ( a ) < ν Q ( f )and his shows (i) .If ν Q ( f ) < ν ( f ), then S Q ( f ) = { } by item (i) of Remark 2.7. Hence, (ii) follows from (i) . (cid:3) Lemma 2.16.
Let h , . . . , h n ∈ K [ x ] be polynomials of degree smaller than deg( Q ) .Let n Y i =0 h i = n X i =0 c i Q i be the Q -expansion of h · · · h n . For every i ∈ { , . . . , n } we have (i): ν Q ( c i Q i ) > ν ( c ) ; and (ii): ν ( c i Q i ) − ν ( c ) > ν ( Q ) − ν ( Q ) .Proof. We will prove (i) by induction on n . For n = 2, let h h = q Q + r and h h = q Q + r the Q and Q -expansions of h h , respectively. By Lemma2.8 we have ν ( q Q ) > ν ( h h ), ν ( q Q ) > ν ( h h ) and ν ( h h ) = ν ( r ) = ν ( r ).Then ν Q ( − q Q + q Q ) = ν Q ( r − h h + h h − r ) = ν Q ( r − r ) = ν ( r − r )= ν ( r − h h + h h − r ) = ν ( − q Q + q Q ) . Since ν Q ( q Q ) = ν ( q Q ) > ν ( r ) = ν ( r ) and ν Q ( − q Q + q Q ) = ν ( − q Q + q Q ) ≥ min { ν ( q Q ) , ν ( q Q ) } > ν ( h h ) = ν ( r ) = ν ( r ) , we must have ν Q ( q Q ) > ν ( r ).Suppose now that the result is true for n , i.e., if n Y i =0 h i = n X i =0 c i Q i is the Q -expansion of h · · · h n , then for each i , 1 ≤ i ≤ n , we have ν Q ( c i Q i ) >ν ( c ). Let h n +1 ∈ K [ x ] be a polynomial of degree smaller than deg( Q ). For i ≤ i ≤ n , write h n +1 c i = q i Q + r i the Q -expansion of h n +1 c i . By the case n = 2,we have ν Q ( q i Q ) > ν ( r i ) = ν ( h n +1 c i ). In particular, ν ( q Q ) > ν ( r ) = ν ( h n +1 c ) . It remains to show that for every i , 1 ≤ i ≤ n , we have ν Q ( q i Q i +1 ) > ν Q ( r i Q i ) > ν ( h n +1 c ) . IMIT KEY POLYNOMIALS ARE p -POLYNOMIALS 9 By assumption we have ν Q ( r i Q i ) = ν ( r i ) + ν Q ( Q i ) = ν ( h n +1 c i ) + ν Q ( Q i )= ν ( h n +1 ) + ν ( c i ) + ν Q ( Q i ) = ν ( h n +1 ) + ν Q ( c i Q i ) > ν ( h n +1 ) + ν ( c ) = ν ( h n +1 c ) , and this shows (i) .For (ii) since ν Q ( Q ) = ν ( Q ) − ( ν ( Q ) − ν ( Q )) e ν ( Q ) ≥ ν ( Q ) the item (i) implies that for every i ≤ i ≤ n we have ν Q ( c i Q i ) = ν ( c i Q i ) − i ( ν ( Q ) − ν ( Q )) > ν ( c ) . Hence ν ( c i Q i ) − ν ( c ) > i ( ν ( Q ) − ν ( Q )) ≥ ν ( Q ) − ν ( Q ) , and this concludes the proof. (cid:3) Lemma 2.17.
Let a ∈ K [ x ] be a polynomial of degree smaller than deg( Q ) andtake n ∈ N . Let aQ n = n X i =0 (cid:18) a (cid:18) ni (cid:19) h i Q n − i (cid:19) = n X i =0 b i Q i be the Q -expansion of aQ n . We have (i): ν Q ( b i Q i ) ≥ ν ( aQ n ) for every i ≤ i ≤ n ; and (ii): ν ( b n ) = ν ( a ) .Proof. Since ν ( h ) ≥ ν Q ( Q ) = ν ( Q ), for every i , 0 ≤ i ≤ n , we have ν Q (cid:18) a (cid:18) ni (cid:19) h i Q n − i (cid:19) ≥ ν (cid:18)(cid:18) ni (cid:19)(cid:19) + ν ( aQ n ) ≥ ν ( aQ n ) . Hence, Lemma 2.16 (i) applied to ah i implies (i) .In order to prove (ii) , for each i , 1 ≤ i ≤ n , let ah i = q i,i Q i + · · · + q i, Q + r i be the Q -expansion of ah i . We have b n Q n = n X i =1 q i,i Q n + aQ n . By Lemma 2.16 (i) , for every i , 1 ≤ i ≤ n , we have ν Q ( q i,i Q i ) > ν ( r i ) = ν ( ah i ).Hence, ν Q ( q i,i Q n ) > ν Q ( ah i Q n − i ) ≥ ν Q ( aQ n )and therefore ν ( b n ) = ν ( a ). (cid:3) Corollary 2.18.
Let f ∈ K [ x ] and δ = δ Q ( f ) . Let f = f n Q n + . . . + f and f = f ′ n Q n + . . . + f ′ be the Q and Q -expansions of f , respectively. We have (i): ν ( f δ ) = ν ( f ′ δ ) ; (ii): δ Q ( f ) ≤ δ Q ( f ) ; and (iii): ν ( f n ) = ν ( f ′ n ) .Proof. In order to prove (i) , for each i , 0 ≤ i ≤ n , let f i Q i = f n,i Q n + . . . + f δ,i Q δ + . . . + f ,i be the Q -expansion of f i Q i . Then f ′ δ = f δ, + . . . + f δ,n . We will analyze the value f δ,i for each i , 0 ≤ i ≤ n .If i < δ the the Q -expansion of f i Q i does not have a term of degree δ . Hence, f δ,i = 0.If i = δ , by Lemma 2.17 (ii) , we have ν ( f δ,δ ) = ν ( f δ ).If i > δ , by Lemma 2.17 (i) , we have ν Q ( f δ,i Q δ ) = ν ( f δ,i Q δ ) ≥ ν ( f i Q i ) . Since δ = δ Q ( f ), we have ν ( f i Q i ) > ν ( f δ Q δ ) and consequently ν ( f δ,i ) > ν ( f δ ).Therefore, ν ( f δ, + . . . + f δ,n ) = ν ( f δ ) and this shows (i) .In order to prove (ii) take r, s , 0 ≤ r < s ≤ n . We will analize the terms f s,i Q i of the Q -expansions of f i Q i for each i , 0 ≤ i ≤ n .If i < s , then the Q -expansion of f i Q i does not have a term of order s , andhence, f s,i Q s = 0.If i ≥ s , by Lemma 2.17 (i) we have ν Q ( f s,i Q s ) ≥ ν ( f i Q i ) > ν ( f δ Q δ ) = ν Q ( f δ Q δ ) . By (i) , the term of degree δ on the Q -expansion of f has values ν ( f r Q r ). Since s > r , we have ν ( f s,i Q s ) > ν ( f r Q r ). Since the term of degree s on the Q -expansionof f is ( f r, + . . . + f r,n ) Q r we conclude (ii) .In order to prove (iii) we observe that for every i , 0 ≤ i ≤ n , the Q -expansionsof f i Q i do not have terms of order n . Hence, we can apply Lemma 2.17 (ii) to themonomial f n Q n to obtain the result. (cid:3) The relation between the derivatives and the Q -truncation. For thissection let Q ∈ K [ x ] be a key polynomial for ν such that I ( Q ) has only one element.Let b Q be this element and set ǫ := ǫ ( Q ). Let h ∈ K [ x ] be a polynomial of degreesmaller than deg( Q ) and take n ∈ N .For b ∈ N , consider the set S b of all tuples γ = ( b , . . . , b r ) where (i): r ∈ { , . . . , n } ; (ii): ≤ b ; (iii): < b ≤ b ≤ . . . ≤ b r ; and (iv): b + . . . + b r = b .By the Leibiniz rule for derivation we have(6) ∂ b ( hQ n ) = X γ ∈S b (cid:18) nr (cid:19) T γ , where T γ := ∂ b ( h ) r Y j =1 ∂ b i ( Q ) Q n − r . IMIT KEY POLYNOMIALS ARE p -POLYNOMIALS 11 Lemma 2.19.
Let b ∈ N and γ = ( b , . . . , b r ) ∈ S b . We have (7) S Q ∂ b ( h ) r Y j =1 ∂ b i ( Q ) = { } , and hence ν Q ( T γ ) = ν ( T γ ) and S Q ( T γ ) = { n − r } .Proof. Since the degrees of ∂ b h and ∂ b i Q , for every i , 1 ≤ i ≤ r , are smaller thandeg( Q ), the first equation follows from Lemma 2.16 (ii) taking Q = Q = Q .Hence, by Remark 2.7 (i) we have ν Q ( T γ ) = ν ( T γ ). Since T γ = ∂ b h · · · ( ∂ b r Q ) Q n − r , the equation (7) gives us S Q ( T γ ) = { n − r } . (cid:3) Lemma 2.20.
Let b ∈ N and γ = ( b , . . . , b r ) ∈ S b . We have (8) ν Q ( T γ ) ≥ ν ( hQ n ) − bǫ. Moreover, the equality is satisfied in (8) if and only if b = 0 and b i = b Q for every i , ≤ i ≤ r . In this case we have r = b/b Q .Proof. We have(9) ν ( ∂ b h ) ≥ ν ( h ) − b ǫ and ν ( ∂ b i Q ) ≥ ν ( Q ) − b i ǫ, for every i , 1 ≤ i ≤ r . Since b + . . . + b r = b we add the above equations to obtain ν Q ( T γ ) = ν ∂ b ( h ) r Y j =1 ∂ b i ( Q ) Q n − r ≥ ν ( hQ n ) − bǫ. The equality is satisfied if and only if all the inequalities in (9) are equalilies. Thishappens if and only if b = 0 and b i ∈ I ( Q ) = { b Q } for every i , 1 ≤ i ≤ r . Thisconcludes the proof. (cid:3) Corollary 2.21.
For every b ∈ N we have (i): ν Q ( ∂ b ( hQ n )) ≥ ν ( hQ n ) − bǫ ; and (ii): ν Q ( ∂ b ( hQ n )) = ν ( hQ n ) − bǫ in and only if b Q | b and p ∤ (cid:18) nb/b Q (cid:19) . Inthis case, S Q ( ∂ b ( hQ n )) = { n − b/b Q } .Proof. Since ∂ b ( hQ n ) = X γ ∈S b (cid:18) nr (cid:19) T γ , Lemma 2.20 gives us ν Q ( ∂ b ( hQ n )) ≥ ν ( hQ n ) − bǫ . This implies (i) .By Lemma 2.20, if γ = ( b , . . . , b r ) ∈ S b , then ν ( T γ ) = ν ( hQ n ) − bǫ if and onlyif γ = (0 , b Q , . . . , b Q ) and r = b/b Q . Assume that ν Q ( ∂ b ( hQ n )) = ν ( hQ n ) − bǫ or b Q | b . Then there exists a unique λ ∈ S b such that ν ( T λ ) = ν ( hQ n ) − bǫ . Since ∂ b ( hQ n ) = X γ ∈S b \{ λ } (cid:18) nr (cid:19) T γ + (cid:18) nb/b Q (cid:19) T λ and ν Q X γ ∈S b \{ λ } (cid:18) nr (cid:19) T γ > ν ( hQ n ) − bǫ we have ν ( ∂ b ( hQ n )) = ν ( hQ n ) − bǫ if and only if ν (cid:16)(cid:0) nb/b Q (cid:1)(cid:17) = 0. This onlyhappens if p ∤ (cid:18) nb/b Q (cid:19) . Since S Q ( T λ ) = { n − b/b Q } , by Remark 2.7 (ii) we have S Q ( ∂ b ( hQ n )) = { n − b/b Q } . This shows (ii) . (cid:3) Corollary 2.22.
Take f ∈ K [ x ] . Then for every b ∈ N we have ν Q ( ∂ b f ) ≥ ν Q ( f ) − bǫ .Proof. Let f = f + . . . + f m Q m be the Q -expansion of f . Since deg( f ) < deg( Q )we have ν Q ( ∂ b f ) ≥ ν ( f ) − bǫ. By Corollary 2.21 we have ν Q ( ∂ b ( f i Q i )) ≥ ν ( f i Q i ) − bǫ for every i , 1 ≤ i ≤ m .Since ∂ b ( f ) = ∂ b ( f ) + . . . + ∂ b ( f m Q m )we have the result. (cid:3) Corollary 2.23.
Let f ∈ K [ x ] such that δ = δ Q ( f ) > . Write δ = p e u where e, u ∈ N and p ∤ u . Set b = p e b Q . Then ν Q ( ∂ b f ) = ν Q ( f ) − bǫ and δ Q ( ∂ b f ) = δ − p e .Proof. Let f = f + . . . + . . . + f m Q m be the Q -expansion of f . We have ∂ b ( f ) = ∂ b f + . . . + ∂ b ( f m Q m ) . Observe that p e = b/b Q .For i = δ , since p ∤ (cid:0) δp e (cid:1) by Corollary 2.21 (ii) we have ν Q ( ∂ b ( f δ Q δ )) = ν ( f δ Q δ ) − bǫ and S Q ( ∂ b ( f δ Q δ )) = { δ − p e } .For i = 0 since deg( f ) < deg( Q ) and b >
0, we have ν Q ( ∂ b f ) > ν ( f ) − bǫ ≥ ν ( f δ Q δ ) − bǫ = ν Q ( ∂ b ( f δ Q δ ))where the second inequality holds because δ = δ Q ( f ).For i < δ and i > ν Q ( ∂ b ( f i Q i )) ≥ ν ( f i Q i ) − bǫ ≥ ν ( f δ Q δ ) − bǫ = ν ( ∂ b ( f δ Q δ ))By Corollary 2.21 (ii) , if the first inequality is an equality, then δ Q ( ∂ b ( f i Q i )) = i − p e < δ − p e . For i > δ we have ν Q ( ∂ b ( f i Q i )) ≥ ν ( f i Q i ) − bǫ > ν ( f δ Q δ ) − bǫ = ν ( ∂ b ( f δ Q δ ))where the second inequality holds because δ = δ Q ( f ).By the case 0 < i < δ and i = δ Remark 2.7 (iii) gives us ν Q ( ∂ b ( f Q + . . . + f δ Q δ )) = ν Q ( ∂ b ( f δ Q δ )) and δ Q ( ∂ b ( f δ Q δ )) = δ − p e . IMIT KEY POLYNOMIALS ARE p -POLYNOMIALS 13 Hence, by the cases i = 0 and i > δ and Remark 2.7 (ii) we have ν Q ( ∂ b f ) = ν Q ( ∂ b ( f δ Q δ )) = ν Q ( f ) − bǫ and δ Q ( ∂ b f ) = δ − p e . This concludes the proof. (cid:3) Proof of Theorem 1.1
For this section let Ψ α be the set of key polynomials of degree α . In Ψ α weconsider the order given by Q ≤ Q if and only if ν ( Q ) ≤ ν ( Q ) (or equivalently,by Lemma 2.12, Q ≤ Q if and only if ǫ ( Q ) ≤ ǫ ( Q )). We will assume that Ψ α does not have a largest element. We will use consider the(10) S α := { f ∈ K [ x ] | ν Q ( f ) < ν ( f ) for every Q ∈ Ψ α } . Proposition 3.1.
Every monic polynomial f ∈ S α with smallest degree amongpolynomials in S α is a key polynomial for ν .Proof. For each i , 1 ≤ i ≤ deg( f ), since deg( ∂ i f ) < deg( f ) there exists Q i ∈ Ψ α such that ν Q i ( ∂ i f ) = ν ( ∂ i f ). Let Q M = max { Q , . . . , Q deg( f ) } and Q ≥ Q M inΨ α . By Lemma 2.14 we have ν Q ( ∂ b i f ) = ν ( ∂ b i f ) for every i , 1 ≤ i ≤ deg( f ). Since ν Q ( f ) < ν ( f ), Remark 2.7 (i) gives us S Q ( f ) = 0. Hence, by Corollary 2.23 thereexists b ∈ N such that ν ( ∂ b f ) = ν Q ( ∂ b f ) = ν Q ( f ) − bǫ ( Q ) < ν ( f ) − bǫ ( Q ) . Hence, ǫ ( f ) > ǫ ( Q ). Since Q ≥ Q M can be chosen arbitrarily we have ǫ ( f ) > ǫ ( Q )for every Q ∈ Ψ α . It remains to show that for g ∈ K [ x ], if deg( g ) < deg( f ),then ǫ ( g ) < ǫ ( f ). Reasoning as before, there exists Q ∈ Ψ α such that for every i ,1 ≤ i ≤ deg( g ), we have ν Q ( ∂ i g ) = ν ( ∂ i g ) and ν Q ( g ) = ν ( g ) . Then by Corollary 2.22 we have ν ( ∂ i g ) = ν Q ( ∂ i g ) ≥ ν Q ( g ) − iǫ ( Q ) = ν ( g ) − iǫ ( Q )for every i , 1 ≤ i ≤ r . This means that ǫ ( g ) ≤ ǫ ( Q ) < ǫ ( f ) which is what we wantedto prove. (cid:3) Definition 3.2. A limit key polynomial for Ψ α is a monic polynomial f ∈ K [ x ]as in Proposition 3.1. Remark 3.3.
In [11], limit key polynomial was defined slightly different. Therewe fix a polynomial Q and say that it is a limit key polynomial if there exists a keypolynomial Q − with some properties. One can easily show that such polynomialwill be a limit key polynomial for the family Ψ α where α = deg( Q − ). Remark 3.4.
Consider a cofinal subset Ψ ⊆ Ψ α of Ψ α . By Lemma 2.15 and thefact that Ψ does not have largest element, we have f ∈ S α ⇐⇒ ν Q ( f ) < ν Q ( f ) for every Q , Q ∈ Ψ with ν ( Q ) < ν ( Q ) . Lemma 3.5.
Assume that Ψ α admits a limit key polynomial. Then ν (Ψ α ) isbounded.Proof. Since Ψ α is infinite, by Corollary 2.18 (ii) there exists Q ∈ Ψ α such thatfor every Q ∈ Ψ α with Q ≥ Q we have δ := δ Q ( f ) = δ Q ( f ). Let f δ be thecoefficient of degree δ in the Q -expansion of f . By Corollary 2.18 (i) , for every Q ∈ Ψ α with Q ≥ Q , the coefficient of degree δ in the Q -expansion f has value ν ( f δ ). Since δ = δ Q ( f ) we have ν Q ( f ) = ν ( f δ ) + δν ( Q ) < ν ( f ) e consequently ν ( Q ) < ( ν ( f ) − ν ( f δ )) /δ . Since Q can be chosen as large as needed in Ψ α , theresult follows. (cid:3) Since rk( ν ) = 1, the fact that Ψ α is limited implies that it admits a supremum.Set B = sup ν (Ψ α ). Assume that Ψ α admits a limit key polynomial Q α . The nextstep is to show that for every Q ∈ Ψ α , δ Q ( Q α ) = deg Q ( Q α ), the leading coefficientin the Q -expansion of Q α is 1 and deg( Q α ) = p r α for some r ∈ N . We start byproving the following useful result. Lemma 3.6.
For g ∈ S α set C = sup Q ∈ Ψ α { ν Q ( g ) } . Let Q ∈ Ψ α , a ∈ K [ x ] a polynomial of degree smaller than deg( Q ) and n ∈ N . If ν ( aQ n ) ≥ C , then g − aQ n ∈ S α .Proof. For every Q ∈ Ψ α with Q ≥ Q we have ν Q ( aQ n ) = ν ( aQ n ) ≥ C andconsequently ν Q ( g − aQ n ) < ν ( g − aQ n ). For Q < Q , by Lemma 2.14 we have ν Q ( g − aQ n ) ≤ ν Q ( g − aQ n ) < ν ( g − aQ n ) . This concludes the proof. (cid:3)
For Q , Q ∈ Ψ α large enough in Ψ α , by Corollary 2.18 (ii) we have δ := δ Q ( Q α ) = δ Q ( Q α ). Let Ψ δ ⊆ Ψ α be the set of all Q ∈ Ψ α such that δ Q ( Q α ) = δ .For a given Q ∈ Ψ δ let Q α = a d Q d + . . . + a δ Q δ + . . . + a be the Q -expansion of Q α . Set γ d = ν ( a d ) and γ δ = ν ( a δ ). By Corollary 2.18 (i) and (iii) , for every Q ′ ∈ Ψ δ , if Q α = a ′ d Q ′ d + . . . + a ′ δ Q ′ δ + . . . + a ′ is the Q ′ -expansion of Q α , then γ d = ν ( a ′ d ) and γ δ = ν ( a ′ δ ). Observe that ν Q ( Q α ) = γ δ + δν ( Q ) < γ δ + δB for every Q ∈ Ψ δ . Hence, by the cofinality of Ψ δ in Ψ α by Lemma 2.14 we have ν Q ( Q α ) < ν ( a δ ) + δB for every Q ∈ Ψ α . Lemma 3.7.
We have δ = d . IMIT KEY POLYNOMIALS ARE p -POLYNOMIALS 15 Proof.
Suppose, aiming for a contradiction, that δ = d and fix Q ∈ Ψ δ . We have γ δ + δν ( Q ) < γ d + dν ( Q ) and consequently γ δ < γ d + ( d − δ ) ν ( Q ) =: k. Take Q ∈ Ψ δ with ν ( Q ) > ν ( Q ) such that δν ( Q ) + k > γ δ + δB. Since d > δ we have γ d + dν ( Q ) > γ d + ( d − δ ) ν ( Q ) + δν ( Q ) = k + δν ( Q ) > γ δ + δB. This means that if Q α = a ′ d Q d + . . . + a ′ is the Q -expansion of Q α , then ν ( a ′ d Q d ) > ν ( a δ ) + δB . By Lemma 3.6 we have ν Q (cid:0) Q α − a ′ d Q d (cid:1) < ν (cid:0) Q α − a ′ d Q d (cid:1) for every Q ∈ Ψ α . Since deg( Q α − a ′ d Q d ) < deg( f ) this is a contradiction to the minimality of thedegree of Q α in S α . (cid:3) Proposition 3.8.
For every Q ∈ Ψ α we have δ Q ( Q α ) = deg Q ( Q α ) .Proof. By Lemma 3.7, for Q ∈ Ψ α large enough we have δ Q ( Q α ) = deg Q ( Q α ).Since, By Corollary 2.18 (ii) , δ Q ( Q α ) ≥ δ Q ( Q α ) for every Q , Q ∈ Ψ α with Q ≤ Q , the result follows. (cid:3) Lemma 3.9.
Let Q , Q ∈ Ψ α such that ν ( Q ) − ν ( Q ) > δ ( B − ν ( Q )) . Then, the Q -expansion of Q α is of the form Q α = Q δ + . . . + f (i.e., a δ = 1 ).Proof. Let Q α = a δ Q δ + . . . + a be the Q -expansion of Q α . Suppose, aimingfor a contradiction, that a δ = 1. Since Q α and Q are monic this implies thatdeg( Q α ) > δα . By Corollary 2.6 Q is irreducible. Hence, by B´ezout’s identitythere exist l, q ∈ K [ x ], with deg( q ) < α , such that la δ = 1 + qQ . Let lQ α = ( q Q + r ) + . . . + ( q δ − Q δ + r δ − Q δ − ) + ( qQ δ +12 + Q δ )be the Q -expansion of lf where lf i = q i Q + r i is the Q -expansion of lf i for each i , 0 ≤ i ≤ δ −
1. Observe that ν Q ( lQ α ) = ν ( l ) + ν Q ( Q α ) < ν ( lQ α ) for every Q ∈ Ψ α . Since ν Q ( lQ α ) < ν ( la δ ) + δB for every Q ∈ Ψ α (by Corollary 2.18 (iii) ), if somemonomial c i Q i in the expansion of lQ α has value larger than ν ( la δ ) + δB , we canuse Lemma 3.6 to obtain that lQ α − c i Q i ∈ S α .By Lemma 2.16 (ii) we have ν ( qQ ) > ν (1) + ( ν ( Q ) − ν ( Q )) = ν ( la δ ) + ( ν ( Q ) − ν ( Q )) > ν ( la δ ) + δ ( B − ν ( Q )) , and consequently ν ( qQ δ +12 ) > ν ( la δ ) + δ ( B − ν ( Q )) + δν ( Q ) > ν ( la δ ) + δB. Observe that ν ( r δ − ) = ν ( la δ − ) ≥ ν ( la δ ) + ν ( Q ) since δ Q ( lQ α ) = δ . Hence,by Lemma 2.16 (ii) we have ν ( q δ − Q ) > ν ( r δ − ) + ( ν ( Q ) − ν ( Q )) ≥ ν ( la δ ) + ν ( Q ) + ( ν ( Q ) − ν ( Q )) > ν ( la δ ) + δB − ( δ − ν ( Q ) . Hence ν ( q δ − Q α ) > ν ( la δ ) + δB + ( δ − ν ( Q ) + ( δ − ν ( Q ) > ν ( la δ ) + δB. Therefore, the polynomial h := lQ α − qQ δ +12 − q δ − Q δ = r + ( q + r ) Q + . . . + ( q δ − + r δ − ) Q δ − + Q δ is monic, deg( h ) = δα < deg( Q α ) and h ∈ S α . This is a contradiction to theminimality of Q α in S α . (cid:3) Proposition 3.10.
For every Q ∈ Ψ α the leading coefficient in the Q -expansionof Q α is .Proof. By Lemma 3.9 the Q -expansion of Q α is of the form(11) Q α = δ − X i =0 a i Q i + Q δ . Let Q ∈ Ψ α and Q = Q + h the Q -expansion of Q . Substituting Q = Q + h in Q α for every i , 0 ≤ i ≤ δ −
1, the Q -expansion of a i ( Q + h ) i does not have aterm of degree δ . On the other hand, in the Q -expansion of ( Q + h ) δ the coefficentof degree δ is 1. This shows the result. (cid:3) It remains to show that δ is a power of p . In order to do this, consider Ψ I acofinal subset of Ψ α with I ( Q ) = I ( Q ) for every Q , Q ∈ Ψ I . This is possible because I ( Q ) ⊂ { , . . . , α } for each Q ∈ Ψ α ). Set I := I ( Q ) forsome (and hence for every) Q ∈ Ψ I . By Corollary 2.13 we have that I = { b ∞ } forsome b ∞ ∈ N . Proposition 3.11.
We have δ = p r for some r ∈ N .Proof. Write δ = p r u with r, u ∈ N and p ∤ u . Set b := p r b ∞ . By Corollary 2.23 forevery Q ∈ Ψ I we have δ Q ( ∂ b Q α ) = δ − p r . If δ − p r >
0, then by Lemma 2.15 (i) we have ν Q ( ∂ b Q α ) < ν Q ( ∂ b Q α )for every Q , Q ∈ Ψ I with Q < Q . Since Ψ I is cofinal in Ψ α we have ∂ b Q α ∈ S α .Since deg( ∂ b Q α ) < deg( Q α ), this is a contradiction to the minimality of the degreeof Q α in S α . Therefore, δ = p r which is what we wanted to prove. (cid:3) IMIT KEY POLYNOMIALS ARE p -POLYNOMIALS 17 Remark 3.12.
By Proposition 3.11 we conclude that if char( Kν ) = 0 (i.e., p = 1),then ν does not admit limit key polynomial.4. Proof of Theorem 1.2
Let Ψ α be the set of key polynomials of degree α and assume that Ψ α has a limitkey polynomial Q ′ α . By Lemma 3.5 ν (Ψ α ) is bounded. Since rk( ν ) = 1 we have that B = sup ν (Ψ α ) exists. For every Q ∈ Ψ α , by Theorem 1.1, the Q decomposition of Q ′ α is of the form(12) Q ′ α = Q p r + a p r − Q p r − + . . . + a and δ := δ Q ( Q ′ α ) = p r . Hence, ν Q ( Q ′ α ) = p r ν ( Q ).In order to prove Theorem 1.2 we will find Q ∈ Ψ α such that ν ( a i Q i ) > sup Q ′ ∈ Ψ { ν Q ′ ( Q ′ α ) } for every i , 1 ≤ i ≤ p r , for which i is not a power of p . Then, by Lemma 3.6 wecan remove these monomials from Q ′ α to obtain Q α . Notation 4.1.
Take Q ∈ Ψ α and Q ′ α a limit key polynomial for Ψ α with Q -expansion as (12). We define γ Q ( Q ′ α ) = max { i | ≤ i ≤ δ, i is not a power of p and ν ( a i Q i ) < δB } ;Λ Q ( Q ′ α ) = { i | γ Q ( Q ′ α ) ≤ i ≤ δ, i = p s and ν ( a i Q i ) < δB } ; ω Q ( Q ′ α ) = max (cid:26) i | ≤ i ≤ γ Q ( Q ′ α ) and ν ( a i Q i ) = min ≤ j ≤ γ Q ( Q ′ α ) (cid:8) ν (cid:0) a j Q j (cid:1)(cid:9)(cid:27) . Observe that γ Q ( Q ′ α ) is well-defined because ν ( a ) ≤ δB . Indeed, if this wouldnot be the case, then by Lemma 3.6 the polynomial Q ′ − a would be a limitkey polynomial for Ψ α . This would contradict the fact that key polynomials areirreducible (Corollary 2.6).Let Ψ ′ ⊆ Ψ α be a cofinal subset of Ψ α such that (i): ν ( p ) > δ ( B − ν ( Q )) for every Q ∈ Ψ ′ ; (ii): for every Q , Q ∈ Ψ ′ we have γ Q ( Q ′ α ) = γ Q ( Q ′ α ) , Λ Q ( Q ′ α ) = Λ Q ( Q ′ α ) and ω Q ( Q ′ α ) = ω Q ( Q ′ α ); and (iii): for every Q , Q ∈ Ψ ′ we have I ( Q ) = I ( Q ).We set γ := γ Q ( Q ′ α ), Λ := Λ Q ( Q ′ α ), ω := ω Q ( Q ′ α ) and I := I ( Q ) for some (and forany) Q ∈ Ψ ′ . By Corollary 2.13, I = { b ∞ } for some b ∞ ∈ N .Let Q , Q ∈ Ψ ′ such that ǫ ( Q ) − ǫ ( Q ) > δ ( B − ν ( Q )) . Let Ψ ⊂ Ψ ′ the subset of elements Q ∈ Ψ ′ for which Q ≥ Q . Observe that(13) ǫ ( Q ) − ǫ ( Q ) > δ ( B − ν ( Q )) , for every Q ∈ Ψ and Ψ is cofinal in Ψ α . By Lemma 2.11 we have ν ( ∂ b ∞ Q ) = ν ( ∂ b ∞ Q ) para todo Q ∈ Ψ . Hence, replacing ǫ ( Q ) = ( ν ( Q ) − ν ( ∂ b ∞ Q )) /b ∞ and ǫ ( Q ) = ( ν ( Q ) − ν ( ∂ b ∞ Q )) /b ∞ in (13) we have ν ( Q ) − ν ( Q ) > δ ( B − ν ( Q )) for every Q ∈ Ψ . If γ = 0, then for every Q ∈ Ψ ν (cid:0) a i Q i (cid:1) > δB for every i, ≤ i ≤ δ, with i = p s . Hence, by Lemma 3.6 we can remove these terms to conclude the proof of Theorem1.2. In order to show that γ = 0, we will suppose that γ = 0 and obtain acontradiction.Let Q ′ α = a + . . . + a δ Q δ be the Q -expansion of Q ′ α . Lemma 4.2.
Let Q ∈ Ψ and n ∈ { , . . . , δ } a power of p . Let a n Q n = n X i =0 b i Q i be the Q -expansion of a n Q n . For every i , ≤ i ≤ n − , we have ν ( b i Q i ) > δB .Proof. We have ν ( a n Q n ) ≥ ν Q ( Q ′ α ) = δν ( Q )e consequently ν ( a n ) ≥ ( δ − n ) ν ( Q ). Let Q = Q + h be the Q -expansion of Q .Since ν ( Q ) < ν ( Q ) we have ν ( h ) = ν ( Q ). Replacing Q + h in a n Q n we obtain a n Q n = n X i =0 (cid:18)(cid:18) ni (cid:19) a n h n − i Q i (cid:19) = n X i =0 b i Q i . In order to show our result, for each i , 0 ≤ i ≤ n we will show that if (cid:18) ni (cid:19) a n h n − i Q i = n X j =0 a ij Q j is the Q -expansion of (cid:18) ni (cid:19) a n h n − i Q i , then ν (cid:0) a ij Q j (cid:1) for every j, ≤ j ≤ n − . The monomial (cid:0) nn (cid:1) a n Q n = a n Q n does not play a role because a nj = 0 for every j ,1 ≤ j ≤ n − i = 0 we have ν ( a ) = ν (cid:18)(cid:18) n (cid:19) a n h n (cid:19) = ν ( a n ) + nν ( Q ) ≥ δν ( Q ) . Hence, by Lemma 2.16 (ii) , for every j , 1 ≤ j ≤ n we have ν ( a j Q j ) > ν ( Q ) − ν ( Q ) + ν ( a ) > δ ( B − ν ( Q )) + δν ( Q ) = δB, and this concludes the case i = 0. IMIT KEY POLYNOMIALS ARE p -POLYNOMIALS 19 For i , 1 ≤ i ≤ n −
1, since p | (cid:18) ni (cid:19) and ν ( Q ) > ν ( Q ) we have ν Q (cid:18)(cid:18) ni (cid:19) a n h n − i Q n (cid:19) ≥ ν ( p )+ ν ( a n h n − i )+ ν ( Q n ) > δ ( B − ν ( Q ))+ δ ( ν ( Q )) = δB, and this concludes the proof. (cid:3) Lemma 4.3.
Let n ∈ { ω, γ } ∪ Λ Q ( f ) . For every Q ∈ Ψ if Q ′ α = Q δ + b δ − Q δ − + . . . + b is the Q -expansion of Q ′ α , then ν ( b n Q n ) = ( δ − n ) B .Proof. For n = δ we have ν (1) = 0 = ( δ − δ ) B . Observe that 0 / ∈ { ω, γ } ∪ Λ.Assume that n = δ . We will first show that ν ( b n ) = ν ( a n ). For each i , let a i Q i = b ii Q i + . . . + b i be the Q -expansion of a i Q i . Then b n = b n, + . . . + b n,δ . For i < n we have b n,i = 0.For i = n , by Lemma 2.17 (ii) we have ν ( b n,n ) = ν ( a n ). For i > n and i = p s , byLemma 4.2 we have ν ( b n,i Q n ) > δB . Hence, ν ( b n,i ) > δB − nν ( Q ).For n = ω , i > n and i = p s we have i > γ and hence ν ( a i Q i ) ≥ δB . Since ν ( Q ) > ν Q ( Q ),b y Lemma 2.17 (i) we have ν ( b n,i Q n ) > ν Q ( b n,i Q n ) ≥ ν ( a i Q i ) ≥ δB. Hence, ν ( b n,i ) > δB − nν ( Q ).For n = ω and i ∈ { ω + 1 , . . . γ } by Lemma 2.17 (i) and the definition of ω wehave ν Q ( b n,i Q n ) ≥ ν ( a i Q i ) > ν ( a n Q n ) . Hence, ν ( b n,i ) > ν ( a n ).Since n ∈ { ω, γ } ∪ Λ Q ( f ) we have ν ( b n ) < δB − nν ( Q ). Consequently, theprevious analysis gives us ν ( b n ) = ν ( b n,n ) = ν ( a n ) . We will show now that ν ( a n ) = ( δ − n ) B . Assume, aiming for a contradiction,that ν ( a n ) = ( δ − n ) B . If ν ( a n ) < ( δ − n ) B we can assume that ( δ − n ) ν ( Q ) > ν ( a n ).Since ν ( b n Q n ) = ν ( a n ) + nν ( Q )we have a contradiction to ν ( a n ) + nν ( Q ) < δν ( Q ) = ν Q ( Q ′ α ). If ν ( a n ) > ( δ − n ) B we can assume that nν ( Q ) + ν ( a n ) > δB . Since ν ( b n Q n ) = ν ( a n ) + nν ( Q )we obtain a contradiction with n ∈ { ω, γ } ∪ Λ Q ( Q ′ α ) and Q ∈ Ψ. (cid:3) Corollary 4.4.
We have ω = γ . Proof.
Let Q ∈ Ψ. By Lemma 4.3 and the definitions of ω and γ we have( δ − ω ) B + ων ( Q ) = ν ( b ω Q ω ) ≤ ν ( b γ Q γ ) = ( δ − γ ) B + γν ( Q ) . Since ν ( Q ) < B and ω ≤ γ we must ω = γ . (cid:3) Set ǫ ( B ) = sup ǫ (Ψ α ). Observe that ν ( ∂ b Q ) = ν ( ∂ b ∞ Q ) for every Q ∈ Ψbecause ǫ ( B ) = ( B − ν ( ∂ b ∞ Q )) /b ∞ . Lemma 4.5.
For every Q ∈ Ψ and b ∈ N we have ν ( ∂ b Q ) ≥ B − bǫ ( B ) .Proof. Take b ∈ N . For every h ∈ K [ x ] with ν ( h ) = ν ( Q ) and deg( h ) < deg( Q )we have ν ( ∂ b h ) > B − bǫ ( Q ) > B − bǫ ( B ) . Indeed, since ǫ ( Q ) − ǫ ( Q ) > δ ( B − ν ( Q )) we have ν ( Q ) − bǫ ( Q ) > B − bǫ ( Q ).Since ν ( h ) = ν ( Q ) and deg( h ) < deg( Q ) we obtain ν ( ∂ b h ) > ν ( h ) − bǫ ( Q ) = ν ( Q ) − bǫ ( Q ) > B − bǫ ( Q ) > B − bǫ ( B ) . If ν ( ∂ b Q ) ≥ B − bǫ ( B ), the for every Q ∈ Ψ we have ν ( ∂ b Q ) ≥ B − bǫ ( B ).Inded, Since Q ≥ Q , if Q = Q + h is the Q -expansion of Q , then ν ( h ) = ν ( Q ).Hence, ν ( ∂ b h ) > B − bǫ ( B ). Since ∂ b Q = ∂ b Q + ∂ b h we obtain ν ( ∂ b Q ) ≥ B − bǫ ( B )and consequently the result.Analogously, if ν ( ∂ b Q ) ≤ B − bǫ ( B ), the ν ( ∂ b Q ) = ν ( ∂ b Q ) for every Q ∈ Ψ.Assume that ν ( ∂ b Q ) ≤ B − bǫ ( B ). Then, for every Q ∈ Ψ we have(14) ν ( ∂ b Q ) = ν ( ∂ b Q ) ≥ ν ( Q ) − bǫ ( Q ) . Taking the limit in (14) we have ν ( ∂ b Q ) ≥ B − ǫ ( B ) and this concludes theproof. (cid:3) Write γ = p e u for e, u ∈ N and p ∤ u . Since γ is not a power of p we have γ − p e >
0. Set b = p e b ∞ . Lemma 4.6.
For every Q ∈ Ψ we have S Q ( ∂ b Q ′ α ) = { } .Proof. Since ∂ b Q ′ α = ∂ b b + . . . + ∂ b ( b δ Q δ )we will study separately ν Q ( ∂ b ( b i Q i )) and δ Q ( ∂ b ( b i Q i )) for each i , 0 ≤ i ≤ δ . Ourmain goal is to show that if ∂ b Q ′ α = c δ Q δ + . . . + c is the Q -expansion of ∂ b Q ′ α , then ν ( c ) ≥ ν (cid:0) c γ − p r Q γ − p r (cid:1) .For i = 0, since ǫ ( Q ) − ǫ ( Q ) > δ ( B − ν ( Q )), ν ( b ) ≥ ν Q ( Q ′ α ) = δν ( Q ) and b > ν Q ( ∂ b b ) ≥ ν ( b ) − bǫ ( h ) ≥ δν ( Q ) − bǫ ( Q ) > δB − bǫ ( Q ) . IMIT KEY POLYNOMIALS ARE p -POLYNOMIALS 21 For 0 < i < γ , since ω = γ by Corollary 4.4, we have ν ( b i Q i ) ≥ ν ( b γ Q γ ). ByCorollary 2.21 (i) we have ν ( ∂ b ( b i Q i )) ≥ ν ( b i Q i ) − bǫ ( Q ) ≥ ν ( b γ Q γ ) − bǫ ( Q ) . By Corollary 2.21 (ii) , if the first inequality is an equality, then δ Q ( ∂ b ( b i Q i )) = i − p r < γ − p r .For i = δ , since p ∤ (cid:18) γp r (cid:19) , by Corollary 2.21 (ii) we have ν Q ( ∂ b ( a γ Q γ )) = ν ( a γ Q γ ) − bǫ ( Q ) and S Q ( ∂ b ( a γ Q γ )) = { γ − p r } . For i > δ and i / ∈ Λ, we have ν ( a i Q i ) > ν ( a γ Q γ ). Hence, by Corollary 2.21 (i) we have ν Q ( ∂ b b i Q i ) > ν ( b γ Q γ ) − bǫ ( Q ).Set g = f − X i ∈ Λ b i Q i . Then ν Q ( g ) = ν ( b γ Q γ ) − bǫ ( Q ) = ( δ − γ ) B + γν ( Q ) − bǫ ( Q ) , where the second equality holds by Lemma 4.3. Moreover, δ Q ( g ) = γ − p r . Itremains to show that if ∂ b ( b i Q i ) = b iδ Q δ + . . . + b i is the Q -expansion of ∂ b ( a i Q i ), then ν ( b i ) > ν ( g ) for every i ∈ Λ . For i ∈ Λ, by Lemma 2.19 we have b i = X α T α where α = ( b , . . . , b r ) and r = i. Let S b,i ⊂ S b the set of uples for which r = i . It remains to show that ν ( T α ) >ν ( g )for each α ∈ S b,i . Let α = ( b , . . . , b i ) ∈ S b,i . Since b j > j ∈{ , . . . , i } , by Lemma 4.5 we have ν ( ∂ b j Q ) ≥ B − b j ( ǫ ( B ) − ǫ ( Q )) − b j ǫ ( Q ) . By Lemma 4.3 we have ν ( a i ) = ( δ − i ) B and consequently ν ( ∂ b a i ) ≥ ( δ − i ) B − b o ǫ ( Q ) ≥ ( δ − i ) B − b ( ǫ ( B ) − ǫ ( Q )) − b ǫ ( Q ) . Therefore, adding the inequalities we obtain ν ( T α ) ≥ δB − b ( ǫ ( B ) − ǫ ( Q )) − bǫ ( Q ) = ( δ − p r ) B + p r ν ( Q ) − bǫ ( Q ) . Since γ > p r we have ν ( T α ) > ν Q ( ∂ b ( ∂ b a γ Q γ )) = ν Q ( g ) which shows the result. (cid:3) Corollary 4.7.
We have γ = 0 .Proof. Suppose that γ >
0. By Lemma 4.6 we have S Q ( ∂ b f ) = { } for every Q ∈ Ψ.Hence, by Lemma 2.15 (i) we have ν Q ( ∂ b f ) < ν Q ( ∂ b f ) for which Q , Q ∈ Ψ with Q < Q . Since Ψ is cofinal in Ψ α we have ν Q ( ∂ b f ) < ν ( ∂ b f ) for every Q ∈ Ψ α . Thisis a contradiction with the minimality of the degree of Q ′ α . Therefore, γ = 0. (cid:3) Proof of Theorem 1.2.
Take Q ∈ Ψ. By Corollary 4.7 we have γ = 0. Hence, everymonomial on the Q -expansion of Q ′ α that are not a power of p nor 0 must havevalue greater than δB . By Lemma 3.6 we can remove these monomials to obtain alimit key polynomial Q α for Ψ α . Therefore, Q α has the desired properties. (cid:3) References [1] V. Alexandru, N. Popescu and A. Zaharescu,
A theorem of characterization of residual tran-scendental extensions of a valuation , J. Math. Kyoto Univ. (1988), 579 – 592.[2] S. D. Cutkosky and H. Mourtada, Defect and local uniformization , RACSAM (2019),4211–4226.[3] M. dos S. Barnab´e and J. Novacoski,
Generating sequences and key polynomials ,arXiv:2007.12293, 2020.[4] J. Decaup, M. Spivakovsky and W. Mahboub,
Abstract key polynomials and comparisontheorems with the key polynomials of MacLane – Vaquie , Illinois J. Math. Vol , Number (2018), 253 – 270.[5] F.J. Herrera Govantes, W. Mahboub, M.A. Olalla Acosta and M. Spivakovsky, Key polyno-mials for simple extensions of valued fields , arXiv:1406.0657, 2014.[6] I. Kaplansky,
Maximal fields with valuations I , Duke Math. Journ. (1942), 303 – 321.[7] F.-V. Kuhlmann, A classification of ArtinSchreier defect extensions and characterizations ofdefectless fields , Illinois J. Math. Volume 54, Number 2 (2010), 397–448.[8] S. MacLane,
A construction for prime ideals as absolute values of an algebraic field , DukeMath. J. (1936), 492 – 510.[9] S. MacLane, A construction for absolute values in polynomial rings , Trans. Amer. Math.Soc. (1936), 363 – 395.[10] J. Novacoski, Key polynomials and minimal pairs , J. Algebra (2019), 1 – 14.[11] J. Novacoski and M. Spivakovsky,
Key polynomials and pseudo-convergent sequences , J. Al-gebra (2018), 199 – 219.[12] M. Vaqui´e,
Extension d’une valuation , Trans. Amer. Math. Soc. (2007), no. 7, 3439 –3481.[13] M. Vaqui´e,
Famille admissible de valuations et defaut d’une extension , J. Algebra (2007),no. 2, 859 – 876.JOSNEI NOVACOSKIDepartamento de Matem´atica–UFSCarRodovia Washington Lu´ıs, 23513565-905 - S˜ao Carlos - SPEmail: [email protected]