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On stable and fixed polynomials

2023, Journal of Pure and Applied Algebra

https://doi.org/10.1016/J.JPAA.2022.107216
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Abstract

Let ν be a rank one valuation on K[x] and Ψn the set of key polynomials for ν of degree n ∈ N. We discuss the concepts of being Ψn-stable and (Ψn, Q)-fixed. We discuss when these two concepts coincide. We use this discussion to present a simple proof of Proposition 8.2 of [3] and Theorem 1.2 of .

On stable and fixed polynomials J. Novacoski, M. Spivakovsky To cite this version: J. Novacoski, M. Spivakovsky. On stable and fixed polynomials. Journal of Pure and Applied Algebra, 2022, 227 (3), pp.107216. �10.1016/j.jpaa.2022.107216�. �hal-03873675� HAL Id: hal-03873675 https://hal.science/hal-03873675v1 Submitted on 27 Nov 2022 HAL is a multi-disciplinary open access archive for the deposit and dissemination of scientific research documents, whether they are published or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers. L’archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires publics ou privés. arXiv:2203.00576v1 [math.AC] 1 Mar 2022 ON STABLE AND FIXED POLYNOMIALS J. NOVACOSKI AND M. SPIVAKOVSKY Abstract. Let ν be a rank one valuation on K[x] and Ψn the set of key polynomials for ν of degree n ∈ N. We discuss the concepts of being Ψn -stable and (Ψn , Q)-fixed. We discuss when these two concepts coincide. We use this discussion to present a simple proof of Proposition 8.2 of [3] and Theorem 1.2 of [5]. 1. Introduction Let ν be a rank one valuation on K[x]. For n ∈ N set Ψn = {Q ∈ K[x] | Q is a key polynomial for ν and deg(Q) = n}. Suppose that Ψn is non-empty and bounded (i.e., there exists a ∈ K[x] such that ν(a) > ν(Q) for every Q ∈ Ψn ) and that ν(Ψn ) does not have a maximum. Set K[x]n = {a ∈ K[x] | deg(a) < n}. For each f ∈ K[x] and Q ∈ Ψn the Qexpansion of f is the expression f = a0 + a1 Q + . . . + ar Qr = l(Q) where l(X) ∈ K[x]n [X]. We denote the value r (which does not depend on the choice of Q ∈ Ψn ) in the previous expression by degX (f ). The truncation of ν on Q is given by νQ (f ) := min {ν(ai Qi )}. 0≤i≤r Let Sn = {f ∈ K[x] | νQ (f ) < ν(f ) for every Q ∈ Ψn }. A polynomial f is said to be Ψn -stable if it does not belong to Sn . A monic polynomial F ∈ K[x] is called a limit key polynomial for Ψn if it belongs to Sn and has the smallest degree among polynomials in Sn . In [4], Kaplansky introduces the concept of pseudo-convergent sequences. These objects are strongly related to the set Ψ1 . For a given such sequence a we can define what it means for a polynomial f ∈ K[x] to be fixed by a. Here, we generalize this concept for any of the sets Ψn . We say that f = l(Q), l(X) ∈ K[x]n [X], is (Ψn , Q)fixed if there exists Q′ ∈ Ψn , ν(Q) < ν(Q′ ) such that ν(f ) = ν(l(Q′ − Q)). 2010 Mathematics Subject Classification. Primary 13A18. Key words and phrases. Key polynomials, stable polynomials, truncations of valuations, fixed polynomials. During the realization of this project the first author was supported by a grant from Fundação de Amparo à Pesquisa do Estado de São Paulo (process number 2017/17835-9). 1 2 J. NOVACOSKI AND M. SPIVAKOVSKY Our first main result (Proposition 3.2) is that we can choose a suitable Q ∈ Ψn such that for any f ∈ K[x] with degX (f ) ≤ degX (F ), we obtain that f is Ψn -stable if and only if it is (Ψn , Q)-fixed. We will fix a suitable Q ∈ K[x] (see (2) and (3)). Then we fix a limit ordinal λ and any cofinal well-ordered (with respect to ν) subset {Qρ }ρ<λ ⊆ {Q′ ∈ Ψn | ǫ(Q) < ǫ(Q′ )}. This means that if ρ < σ < λ, then ν(Qρ ) < ν(Qσ ) and that for every Q′ ∈ Ψn , there exists ρ < λ such that ν(Q′ ) < ν(Qρ ). For each ρ < λ set hρ := Q − Qρ ∈ K[x]n and γρ = ν(Qρ ). It follows from the definition that {hρ }ρ<λ is a pseudoconvergent sequence for ν. For simplicity, we will denote νQρ by νρ . Let p = char(Kν) and I be the set of all non-negative powers of p. As an application of Proposition 3.2 we can prove the following. Theorem 1.1. Let F be a limit key polynomial for Ψn and write F = L(Q) for some L(X) ∈ K[x]n [X]. Then we have the following. (i): There exists σ < λ such that for every θ > σ the polynomial X Fp = L(hθ ) + ∂i L(hθ )Qiθ i∈I is a limit key polynomial for Ψn . Here ∂i L denotes the Hasse derivative of L(X) (as a polynomial in K(x)[X]) of order i. (ii): For each i ∈ I ∪ {0} there exists ai ∈ K[x]n such that X ai Qiθ Fp = i∈I∪{0} is a limit key polynomial for Ψn . Kaplansky proved the above result in the case n = 1. In that case, (ii) follows trivially from (i). Our proof of Theorem 1.1 follows Kaplansky’s proof. If n > 1, then (ii) was proven in [3] (Proposition 8.2). An alternative proof of it was presented in [5] (Theorem 1.2). The advantage of our proof is that it is much simpler and presents as algorithm on how to construct the limit key polynomial of this form (from a given limit key polynomial). Also, our proof does not require that are in the equicharacteristic case. 2. Preliminaries Throughout this paper ν will denote a rank one valuation on K[x]. For f ∈ K[x] we denote ν(f ) − ν(∂b f ) (1) ǫ(f ) = max , b 1≤b≤deg(f ) where ∂b f denotes the Hasse derivative of f of order b. We denote ν(f ) − ν(∂b f ) I(f ) = b, 1 ≤ b ≤ deg(f ) | ǫ(f ) = . b ON STABLE AND FIXED POLYNOMIALS 3 A monic polynomial Q ∈ K[x] is said to be a key polynomial for ν if for every f ∈ K[x], if ǫ(f ) ≥ ǫ(Q), then deg(f ) ≥ deg(Q). For a polynomial f ∈ K[x] let f = f0 + f1 Q + . . . + fr Qr be the Q-expansion of f . We set SQ (f ) = {i, 0 ≤ i ≤ r | νQ (f ) = ν(fi Qi )} and δQ (f ) = max SQ (f ). Throughout this paper we will fix a limit key polynomial F for Ψn and denote d := degX (F ). Set B = lim ν(Q) and B = lim νQ (F ). Q∈Ψn Q∈Ψn Take Q0 ∈ Ψn and choose Q ∈ Ψn such that (2) ǫ(Q) − ǫ(Q0 ) > d(B − ν(Q)) and (3) ǫ(Q) − ǫ(Q0 ) > B − νQ (F ). Write F = L(Q) for L(X) ∈ K[x]n [X]. The next result is well-known. We will reprove it here because we need this slightly stronger statement. Lemma 2.1. Let Q be a key polynomial and take f such that f = qQ + r with γ = max{ǫ(f ), ǫ(r)} < ǫ(Q). Then we have νQ (qQ) − (ǫ(Q) − γ) ≥ ν(f ) = ν(r). Proof. Take b ∈ I(qQ). Since ǫ(qQ) = max{ǫ(Q), ǫ(q)} ≥ ǫ(Q) (Corollary 4.4 of [1]) we have ν(qQ) − bǫ(Q) ≥ ν(∂b (qQ)) ≥ min{ν(∂b (f )), ν(∂b (r))} ≥ min{ν(f ), ν(r)} − bγ. Consequently, ν(f ) = ν(r) and ν(qQ) − ν(f ) ≥ b(ǫ(Q) − γ) ≥ ǫ(Q) − γ. Applying the above discussion to νQ instead of ν we obtain the result. Take f ∈ K[x] and Q′ ∈ Ψn with ǫ(Q) ≤ ǫ(Q′ ) with ǫ(f ) < ǫ(Q0 ) and write f = qQ′ + r with deg(r) < deg(Q′ ) = deg(Q0 ). By (2), (3), Lemma 2.1 and the fact that ǫ(r) < ǫ(Q0 ) we have (4) νQ′ (qQ′ ) > ν(f ) + d(B − ν(Q)) and (5) νQ′ (qQ′ ) > ν(f ) + B − νQ (F ). The next result is a well-known result about key polynomials. 4 J. NOVACOSKI AND M. SPIVAKOVSKY Lemma 2.2. Take Q, Q′ ∈ Ψn be such that ν(Q) < ν(Q′ ). For f ∈ K[x] let r X fi Q′i be the Q′ -expansion of f . Then f= i=0 νQ (f ) = min {νQ (fi Q′i )}. 0≤i≤r ′ Proof. Since Q is monic and has the smallest degree among all polynomials f such that νQ (f ) < ν(f ), it is a (Mac Lane-Vaquié) key polynomial for νQ (Theorem 31 of [2]). In particular, Q′ is νQ - minimal and the result follows from Proposition 2.3 of [6]. Lemma 2.3. Let Q′ ∈ Ψn such that ǫ(Q) < ǫ(Q′ ). For any f ∈ K[x] let f = a0 + a1 Q + . . . + ar Qr and f = b0 + b1 Q′ + . . . + br Q′r be the Q and Q′ -expansions of f , respectively. For l = δQ (f ) we have ν(al − bl ) > ν(al ). In particular, ν(al ) = ν(bl ). Proof. Let h := Q − Q′ so that Q = Q′ + h. Then i X i i ai hj Q′i−j . ai Q = j j=0 For each i, 0 ≤ i ≤ r, and j, 0 ≤ j ≤ i, let i (6) ai hj = aij0 + aij1 Q′ + . . . + aijn Q′n j be the Q′ expansion of ai hj . Then bl = X aijk . i−j+k=l For i, 0 ≤ i ≤ r, and j, 0 ≤ j ≤ i, if k := l + j − i > 0, then by (4), we have ν(aijk ) + kν(Q′ ) > ν(ai hj ) + d(B − ν(Q)) ≥ ν(ai Qi ) + (j − i)ν(Q) + k(B − ν(Q)) ≥ ν(al Ql ) + (j − i)ν(Q) + k(B − ν(Q)) ≥ ν(al ) + (l + j − i)ν(Q) + k(B − ν(Q)) = ν(al ) + kB. Since B > ν(Q′ ) we have ν(aijk ) > ν(al ). Suppose now that k := l+j−i = 0 (i.e., that i = l+j). If j = 0, then by definition aijk = al . If j > 0, then i > l. Since l = δQ (f ) we have ν(ai Qi ) > ν(al Ql ). Then by Lemma 2.2, applied to (6), we have νQ (aijk Q′k ) ≥ νQ (ai hj ) = ν(ai Qi ) + (j − i)ν(Q) > ν(al ) + (l + j − i)ν(Q). Since νQ (Q′ ) = ν(Q) we obtain that ν(aijk ) > ν(al ) and the result follows. ON STABLE AND FIXED POLYNOMIALS 5 For each ρ < λ and f ∈ K[x], let f = aρ0 (f ) + aρ1 (f )Qρ + . . . + aρr (f )Qrρ be the Qρ -expansion of f . The value of aρ0 (f ) will be very important in what follows. Proposition 2.4. For f ∈ K[x] with degX (f ) ≤ d, write f = l(Q) for l(X) ∈ K[x]n [X]. For every ρ < λ we have (7) νρ (l(hρ )) ≥ νρ (f ). Moreover, the equality holds in (7) if and only if ν (aρ0 (f )) = νρ (f ) = ν(l(hρ )). Proof. By definition f = l(Q) = l(Qρ + hρ ) = Qρ p(x) + l(hρ ) for some p(x) ∈ K[x]. Hence aρ0 (f ) = aρ0 (l(hρ )). Let l(hρ ) = aρ0 (f ) + b1 Qρ + . . . + bl Qlρ (8) be the Qρ -expansion of l(hρ ). We will show that ν(bi Qiρ ) > νρ (f ) for every i, 1 ≤ i ≤ l, and this will imply our result. Let f = a0 + a1 Q + . . . + ar Qr be the Q-expansion of f , so that l(hρ ) = a0 + a1 hρ + . . . + ar hrρ . (9) For each j, 1 ≤ j ≤ r, consider the Qρ -expansion aj hjρ = aρ0j + aρ1j Qρ + . . . + aρlj Qlρ (10) of aj hjρ . Comparing (8), (9) and (10), it is enough to show that ν aρij Qiρ > ν (aρ0 (f )) for every i, j, 1 ≤ i ≤ l and 1 ≤ j ≤ r. For a fixed j, 1 ≤ j ≤ r, by (4) applied to (10) we have (11) ν aρij Qjρ > ν aj Qj + s(B − ν(Q)). Since ν(Q) = ν(hρ ) = νρ (Q), if ν(a0 ) < ν(ai Qi ) for every i, 1 ≤ i ≤ r, then νρ (f ) = ν(a0 ) = ν(l(hρ )) and we are done. Suppose not and take l = δQ (f ) > 0. By (11) and the fact that ν(aρl (f )) = ν(al ) (Lemma 2.3), we have > ν(aj Qj ) + l(B − ν(Q)) ≥ ν(al ) + lν(Q) + l(B − ν(Q)) ν aρij Qjρ = ν(al ) + lB > ν(al ) + lν(Qρ ) ≥ νρ (f ). This completes the proof. 6 J. NOVACOSKI AND M. SPIVAKOVSKY Corollary 2.5. If deg(f ) < deg(F ), then there exists ρ such that ν(l(hσ )) = ν(f ) = νσ (f ) = ν(aσ0 (f )) for every σ, ρ < σ < λ. Proof. It is well-known that if f is Ψn -stable, then there exists Q′ ∈ Ψn such that 0 ∈ SQ′ (f ). The result follows immediately. 3. The Taylor expansion of a polynomial We will consider the ring K(x)[X] where X is an indeterminate and let ∂i denote the i-th Hasse derivative with respect to X. Then, for every l(X) ∈ K(x)[X] and a, b ∈ K[x] we have the Taylor expansion degX l l(b) = l(a) + X ∂i l(a)(b − a)i . i=1 Lemma 3.1 (Lemma 4 of [4]). Let Γ be an ordered abelian group, β1 , . . . , βn ∈ Γ and {γρ }ρ<λ an increasing sequence in Γ, without a last element. If t1 , . . . , tn are distinct positive integers, then there exist b, 1 ≤ b ≤ n, and ρ < λ, such that βi + ti γσ > βb + tb γσ for every i, 1 ≤ i ≤ n, i 6= b and σ > ρ. Proposition 3.2. Take f ∈ K[x] such that degX (f ) ≤ d and write f = l(Q) for l(X) ∈ K[x]n [X]. Then there exists ρ < λ such that νσ (f ) = ν(l(hσ )) for every σ, ρ < σ < λ. In particular, f is Ψn -stable if and only if it is (Ψn , Q)-fixed. Proof. Since for every j, 1 ≤ j ≤ degX (l), we have deg(∂j l(Q)) < deg(F ) we can use Corollary 2.5 to obtain ρ < λ such that (12) βj := νσ (∂j l(hσ )) = ν (∂j l(hσ )) for every j, 1 ≤ j ≤ degX (l) and ρ < σ. By Lemma 3.1, there exist b, 1 ≤ b ≤ degX (l), and ρ < λ such that for every λ, ρ < σ < λ and i 6= b we have (13) βb + bγρ < βi + iγρ and (12) is satisfied. For σ, ρ < σ < λ, since degX (f ) f − l(hσ ) = X ∂i l(hσ )Qiσ i=1 we have (14) νσ (f − l(hσ )) = βb + bγσ and (15) ν (f − l(hσ )) = βb + bγσ . ON STABLE AND FIXED POLYNOMIALS 7 By Proposition 2.4 we have νσ (l(hσ )) ≥ νσ (f ). This and (14) imply that (16) ν(l(hσ )) ≥ νσ (l(hσ )) ≥ βb + bγσ . If νσ (f ) = ν(f ), then ν(l(hσ )) ≥ ν(f ). Suppose, aiming for a contradiction, that ν(l(hσ )) > ν(f ). Then by (15) we have ν(f ) = βb + bγσ . For any σ ′ > σ, by (16) we would obtain ν(l(hσ′ )) ≥ βb + bγσ′ > βb + bγσ = ν(f ) and this contradicts (15) (with σ replaced by σ ′ ). Hence, ν(l(hσ )) = ν(f ). Suppose now that νσ (f ) < ν(f ). If νσ (l(hσ )) > νσ (f ), then by (14) we have νσ (f ) = βb + bγσ . Consequently, ν(f − l(hσ )) ≥ min{ν(f ), ν(l(hσ ))} > νσ (f ) = βb + bγσ , what is a contradiction to (15). Hence, νσ (l(hσ )) = νσ (f ). By the second part of Proposition 2.4 we obtain that νσ (f ) = ν(l(hσ )). This, (14)–(16) and the fact that νσ (f ) < ν(f ) imply that νσ (f ) = ν(l(hσ )) = βb + bγσ . Remark 3.3. In the proof of Theorem 1.1 we will use the explicit calculation of ν(l(hσ )) obtained in the previous proposition. 4. Proof of Theorem 1.1 We will adapt the proof by Kaplansky in [4]. For each i, 1 ≤ i ≤ d, the polynomial ∂i L(Q) has degree smaller than deg(F ), hence by Corollary 2.5 there exists ρ0 < λ such that (17) βi := ν(∂i L(Q)) = ν(∂i L(hρ )) for every ρ, ρ0 < ρ < λ. Lemma 4.1. If i = pt and j = pt r with r > 1 and p ∤ r, then there exists ρ < λ such that βi + iγσ < βj + jγσ for every σ, ρ < σ < λ. Moreover, if C in the value group of ν is such that C > γρ for every ρ < λ, then βi + iC < βj + jC. 8 J. NOVACOSKI AND M. SPIVAKOVSKY Proof. From the Taylor formula (applied to ∂i L) we have ∂i L(hσ ) − ∂i L(hρ ) = = n−i X ∂k ∂i L(hρ )(hσ − hρ )k k=1 n−i X k=1 i+k ∂i+k L(hρ )(hσ − hρ )k . i By Lemma 3.1, for ρ < σ large enough (18) i+k k . ν ν (∂i L(hσ ) − ∂i L(hρ )) = min ∂i+k L(hρ )(hσ − hρ ) 1≤k≤n−i i In particular, taking k = j − i, this gives j j−i (19) ν (∂i L(hσ ) − ∂i L(hρ )) ≤ ν ∂j L(hρ )(hσ − hρ ) i By (17) and (19) we have βi ≤ ν(∂i L(hσ ) − ∂i L(hρ )) j j−i ≤ ν ∂j L(hρ )(hσ − hρ ) i j = ν + βj + (j − i)γρ . i j j Since p ∤ and char(Kν) = p we have ν = 0. Consequently, i i βi ≤ βj + (j − i)γρ . This means that for every σ, ρ < σ < λ, we have βi + iγσ < βj + jγσ . Take C > γρ for every ρ < λ. If βi + iC ≥ βj + jC, then βi − βj ≥ (j − i)C > (j − i)γσ for every ρ < λ and this contradicts the first part. The proof of the next result is very similar to the proof of Proposition 2.4. Lemma 4.2. Fix θ < λ and for each i, 0 ≤ i ≤ r, set ai0 := aθ0 (∂i L(hθ )). Then νθ (∂i L(hθ ) − ai0 ) + iν(Qθ ) > B. Proof. Since F = a0 + a1 Q + . . . + ar Qr we have ∂0 L(hθ ) = a0 + a1 hθ + . . . + ar hrθ ∂1 L(hθ ) = .. . ∂r L(hθ ) = a1 + . . . + rar hr−1 θ ar . If we write ∂i L(hθ ) = bi0 + bi1 hθ + . . . + bis hsθ , ON STABLE AND FIXED POLYNOMIALS 9 then ν bij hjθ + iν(Q) ≥ ν(ai+j Qi+j ) ≥ νQ (F ). For each j > 1, write bij hjθ = ai0j + ai1j Qθ + . . . + aisj Qsθ . For every i, j and k > 0, by (5) we have ν aikj Qkθ + iν(Qθ ) > ν(bij hjθ ) + B − νQ (F ) + iν(Qθ ) > νQ (F ) + B − νQ (F ) = B. For every i, we have ∂i L(hθ ) − ai0 = X aikj Qjθ . k,j>0 The result follows. We proceed now with the proof of Theorem 1.1. Proof of Theorem 1.1. For each i = ps with 1 ≤ i ≤ deg(F ) and j = ps r, with p ∤ r, by Lemma 4.1 we have (20) βi + iB < βj + jB for ρ large enough. Then there exists ρij such that for every σ > ρij we have (21) βi + iγρ < βj + jγσ for every ρ < λ. Take σ such that (21) is satisfied for every i = ps and j = ps r, with p ∤ r. Write I = {l | 1 ≤ l ≤ d such that l = pi for some i ∈ N} and J := {1, . . . , d} \ I. Then, for every j ∈ J there exists i ∈ I such that βi + iγρ < βj + jγσ for every ρ < λ. This means that for every ρ > σ we have   X (22) ν  ∂j L(hσ )(hρ − hσ )j  ≥ min{βj + jγσ } > min{βi + iγρ } = βb + iγb . j∈J j∈J i∈I In order to prove (i), take θ > σ and consider the polynomial X Fp (x) = L(hθ ) + ∂i L(hθ )(Q − hθ )i =: Lp (Q). i∈I Then L(hρ ) − Lp (hρ ) = X ∂j L(hθ )(hρ − hθ )j . j∈J Consequently, ν (L(hρ ) − Lp (hρ )) > βb + bγρ = ν(L(hρ )). 10 J. NOVACOSKI AND M. SPIVAKOVSKY The last equality follows from the proof of Proposition 3.2 (as observed in Remark 3.3). Hence, ν(Lp (hρ )) = βb + bγρ . If r ∈ / I, then degX (Fp ) < d so we apply Lemma 3.2 to obtain that Fp ∈ Sn . This is a contradiction to the minimality of the degree of F in Sn . Hence, degX (Fp ) = d and Fp is monic. Since degX (Fp ) ≤ d, by Lemma 3.2 we obtain that Fp ∈ Sn and so is a limit key polynomial for Ψn . In order to prove (ii), for each i ≥ 0 take ai = aθ0 (∂i L(hθ )) and F p := X ai Qiθ . i∈I∪{0} By Lemma 4.2 we have νρ Fp − F p > B > νρ (F ) for every ρ, θ < ρ < λ. As before, we conclude that F p is a limit key polynomial for Ψn and this completes the proof. Remark 4.3. One can prove (Proposition 3.5 of [5]) that ar = 1 and in particular deg(F ) = n degX (F ). References [1] M. dos S. Barnabé and J. Novacoski, Generating sequences and key polynomials, to appear in Michigan Mathematical Journal, arXiv:2007.12293, 2020. [2] J. Decaup, W. Mahboud and M. Spivakovsky, Abstract key polynomials and comparison theorems with the key polynomials of Mac Lane-Vaquié, Illinois Journal of Mathematics 62 (2018), 253–270. [3] F.J. Herrera Govantes, W. Mahboub, M.A. Olalla Acosta and M. Spivakovsky, Key polynomials for simple extensions of valued fields, arXiv:1406.0657, 2014. [4] I. Kaplansky, Maximal fields with valuations I, Duke Math. Journ. 9 (1942), 303 – 321. [5] M. Moraes and J. Novacoski, Limit key polynomials as p-polynomials, J. Algebra 579, 152– 173 (2021). [6] E. Nart, Key polynomials over valued fields, Publ. Mat. 64 (2020), 195–232.

References (6)

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