Automorphisms of a Polynomial Ring Which Admit Reductions of Type I

45(2009), 907–917

Automorphisms of a Polynomial Ring Which Admit Reductions of Type I

By

ShigeruKuroda∗

Abstract

Recently, Shestakov-Umirbaev solved Nagata’s conjecture on an automorphism of a polynomial ring. To solve the conjecture, they defined notions called reductions of types I–IV for automorphisms of a polynomial ring. An automorphism admitting a reduction of type I was first found by Shestakov-Umirbaev. Using a computer, van den Essen–Makar-Limanov–Willems gave a family of such automorphisms. In this paper, we present a new construction of such automorphisms using locally nilpo- tent derivations. As a consequence, we discover that there exists an automorphism admitting a reduction of type I which satisfies some degree condition for each possible value.

§1. Introduction

Let k be a field of characteristic zero, and k[x] = k[x₁, . . . , x_n] the poly- nomial ring in n variables over k. We will identify an endomorphism F ∈ End_kk[x] with then-tuple (f₁, . . . , f_n) of elements of k[x], where f_i =F(x_i) for eachi. Then,F is invertible if and only ifk[f₁, . . . , f_n] =k[x]. If this is the case, the sum degF := _n

i=1degf_i of the total degrees of f_i’s is necessarily at leastn. An automorphismF ∈Aut_kk[x] is said to beaffineif degF =n, and elementary if there exist i ∈ {1, . . . , n} and a polynomial φ ∈ k[x] not depending onx_i such that f_i = x_i+φ and f_j = x_j for eachj = i. We say thatF admits anelementary reductionif there exists an elementary automor- phism G such that deg(F ◦G) < degF. Note thatF admits an elementary

Communicated by S. Mori. Received August 12, 2008.

2000 Mathematics Subject Classification(s): Primary 14R10; Secondary 13N15.

Partly supported by the Grant-in-Aid for Young Scientists (Start-up) 19840041, The Ministry of Education, Culture, Sports, Science and Technology, Japan.

∗Department of Mathematics and Information Sciences, Tokyo Metropolitan University, 1-1 Minami-Osawa, Hachioji, Tokyo 192-0397, Japan.

c 2009 Research Institute for Mathematical Sciences, Kyoto University. All rights reserved.

reduction if and only if there exists φ∈k[f₁, . . . , f_i−1, f_i+1, . . . , f_n] such that deg(f_i+φ)< degf_i for some i. The subgroup T_kk[x] of the automorphism group Aut_kk[x] generated by affine automorphisms and elementary automor- phisms is called thetame subgroup, and each element of Tkk[x] is called atame automorphism.

By Jung [3], it follows that T_kk[x] = Aut_kk[x] whenn = 2. In fact, he showed that each F ∈ Aut_kk[x] for n = 2 admits an elementary reduction whenever degF >2. Thereby,

degF >deg(F◦G₁)>· · ·>deg(F◦G₁◦ · · · ◦G_r) = 2

for some elementary automorphisms G₁, . . . , G_r of k[x]. Note that a similar result for a fieldkof an arbitrary characteristic was given by van der Kulk [4].

Now, assume thatn= 3. Nagata [8] conjectured that the automorphism (1.1) (x₁−2(x₁x₃+x²₂)x₂−(x₁x₃+x²₂)²x₃, x₂+ (x₁x₃+x²₂)x₃, x₃) of k[x] is not tame. In 2003, this well-known conjecture was finally solved in the affirmative by Shestakov-Umirbaev [9], [10]. They defined four types of reductions, said to be of types I, II, III and IV, for elements of Aut_kk[x].

Then, they showed that each non-affine element of T_kk[x] admits an elementary reduction or one of these four types of reductions. One can easily check that Nagata’s automorphism admits none of these reductions. Therefore, Nagata’s conjecture holds true.

In the present paper, we study automorphisms admitting reductions of type I.

Definition 1.1[10, Definition 1]. Assume that n = 3. We say that an automorphism F = (f₁, f₂, f₃) of k[x] admits a reduction of type I if the following conditions hold:

(i) There exists an odd numbers≥3 such that degf₁: degf₂= 2 :s.

(ii) degf₁<degf₃≤degf₂.

(iii) ¯f₃ does not belong tok[ ¯f₁,f¯₂], where ¯f denotes the highest homoge- neous part off for eachf ∈k[x].

(iv) There existα∈k\{0}andφ∈k[f₁, f₂−αf₃] such that deg(f₃+φ)<

degf₃ and deg[f₁, f₃+φ]<degf₂+ deg[f₁, f₂−αf₃]. Here, we define (1.2) deg[f, g] = max

deg

∂f

∂x_i

∂g

∂x_j − ∂f

∂x_j

∂g

∂x_i 1≤i < j≤3

+ 2 for eachf, g∈k[x].

We also say thatF admits a reduction of type I if (f_σ(1), f_σ(2), f_σ(3)) sat- isfies (i)–(iv) for some permutationσof{1,2,3}.

We mention that the conditions listed above imply that deg(f₂−αf₃) = degf₂and thatf₁andf₂−αf₃form a “∗-reduced pair”, which are also included in the definition of reduction of type I by Shestakov-Umirbaev.

Note that (f₁, f₂−αf₃, f₃) admits an elementary reduction by (iv), while (f₁, f₂, f₃) does not (cf. [10, Proposition 1]). Shestakov-Umirbaev [10, Exam- ple 1] gave the first example of a tame automorphism which admits a reduction of type I in case of s = 3. Van den Essen–Makar-Limanov–Willems [1] con- structed a family of such automorphisms when s = 3,5,7 using a computer.

Reductions of types II, III and IV are also defined theoretically [10], but no automorphisms admitting these reductions are found. To study the structures of Aut_kk[x] and T_kk[x], it is of great importance to investigate automorphisms admitting reductions of these four types.

The purpose of this paper is to construct new automorphisms ofk[x] which admit reductions of type I by employing the theory of locally nilpotent deriva- tions. As a consequence, we discover that there exists a tame automorphism admitting a reduction of type I such that degf₁ : degf₂ = 2 :s for each odd numbers≥3.

The author would like to thank Prof. Hiraku Kawanoue for informing him of an example of polynomials discussed in Section 4, to Prof. Eric Edo for many invaluable comments, and to Prof. Gene Freudenburg for helpful discussions.

§2. Tame Automorphisms Admitting Reductions of Type I Before stating our main result, we prove a lemma. In what follows, we assume thatn= 3.

Lemma 2.1. Lets≥3 be an odd number, andH = (h₁, h₂, h₃)a tame automorphism ofk[x]such that

degh₁: degh₂: degh₃= 2 :s: 1, s−1

2 degh₁<deg(ch^s₁+h²₂)<degh₂ for somec∈k\{0}. Then,H= (h₁, h₂, h₃)is a tame automorphism admitting a reduction of type I for whichdegh₂ = degh₂ and degh₃ = deg(ch^s₁+h²₂).

Here,h₂=h₂+h₃+ch^s₁+h²₂ andh₃=h₃+ch^s₁+h²₂.

Proof. LetG₁ andG₂ be elementary automorphisms ofk[x] defined by G₁(x₃) = x₃ +cx^s₁ +x²₂, G₂(x₂) = x₂+x₃ and G_i(x_j) = x_j for (i, j) = (1,3),(2,2). Then,H =H◦G₁◦G₂. Hence,His a tame automorphism ofk[x], since so isH. By assumption, degh₂is greater than degh₃and deg(ch^s₁+h²₂),

while degh₃ is less than deg(ch^s₁+h²₂). Hence, degh₂= degh₂ and degh₃ = deg(ch^s₁+h²₂). It follows that

ldegh₁<degh₃<degh₂= s

2degh₁<(l+ 1) degh₁, where l=s−1 2 . This implies that ¯h₃ does not belong to k[¯h₁,h¯₂], since degh₃ < degh₂. If α= 1, thenh₂−αh₃=h₂, soφ:=−ch^s₁−h²₂ is contained ink[h₁, h₂−αh₃].

The total degree ofh₃+φ=h₃is less than degh₃. In addition, deg[h₁, h₃+φ] = deg[h₁, h₃]≤degh₁+ degh₃≤degh₂

= deg(h₂−αh₃)<deg(h₂−αh₃) + deg[h₁, h₂−αh₃].

Therefore, (h₁, h₂, h₃) satisfies all the conditions of Definition 1.1.

Now, letpand qbe natural numbers, and consider triangular derivations DandE ofk[x] defined by

(2.1) D(x₁) =x^q+1₂ , D(x₂) = 0, D(x₃) = (p+ 1)x^p₁x^q₂, E(x₁) = 2x₃, E(x₂) = 2(p+ 1)x^p₁, E(x₃) = 1.

Here, we say that ak-derivation Δ ofk[x] istriangular if Δ(x_σ(i)) belongs to k[x_σ(1), . . . , x_σ(i−1)] for eachi for some permutationσ of{1, . . . , n}. If this is the case, Δ is locally nilpotent, i.e., Δ^l(f) = 0 for sufficiently large l for each f ∈k[x]. In particular,

(2.2) f_i=

∞ l=0

D^l(x_i)

l! , g_i= ∞

l=0

E^l(x_i) l! (−x₃)^l

are elements ofk[x] for each i. We setF = (f₁, f₂, f₃),G = (g₁, g₂, x₃), and defineh₁=F(g₁),h₂=F(g₂) andh₃=f₃. Namely, (h₁, h₂, h₃) =F◦G. Put (2.3) m=pq+p+q, c= (−2)^p+1

p i=1

i+ 1 2i+ 1. Here is our main result.

Theorem 2.1. Let pandqbe natural numbers. Then, (h₁, h₂, h₃)is a tame automorphism ofk[x]forn= 3 such that

(2.4) degh₁= 2m, degh₂= (2p+ 1)m, degh₃=m, deg(c²h^2p+1₁ +h²₂) = 2pm+p+ 1.

Note that (h₁, h₂, h₃) satisfies the assumptions of Lemma 2.1 fors= 2p+1.

Actually, (s−1)/2 =pand

pdegh₁<2pm+p+ 1 = (2p+ 1)m−(p+ 1)q+ 1<degh₂. Therefore, we obtain the following corollary to Theorem 2.1.

Corollary 2.1. There exists a tame automorphism (h₁, h₂, h₃) of k[x]

admitting a reduction of type I such that

degh₁= 2m, degh₂= (2p+ 1)m, degh₃= 2pm+p+ 1 for eachp, q∈N, wherem=pq+p+q.

For a triangular derivation Δ, it is known that the exponential map exp Δ : k[x]→k[x] is a tame automorphism ofk[x]. If furthermore Δ(x_n) = 1, then ker Δ =k[g₁, . . . , g_n−1 ], and (g₁, . . . , g_n−1, x_n) is a tame automorphism ofk[x].

Here, we define

(exp Δ)(f) = ∞ l=0

Δ^l(f)

l! , g_i= ∞ l=0

Δ^l(x_i) l! (−x_n)^l

for each f ∈ k[x] and i = 1, . . . , n−1 (cf. [2, Sections 1.3 and 6.1]). Hence, F and Gare tame automorphisms of k[x], and so is F ◦G= (h₁, h₂, h₃). In addition,E(g_i) = 0 fori= 1,2. In Section 3, we will consider the polynomial

I=x₂x₃−x^p+1₁ . SinceD(I) = 0, it follows thatF(I) = (expD)(I) =I.

To conclude this section, we give explicit descriptions of f_i and g_j for i= 1,2,3 andj = 1,2. By a straightforward computation, we get

(2.5) f₁=x₁+x^q+1₂ , f₂=x₂, g₁=x₁−x²₃. We show that

(2.6) f₃=x₃+ p i=0

p+ 1 i+ 1

x^p−i₁ x^(q+1)i+q₂ , g₂=x₂+ p i=0

c_ix^p−i₁ x²ⁱ⁺¹₃ ,

where

c_i= (−2)ⁱ⁺¹

i l=0

p−l+ 1 2l+ 1

for eachi≥0. For the first equality of (2.6), it suffices to verify that

(2.7) Dⁱ⁺¹(x₃)

(p+ 1)! = x^p−i₁ x^(q+1)i+q₂ (p−i)!

fori= 0, . . . , p. We prove (2.7) by induction oni. The casei= 0 follows from the definition ofD. Assume that (2.7) is true ifi=lfor some 0≤l < p. Then,

D^l+2(x₃) (p+ 1)! =D

D^l+1(x₃) (p+ 1)!

=D

x^p−l₁ x^(q+1)l+q₂ (p−l)!

= (p−l)x^p−l−1₁ x^(q+1)l+q₂ D(x₁)

(p−l)! =x^p−(l+1)₁ x(q+1)(l+1)+q

(p−(l2+ 1))! .

Hence, (2.7) holds fori=l+ 1, and thus holds for any 0≤i≤p. Therefore, we have proved the first equality of (2.6). Next, letg₂be the right-hand side of the second equality of (2.6). Then,g₂−g₂ =x₃ψfor someψ∈k[x]. To conclude thatg₂=g₂, it suffices to show thatE(g₂−g₂) = 0, sinceEis locally nilpotent and E(x₃) = 0 by definition. In fact, for a locally nilpotent derivation Δ of k[x], the condition Δ(φψ) = 0 impliesψ = 0 for φ, ψ ∈ k[x] with Δ(φ) = 0 (cf. [2, Proposition 1.3.32]). It follows thatE(g₂) = 0 as mentioned. Note that c₀=−2(p+ 1), and 2(p−i)c_i=−(2i+ 3)c_i+1 fori= 0, . . . , p. Hence, we have

E(g₂) =E(x₂) + p i=0

c_i

(2i+ 1)x^p−i₁ x²ⁱ₃E(x₃) + (p−i)x^p−i−1₁ x²ⁱ⁺¹₃ E(x₁)

= 2(p+ 1)x^p₁+ p i=0

(2i+ 1)c_ix^p−i₁ x²ⁱ₃ + 2(p−i)c_ix^p−i−1₁ x²ⁱ⁺²₃

= 2(p+ 1)x^p₁+ p i=0

(2i+ 1)c_ix^p−i₁ x²ⁱ₃ −(2i+ 3)c_i+1x^p−(i+1)₁ x²⁽ⁱ⁺¹⁾₃

= 2(p+ 1)x^p₁+ p i=0

(2i+ 1)c_ix^p−i₁ x²ⁱ₃ −

p+1

i=1

(2i+ 1)c_ix^p−i₁ x²ⁱ₃ = 0.

Thus,E(g₂−g₂) =E(g₂)−E(g₂) = 0. Therefore,g₂ is expressed as in (2.6).

§3. Proof of the Main Result In this section, we prove the four equalities of (2.4).

Letf =

α∈Zⁿc_αx^αbe a Laurent polynomial inx₁, . . . , x_n overk, where c_α∈kandx^α=x^α₁¹· · ·x^α_nⁿ for each α= (α₁, . . . , α_n). Then, we set

|f|={α∈Zⁿ |c_α= 0}.

Forη ∈Rⁿ, we define deg_ηf to be the maximum among the inner products α·η forα∈ |f|, and put

f^η=

α∈Zⁿ

c_αx^α, where c_α=

c_α ifα·η= deg_ηf 0 otherwise.

Clearly, deg_ηf = deg_ηf^η for each f ∈ k[x]. We note that (f +g)^η is equal to one off^η, g^η and f^η+g^η for each f, g ∈k[x] with|f^η| ∩ |g^η| =∅. For a k-derivation Δ ofk[x], we define ak-derivation Δ^η ofk[x] by setting

Δ^η(x_i) =

(Δ(x_i))^η if deg_η(Δ(x_i)x⁻¹_i ) = deg_ηΔ

0 otherwise

for each i, where deg_ηΔ denotes the maximum among deg_η(Δ(x_i)x⁻¹_i ) for i = 1, . . . , n. Then, we have Δ^η(f^η) = 0 for each f ∈ ker Δ, for otherwise 0= Δ^η(f^η) = (Δ(f))^η, a contradiction.

Now, we setω_i= degf_i fori= 1,2,3, andω= (ω₁, ω₂, ω₃). Then, ω₁=q+ 1, ω₂= 1, ω₃=pq+p+q=m.

For eachα= (α₁, α₂, α₃), we have

degF(x^α) = degf₁^α1f₂^α2f₃^α3=α₁ω₁+α₂ω₂+α₃ω₃= deg_ωx^α. Hence, degF(f) ≤ deg_ωf for each f ∈ k[x]. The equality holds when f^ω is a term. By (2.5) and (2.6), we see thatg^ω₁ = −x²₃ and g₂^ω = c_px^2p+1₃ are terms, so degh_i = degF(g_i) = deg_ωg_i fori = 1,2. Hence, degh₁ = 2m and degh₂ = (2p+ 1)m. In addition, degh₃= degf₃=m. Thus, we have proved the first three equalities of (2.4).

Next, we consider the polynomial P := c²g₁^2p+1 +g₂². Our goal is to establish that degF(P) = 2pm+p+ 1, which immediately implies the last equality of (2.4). WriteP=P₁+P₂, whereφ=g₂−x₂,

P₁=c²g₁^2p+1+φ² and P₂=x²₂+ 2φx₂.

Set = (1,0,−2). Then,g₁^2p+1 and φ² belong to x^2(2p+1)₃ k[x], sinceg₁and φ are inx²₃k[x] and x^2p+1₃ k[x] by (2.5) and (2.6), respectively. Hence,

P₁^ω=cx^ux^2(2p+1)₃ =cx^u₁x^2(2p−u+1)₃ , whereu≥0, c∈k\ {0}. We claim that u = 0. In fact, the monomial x^2(2p+1)₃ appears in g^2p+1₁ and φ² with coefficients −1 and c²_p, respectively. By definition, c_p = c. Hence,

x^2(2p+1)₃ does not appear in P₁, so we get u= 0. On the other hand, P₂^ω = x₂(x₂+ 2φ)^ω= 2x₂φ^ω= 2cx₂x^2p+1₃ . Clearly,|P₁^ω| ∩ |P₂^ω|=∅. Hence,P^ω must be equal toP₁^ω or P₂^ω or P₁^ω+P₂^ω. Recall that E(g_i) = 0 for i = 1,2. So, E(P) = 0. This implies that E^ω(P^ω) = 0 as mentioned. A straightforward computation shows that

deg_ω(E(x₁)x⁻¹₁ ) = deg_ωx⁻¹₁ x₃=−(q+ 1) +m=pq+p−1, deg_ω(E(x₂)x⁻¹₂ ) = deg_ωx^p₁x⁻¹₂ =p(q+ 1)−1 =pq+p−1, deg_ω(E(x₃)x⁻¹₃ ) = deg_ωx⁻¹₃ =−m=−pq−p−q < pq+p−1.

Accordingly, we getE^ω(x_i) = (E(x_i))^ω =E(x_i) for i= 1,2 and E^ω(x₃) = 0.

Then, it follows thatE^ω(P_i^ω)= 0 fori= 1,2. Consequently, P^ω=P₁^ω+P₂^ω=cx^u₁x^2(2p−u+1)₃ + 2cx₂x^2p+1₃ , 0 =E^ω(P^ω) = 2ucx^u−1₁ x^4p−2u+3₃ + 4c(p+ 1)x^p₁x^2p+1₃ , sou=p+ 1 and c =−2c. Therefore, we get

P^ω=−2cx^p+1₁ x^2p₃ + 2cx₂x^2p+1₃ = 2cx^2p₃ (x₂x₃−x^p+1₁ ) = 2cx^2p₃ I.

Hence, deg_ωP = 2pm+m+ 1. Since F(I) =Ias mentioned, (3.1) degF(P^ω) = deg(2cf₃^2pI) = 2pm+p+ 1.

Finally, let Q = P −P^ω. Since F(P) = F(P^ω) +F(Q), it remains only to verify that degF(Q)<2pm+p+ 1 by (3.1). Observe thatP andQbelong to x²₂k[x⁻¹₂ x^2p+1₃ ,x]. Furthermore, deg_ωx=q+ 1−2m <0, and

deg_ωx⁻¹₂ x^2p+1₃ = (2p+ 1)m−1 =−(p+ 1) deg_ωx.

Hence, deg_ωQ≡deg_ωP (mod deg_ωx). Since deg_ωQ <deg_ωP, we get deg_ωQ≤deg_ωP+ deg_ωx= 2pm+m+ 1 +q+ 1−2m

= 2pm+p+ 1−p(q+ 2) + 1<2pm+p+ 1.

Thus, degF(Q)≤deg_ωQ <2pm+p+ 1, and thereby proving the last equality of (2.4).

§4. Remarks

As far as we know, the answer to the following simple question is not known.

Question 4.1. Do there exist polynomials f, g ∈ k[x] for n = 3 as follows?

(i)k[f, g, h] =k[x] for someh∈k[x].

(ii) deg(f³+g²)≤degf.

This question is closely related to the study of Aut_kk[x] for n = 3. In fact, no automorphism ofk[x] admits a reduction of type II or III or IV if the answer to Question 4.1 is negative. The reason is as follows.

Suppose that there exists an automorphism ofk[x] admitting a reduction of type II or III or IV. Then, it follows from [10, Definitions 2, 3 and 4] that there exists an automorphism (g₁, g₂, g₃) as follows:

(1) degg₁= 2l and degg₂= 3l for somel∈N.

(2) There exists φ ∈k[g₁, g₂]\k with degφ≤2l such that ¯φand ¯g₁ are linearly independent overk.

Since degφ ≤ degg₁ and degφ < degg₂, the condition (2) implies that φ¯∈k[¯g₁,g¯₂]. Writeφ=

i,jc_i,jgⁱ₁g₂^j, where c_i,j ∈k for eachi andj. Letu₁ andu₂ be the maximal numbers such thatc_u1,j = 0 andc_i,u2= 0 for somej andi, and letq_i andr_i respectively be the quotient and residue ofu_i divided bye_i fori= 1,2, where

e₁= degg₂

gcd(degg₁,degg₂) = 3, e₂= degg₁

gcd(degg₁,degg₂) = 2.

Then, due to [9, Theorem 3] (see also [5], [7] and [11]), it follows that degφ≥q_i(lcm(degg₁,degg₂)−degg₁−degg₂+ deg[g₁, g₂]) +r_idegg_i

≥q_i(l+ 2) +r_idegg_i

fori= 1,2. Since degg₁= 2l and degφ≤2l by assumption, (q₁, r₁) must be (0,1) or (1,0). Hence,u₁= 3q₁+r₁is equal to 1 or 3. Similarly, (q₂, r₂) = (1,0), and so u₂ = 2. In particular, u₁ ≤ 3 and u₂ = 2. The polynomials g₁ⁱg₂^j for i= 0,1,2,3 andj= 0,1,2 with (i, j)= (3,0),(0,2) have distinct total degrees.

This implies that c_i,j = 0 for each (i, j) with 2i+ 3j > 6, while c_3,0 = 0 andc_0,2= 0, for otherwise ¯φ=c_i,jg¯ⁱ₁g¯₂^j for some (i, j), which contradicts that φ¯∈k[¯g₁,¯g₂]. Hence,

(4.1) φ=c_3,0g₁³+c_0,2g²₂+c_1,1g₁g₂+c_2,0g₁²+c_0,1g₂+c_1,0g₁+c_0,0, in whichc_3,0= 0 andc_0,2= 0. Without loss of generality, we may assume that c_0,2= 1. Then, (4.1) is expressed as

(4.2) φ=c_3,0fˆ³+ ˆg²+bfˆ+c, where ˆf =g₁+a, gˆ=g₂+c_1,1

2 g₁+c_0,1 2

and a, b, c ∈ k. Indeed, φ = c_3,0g₁³+ ˆg²+c_2,0g²₁ +c_1,0g₁ +c_0,0 for some c_2,0, c_1,0, c_0,0 ∈k. Then, we have (4.2) for a=c_0,2/(3c_3,0) and some b, c∈k.

Finally, putf = c_3,0fˆand g = c_3,0g. Clearly, degˆ f = degg₁ = 2l, degg = degg₂= 3l, andk[f, g, g₃] =k[x]. Moreover,

deg(f³+g²) = degc²_3,0(c_3,0fˆ³+ ˆg²)≤2l= degf

by (4.2), since the total degrees ofφ andbfˆ+c are at most 2l. Therefore,f andgsatisfy the conditions of Question 4.1.

It is worthwhile to mention that, if there exists a tame automorphism (h₁, h₂, h₃) with degh₁: degh₂: degh₃= 2 : 3 : 1 and deg(ch³₁+h²₂)≤degh₁ for somec∈k\ {0}, then we can construct a tame automorphism admitting a reduction of type II or III. On the other hand, ifp= 1, then (2.4) gives that degh₁ = 2m, degh₂= 3m, degh₃ =mand deg(c²h³₁+h²₂) = 2m+ 2. In this case, we havem= 2q+ 1 and

degh₁

deg(c²h³₁+h²₂)= 2m

2m+ 2 →1 (q→ ∞), although deg(c²h³₁+h²₂)>degh₁.

Assume that f, g ∈ k[x] are algebraically independent over k for which degf : degg=r:s, wherer, s∈Nwith 2≤r < sand gcd(r, s) = 1. Then, it easily follows from [9, Theorem 3] (see also [5, Theorem 2.1]) that

deg(f^s+g^r)>

degg ifr≥3

degf ifr= 2 and s≥5.

Hence, deg(f^s+g^r) ≤ degf is possible only if (r, s) = (2,3). We define f, g∈k[x₁, x₂] by

(4.3) f =−x^4l₁x^2(2m−1)₂ −2x^l₁x^m₂, g=x^6l₁x^3(2m−1)₂ + 3x^3l₁x^3m−1₂ +3 2x₂, wherel, m∈N. Then, degf : degg= 2 : 3. Moreover,fandgare algebraically independent overk, and

f³+g²=x^3l₁x^3m₂ +9 4x²₂,

sincef =−x^l₁x^m₂(x^α+ 2) andg=x₂(x^2α+ 3x^α+ 3/2), whereα= (3l,3m−2).

In particular, deg(f³+g²) = degf if l =m = 1, and deg(f³+g²) < degf otherwise. Ifkis of characteristicr >0, thenf =x^rl₁ andg=x₂+x^sl₁ satisfy f^s−g^r=−x^r₂for anyl, s∈N. Hence, deg(f^s−g^r)≤degf in this case.

Note: Instead of Question 4.1, the author first asked a question whether there existf, g∈k[x] with deg(f³+g²)≤degf which are algebraically inde- pendent overk. In answer to the question, Prof. Hiraku Kawanoue informed him of an example satisfying deg(f³+g²) = degf. The example (4.3) is a modification of Kawanoue’s example by the author.

Recently, the author [6] showed that no tame automorphism of k[x] for n= 3 admits a reduction of type IV.

References

[1] A. van den Essen, L. Makar-Limanov and R. Willems, Remarks on Shestakov-Umirbaev, Report 0414, Radboud University of Nijmegen, Toernooiveld, 6525 ED Nijmegen, The Netherlands, 2004.

[2] A. van den Essen, Polynomial automorphisms and the Jacobian conjecture, Progr.

Math., 190, Birkh¨auser, Basel, 2000.

[3] H. W. E. Jung, ¨Uber ganze birationale Transformationen der Ebene, J. Reine Angew.

Math.184(1942), 161–174.

[4] W. van der Kulk, On polynomial rings in two variables, Nieuw Arch. Wiskunde (3)1 (1953), 33–41.

[5] S. Kuroda, A generalization of the Shestakov-Umirbaev inequality, J. Math. Soc. Japan 60(2008), no. 2, 495–510.

[6] , Shestakov-Umirbaev reductions and Nagata’s conjecture on a polynomial au- tomorphism, arXiv:math.AC/0801.0117.

[7] L. Makar-Limanov and J.-T. Yu, Degree estimate for subalgebras generated by two elements, J. Eur. Math. Soc. (JEMS)10(2008), no. 2, 533–541.

[8] M. Nagata, On automorphism group of k[x, y], Department of Mathematics, Kyoto University, Lectures in Mathematics, No. 5, Kinokuniya Book Store, Tokyo, 1972.

[9] I. P. Shestakov and U. U. Umirbaev, Poisson brackets and two-generated subalgebras of rings of polynomials, J. Amer. Math. Soc.17(2004), no. 1, 181–196 (electronic).

[10] , The tame and the wild automorphisms of polynomial rings in three variables, J. Amer. Math. Soc.17(2004), no. 1, 197–227 (electronic).

[11] S. V´en´ereau, A parachute for the degree of a polynomial in algebraically independent ones, arXiv:math.AC/0704.1561.