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On some nice polynomial automorphisms ˙ ELZBIETA ADAMUS 6 Faculty of Applied Mathematics, 1 AGH University of Science and Technology 0 2 al. Mickiewicza 30, 30-059 Kraków, Poland n e-mail: [email protected] a J PAWEL BOGDAN 6 Faculty of Mathematics and Computer Science, ] G Jagiellonian University A ul. L ojasiewicza 6, 30-348 Kraków, Poland . e-mail: [email protected] h t a m TERESA CRESPO Departament d’A`lgebra i Geometria, [ Universitat de Barcelona 1 v Gran Via de les Corts Catalanes 585, 08007 Barcelona, Spain 0 e-mail: [email protected] 1 2 1 ZBIGNIEW HAJTO 0 Faculty of Mathematics and Computer Science, . 1 Jagiellonian University 0 6 ul. L ojasiewicza 6, 30-348 Kraków, Poland 1 e-mail: [email protected] : v i X r Abstract a Given a polynomial endomorphism F of the n-dimensional affine space An K over a field K, we define a sequence of polynomial endomorphisms (Pk)k∈N of AnK associated to F. We call F nice if there exists an integer m such that P = 0 for m some m. Then F is invertible and F−1 may be computed in terms of the endomorphisms P ,k = 0,...,m−1. In this paper we study the class of nice polyno- k mial automorphisms and obtain that this class is large and includes triangulable automorphisms and all linear cubic homogeneous polynomial automorphisms of nilpotence index up to 3. We prove as well that the nicety property is invariant under linear conjugation and determine that seven of the eight forms in Hubbers’ 1 classification of cubic homogeneous automorphisms in dimension 4 are nice and moreover the eighth one is a composition of nice maps. 1 Introduction The Jacobian Conjecture originated in the question raised by Keller in [7] on the invertibility of polynomial maps with Jacobian determinant equal to 1. The question is still open in spite of the efforts of many mathematicians. We recall in the sequel the precise statement of the Jacobian Conjecture and other results we shall use. We refer to [5] for a detailed account of the research on the Jacobian Conjecture and related topics. Let K be a field and K[X] = K[X ,...,X ] the polynomial ring in the variables 1 n X ,...,X over K. A polynomial map is a map F = (F ,...,F ) : Kn → Kn of the 1 n 1 n form (X ,...,X ) 7→ (F (X ,...,X ),...,F (X ,...,X )), 1 n 1 1 n n 1 n where F ∈ K[X],1 ≤ i ≤ n. The polynomial map F is invertible if there exists a i polynomial map G = (G ,...,G ) : Kn → Kn such that X = G (F ,...,F ),1 ≤ i ≤ 1 n i i 1 n n. We shall call F a Keller map if the Jacobian matrix ∂F i J = (cid:18)∂Xj(cid:19)11≤≤ji≤≤nn has determinant equal to a nonzero constant. Clearly an invertible polynomial map F is a Keller map. Jacobian Conjecture. Let K be a field of characteristic zero. A Keller map F : Kn → Kn is invertible. In the sequel, K will always denote a field of characteristic 0. For F = (F ,...,F ) ∈ 1 n K[X]n, we define the degree of F as degF = max{degF : 1 ≤ i ≤ n}. It is known that i if F is a polynomial automorphism, then degF−1 ≤ (degF)n−1 (see [2] or [8]). The Jacobian conjecture for quadratic maps was proved by Wang in [9]. We state now the reduction of the Jacobian conjecture to the case of maps of third degree (see [2], [10], [3] and [4]). Proposition 1. a) (Bass-Connell-Wright-Yagzhev) Given a Keller map F : Kn → Kn, there exists a KellermapF : KN → KN, N ≥ n ofthe formF = Id+H, where H(X) e e 2 is a cubic homogeneous map and having the following property: if F is invertible, then F is invertible too. e b) (Druz˙kowski) The cubic part H may be chosen of the form N N ( a X )3,...,( a X )3 1j j Nj j ! j=1 j=1 X X and with the matrix A = (a ) satisfying A2 = 0. ij 1≤i≤N 1≤j≤N We recall now the sufficient condition for the invertibility of a polynomial map we gave in [1] Corollary 5. Proposition 2. Given a polynomial map F : Kn → Kn, we define for i = 1,...,n the following sequence of polynomials in K[X], Pi(X ,...,X ) = X , 0 1 n i Pi(X ,...,X ) = F −X , 1 1 n i i and, assuming Pi is defined, k−1 Pi(X ,...,X ) = Pi (F ,...,F )−Pi (X ,...,X ). k 1 n k−1 1 n k−1 1 n If there exists an integer m such that Pi = 0, for all i = 1,...,n, then F is invertible m and the inverse G of F is given by m−1 G (Y ,Y ,...,Y ) = (−1)lPi(Y ,Y ,...,Y ), 1 ≤ i ≤ n. i 1 2 n l 1 2 n l=0 X We shall say that the polynomial map F is nice if there exists an integer m such that Pi = 0, for all i = 1,...,n. The aim of this paper is to prove several properties of nice m polynomial automorphisms, namely that nicety is invariant under linear conjugation, that all triangulable polynomial automorphisms are nice and that the inverse of a nice polynomial automorphism is nice. We also prove that if f and F form a Gorni-Zampieri pairing and F is nice, then f is nice and that all linear cubic polynomial maps of nilpotence index ≤ 3 are nice. 3 2 Triangulable polynomial automorphisms Theorem 3. Let F : Kn → Kn be a polynomial map of the form F (X ,...,X ) = X +H (X ,...,X ) 1 1 n 1 1 1 n F (X ,...,X ) = X +H (X ,...,X )  2 1 n 2 2 1 n .  ..   F (X ,...,X ) = X +H (X ,...,X ), n 1 n n n 1 n  where Hi(X1,...,Xn) is a polynomial in X1,...,Xn of lower degree ≥ 2, 1 ≤ i ≤ n. Let T ∈ GL(n,K) and F := T−1 ◦F ◦T. If F is nice, then F is nice. Proof. We have clearly F := T−1 ◦ F ◦ T = T−1 ◦ (I + H) ◦ T = I + H, where e e H = T−1 ◦ H ◦ T has lower degree ≥ 2. Let us consider the polynomial sequences Pi k and Pi defined by e e k e Pi(X ,...,X ) = X , e 0 1 n i Pi(X ,...,X ) = F −X , 1 1 n i i . . . Pi(X ,...,X ) = Pi (F ,...,F )−Pi (X ,...,X ) k 1 n k−1 1 n k−1 1 n . . . and Pi(X ,...,X ) = X , 0 1 n i Pi(X ,...,X ) = F −X , 1 1 n i i e .. . e e Pi(X ,...,X ) = Pi (F ,...,F )−Pi (X ,...,X ) k 1 n k−1 1 n k−1 1 n . . . e e e e e and the polynomial maps P := (P1,...,Pn) and P := (P1,...,Pn). We shall prove k k k k k k by induction on k e e e P = T−1 ◦P ◦T, for all integer k ≥ 1. k k Then, if P = 0 for some integer m, also P = 0. For k = 1, we have P = H = m m 1 e T−1 ◦H ◦T = T−1 ◦P ◦T. Let us assume P = T−1 ◦P ◦T. We have then 1 k k e e e e 4 P = P ◦F −P k+1 k k = (T−1 ◦P ◦T)◦(T−1 ◦F ◦T)−T−1 ◦P ◦T k k e = Te−1 ◦eP ◦eF ◦T −T−1 ◦P ◦T k k = T−1 ◦(P ◦F −P )◦T k k = T−1 ◦P ◦T. k+1 ✷ Theorem 4. Let F be a triangular polynomial map in dimension n, i.e. a polynomial map of the form F (X ,...,X ) = X +H (X ,...,X ) 1 1 n 1 1 2 n F (X ,...,X ) = X +H (X ,...,X )  2 1 n 2 2 3 n .  ..    Fn−1(X1,...,Xn) = Xn−1 +Hn−1(Xn) F (X ,...,X ) = X . n 1 n n    Then F is nice.   Proof. Let H ∈ K[X ,...,X ] have degree d in X and define 2 n 2 P (X ,...,X ) = H, 1 2 n P (X ,...,X ) = P (F ,...,F )−P (X ,...,X ). k 2 n k−1 2 n k−1 2 n Since the polynomials H , for j > 1, have degree 0 in X , we obtain that the degree in j 2 X of P is d−k+1, hence the degree in X of P is 0. Taking H = H , we obtain that 2 k 2 d+1 1 P1 is a polynomial in X ,...,X . If P1 has degree d′ in X , taking H = P1 and d+1 3 n d+1 3 d+1 the polynomial map (F ,...,F ), we obtain that P1 is a polynomial in X ,...,X . 2 n d+d′+1 4 n By iteration, we obtain P1 = 0, for some m, hence F is nice. ✷ m Theorems 3 and 4 together give the following corollary. Corollary 5. A triangulable polynomial automorphism is nice. Example 6. We consider the Nagata automorphism defined by F = X −2pX −p2X 1 1 2 3 F = X +pX 2 2 3  F = X  3 3 where p = X X +X2. It is known that the Nagata automorphism is not tame. However 1 3 2  it is nice. Indeed, by calculation, we obtain P1 = 0,P2 = 0. 3 2 5 3 Products of nice automorphisms Theorem 7. If F is a nice polynomial automorphism of Kn, then F−1 is nice. Proof. Let F(X) be nice and G(Y) be its inverse. Let the polynomials Pi be defined by k Pi(X ,...,X ) = X , 0 1 n i Pi(X ,...,X ) = F −X , 1 1 n i i Pi(X ,...,X ) = Pi (F ,...,F )−Pi (X ,...,X ) k 1 n k−1 1 n k−1 1 n and assume Pi = 0. Then by Proposition 2, we have m m−1 X = (−1)lPi(F) i l l=0 X hence m−1 G (Y) = (−1)lPi(Y). (1) i l l=0 X Let now the polynomials Qi be defined by k Qi(Y ,...,Y ) = Y , 0 1 n i Qi(Y ,...,Y ) = G −Y , 1 1 n i i Qi(Y ,...,Y ) = Qi (G ,...,G )−Qi (Y ,...,Y ). k 1 n k−1 1 n k−1 1 n We shall compute the polynomials Qi using (1). We have k Qi(Y) = G −Y = m−1(−1)lPi(Y) 1 i i l=1 l Qi(Y) = Qi(G)−Qi(Y) = m−1(−1)l(Pi(G)−Pi(Y)) 2 1 P1 l=1 l l Now Pi(X) = Pi (F)−Pi (X) implies Pi(G) = Pi (Y)−Pi (G), therefore k k−1 k−1 Pk k−1 k−1 m−1 Qi(Y) = (−1)lPi(G) 2 l l=2 X and, by induction, we obtain m−1 Qi(Y) = (−1)lPi(Gk−1), k l l=k X where Gk−1 denotes composition of G with itself k − 1 times. We conclude then that Qi (Y) = 0. ✷ m 6 Remark 8. The set of nice polynomial automorphisms is not a subgroup of the group of polynomial automorphisms as can be shown by considering the following example. Let F and F be the polynomial automorphisms of K2 defined by X 7→ (X +X3,X ) and 1 2 1 2 2 X 7→ (X ,X +X2) respectively. It is easy to check that F and F are nice. However 1 2 1 1 2 F := F ◦F is not nice. We have F(X) = (X +(X2+X )3,X +X2). Let Q denote 1 2 1 1 2 2 1 k the term of lower degree of P1. Taking into account that Q is the term of lower k k+1 degree of ∂Q ∂Q k(X2 +X )3 + kX2 ∂X 1 2 ∂X 1 1 2 and that, if Q is a homogeneous polynomial, the term of lower degree of ∂Q ∂Q (X2 +X )3 + X2 is ∂X 1 2 ∂X 1 1 2 ∂Q ∂Q X2 if 6= 0 ∂X 1 ∂X 2 2 ∂Q ∂Q X3 if = 0 ∂X 2 ∂X 1 2 we obtain the following recursive formula for Q , which shows P1 6= 0 for all k. Let k k us write k = 4m+ r, with r = 1,2,3,4 and define µ(r) by µ(1) = 1,µ(2) = 3,µ(3) = µ(4) = 6. Then m Q = µ(r)6m (6+5(j −1))X2r−2+5mX4−r. k 1 2 j=0 Y Remark 9 (Hubbers’ classification). Hubbers [6] classified up to conjugation all cubic homogeneous polynomial maps in dimension 4. He obtained eight conjugation classes. We have checked that all but the last are nice automorphisms. Let us now consider the eighth class in Hubbers classification. It is represented by F = X 1 1 F = X −X3/3  2 2 1 F = X −X2X −e X X2 +g X X X −k X3 +m X2X +g2X2X .  3 3 1 2 3 1 2 4 1 2 3 3 2 4 2 3 4 2 4  F = X −X2X −e X X2 −2m X X X /g −g X X X −k X3  4 4 1 3 4 1 2 4 1 2 3 4 4 1 2 4 4 2 −m2X2X /g2−m X2X 4 2 3 4 4 2 4     We have F = G◦H, where H and G defined by 7 H = X G = X 1 1 1 1 H = X G = F  2 2 and  2 2 H = F G = X  3 3  3 3     H = F G = X 4 4 4 4 are both nice.     Remark 10. It is known that all cubic homogeneous polynomial automorphisms in dimension ≤ 3 are triangulable (see [5], Proposition 7.1.1), hence they are nice. We obtain thenthat thesubset ofnice cubic homogeneous polynomial automorphisms indimension ≤ 4 generate the whole set of cubic homogeneous polynomial automorphisms. Question 11. The above remark suggest the following question: Is every polynomial automorphism a product of nice ones? 4 Gorni-Zampieri pairing We shall now prove that if f and F form a Gorni-Zampieri pair (see [5] §6.4) and F is nice then f is nice. Let f = Kn → Kn be a cubic homogeneous map and F : KN → KN a cubic linear map, with N > n. Let x = (x ,...,x ) ∈ Kn and X = (X ,...,X ) ∈ KN. 1 n 1 N For f = (f ,...,f ), where f ∈ K[x] we define p (x) = (p1(x),...,pn(x))T in the 1 n i t t t following way pi(x) = x 0 i pi(x) = f (x)−x 1 i i pi(x) = pi(f(x))−pi(x) 2 1 1 . . . pi (x) = pi(f(x))−pi(x) t+1 t t For F = (F ,...,F ), where F ∈ K[X] we define P (X) = (P1(X),...,Pn(X))T in 1 N i t t t the following way Pi(X) = X 0 i Pi(X) = F (X)−X 1 i i Pi(X) = Pi(F(x))−Pi(X) 2 1 1 . . . Pi (X) = Pi(F(X))−Pi(X) t+1 t t 8 Lemma 12. Let f and F be as above. Assume that f and F are paired through the matrices B ∈ M (K) and C ∈ M (K). Then p and P are paired through the n×N N×n t t same matrices B and C, for every t ≥ 0. Proof. Our main assumption is that F and f arepaired, so in particular f(x) = BF(X), where X = Cx. Note that since BC = I , then X = Cx implies x = BX. n The proof goes by induction on t. For t = 0 we have p (x) = x = BX and P (X) = 0 0 X, so p (x) = BP (X). Similarly for t = 1 we have 0 0 p (x) = p (f(x))−p (x) = f(x)−x, 1 0 0 P (X) = P (F(X))−P (X) = F(X)−X. 1 0 0 So p (x) = f(x)−x = BF(X)−BX = B(F(X)−X) = BP (X). 1 1 Let us assume that for a given t we have p (x) = BP (X) (2) t t or equivalently p (BX) = BP (X) ∀X ∈ KN. (3) t t We will prove that p (x)−BP (X) = 0. t+1 t+1 We have that p (x)−BP (X) = p (f(x))−p (x)−[BP (F(X))−BP (X)] =(2) t+1 t+1 t t t t = p (f(x))−BP (F(X)) = p (BF(X))−BP (F(X)) =(3) 0 t t t t ✷ Corollary 13. Let f and F be as above. If f and F are paired, then if F is nice, f is also nice. Proof. If f and F are paired through the matrices B and C, Lemma 12 gives p (x) = t BP (X), for all t ≥ 0. Then if P = 0, for some integer m, we have p = 0. ✷ t m m 9 Example 14. In [1] Example 7, we proved that the polynomial automorphism f of K4 defined by f = x +px 1 1 4 f = x −px  2 2 3 , f = x +x3  3 3 4  f = x 4 4 where p := x1x3 + x2x4, is not nice. We consider the polynomial automorphism F of K16 defined by F = X +(L +X +X )3 1 1 1 15 16 F = X +(L −X −X )3 2 2 1 15 16  F = X +(L +X −X )3  3 3 1 15 16   F = X +(L −X +X )3  4 4 1 15 16   F = X +(L +X )3  5 5 2 16   F = X +(L −X )3  6 6 2 16   F = X +L3  7 7 2  F = X +(L +X )3  8 8 1 15   F = X +(L −X )3  9 9 1 15   F = X +L3 10 10 1 F = X +(L +X +X )3  11 11 2 15 16  F = X +(L −X −X )3  12 12 2 15 16  F = X +(L +X −X )3  13 13 2 15 16  F = X +(L −X +X )3  14 14 2 15 16  F = X +X3  15 15 16   F = X  16 16   where   L = (X +X −X −X +4X +4X −8X )/24 1 1 2 3 4 5 6 7 L = (−4X −4X +8X −X −X +X +X )/24 2 8 9 10 11 12 13 14 which is paired with f through the matrices 1 1 −1 −1 4 4 −8 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 −4 −4 8 −1 −1 1 1 0 0 B =   24 0 0 0 0 0 0 0 0 0 0 0 0 0 0 24 0  0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 24      10