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2 A new class of invertible polynomial maps Arno van den Essen and Engelbert Hubbers Abstract In this paper we present a new large class of polynomial maps F = X + H : A n A n (denition.) on every commutative ringa for which the Jacobian Conjecture is true. In particular H does not need to be homogeneous. We alsoshow that for all H in this class satisfying H() = the n-th iterate H H =. Introduction In [] it was shown that it suces to prove the Jacobian Conjecture for cubic homogeneous polynomial maps, i.e. maps of the form F = X + H : C n C n where H =(H ::: H n )andeach H i is either zero or a homogeneous polynomial map of degree three. In this case the Jacobian condition det(jf) C is equivalent tojh is nilpotent. (JF and JH are the Jacobian matrices of F and H.) So understanding nilpotent Jacobian matrices is crucial in the study of the Jacobian Conjecture. In [4] Wright showed that if n =3allJH where H is cubic homogeneous are linearly triangularizable. In [] the second author gave a complete description of all cubic homogeneous Jacobian matrices in case n = 4. They are no longer linearly triangularizable. However it turns out that the rows of the Jacobian matrices are linearly dependent over C (or equivalently that H H H 3 and H 4 are linearly dependent over C ). Already in [4] Dru_zkowski and Rusek conjectured that if H = ` ::: H n = ` n, where each `i is a linear form, then the nilpotence of JH implies the linear dependence of H ::: H n. The same question of linear dependence of H ::: H n was raised by Olech in [3] and Meisters in [], in case H ::: H n are cubic homogeneous. Then it was observed by the authors that the following more general dependence problem would imply the Jacobian Conjecture: does JH nilpotent (not necessarily homogeneous) and H() = imply that H ::: H n are linearly dependent over C? (Recently in [3] this dependence problem appeared as a Conjecture, the Nilpotent Conjecture, where it was shown that an armative answer would imply the Jacobian Conjecture.) Our aim was to investigate what consequences could be deduced assuming that the dependence question had an armativeanswer.

3 The class H n (A) The result is that for every commutativeringa we dened a large class, denoted H n (A), of polynomial maps H A[X ::: X n ] n such that the Jacobian matrix JH is nilpotent. It is shown that for all H H n (A) the map F := X + H is invertible with det(jf) = and that the inverse is of the form X + G with G H n (A). Furthermore we show that H n = H H =forallh H n (A), with H() =, a phenomenon rst observed by Meisters in []. Then in section 4 we consider the question if every H with JH nilpotent belongs to H n (A) (which, if true, would imply the Jacobian Conjecture). We show that the answer is yes if n = and A is a Q-algebra which is a U.F.D. (this result was already obtained by the second author in []), and that the answer is no for all n 3 and every domain A, whichisaq-algebra. This last result is based on recent counterexamples to the dependence problem for all n 3 obtained by the rst author in [7]. Finally in section 5 we show that all counterexamples found in [5], [8] and [] belong to H n (C ) (which is a subclass of H n (C )). In a subsequent paper [9] we make a detailed study of the class H n (A) and show thatallf of the form X + H with H H n (A) are stably tame. This implies that all cubic homogeneous maps in dimension 4, obtained in [], are stably tame. The class H n (A) Throughout this section A denotes an arbitrary commutative ring and let us denote by A[X] :=A[X ::: X n ] the polynomial ring in n-variables over A. Let F =(F ::: F n ) A[X] n. Such anf is called invertible over A or F ::: F n is called a coordinate system of A[X] ifa[f ::: F n ]=A[X ::: X n ]. In other words, if there exist G ::: G n A[X] suchthatx i = G i (F ::: F n ) for all i. It is an immediate consequence of the formal inverse function theorem that G =(G ::: G n ) is uniquely determined and satises F G = X. Now we come to the main denition of this paper. First we put N n (A) :=fh A[X] n : JH is nilpotentg: For each n N, n and each commutative ringa we are going to dene a set H n (A) A[X] n which will turn out to be a subset of the set N n (A). (cf. Theorem.3.) Denition. Put H (A) =A, for each commutative ringa and inductively for n and H A[X] n we dene that H H n (A) if and only if there exist T M n (A), c A n and ~ H H n; (A[X n ]) such that H =Adj(T ) jtx + c () where Adj(T ) denotes the adjoint matrix of T and jtx the `evaluation at the vector TX'.

4 Arno van den Essen and Engelbert Hubbers 3 Example. Let H = H H there exist T = t t a a such that H a = ;t H ;a t A[X X ]. Then H H (A) if and only if M (A), c c A and f(x ) H (A[X ]) = A[X ] f(x ) + t X + t X a X + a X In other words: if and only if H and H are of the form c c : H = a f(a X + a X )+c H = ;a f(a X + a X )+c for some a a c c A and f A[X ]. Remark.3 It was shown in [, Theorem 3.] that if A is a Q-algebra and a unique factorization domain then H (A) =N (A). We willgive a short proof in section 4 (theorem 4.3). However if A is a domain which is not a unique factorization domain it can happen that H (A)&N (A) (see section 3 below). Properties of H n (A) Lemma. Let H H n (A), r A, c A n. Then i). rh + c H n (A). ii). If S M n (A) then Adj(S)H jsx H n (A). iii). If ' : A S is a ringhomomorphism then '(H) H n (S) where '(H) is obtained by applying ' to the coecients of H. Proof. i) and iii) follow readily by induction on n. It therefore remains to prove ii). So let M n (A) andh H n (A). Then according to denition. we get Adj(S)H jsx = ) = Adj(S)Adj(T = Adj(TS) ~ H j(t S)X jtx jtx + ca and this is of the desired form. + c A jsx jsx + c

5 4 Properties of H n (A) Corollary. Let A[Y ] be the polynomial ring in one variable over A. Let a A. If H =(H ::: H n ) H n (A[Y ]), then H(Y = a) H n (A). Proof. Apply lemma..iii) to the substitution homomorphism ' : A[Y ] A sending Y to a. To simplify notations we abbreviate () by H = H[T c]orby ~ H = H[T ~ ]incase c =. As before we denote the Jacobian matrix with respect to X ::: X n of an element H A[X] n by JH or if confusion is possible by J n H. One then easily veries that J n H[T c] ~ (J = Adj(T ) n; H)jTX ~ T: () Theorem.3 For all rings A and n N, n we have H n (A) N n (A). Proof. Induction on n. The case n = is obvious. So let n and let H = ~ H[T c] for some T Mn (A), c A n and ~ H Hn; (A[X n ]). By the induction hypothesis we have that J n; ~ H is nilpotent. Hence so is (Jn; ~ H)jTX. From remark () it then follows that J n ~ H[T c]=jn H is nilpotent. Theorem.4 Let H H n (A). PutF := X + H. Then i). det(jf)=and ii). F is invertible over A. Furthermore F ; = X + G with G H n (A). Before we can prove this theorem we need some preliminaries. Therefore consider the polynomial ring A[T ij ]:=A[T ij i j n] in n indeterminates over A. Put T u =(T ij ), d := det(t u ) and consider the ring S := A[T ij ][d ; ]. We claim that A[T ij ] S. This follows immediately from Lemma.5 d is not a zero-divisor in A[T ij ]. Proof. We use induction on n the case n = is obvious, so let n. Write d n; := det(t ij ) i jn;.puta := A[T ni T in i n;] and B := A [T ij i j n ; ]. So B = A[T ij (i j) 6= (n n)]. If we nowdevelop d with respect to the n th column we get d = d n; T nn + b for some b B. In particular b does not contain any T nn. Now supposed is a zero-divisor in A[T ij ]. Then there exists an element 6= g A[T ij ] with dg =. Now develop g after powers of T nn,i.e. g = g m T m nn + + g with m, g m 6=andg i B for all i. Looking at the coecient oftnn m+ in the equation dg =wegetd n; g m =. But if we apply the induction hypothesis to the ring A we get that d n; is no zero-divisor in A [T ij i j n ; ]. Consequently g m =,acontradiction. Hence d is no zero-divisor in A[T ij ].

6 Arno van den Essen and Engelbert Hubbers 5 Proof of theorem.4. By induction on n. Again the case n = is clear. So let n and let H = ~ H[T c] for some T =(tij ) M n (A), c A n and ~ H H n; (A[X n ]). Since the transformation T c : x 7 x+c is bijectivewithinverse T ;c, we get that T ;c H = ~ H[T ] and hence we may assume that c = without loss of generality. i). Let S = A[T ij ][d ; ]asabove andputs := A[T ij ]. By lemma.5 we have that S is a subring of S. By lemma. we canview H ~ as an element of H n; (S [X n ]) H n; (S[X n ]). Now dene the universal H u := H[Tu ~ ] and F u := X + H u.notethat d H ~ H u = T ; u So we get jtux det(j n F u ) = det(j n (X + H u )) = ; u J n X + d H ~ jtux T u A = det(j n; (X + d ~ H))jT ux where X =(X ::: X n; ) t. However since d ~ H Hn; (S[X n ]), the last determinant equals by the induction hypothesis. So also det(jf u )=. Finally making the substitutions T ij t ij we obtain det(jf)=. ii). Since H ~ H n; (S [X n ]) and d S we get d H ~ H n; (S [X n ]). So by the induction hypothesis we get that X +d H ~ is invertible over S [X n ] with inverse X + G,where ~ G ~ Hn; (S [X n ]). The equation (X + d H) ~ (X + G)=X ~ implies that X + G ~ + d H(X ~ + G)=X ~ so G ~ = ;d H(X ~ + G). ~ Now observe that F u = T ; u (X + d ~ H ) jt ux and that its inverse over S is given by T ; u (X + ~ G ) jt ux = X + Adj(T u ) G ~ d jtux : Since d ~ G = ; ~ H(X + ~ G) belongs to S [X n ] n;, it follows that F u is in fact invertible over S. Asini)we conclude the proof by making the substitutions T ij t ij for all i j. The next result shows a remarkable nilpotence property of the elements of H n (A). For special examples this property was discovered by Meisters in [].

7 6 Properties of H n (A) Theorem.6 Let H H n (A). Then H n = H H A n for all n. In particular if H() =, then H n =. Before we can give the proof of this theorem we rst present two lemmas. Lemma.7 Let H =(H ::: H n ) A[X] n and assume H n = c n A. Now write for each i n ; H i = H i (X n = c n ).Put H := (H ::: H (n;) ) A[X ::: X n; ] n; : Then for all p H p = H p; (H ::: H n; ) c n : Proof. We use induction on p. First observe that for all i n ; H i (H ::: H n )=H i (H ::: H n; ) (3) which proves the case p =. Nowletp 3. Then by the induction hypothesis H p = H p; H = H p; (H ::: H n; ) c n H So by (3) we get H p = = H p; (H (H ::: H n ) ::: H (n;) (H ::: H n )) H p (H ::: H n; ) c n c n Proof of theorem.6. Let H = H[T c]forsomec ~ A n, T =(t ij ) M n (A) and H ~ H n; (A[X n ]). As in the proof of theorem.4 consider the ring S and H u S[X] n. It suces to prove that Hu n S n, for then Hu n S n \ A[T ij ][X] n = A[T ij ] n, so making the substitutions T ij 7 t ij gives H n A n as desired. Now observe that where a deduce that H u = T ; u = d ~ H d ~ H jtux + T ; (T u uc) =T ; u a jtux + T u c. Observe thata S and ~ H Hn; (S[X n ]). We H n u = T ; u a n jtux :

8 Arno van den Essen and Engelbert Hubbers 7 So it suces to show that H = H. H n a n S n. Therefore we may assume that C A A[X]n with H ~ = H. H n; C A H n;(a[x n ]) and H n A. Write c n instead of H n.soweneed to show that H n A n.we use induction on n. First write H i = H i +(X n ; c n )Hi as in lemma.7 above and put H =(H ::: H (n;) ). Then lemma.7gives H n = H n; (H ::: H n; ) c n Furthermore by corollary. we have H H n; (A). So if n = then H A. Finally if n 3 then the induction hypothesis, applied to H, gives that H n; A n;, whence H n A n by(4). (4) 3 A domain A with H (A) & N (A) Throughout this section A denotes the domain Z[X Y Z]=(X + YZ). Theorem 3. Let H = c X + c X, H = d X + d X in A[X X ] where c = X, c = Y, d = Z and d = ;X. Then i). H =(H H ) N (A). ii). H 6 H (A). iii). YH H (A). Proof. i). JH = X Y. Since Tr(JH) = and det(jh)=;(x + Y Z) =we Z ;X deduce that H N (A). ii). Suppose H H (A). Then by example. there exist a a A and f A[T ] with f() = such that H = a f(a X + a X ) H = ;a f(a X + a X ) Now since both deg(h ) = deg(h )=we deduce that f(t )= bt for some b A nfg. Consequently X = ba a and Y = ba. Let A A B Z[X Y Z]such that a = A, a = A and b = B. Thenmultiplying X by a and Y by a we obtain a X = a Y,i.e.A X ; A Y = c(x + YZ) for some

9 8 The class H n (A) c Z[X Y Z]. Consequently X(A ; cx) =Y (A + cz). So A ; cx = dy for some d Z[X Y Z] and hence A + cz = dx. Summarizing A = dx ; cz and A = cx + dy with c d Z[X Y Z]. Consequently the equation X = ba a,i.e. X ; BA A (X + YZ), implies X (X Y Z), a contradiction. So H 6 H (A). Y XX iii). YH = + Y X. Since Y Z = ;X,we see that we can take Y ZX ; Y XX a = X, a = Y and f(t )=T to get the desired form of example.. 4 The class H n (A) In the previous section we saw that there exists a commutative domain A such that H A[X], H 6 H (A) butrh H (A), for some 6= r A. This leads us to the following denition, where we take the closure of H n (A) with respect to this property. Throughout this section: A is a commutative domain. Denition 4. First dene H (A) =A. NowletnandH A[X] n. Then H H n (A) if and only if there exist 6= r A, T M n (A), c A n and H n; (A[X n ]) such that rh = Adj(T ) As in section we have the following result. jtx Theorem 4. i). H n (A) N n (A), for all n. ii). Let H H n (A) and put F := X + H. Then det(jf)=and F is invertible with inverse F ; equal to X + G where G H n (A). iii). Let H H n (A). Then H n A n, for all n Proof. (Sketch.) The proofs of these theorems are obtained from the proofs of theorems.3,.4 and.6 given in section by replacing H n (A) by H n (A) and using localizations. Finally we consider the question whether H n (A) =N n (A)? As already observed earlier, it was proved in [] that in case A is a U.F.D., then H (A) =N (A), hence H (A) =N (A). Since the paper [] is not easy available we give a short proof of this result. + c:

10 Arno van den Essen and Engelbert Hubbers 9 Theorem 4.3 ([], 994) Let A be a U.F.D., then H (A) =N (A). Proof. i). First assume that A = k is a eld. Then the result is proved in []. ii). Now let A be a U.F.D. and let H =(H H ) N (A). Then H N (K) where K is the quotient eld of A. So by i) there exist g(t ) K[T ] with g() = and d d K such that H = g( X + X )+d H = ; g( X + X )+d (see example.). So clearing denuminators we get: there exist a A, a 6=, f(t ) A[T ] with f() = and c c A such that ah = g( X + X )+c ah = ; g( X + X )+c (5) Substituting X = X =in(5)we obtain that c = a ~c and c = a ~c for some ~c ~c A. So replacing H i by H i ; ~c i wemay assume that c = c =. iii). Nowweshowthatwemay assume that gcd( ) = : therefore let = ~ d, = ~ d where d = gcd( ). So gcd( ~ ~ )=and i f( X + X )= ~ i df(d( ~ X +~ X )). Hence if we put ~ f(t )=df(dt )weget i f( X + X )= ~ f(~ X +~ X ): iv). Consequently we may assume that gcd( )=. Write f = P N i= f i T i, with f i A. From (5) we see that wemay assume that gcd(a f ::: f N )=. Claim: a is a unit in A (and hence we are done). Suppose that p is a prime factor of a. Then (5) implies that p divides f( X + X ) (since gcd( ) = ). So in particular p divides both f( X ) and f( X ), so p divides f i i and f i i for all i and hence p divides f i for all i which contradicts gcd(a f ::: f N ) =. So a is a unit. In the remainder of this section we will show that such aresultisnolonger true if n 3. More precisely we have: Theorem 4.4 Let A be any Q-algebra. Then H n (A)&N n (A), for all n 3. To prove this result we need the following lemma. Lemma 4.5 Let A be a domain, n and H H n (A) with H() =. Then there exist ::: n A, not all zero, such that H + + n H n =.

11 Final remarks Proof. If H = we are done, so let H 6=. Then there exist 6= r A, T M n (A), c A n and H ~ Hn; (A[X n ]) such that rh =Adj(T ) + c So, multiplying by T we get rth = det(t ) jtx jtx + Tc (6) i). If det(t ) = it follows from (6) that rth = Tc. Since H() = and A is a domain we deduce that TH =. Since T 6= (otherwise H = ) there exists a non-zero row, say thei-th, whence t i H + + t in H n =, as desired. ii). If det(t ) 6=, then equating the n-th components of the vectors in (6) we get r(th) n =(Tc) n. Since H() =, we get (Tc) n =,so(th) n = i.e. t n H ++t nn H n =.Obviously t nj 6= for some j (otherwise det(t ) = ). Now let n 3 and A be a Q-algebra. It was shown in [7] that the following H =(H ::: H n ) A[X] n belongs to N n (A): let (X ) := X n; H := X ; (X ) H i := X i+ + (;)i (i ; ) (i;) (X ; (X )) i; for i n ; H n := (;)n (n ; ) (n;) (X ; (X )) n; : Proof of theorem 4.4. Let n 3andH be as dened above. Then, as observed H N n (A). However if H H n (A), then by lemma 4.5 there exist ::: n A, not all zero, such that H + + n H n =. It follows readily that = = n = (look at the monomials X 3 X 4 ::: X n respectively). So H + n H n =, which easily implies that also = n =(n 3), contradiction. So H 6 H n (A). In particular this proof shows that the dependence problem, or the Nilpotent Conjecture from [3], is false. (See [7] for more details.) 5 Final remarks To conclude this paper we explain the counterexamples found in [5], [8] and []. They all belong to H n (C ). To see this consider example.. First take A = C [X 3 X 4 ], a = X 3, a = X 4, c = c =andf(t )=T. Then (H H )=(X 4 (X 3 X + X 4 X ) ;X 3 (X 3 X + X 4 X ))

12 References belongs to H (C [X 4 ][X 3 ]). Consequently (H H ) belongs to H 3 (C [X 4 ]) and hence (H H X 3) belongs to H 4 3(C [X 4 ]). This implies that (H H X 3 4 ) belongs to H 4 (C ) and hence that H := (H H X 3 ::: ) belongs to H 4 n(c ) for all n 4. Then X + H is exactly the counterexample to Meisters' Linearization Conjecture given in [5]. Similarly, taking f(t )=T we nd the counterexamples to the Deng-Meisters- Zampieri Conjecture and the Discrete Markus-Yamabe problem given in [8]. Finally the counterexamples to the Markus-Yamabe Conjecture and the Discrete Markus-Yamabe problem in []: take A = C [X 3 ], a =,a = X 3, c = c = and f(t )=T in example.. Then (H H )=(X 3 (X + X 3 X ) ;(X + X 3 X ) ) belongs to H (C [X 3 ]), hence (H H ) belongs to H 3 (C ) and consequently the n- dimensional map (H H ::: ) belongs to H n (C )foralln3. Then ;X + H resp. X + H are exactly the counterexamples to the Markus-Yamabe Conjecture resp. the Discrete Markus-Yamabe problem given in []. References [] H. Bass, E. Connell, and D. Wright, The Jacobian Conjecture: Reduction of Degree and Formal Expansion of the Inverse, Bulletin of the American Mathematical Society 7 (98), 87{33. [] A. Cima, A.R.P. van den Essen, A. Gasull, E.-M.G.M. Hubbers, and F. Ma~nosas, A polynomial counterexample to the Markus-Yamabe Conjecture, Report 955, University of Nijmegen, Toernooiveld, 655 ED Nijmegen, The Netherlands, 995. [3] A. Cima, A. Gasull, and F. Ma~nosas, The discrete Markus-Yamabe problem, Prepublicacions Num. 6/995, Universitat Autonoma de Barcelona, 995. [4] L.M. Dru_zkowski and K. Rusek, The formal inverse and the Jacobian conjecture, Annales Polonici Mathematici 46 (985), 85{9. [5] A.R.P. van den Essen, A counterexample to a conjecture of Meisters, In Automorphisms of Ane Spaces [6], Proceedings of the conference `Invertible Polynomial maps', pp. 3{34. [6] A.R.P. van den Essen (ed.), Automorphisms of Ane Spaces, Curacao, Caribbean Mathematics Foundation, Kluwer Academic Publishers, July 4{ 8 994, 995, Proceedings of the conference `Invertible Polynomial maps'. [7] A.R.P. van den Essen, Nilpotent Jacobian matrices with independent rows, Report 963, University of Nijmegen, Toernooiveld, 655 ED Nijmegen, The Netherlands, 996.

13 References [8] A.R.P. van den Essen and E.-M.G.M. Hubbers, Chaotic Polynomial Automorphisms counterexamples to several conjectures, Report 9549, University of Nijmegen, Toernooiveld, 655 ED Nijmegen, The Netherlands, 995, To appear in Advances of Applied Mathematics. [9] A.R.P. van den Essen and E.-M.G.M. Hubbers, D n (A) for polynomial automorphisms, Report 964, University of Nijmegen, Toernooiveld, 655 ED Nijmegen, The Netherlands, 996. [] E.-M.G.M. Hubbers, The Jacobian Conjecture: Cubic Homogeneous Maps in Dimension Four, Master's thesis, University of Nijmegen, Toernooiveld, 655 ED Nijmegen, The Netherlands, February 7 994, directed by A.R.P. van den Essen. [] G.H. Meisters, Inverting polynomial maps of n-space by solving dierential equations, Delay and Dierential Equations, Proceedings in Honor of George Seifert, Ames, Iowa, Oct. 8{9, 99 (A.M. Fink, R.K. Miller, and W. Kliemann, eds.), World Sci. Pub. Singapore Teaneck NJ London Hong Kong, 99, ISBN 98{{89-X, pp. 7{66. [] G.H. Meisters, Polyomorphisms conjugate to Dilations, In van den Essen [6], Proceedings of the conference `Invertible Polynomial maps', pp. 67{88. [3] C. Olech, On Markus-Yamabe stability conjecture, International Meeting on Ordinary Dierential Equations and Their Applications, University of Florence, Florence, Italy, September {4 993, 995, pp. 7{37. [4] D. Wright, The Jacobian Conjecture: linear triangularization for cubics in dimension three, Linear and Multilinear Algebra 34 (993), 85{97.