On the lifting of the Nagata automorphism
aa r X i v : . [ m a t h . A C ] F e b ON THE LIFTING OF THE NAGATAAUTOMORPHISM
ALEXEI BELOV-KANEL AND JIE-TAI YU
Abstract . It is proved that the Nagata automorphism (Nagata coor-dinates, respectively) of the polynomial algebra F [ x, y, z ] over a field F cannot be lifted to a z -automorphism ( z -coordinate, respectively)of the free associative algebra F h x, y, z i . The proof is based on thefollowing two new results which have their own interests: degree esti-mate of Q ∗ F F h x , . . . , x n i and tameness of the automorphism groupAut Q ( Q ∗ F F h x, y i ).1. Introduction and main results
The long-standing famous Nagata conjecture for characteristic 0 wasproved by Shestakov and Umirbaev [12, 13], and a strong version of theNagata conjecture was proved by Umirbaev and Yu [14]. That is, theNagata automorphism ( x − y ( y + xz ) − ( y + xz ) z, y + ( y + xz ) z, z )(Nagata coordinates x − y ( y + xz ) − ( y + xz ) z and y + ( y + xz ) z re-spectively) is (are) wild. In [11, 14], a stronger question (which impliesthe Nagata conjecture and the strong Nagata conjecture) was raised:whether the Nagata automorphism (coordinates) of the polynomial al-gebra F [ x, y, z ] can be lifted to an automorphism (coordinates) of thefree associative F h x, y, z i over a field F ? We can also formulate The General Lifting Problem.
Let φ = ( f , . . . , f n ) be an auto-morphism of the polynomial algebra F [ x , . . . , x n ] over a field F . Doesthere exists an F -automorphism φ ′ = ( f ′ , . . . , f ′ n ) of the free associativealgebra F h x , . . . , x n i such that each f i is the abelianization of f ′ i ? Mathematics Subject Classification.
Primary 13S10, 16S10. Secondary13F20, 13W20, 14R10, 16W20, 16Z05.
Key words and phrases.
Automorphisms, coordinates, degree estimate, subal-gebras generated by two elements, polynomial algebras, free associative algebras,commutators, Jacobian, lifting problem.The research of Jie-Tai Yu was partially supported by an RGC-GRF Grant.
For n = 2, the answer of the above problem is positive, as due to Jung[5] and van der Kulk [6] every automorphism of F [ x, y ] is composationof linear and elementary automorphisms which are liftable to automor-phisms of F h x, y i . Moreover, Makar-Limanov [9] and Czerniakiewicz[7] proved independently that Aut( F h x, y i ) is actually isomorphic toAut( F [ x, y ]), which implies that the lifting is unique.In this paper we prove the following new result, which partially answersthe question raised in [14] negatively. The result can be viewed as thefirst step to attack the general lifting problem. In a forthcoming paper[1], we will deal the general lifting problem. Theorem 1.1.
Let ( f, g ) be an wild F [ z ] -automorphism of F [ x, y, z ] = F [ z ][ x, y ] . Then ( f, g, z ) , as an F -automorphism of F [ x, y, z ] , cannotbe lifted to an automorphism of F h x, y, z i fixing z . The crucial step to prove Theorem 1.1 is the following
Theorem 1.2.
Let ( f, g ) be a wild F [ z ] -automorphism of F [ x, y, z ] = F [ z ][ x, y ] , which can be effectively obtained as the product of the canon-ical sequence of uniquely determined alternative operations (elementary F ( z ) -automorphisms), and the sequence contains an elementary F ( z ) -automorphism of the type ( x, y + z − k x l + . . . ) or ( x + z − k y l + . . . , y ) where l > . Then ( f, g, z ) , as an F -automorphism of F [ x, y, z ] , cannotbe lifted to an automorphism of F h x, y, z i fixing z . Corollary 1.3.
The Nagata automorphism cannot be lifted to an au-tomorphism of F h x, y, z i fixing z . Corollary 1.4.
Let ( f, g ) be a wild F [ z ] -automorphism of F [ x, y, z ] = F [ z ][ x, y ] . Then neither f nor g can be lifted to a z -coordinate of F h x, y, z i . In particular, the Nagata coordinates x − y ( y + xz ) − ( y + xz ) z and y + ( y + xz ) z cannot be lifted to any z -coordinate of F h x, y, z i .Proof. Suppose ( f, h ) is an F [ z ]-automorphism, then obviously ( f, h ) isthe product of ( f, g ) and an elementary F [ z ]-automorphism of the type( x, h ). Therefore ( f, h ) is liftable if and only if ( f, g ) is liftable. Henceany F [ z ]-automorphism of the type ( f, h ) is not liftable. Therefore f cannot be lifted to a z -coordinate of F h x, y, z i . Same for g . (cid:3) N THE LIFTING OF THE NAGATA AUTOMORPHISM 3
Crucial to the proof of Theorem 1.2 is the following new result, that im-plies that the automorphism group Aut Q ( Q ∗ F F h x, y i ) is tame, whichhas its own interests. Theorem 1.5 (on degree increasing process) . Let Q be an extensionfield over a field F . A Q -automorphism of Q ∗ F F h x, y i can be effectivelyobtained as the product of a sequence of uniquely determined alternatingoperations (elementary automorphisms) of the following types: • x → x, y → ryr ′ + P r xr x · · · r k xr k +1 , • x → qxq ′ + q P yq y · · · q k yq k +1 , y → y where r, q, r j , q j ∈ Q . The following new result of degree estimate is also essential to the proofof Theorem 1.2.
Theorem 1.6 (Degree estimate) . Let Q be an extension field of a field F . Let A = Q ∗ F F h x , . . . , x n i be a co-product of Q and the freeassociative algebra F h x , . . . , x n i over F . Suppose f, g ∈ A are alge-braically independent over Q , f + and g + are algebraically independentover Q ; or f + and g + are algebraically dependent, and neither f + is Q -proportional to a power g + , nor g + is Q -proportional to a power f + .Let P ∈ Q ∗ F F h x, y i\ Q . Then deg( P ( f, g )) ≥ deg([ f, g ])deg( f g ) w deg( f ) , deg( g ) ( P ) , where the degree is the usual homogeneous degree with respect to x , . . . , x n and w r,s is the weight degree with respect to r, s. Note that u is proportional to v for u, v ∈ Q ∗ F F h x , . . . , x n i meansthat there exist p , . . . , p m ; q , . . . , q m ∈ Q such that u = Σ mi =1 p i vq i (itis important that ‘proportional’ is not reflexive, i.e. u is proportionalto v does not imply v is proportional to u ), and that f + is the highesthomogeneous form of f . Remark 1.7.
Theorem 1.6 is still valid for an arbitrary division ring Q over a field F . The proof is almost the same. When Q = F ( z )then the result can be directly deduced from the degree estimate in[10, 8] via substitution ψ : x → P ( z ) xP ( z ); y → R ( z ) yR ( z ) forappropriate P i , R i ∈ F [ z ]. For any element τ in Q ∗ F F h x, y i there A.BELOV-KANEL AND JIE-TAI YU exist P i , R i ∈ F [ z ] such that ψ ( τ ) ∈ F h x, y, z i . In the sequel we onlyuse this special case regarding the lifting problem.2. Proofs
Proof of Theorem 1.6
Similar to the proof of the main result of Liand Yu [8], where Bergman’s Lemma [3, 4] on centralizers is used. Seealso Makar-Limanov and Yu [10] for the special case of characteristic0, where Bergman’s Lemma on radical [3, 4] is used. (cid:3)
Proof of Theorem 1.5.
Let φ = ( f, g ) be a Q -automorphism inAut Q ( Q ∗ F F h x, y i ) which is not linear, namely,deg x,y ( f ) + deg x,y ( g ) ≥ . By Theorem 1.5, we obtain that either a power of f + is proportionalto g + , or a power of g + is proportional to f + . Now the proof is doneby induction. (cid:3) To prove the main result, we need a few more lemmas.
Definition.
Let D be a domain containing a field K , E the field offractions of D . A monomial ∈ E ∗ K K h x, y i of the following form, ......ptq...... where t ∈ E \ D , p, q ∈ { x, y } , is called a sandwich monomial , or just a sandwich for short. Lemma 2.1 (on sandwich preserving) . In the constructive decomposa-tion in Theorem 1.5, suppose a sandwich ...ptq... (where p, q ∈ { x, y } , t ∈ F ( z ) \ F [ z ] , appears on some step during the process of the effectivedecomposation, then there will be some sandwich in any future step.Proof. Let f be the polynomial obtained in the ( n − − th step of theeffective operation in Theorem 1.5, k = deg( f ). Take all sandwiches s α of the maximum total degree with respect to x and y . Let S = P s α be their sum. Let T = P t β be the sum of components (monomials) t β of f maximum total degree respect to x and y . It is possible that s α = t β for some α, β , then deg( s α ) = deg( t β ) for all α, β . In this case T = S + D . N THE LIFTING OF THE NAGATA AUTOMORPHISM 5
Suppose the n − th step has the following form x → x, y → y + G ( x ).Let ¯ G be the sum of monomials in G with the maximum degree. Then¯ G ( x ) = e G ( x, . . . , x ) where e G ( x , . . . , x n ) = X i q i, x q i, x · · · x m q i,m +1 , q ij ∈ F ( z ) ,m be the degree of the n − th step operation (elementary automorphism).Let deg( S ) < deg( T ). Consider elements of the form e G ( S, T, . . . , T ).It is a linear combination of sandwiches. All of them have the followingform q s i q t · · · t m − q m , q i ∈ Q. Their sum is not zero, because for any polynomial of the form H = P i q i x q i x · · · x m q i m +1 such that H ( x, . . . , x ) = 0 and for any S, T / ∈ Q , H ( S, T, . . . , T ) = 0.If deg( S ) = deg( T ), we consider elements of the form e G ( S, S, . . . , S ).It is a linear combination of sandwiches. All of them have the followingform q s q · · · s m q m , q i ∈ Q.s i are monomials from S . Their sum is not zero, because for any polyno-mial of the form H = P i q i x q i x · · · x m q i m +1 such that H ( x, . . . , x ) =0 and for any S ∈ Q , H ( S, . . . , S ) = 0.Now we are going to prove (via degree estimate) that they cannotcancel out by other monomials (which must be sandwiches). That is,there are no other sandwiches which are in this form. They cannot beproduced by H ( R , . . . , R m ) where R i are monomials either from S or T . This can be easily seen from the following argument:Suppose deg( S ) = deg( T ) and D = 0. Then if we substitute mono-mials forming S and D in different ‘words’, the outcomes would bedifferent. Similarly suppose deg( S ) < deg( T ) and D = 0, then sub-stituting monomials forming S and T in different ‘words’, the out-comes would be different. Suppose D = 0, i.e. S = T . Then H ( T, . . . , T ) = H ( S, . . . , S ). Obviously in this case we need to donothing.
A.BELOV-KANEL AND JIE-TAI YU
Suppose we get such a sandwich because an element from F ( z ) \ F [ z ]appears between summands of polynomials, obtained by the the (previ-ous) ( n − − th step. It means that ‘fractional coefficient’ in F ( z ) \ F [ z ]appears in the position in some term, between two monomials obtainedon the ( n − x → X v i , y → X u i be an automorphism, obtained on the ( n − n -thstep: x → x, y → y + X i q i xq i · · · xq in i . Let v i = a i ¯ v i b i where a i , b i ∈ Q, ¯ v i begins with either x or y and alsoends with either x or y .Suppose the leftmost factor q ∈ F ( z ) \ F [ z ] corresponding to the left-most factor q in monomial s i in s appears in the corresponding sandwich w . Then it has the form w = q i v α q i · · · a α k ¯ v α k b α k q ik a α k +1 ¯ v α k +1 b α k +1 · · · a α ni ¯ v α ni b α ni q in i and the position ¯ v α k b α k q ik a α k +1 ¯ v α k +1 corresponds to the position of frac-tional coefficient in the sandwich s · Q n − i =1 v i living inside s = s qs , s ends with x or y , s begins with x or y . Then q = b α k q ik a α k +1 , s = q i v α q i · · · a α k ¯ v α k , s T i · · · T i n == ¯ v α k +1 b α k +1 · · · a α ni ¯ v α ni b α ni q in i . Only in that case cancellation is possible. Here T i are monomial sum-mands of T .Now let us compare the degrees. deg( s ) < deg( s ), P ni = k deg( v i ) ≤ ( n i − k + 1) deg( T ) ≤ ( m −
1) deg( T ). Hencedeg( W ) < deg( s ) + ( m −
1) deg( T ) = deg( e G ( s, T, . . . , T ))so any cancellation is impossible. (cid:3) Lemma 2.2 (on coefficient improving) . a) Let x ′ = pxq ; p, q ∈ F ( z ) , M ~q ( x ) = xq xq · · · x , q ′ i = q − q i p − . Then M ~q ( x ′ ) = pM ~q ′ ( x ) q . N THE LIFTING OF THE NAGATA AUTOMORPHISM 7 b) Take the process in Theorem 1.5 without sandwiches. Then aftereach step (except the last step), the outcome ( f, g ) has the followingproperties: • Both f and g are sandwich-free. • The left coefficients of f and g belong to F [ z ] . Moreover, thetwo coefficients are relatively prime. • The right coefficients of f and g belong to F [ z ] . Moreover, thetwo coefficients are relatively prime.Now we can clearly see that the outcome of the last step also has theabove property.Proof. a) is obvious; b) is a consequence of a). (cid:3) Lemma 2.3.
Let f ∈ F ( z ) ∗ F F h x, y i , P ( u ) ∈ F ( z ) ∗ F [ u ] such thateach monomial has degree ≥ respect to x and y . Suppose that one ofthe coefficients of P has zero right z -degree and one of the coefficientsof f has zero right z -degree, and there is no coefficients of P and f withnegative right z -degree. Then P ( f ) has one of the coefficients with zeroright z -degree and the degree (respect to x and y ) of corresponding termis strictly more then deg( f ) .Proof. Consider the highest degree monomials of P and f with zeroright z -degree, let e P , e f will be their sums. Let g be sum of terms of f with zero right z -degree, h be the sum of terms of f of maximal degree.Now consider again the highest degree monomials in P ( u ) with zeroright z -degree and substitute e f on the rightmost position instead of u and h on other positions of u . We shall get some terms with non-zerosum T (same argument as in the proof of sandwich lemma). All suchterms have zero right z -degree.It remains to prove that such terms cannot cancel out from the otherterms. First of all, we need to consider only terms of P with zero right z -degree, other terms can not make any influence. Second, we haveto consider substitutions only of terms with zero right z -degree on therightmost positions of u . Let V be their sum. A.BELOV-KANEL AND JIE-TAI YU
But the sum of highest terms satisfying this conditions is equal to T and T is the highest homogeneous component of V , hence V = 0. (cid:3) Corollary 2.4.
Let f be a polynomial, P ∈ F ( z ) ∗ F [ x ] such that eachmonomial has degree ≥ . Suppose that one of the coefficients of P ( f ) has (zero)negative right z -degree. Then P ( f ) has one of the coefficientswith (zero) negative right z -degree and degree of corresponding term isstrictly more then deg( f ) . There is just the ‘dual’ left version of lemma 2.3 and corollary 2.4.As a consequence of the above corollary, we get
Lemma 2.5.
In the step x → x , ( z → z because we are working with z -automorphisms) y → y + x k z − l , k > of the degree-strictly-increasingprocess, applied to the automorphism x → x + higest terms , y → y + higest terms causes some negative power(s). In order to prove Theorem 1.1 we need a similar statement which isalso a consequence of the Corollary 2.4.
Lemma 2.6.
In the step x → x , ( z → z because we are working with z -automorphisms) y → y + P ( x ) , such that P has negative powers of z asleft coefficients of some monomial of degree ≥ in the degree-strictly-increasing process causes some negative power(s) on any succeedingstep. Lemma 2.3, Lemma 2.6 and Corollary 2.4 says that any further stepof non-linear operation either contains terms of negative power withbigger degree, or does not interfere in the process. Hence they implythe following
Lemma 2.7. a) Consider stage in strictly increasing process of follow-ing form. x → T + h , y → T + h where T i are sums of the terms with negative powers of z to the right, h i – are sums of the terms without negative powers of z to the right.If T i are F ( z ) -linear independent, then the negative powers can not becancelled in the strictly increasing process. N THE LIFTING OF THE NAGATA AUTOMORPHISM 9 b) Suppose T is the sum of the terms with negative powers of z to theright, h – is the sum of the terms without negative powers of z to theright, T is the sum of the terms with zero powers of z to the right, h is the sum of the terms with positive powers of z to the right.If T i are F ( z ) -linear independent, then the negative powers can not becancelled in the strictly increasing process. In order to prove Theorem 1.1 we need slight generalization of theprevious lemma, which also follows from the Lemma 2.3 and Corollary2.4.
Proposition 2.8.
Consider stage in strictly increasing process of fol-lowing form: x → T + h + g , y → T + h + g where T i are sums of the terms with negative powers of z to the right, h i – are sums of the terms with zero powers of z to the right, g i aresums of the terms with positive powers of z to the right.If T i are F ( z ) -linear independent, or wedge product of vectors ( T , T ) ^ F [ z l ,z r ] ( h , h ) = 0 , then the negative powers can not be cancelled in the strictly increasingprocess. Wedge product is taken respect to left and right F ( z ) -actions,i.e. as F [ z l , z r ] -modula, monomial (respect to x, y , and inner positionsof z ) are considered as basis vectors.Proof. Pbviously, any linear operation cannot cancel the negative pow-ers of z , but Lemma 2.3 and corollary 2.4 allows us to consider onlysuch operations. (cid:3) Remark 2.9.
Considering the substitutions z → z + c one can getsimilar results for negative powers of z + c (or via considering othervaluations of F ( z )). Lemma 2.10.
Consider the step in the strictly increasing process offollowing form. x → T + h ′ , y → U + h ′ where T is the sum of the terms with negative powers of z to the right, U the sum of the terms with negative powers of z to the right, h is the sum of the terms without negative powers of z to the right, h is thesum of the terms with positive powers of z to the right.If T and U are F ( z ) -linear independent, then the negative powers can-not be cancelled in the strictly increasing process.Proof. By induction. Input of composition with polynomials with x -degree ≥ F ( z )-independence preserves). But the x -linear term action only produces the F ( z )-linear combinations. (cid:3) Consider, for instance, the elementary automorphism x → x, y → y + z n x k . It can be lifted to an Q -automorphism x → x, y → y + z n x k z n · · · x k s z n s , X k i = k, X n i = n. Though n < n and n s can still be non-negative. It is necessary todeal with that kind of situation by the next lemma. Lemma 2.11.
Consider a elementary mapping x → x ; y → y + P ( x ) such that P ( x ) has a monomial of the following form: z k xz k x · · · xz k s where one of k i < for some i such that < i < s . Then if such anelementary transformation occurs in the strictly increasing process, itmust produce some sandwich.Proof. First of all, due to the Lemma 2.1, we may assume without lossof generality that there exists no sandwiches before this step.Consider z k i , the minimum power of z , lying before the variables for allmonomials in P . Next, consider the monomials in P of the minimumdegree containing z k i between x ’s and among them, i.e. the monomialssuch that z k i positioned on the left-most possible position (but then i >
1, it should be a sandwich position). Let us denote such terms T i .Let ϕ ( x ) = X u i , ϕ ( y ) = X v i N THE LIFTING OF THE NAGATA AUTOMORPHISM 11 will be an automorphism, obtained by the previous step. Due to theLemma 2.1 we may assume that no terms come from u i , v i are sand-wiches.Now we consider u i with minimal right z -degree n r , and among them– terms with minimal degree (respect to x and y ). Let u rj will be suchterms, u r = P u rj . Because x is one of the u i , n r ≤
0. Similarlywe consider u i with minimal left z -degree n l , terms u lj and their sum u l = P u lj . We also get n l ≤ T j , consider the element E T j = q ( j )0 x · · · u r z n i u l · · · xq ( j ) s obtained by replacement of u r and u l into the positions of x surroundingoccurrence of z n i as discussed previously, the resulting power of z wouldbe equal to n r + n l + n i ≤ n i < E T j can be presented as a sum E T j = P M E Tj , where M E Tj aremonomials. Monomials M E Tj are sandwiches, they may appear onlythat way which was described previously and hence cannot cancell byother monomials. Hence we must have a sandwich. (cid:3) Proofs of the main theorems
Proof of Theorem 1.2.
Suppose the automorphism ( f, g ) can be lifted to a z -automorphism of F h x, y, z i . Then it induces an automorphism of F ( z ) ∗ F F h x, y i andcan be obtained by the process described in the Lemma on coefficientimproving.Then at some steps some negative powers of z appear either betweenvariables or on the right or on the left and it will be preserved to theend, due to Lemma 2.1 and Lemma 2.11, or Lemmas 2.10, 2.7, 2.5.Hence in the lifted automorphism, there exists some negative power of z . A contradiction. (cid:3) Proof of Theorem 1.1.
Let ( f, g ) be a wild F [ z ]-automorphism of F [ x, y, z ] such that it is notof the type in Theorem 1.2. Consider corresponding strictly increasingprocess. We shall need few more statements.The following lemma is a consequence of Proposition 2.8. Lemma 3.1.
In the strictly increasing process. Consider the steps withnegative powers of z appearing to the right. ϕ : x → x + P ( y ) , y → y Let ψ : x → x, y → y + Q ( x ) + Q ( x ) where deg( Q ) = 1 , each term of Q has degree ≥ and does notcontain negative powers of z . Then ψ = ψ ◦ ψ where ψ : x → x, y → y + Q ( x ) , ψ : x → x, y → y + Q ( x ) and ϕψ ϕ − has no negativepowers of z to the right. Lemma 3.1 together with its left analogue and remark 2.9 imply fol-lowing statement:
Proposition 3.2.
In the strictly increasing process, consider the stepwith appearing coefficients not in F [ z ] . ϕ : x → x + P ( y ) , y → y N THE LIFTING OF THE NAGATA AUTOMORPHISM 13
Let ψ : x → x, y → y + Q ( x ) + Q ( x ) where deg( Q ) = 1 , each term of Q has degree ≥ and does notcontain negative powers of z . Then ψ = ψ ◦ ψ where ψ : x → x, y → y + Q ( x ) , ψ : x → x, y → y + Q ( x ) and ϕ ◦ ψ ◦ ϕ − is a z -automorphism of F h x, y, z i .Proof. Consider set of elements from F ( z ) which are coefficients ofour monomials. If all valuations of F ( z ) centered in finite points arepositive, then they belong to F [ z ] and we are done. Due to symmetry, itis enough to consider right coefficients and due to substitution z → z + a just valuation centered in zero. Then by Lemma 3.1, we are done. (cid:3) Proof.
It is easy to see that ψ ◦ ϕ ◦ ψ − has following form: x → x + c R ( a ′ x + a ′ y ) , y → c R ( a ′ x + a ′ y ), where a ′ ij = αa ij ∈ F [ z ]are relatively prime, α ∈ F [ z ] is the least common multiple of thedenominators of a , a ∈ F [ z ] and c , c ∈ F [ z ] such that c a + c a = 0. Choose r, s ∈ F [ z ] such that ra ′ + sa ′ = 1.Acting the linear automorphism x → rx + sy, y → a ′ x + a ′ y over F [ z ] to ψ ◦ ϕ ◦ ψ − , we get an automorphism of the following form: x → rx + sy + tR ( a ′ x + a ′ y ) , y → a ′ x + a ′ y , which is elementarilyequivalent to x → rx + sy, y → a ′ x + a ′ y . Hence ψ ◦ ϕ ◦ ψ − istame. (cid:3) The next proposition is well-known from linear algebra.
Proposition 3.3.
Let ( f, g ) is a z -automorphism of F [ z ][ x, y ] linearin both x and y . Then it is a tame z -automorphism. Now we are ready to complete the proof of Theorem 1.1. Suppose a z -automorphism ϕ = ( f, g ) of F [ z ][ x, y ] can be lifted to an automorphismof F [ z ] ∗ F F h x, y i (i.e. an automorphism of F h x, y, z i fixing z ), whichis decomposed into product of elementary one according to strictly in-creasing process. The coefficients of elementary operation can be in F ( z ) \ F [ z ] only for linear terms (see Lemma 2.6 and Remark 2.9) andconjugating non-linear elementary step with respect to the automor-phisms corresponding to these terms are z -tame. Hence ϕ is a productof z -tame automorphisms and z -automorphisms linear in both x and y . Now we are done by Proposition 3.3. By carefully looking through the above proofs, we actually obtainedthe following
Theorem 3.4.
An automorphism ( f, g ) in Aut F [ z ] F h x, y, z i , can becanonically decomposed as product of the following type of automor-phisms:i) Linear automorphisms in Aut F [ z ] F h x, y, z i ;ii) Automorphisms which can be obtained by an elementaty automor-phism in Aut F [ z ] F h x, y, z i conjugated by a linear automorphism inAut F ( z ) F ( z ) ∗ F F h x, y i . Theorem 3.4 opens a way to obtain stably tameness of Aut F [ z ] F h x, y, z i ,which will be done in a separate paper [2].4. Acknowledgements
Jie-Tai Yu would like to thank Shanghai University and Osaka Uni-versity for warm hospitality and stimulating atmosphere during hisvisit, when part of the work was done. The authors thank VesselinDrensky and Leonid Makar-Limanov for their helpful comments andsuggestions.
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Department of Mathematics, Bar-Ilan University Ramat-Gan, 52900Israel
E-mail address : [email protected], [email protected] Department of Mathematics, The University of Hong Kong, HongKong SAR, China
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