Automorphisms of the polynomial ring in two variables

Pub . Mat . UAB Vol . 27 Ne 1 AUTOMORPHISMS OF THE POLYNOMIAL RING IN TWO VARIABLES` Warren Dicks Let k be a field, k[x,y] the polynomíal ring in two variables, and Aut k[x,y] the group of all its k-algebraautomorphisms . Such an automorphísm will be denoted by the ordered pair (p,q) where p,q E k[x,y]are the respective images of x,y . THEOREM . The group Aut k[x,y] is generated by (y,x), (x,y -uxn ) u E k, n ; 0 . Moreover Aut k[x,y] = A* U B where A = {(X11x+ñ12y+ñ1'X21x+a22y+ñ2)IX11x22#~21>~12}, B = {(a11x+Xl,a22y+f(x))IXlla22#O,f(x) E k[x]}, 0 = An B = {(X 11 x+a1', 21 x+a22y +a 2)I a 11 a 22~0} . The elements of A are called affine automorphisms, the elements of B de Jonquilres automorphisms, and the elements of the subgroup generated by AUB are called tame automorphisms . The fact that all k-algebra auto morphisms of k[x,y]are tame was provedby Jung [2] for char k = 0, and then by Van der Kulk [8] in the general case . From their work the coproduct decomposition follows fairly easily, but it is not clear who first made the observation . (Kambayashi [3] gives the credit to Shafarevitch [7] .) Rentschler [5] gave a very simple proof of tameness for char k = 0, and then along slightly different lines Makar-Límanov [4] gave a fairly simple proof for arbitrary characterístic .  (News of Van der Kulk'sresult seems not to have reached Moscow at that time, for Makar-Limanov refers to the result as Seminar given at UniversitatAutónoma de Barcelona, July 1981 . 155 unpublished work of Shafarevitch .) In Che spiritof Serre [6l, Roger Alperin El] gave an explicit example of a tree acted on by Aut k[x,yl from which Che coproduct decomposition can be read off . In §l below we give a modified version of Makar-Limanov's proof, and in §2 recall Alperin's example . I am very grateful to P .M . Cohn for providíng me with his translation of Makar-Límanov's thesís . §l  The support of a primitive element Let (f,g) be an automorphism of k[x,yl .  Ide can write f = Ix i jx l y j , aij e k and define supp(f) _ {x y l Iaij ¢ 0} _ <x,y>, where <x,y> is Che free abelían group generated by x,y . Let m = x-deg(f), n = y-deg(f), that is, m is Che highest exponent of x occurríng in supp(f), and similarly for n . Set o = {x l y j lni+mj < mn, i ? 0, j 3 0} S <x,y> . Geometrically, supp(f) lies in Che rectangle determined by l,x m ,x m yn,yn and A occupies Che triangle m determinedby l,x~,y n . The objective of Chis section is to show x m ,y n e supp(f) c Q and min or nlm . If mn = 0 Chis is clear . Thus we may assume mn >'O . Let m' = m/(m,n), n' = n/(m,n) . These are coprimenatural numbers, so we can choose natural numbers s,t such that v sm'-tn' = 1 . Let u = xm /yn , v = y s /x t in <x,y> so x = usv n , y = u t vm . 15 6 y s /x t =v n m' n' u=x /y Thus k[x,y] c k[u,v] and we can write £ = S ij u1 v ] so supp(f) _ {ulv]Ipij 0} . We define the leading v-component of f to be Ifl = (~ u . .U' )V ie k[u] > ' x <v> i 1J where j = v-deg(f) .  If then u-deg(Ifl) = i we definejIfll =ul v i E <u,v> called the leading term of f . This extends to a group homomorphism 11 II : k(u,v) x -> <u,v> .  (Notice the superscrípt x is being used to denote the set of nonzero elements .) The following statement indícates the steps in Makar-Limanov'sargument . TRE OREM 1 .  (í)  Thereexist a,s E k(u) x x <v> c k(u,v)x such that ¡f¡  =  aa a (a  E k x ,  a  EIN + )  and x,y  e k[a +l, o] . (ii) There then exist w,z E <u,v> such that <w> _ <~~II> or <IIaII . ~Ia~I> and +l x,y e semígp<w ,z> . (iíi) Then xm ,yn E supp(f) c p and ¡¡f¡¡ = xm and <w> = <x> . (ív) If <w> = <Ilaib then mln . (v)  If <w> = ~II«II, IISIb then nlm . PROOF .  (i) Let K = k(u) and consider the Laurent series field K((v -1 )) . In a natural way k(u,v c K_ ((v-1 )) and there are maps v-deg : K((v 1 )) x -~ 7L, 1  1 : K((v -1 )) -> Kx x <v> extending the correspondíng maps on k[u,v] .  We view k x as a subgroup of K x x <v> c K((v 1 )) x .  Since v-deg(f)  > 0 there exists aE K x x <v> such that the ímage of a in (Kx x <v>)/k x generates a x maximal cyclic subgroupcontaining the ímage of Ifl, say If¡ = aaa a Ek , a c ]N .  By ínductíon on a we shall show that for any f,g e K((v -1)) with If l = aaa a E k x , aE IN " ' there exists R E K x x<v> such that Ik[f ±l ~g]I c k[a±l,o] . The case a = 0 is vacuous . Let us now define a (possibly finite) sequence inductively . Let an í 9, = g . Suppose we have gí for some i >, 1 . If Igil = a l a)  for some n i a l E k', ni E 7i we set gí+1 = gi-xíf ; íf g í = 0 or g i ¢ 0 and Ig i l is not of this form we let the sequence end at the sequence g 1 9 2 , . . . has a limit g * in K((v -1 )), If g * = 0 then k[f±l,g] c k((f -l ))  so  Ik[f 1 c k[Ifltl] c k[ .±1] and we can take R arbitrary . Thus we may assume g * ¢ 0 so the sequence is finite and k[f ±l . g] l~ k[f ±l .g * ] " If If1,Ig*) are algebraically independent over k then it is easy to see Ik[f ±l ,g * ] x i c k[IfI±1,Ig*I] and we can take S = Ig*I . Thís leaves the case where Ifj,Ig*I are algebraically dependent over k . If c = v-deg(f), d = v-deg(g * ) then Ifid,Ig*I, are algebraicallydependent over k and are v-homogeneous with the same v-degree . It follows that Ifld/ig*Ic lies in K and is algebraic over k so lies in k . Thus i Ig*I c -  Ifl daad (mod k x ) .  But  (K xx <v>)/k x is a torsion-free abelian group, and the ímage of a generates a maximal cyclic subgroup, so ciad and Ig*I = a b (mod k x ) where b = ad/c . Saylg * 1= pab , p e kx . By the definition of g* we know alb, say b = aq+r  0<r<a .  Let h = g* /f q . Then Ihi - ar (mod k x ) and the induction hypothesis applies to the pair (h,f) . _ Hence there exists B s Kx x <v> such that  Ik[h +1 ,f] x I ck[a ±1 ,B] .  Now Ik[f ±1 ,g]x i  Ik[f+l,h]x1 cIk[f,h]xI<IfI> ck[a l ,s] . By induction Ik[f±l,g]xj c k[aS] for some R e k(u) x x < v >, and (i) is proved Since x .y e Ik[f,g]x i . (ii) Recall that two elements of <u,v> are saíd to be dependent if they generate a cyclic subgroup, and otherwise they are independent, that is, freely generate a free abelian subgroup . If IIaII, IISIIare independent then it is clear that x,y e IIk[a+l,s_  ]x 1I c  semigp<IIal~1,IISII> and we can take w = ¡¡al¡, z =11011 . This leaves the, case where IIaII , IISIIare dependent .  Let w be a generator of 11 .11, 1101¡ : , say 11-11 = w l . 11611 = w j , w = IIairil0Ih " Here Since v-deg(g 1 ) > v-deg(g 2 ) > n g* = g-a l f1-a 2 f n2 g] x i, Ik((f -1 )) x 1 Ila~ i~ = IIPlIl = w lj so there is a uníque V e kxsuch that z = Ila'-PPijj  w íj But z and w lj have the same v-degree so w,z are independent . Let a' = acPd ., P'  = a j /P 1 - u .  Then II k[ a+l ,P +1 7 x ,1 = llk[a' +l . (P'+u)+l]xll c Iik[a' .P'] x Il < w > c semigp<w ±l ,z> . Thus x,y e semigp<w ± l,z> and <w> _ < llalh llPib . (iii) Geometrically x,y e semigp<w ±1, z> means that one of the two half-planes determined by w contains both x and y . Ngw by (íi) llall = w l for some integer i and on replacing w with w 1 _if necessary we may assume i 3 0 .  By (i), li f Il = h irlr =wia and li f li E  semigp<x,y>  so wE  semigp<x,y> .  The only way' this can happen is for w to líe along the x or y axis, that is, w ís a power of x or y . But <w,z> ? <x,y> so w ís x or y . Thus llfljis a power of x or y . But the only place supp(f) meets the x or y axes is in A so llflle Aand this (orces supp(f) S A . The only way x-deg(f) can be m ís for x m to be ín supp(f), and similarly yn e supp(f) .  Thus1Ifll = x m or y n .  But u-deg(x m ) = u-deg(u ms v mn' ) = ms ., udeg(y n ) . = u-deg(untvm'n) = n t = ms-(m,n) <ms  so li f Il = x m .  Hence <w>  =-<x> . (iv)  If ~lall>  = < w >  =<x>  then II a II =x .  But  by  (i) llfll = llaIi a =  x a and  by (íii)llfll =x m so a = m .  Thus Ifl = aa m ín k(x,y) so y-deg(lf1) = m(y-deg(a)) . And y-deglfl = n sínce yn EsupPIfI, so min . (v) If < Ilall, IIPII> _ < w > = <x> then n'7l= v-deg(<x>) = v - deg«JIali, IIPII' .) = v-deg(<a,P>,) . By (i) y c k[n 1 ,P] and this .is v-homogeneous so v-deg(y) E v-deg(<a,P>), that is, - m' is a multíple of n' so nim . . §2  The Automorphi sm Group C For any p =  ijx1yj  E k[x,y] x , we define deg(p) = max{i+jluij ¢ 0} ;  if deg p =d we define p 0 = "id-íxiyd-i called the leading component of p . THEOREM 2 ([2],[8]) . Let (p,q) be a-k-algebra automorphísm of k[x,y] with deg p  :5 deg q .  Then . . either  (p, q)  ís affine or there is a uníque y  E kx  and Positive ínteSer r such that deg(q - uPr ) <deg(q) " PROOF . Let (f,g) be the inverse of (p,q) and let f be as in §l . li deg(p m ) ¢ deg(g n ) then deg(f(p,q)) = max{deg(p m ),deg(g n)} . But f(p,q) = x so p or q is a polynomial in x of degree 1 and the desired conclusion follows easily . This leaves the case where deg(p m ) = deg(g n ) . Here m >. n so nlm and deg(p r ) = deg(q) for r = m'  We may assume (p,q) is not affine so deg q > 1 . n Since f(p,q) = x it follows that p0,g 0 are algebraically dependent over k . Hence q0/pro is algebraic over kso lies in k, say q 0 0 = u . Then deg(q -upr ) < deg(q) as désired . By induction on deg(q) it follows easily from Theorem 2 that all k-algebra automorphisms of k[x,y] are tame . It is even a simple matter to obtain the decomposition . THEOREM 3 .  Aut k[x,y] = A *C B . PROOF . Let P be the orientedgraphwhosevertices are the k-subspaces of k[x,y] and whose edges are the inclusion maps .  Then Aut k[x,y] acts in a natural_ way on r_he . -raph P_ Let T be the orbit of k+kx 3 1-1-1-- , Í . L'-- elaim that T is a tree . Any vertex of T is of the form k+kp or k+kp+kq where (p,q) ís some automorphísm . We define deg(k+kp) = deg(p) and deg(k+kp+kq) _ max{deg(p),deg(q)} - } . It is easy to see these are well-defined . Consider a vertex of the form k+kp . We can find an automorphísm (p,q) with deg(q) minimal, so deg(q) < deg(p) or (p,q) is affine . All the neighbours of k+kp are of the form k+kp+k(q+h) where h e k[p] .  The only neighbour of k+kp with smaller degree is k+kp+kq ; all the others have greater degree . Consider a vertex of the form k+kp+kq where deg(q) < deg(p) .  The neighbours are of the form k+k(ap+Sq) where a,R e kX are not both zero ; only k+kq has smaller degree, all the others have greater degree . Finally, the vertex k+kx+ky has smaller degree than all its neighbours . Thus every path from k+kx+ky is strictly increasing (so T has no circuits) and from each vertex there is a stríctlydecreasing path which must necessarily arrive at k+kx+ky (so T is connected) . Hence T is a tree . Now k+kx ; k+kx+ky is a transversal in T for the actionof Aut k[x,y] and the stabilizer of k+kx ís B while the stabilize_of k+kx+ky ís A . Tb' .s implies G = A *c B . c_° C6] . REFERENCES 1 . R .C . ALPERIN, Homology of the group of automorphísms of k[x,y], J . Pure and Appl . Algebra 15 (1979) 109-115 . 2 . H .W .E . JUNG, Uber ganzebirationaletransformationen der Ebene, J . reine angew . Math . 184 (1942) 161-174 . 3 . T . KAMBAYASHI, On the absence of nontrivial separable forms on the affine plane, J . Algebra 35 (1975) 449-456 . 4 . L .G . MAKAR-LIMANOV, On automorphisms of certain algebras (Russian), Ph .D . Thesís, Moscow, 1970 . 5 . R . RENTSCHLER, Opérations du groupe additif sur le plan affíne, C .R . Acad . Sc . Paris, Ser .A, 267 (1968) 384-387 . 6 . J .-P . SERRÉ, Arbres, amalgames et SL 2 , Astérísque No .46, Société Math . de France, 1977 . 7 . I .R . SHAFAREVITCH, On some infíníte dimensional groups, pp .208-212, Attí-Simposío Intecnaz . di Geom . Alg . Roma, 1965 . 8 . W . VAN DER KULK, On polynomial rings in two variables, Nieuw Archief voor Wisk . (3) I (1953) 33-41 . Rebut el 15juliol 1982 Revisat el 20 abril 1982 BedfordCollege Regent's Park London, NW1 4NS