Some Stably Tame Polynomial Automorphisms

We study the structure of length three polynomial automorphisms of R[X, Y ] when R is a UFD. These results are used to prove that if SLm(R[X1, X2, . . . , Xn]) = Em(R[X1, X2, . . . , Xn]) for all n,≥ 0 and for all m ≥ 3 then all length three polynomial automorphisms of R[X, Y ] are stably tame. 1. Introducton Unless otherwise specified R will be a commutative ring with 1 and R = R[X ] = R[X1, ..., Xn] is the polynomial ring in n variables. A polynomial map is a map F = (F1, ..., Fn) : A n R → A n R where each Fi ∈ R . Such an F is said to be invertible if there exists G = (G1, ..., Gn), Gi ∈ R [n] such that Gi(F1, ..., Fn) = Xi for 1 ≤ i ≤ n. Invertible polynomial maps are in one to one correspondence with Rautomorphisms of the polynomial ring R via the map F → F , F (g) = g(F ), g ∈ R. So we identify the group of R-automorphisms of R with the group of all invertible polynomial maps in n variables. Notice that this identification is not an isomorphism but rather an anti isomorphism. We would like to understand the structure of • GAn(R) = {F = (F1, . . . , Fn) : F is invertible }. Some subgroups of GAn(R) are the following. • The affine subgroup: Afn(R) = {(a11X1+a12X2+. . .+a1nXn+b1, . . . , an1X1+ ..annXn + bn) : (aij) ∈ GLn(R) and bi ∈ R} • The elementary subgroup: EAn(R) = The subgroup generated by automorphisms of the form (X1, X2, . . . , Xi−1, Xi+f(X1, . . . , Xi−1, X̂i, Xi+1, . . . , Xn), . . . , Xn) where f ∈ R[X1, X2, . . . , X̂i, . . . , Xn], i ∈ {1, . . . , n}. • The triangular subgroup: BAn(R)= The subgroup of all R-automorphisms of the form F = (a1X1+f1(X2, . . . , Xn), a2X2+f2(X3, . . . , Xn), . . . , anXn+ fn) where each ai ∈ R ∗ and fi ∈ R[Xi+1, . . . , Xn] for all 1 ≤ i ≤ n− 1 and fn ∈ R. • Tame subgroup: Tn(R) = 〈Afn(R),EAn(R)〉. It is easy to see that GA1(R) = Af1(R) when R is a domain. The structure of GA2(R) when R is a field k is well known and is the so-called Jung-van der Kulk theorem or the Automorphism Theorem.[Jun42], [vdK53] This is part of the author’s doctoral thesis, written at Washington University under the direction of David Wright. 1 2 SOORAJ KUTTYKRISHNAN Theorem 1.1. (Jung, van der Kulk) If k is a field then GA2(k) = T2(k). Further, T2(k) is the amalgamated free product of Af2(k) and BA2(k) over their intersection. However, not much is known about GA3(k). A natural question is whether T3(k) the whole group GA3(k)? Nagata [Nag72] conjectured that the answer is no and gave a candidate counterexample. Example 1.1. (Nagata) N = (X + t(tY +X), Y − 2(tY +X)X − t(tY +X), t) ∈ GA 3(k) Let R be a domain. Then the following algorithm from [vdE00] will determine if F = (P (X,Y ), Q(X,Y )) ∈ GA2(R) is in T2(R). Let tdeg(F ) = deg(P ) + deg(Q) and h1 be the highest degree term of P and h2 that of Q. Algorithm 1.1. Input: F = (P,Q). 1) Let (d1, d2) = (deg(P ), deg(Q)). 2) If d1 = d2 = 1, go to 7. 3) If d1 6= d2, go to 5. 4) If there exists τ ∈ Af2(R) with tdeg(τ ◦ F ) < tdeg(F ), replace F by τ ◦ F and go to 1, else stop : / ∈ T2(R). 5) If d2 < d1, replace F by (Q,P ). 6) If d1 | d2 and there exists c ∈ R with h2 = ch d2/d1 1 , replace F by (X,Y − cX21) ◦ F and go to 1, else stop : F / ∈ T2(R). 7) If detJF ∈ R, stop: F ∈ T2(R), else stop : F / ∈ T2(R). Using this algorithm we can easily conclude that N / ∈ T2(k[t]). We say that N is ‘t wild. Shestakov and Umirbaev in 2002 [SU03] proved that N / ∈ T3(k) and thus proved Nagata’s conjecture. We can extend N from the Example 1.1 naturally as Ñ = (N,W ) ∈ GA4(k). Martha Smith proved [Smi89] that Ñ ∈ T4(k). Definition 1.1. Let F,G ∈ GAn(R). Then (1) F is stably tame if there exists m ∈ N and new variables Xn+1, . . . , Xn+m such that the extended map F̃ = (F,Xn+1, . . . , Xn+m) is tame. i.e (F,Xn+1, . . . , Xn+m) ∈ Tn+m(R) (2) F is tamely equivalent(∼) to G if there exists H1, H2 ∈ Tn(R) such that H1 ◦ F ◦H2 = G. (3) F is stable tamely equivalent(∼st) toH ∈ GAn+m(R) if there exists H̃1, H̃2 ∈ Tn+m(R) such that H̃1 ◦ F̃ ◦ H̃2 = H where F̃ = (F,Xn+1, . . . , Xn+m) So N from Nagata’s example is stably tame with one more variable. Also, N fixes ‘t’ and so N ∈ GA2(k[t]). Viewed this way, by the automorphism theorem N is a tame k(t)-automorphism. In fact this phenomenon occurs in a more general situation as described in the next section. 2. Length Of An Automorphism Proposition 2.1. Let R be a domain K its fraction field and F ∈ GA2(R). Then F = L ◦Da,1 ◦Fm ◦Fm−1 ◦ ... ◦F1 where L = (X + c, Y + d), Da,1 = (aX, Y ), Fi = (X,Y + f(X)) or Fi = (X + g(Y ), Y ) for some c, d ∈ R, a ∈ R , f(X), g(X) ∈ K[X ] SOME STABLY TAME POLYNOMIAL AUTOMORPHISMS 3 Proof. Let F = (P (X,Y ), Q(X,Y )), where P (X,Y ), Q(X,Y ) ∈ R[X,Y ] and L = (X + c, Y + d), with c = P (0, 0) and d = Q(0, 0). Let G = L ◦ F ∈ GA2(R). Viewed as an element of GA2(K), by the Automorphism Theorem G ∈ T 0 2(K). When R is a domain, by the results of Wright [Wri76], the group T2(K) of tame automorphisms of K[X,Y ] preserving the augmentation has a similar description as a free amalgamated product as GA2(k) where k is any field. In particular, T 0 2(K) is generated by the automorphisms F1 = (X,Y + f(X)), F2 = (X + g(Y ), Y ), Da,b = (aX, bY ) where f(X) ∈ K[X ], g(Y ) ∈ R[Y ], f(0) = g(0) = 0, a, b ∈ K. Since Da,b = Dab,1◦Db−1,b and SL2(K) = E2(K) we have that Db−1,b is a product of elementary linear automorphisms and hence we can assume that b = 1. We also have the following equalities. F1 ◦Da,1 = (aX, Y + f(aX)) = Da,1 ◦ F ′ 1 where F ′ 1 = (X,Y + f(aX)). F2 ◦Da,1 = (aX + g(Y ), Y ) = Da,1 ◦ F ′ 2 where F ′ 2 = (X + a g(Y ), Y ). So if G ∈ GA2(R) then G = Da,1◦Fm◦Fm−1◦ . . .◦F2◦F1 where each Fi is either of the type (X,Y +fi(X)) or (X+gi(Y ), Y ), fi(X), gi(X) ∈ K[X ] and a ∈ K ∗ The linear components of G and Fm ◦ Fm−1 ◦ . . . ◦ F2 ◦ F1 are in GL2(R) and SL2(K), respectively. This implies that a ∈ R and both Da,1, Fm ◦ Fm−1 ◦ ... ◦ F2 ◦ F1 ∈ GA2(R). Definition 2.1. (1) Length of F ∈ GA2(R) is the smallest natural number m such that F = Da,1 ◦Fm ◦ Fm−1 ◦ . . . ◦F2 ◦F1 where each Fi is either of the type (X,Y + fi(X)) or (X + gi(Y ), Y ) with fi(X), gi(X) ∈ K[X ], a ∈ R ∗ and fi(0) = gi(0) = 0. (2) L(R) = {F ∈ GA2(R) : F is of length m} Remark 2.1. If F ∈ L(R) as above and F = Da,1 ◦Fm ◦Fm−1 ◦ . . . ◦F2 ◦F1 ∈ L(R) then F is tamely equivalent to G = Fm ◦ Fm−1 ◦ . . . ◦ F2 ◦ F1. Thus F is stably tame iff G is stably tame. Clearly if F ∈ L(R) then F ∈ T2(R). Suppose F ∈ L (R). Then F = Da,1◦F2 ◦F1 with F1 = (X,Y +f1(X)) and F2 = (X+g(Y ), Y ) as in the definition above. G = D a,1 ◦ F = (X + g(Y + f(X)), Y + f(X)) ∈ GA2(R) ⇒ f(X) ∈ R[X ]. Putting X = 0 in the first coordinate of G we get that g(Y ) ∈ R[Y ]. So F ∈ T2(R). Thus the first non trivial case is of length three. Now lets go back to Nagata’s example. Let F1 = (X,Y + X t , t) and F2 = (X + t Y, Y ) Then N = F 1 ◦ F2 ◦ F1. So Nagata’s example is of length three and it is stably tame with one more variable. Drensky and Yu [DY01] began a systematic study of length three automorphisms and proved the following result. Theorem 2.1. (Drensky, Yu) Let k be a field of characteristic zero and F ∈ L (k[t]) such that F = F 1 ◦G◦F1 where F1 = (X,Y +f(X)), G = (X+g(Y ), Y ) with f(X), g(X) ∈ k[t][X ]. Then F is stably tame with one more variable. 4 SOORAJ KUTTYKRISHNAN 3. Stable Tameness Of Polynomials Another important notion is the stable tameness of polynomials. This was studied by Berson in [Ber02], Edo and Vénéreau in [EV01] and Edo in [Edo05]. We’ll give some relevant results from these papers below. Let A be any commutative ring with 1. A polynomial P (X) ∈ A is said to be a variable if there exists F ∈ GAn(A) such that F = (F1, F2, . . . Fn) and F1(X) = P (X). Definition 3.1. VAn(A) = {P ∈ A [n] : There exists F ∈ GA n (A)F = (F1, F2, . . . Fn) and F1(X) = P (X).} TVn(A) = {P ∈ A [n] : There exists F ∈ T n (A)F = (F1, F2, . . . Fn) and F1(X) = P (X).} Following definition is due to Berson [Ber02]. Definition 3.2. (Berson’s Class) l ∈ N, p0 ∈ A , g0, p1, ...pl ∈ A and Q1, ..., Ql ∈ A, we define Pl ∈ A [2] by induction on l. P0 =p0X + g0, P1 =p1Y +Q1(X), P2 =p2X +Q2(p1Y +Q1(X)), Pl =plPl−2 +Ql(Pl−1) for l ≥ 3. B(A) = {Pl : p0 ∈ A , g0, p1, . . . , pl ∈ A,Q1, . . . , Ql ∈ A } B(A) = ⋃ l∈N B(A) (Berson’s polynomials) BV2(A) = VA2(A) ∩ B(A) (Berson’s variables) BV l 2 (A) = VA2(A) ∩ ⋃