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Length Four Polynomial AutomorphismsSooraj Kuttykrishnan aa Department of Computer Science, Center for Genome Sciences , Washington University inSt. Louis , Saint Louis, Missouri, USAPublished online: 16 Aug 2011.

To cite this article: Sooraj Kuttykrishnan (2011) Length Four Polynomial Automorphisms, Communications in Algebra, 39:8,2953-2962, DOI: 10.1080/00927871003591827

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Communications in Algebra®, 39: 2953–2962, 2011Copyright © Taylor & Francis Group, LLCISSN: 0092-7872 print/1532-4125 onlineDOI: 10.1080/00927871003591827

LENGTH FOUR POLYNOMIAL AUTOMORPHISMS

Sooraj KuttykrishnanDepartment of Computer Science, Center for Genome Sciences,Washington University in St. Louis, Saint Louis, Missouri, USA

We study the structure of length four polynomial automorphisms of R�X� Y� when R

is a unique factorization domain. The results from this study are used to prove that, ifSLm�R�X1�X2� � � � � Xn�� = Em�R�X1�X2� � � � � Xn�� for all n�m ≥ 0, then all lengthfour polynomial automorphisms of R�X� Y� that are commutators are stably tame.

Key Words: Polynomial automorphisms.

2000 Mathematics Subject Classification: 14R10; 13B10.

1. INTRODUCTION

Throughout this article R will be a unique factorization domain (UFD).Amongst the many unanswered questions about the structure of GA2�R�, the groupof polynomial automorphisms of the polynomial algebra R�X� Y�, stable tamenessconjecture is a long standing one. In this article, we will prove that certain lengthfour automorphisms are stably tame. We will also give an intriguing example of alength four automorphism, which is not a commutator and stably tame.

First we need a few definitions. A polynomial map is a map F = �F1� � � � � Fn� �

�nR → �n

R where each Fi ∈ R�n�. Such an F is said to be invertible if there existsG= �G1� � � � �Gn��Gi ∈ R�n� such that Gi�F1� � � � � Fn� = Xi for 1 ≤ i ≤ n. The groupof all polynomial automorphisms, GAn�R� is defined as

GAn�R� = �F = �F1� � � � � Fn� � F is invertible�

An important goal in the study of polynomial automorphisms is to understandthe structure of this group in terms of some of its well-understood subgroups.An example of such a subgroup is

Tame subgroup: Tn�R� = �Afn�R��EAn�R���

Received August 10, 2008. Communicated by J.-T. Yu.Address correspondence to Dr. Sooraj Kuttykrishnan, Department of Computer Science, Center

for Genome Sciences, Washington University in St. Louis, 4444 Forest Park Parkway, Saint Louis,MO 63108, USA; E-mail: sooraj@cse.wustl.edu

2953

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2954 KUTTYKRISHNAN

where

Afn�R� = ��a11X1 + a12X2 + · · · + a1nXn + b1� � � � � an1X1 + · · · + annXn + bn� �

�aij� ∈ GLn�R� and bi ∈ R

is the subgroup of affine automorphisms of �nR and the elementary subgroup

EAn�R� =⟨�X1� X2� � � � � Xi−1� Xi + f�X1� � � � � Xi−1� X̂i� Xi+1� � � � � Xn�� � � � � Xn� �

i ∈ �1� � � � � n� f ∈ R�X1� X2� � � � � X̂i� � � � � Xn��⟩�

Another well-studied subgroup of GAn�R� is the triangular subgroup,

BAn�R� =⟨�a1X1 + f1�X2� � � � � Xn�� a2X2 + f2�X3� � � � � Xn�� � � � � anXn + fn� �

ai ∈ R∗� fi ∈ R�Xi+1� � � � � Xn�� 1 ≤ i ≤ n− 1� fn ∈ R⟩�

If R is a domain, then GA1�R� = Af1�R�. When R is a field k, the followingwell-known theorem gives us the structure of GA2�k� [4, 10].

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.

A natural question that arises from Theorem 1.1 is whether T3�k� is the wholegroup GA3�k�? Nagata [6] conjectured that the answer is no and gave a candidatecounterexample.

Example 1.1 (Nagata). Let F1 = �X� Y + X2

t� t� and F2 = �X + t2Y� Y� t�. Then

N = F−11 � F2 � F1�

= �X + t�tY + X2�� Y − 2�tY + X2�X − t�tY + X2�2� t� ∈ GA3�k��

Using the following algorithm from [9], we can conclude that N � T2�k�t��.Let F = �P�X� Y��Q�X� Y�� ∈ GA2�R� and tdeg�F� = deg�P�+ deg�Q� and h1

be the homogeneous part of highest degree of P and h2 that of Q.

Algorthm 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 = d2, go to 5.4) If there exists ∈ Af2�R� with tdeg� � F� < tdeg�F�, replace F by � F and go

to 1, else stop: F � T2�R�.5) If d2 < d1, replace F by �Q� P�.6) If d1 d2 and there exists c ∈ R with h2 = ch

d2/d11 , replace F by �X� Y − cXd2/d1� �

F and go to 1, else stop: F � T2�R�.7) If det JF ∈ R∗, stop: F ∈ T2�R�, else stop: F � T2�R�.

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LENGTH FOUR POLYNOMIAL AUTOMORPHISMS 2955

Shestakov and Umirbaev in 2002 [8] proved that N � T3�k� and thus provedNagata’s conjecture.

Definition 1.1. Let F�G ∈ GAn�R�. Then:

(1) F is stably tame if there exist m ∈ � and new variables Xn+1� � � � � Xn+m such thatthe 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� to H ∈ GAn+m�R� if there exists H̃1� H̃2 ∈Tn+m�R� such that H̃1 � F̃ � H̃2 = H , where F̃ = �F�Xn+1� � � � � Xn+m�.

Martha Smith proved [7] that N from Nagata’s example is stably tame withone more variable. This result led to the formulation of the following conjecture.

Conjecture 1. If k is a field and F ∈ GAn�k�, then F is stably tame.

In her proof of the stable tameness of Nagata’s example, Martha Smithexploited the decomposition of N in Example 1.1 into certain special type ofelementary automorphisms as shown in the example. This led to further study ofsuch decompositions and the notion of the length of an automorphism, which wediscuss below. The following proposition due to Wright [11] is well known and aproof is given in [5, Proposition 2.1].

Proposition 1.1. Let R be a domain K its fraction field and F ∈ GA2�R�. ThenF = L �Da�1 � Fm � Fm−1 � · · · � F1 where L = �X + c� Y + d�, Da�1 = �aX� Y�, Fi =�X� Y + fi�X�� or Fi = �X + gi�Y�� Y� for some c� d ∈ R, a ∈ R∗, fi�X�, gi�X� ∈ K�X�such that fi�0� = gi�0� = 0.

Definition 1.2.

(1) Length of F ∈ GA02�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�m��R� = �F ∈ GA02�R� � F is of length m

Remark 1.1. If F ∈ L�m��R� as above and F = Da�1 � Fm � Fm−1 � · · · � F2 � F1 ∈L�m��R� then F is tamely equivalent to G = Fm � Fm−1 � · · · � F2 � F1. Thus F is stablytame iff G is stably tame.

It is easy to see that Nagata’s example is of length three and it is stably tamewith one more variable. Drensky and Yu [2] began a systematic study of length threeautomorphisms and proved the following result.

Theorem 1.2 (Drensky, Yu). Let k be a field of characteristic zero and F ∈L�3��k�t�� such that F = F−1

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.

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2956 KUTTYKRISHNAN

The following theorem was proved in [5]. Let SLn�R� denote the set of all n× nmatrices with entries from R and determinant equal to 1, and En�R� denote thegroup generated by the set of all n× n elementary matrices with entries from R.

Theorem 1.3. Suppose R is a UFD such that SLm�R�X1� X2� � � � � Xn�� =Em�R�X1� X2� � � � � Xn�� for all n�m ≥ 0. Then F ∈ L�3��R� ⇒ F is stably tame.

This theorem was also claimed by Edo in [3] without the assumption thatSLm�R�X1� X2� � � � � Xn�� = Em�R�X1� X2� � � � � Xn�� for all n�m ≥ 0. However, it is theauthor’s contention that this assumption is required for the proof provided in [3]to hold. So a natural question at this point to ask is if F ∈ L�4��R� is stably tame?As an evidence to an affirmative answer to this question, we prove the followingtheorem in this article.

Theorem 1.4 (Main Theorem). Let R be a UFD that is a �-algebra suchthat SLm�R�X1� X2� � � � � Xn�� = Em�R�X1� X2� � � � � Xn�� for all n�m ≥ 0. Let F ∈L�4��R� and F = G−1

1 � F−11 �G1 � F1, where F1 = �X� Y + f�X���G1 = �X +

g�Y�� Y�� f�X�� g�X� ∈ K�X� with f�0� = g�0� = 0. Then F is stably tame.

Remark 1.2. In [1] Berson, van den Essen, and Wright recently proved that ifF ∈GA2�R�, where R is a regular ring then F is stably tame. This is a much strongerresult. However, our result does not require the ring to be regular.

Before we present the proof of the main theorem, here is an example of anontame automorphism of length four.

Example 1.2. Let R be a UFD and t ∈ R\�0. Let

F1 =(X� Y + �t + 1�3X2

t

)�

G1 =(X + t2Y

�t + 1�� Y

)� and

F = G−11 � F−1

1 �G1 � F1

= (X + t�t + 1�X2 − t5Y 2 − t3�t + 1�6X4 − 2t3�t + 1�XY − 2t2�t + 1�4X3

− 2t3�t + 1�4X2Y� Y − t3�t + 1�Y 2 − t�t + 1�7X4

− 2t�t + 1�2XY − 2�t + 1�5X3 − 2t�t + 1�5X2Y)�

Using the Algorithm 1.1, we can see that F � T2�R�.

2. STRUCTURE OF LENGTH FOUR AUTOMORPHISMS

Lemma 2.1. Let R be a UFD and a �-algebra, K its fraction field, and A�X�� B�X� ∈R�X�� b ∈ R\0 be such that A�0� = B�0� = 0, gcd�B� b� = 1. Then A

(B�X�

b

) ∈ R�X� ⇒A�X� ∈ R�bX�.

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LENGTH FOUR POLYNOMIAL AUTOMORPHISMS 2957

Proof.

A

(B�X�

b

)= A

(0+ B�X�

b

)

=n∑

i=0

A�i��0�B�X�i

i! bi =n∑

i=1

A�i��0�B�X�i

i! bi

= B�X�

( n∑i=1

A�i��0�B�X�i−1

i! bi)∈ R�X�

⇒ B�X�n∑

i=0

A�i��0�bn−i B�X�i−2

i! ≡ 0 mod bn�

Since gcd�B� b� = 1, we get

n∑i=1

A�i��0�bn−i B�X�i−1

i! ≡ 0 mod bn� (*)

Putting X = 0 in (*) gives us A′�0� ≡ 0 mod b. That is, coefficient of X in A�X� isdivisible by b. So (*) becomes

n∑i=2

A�i��0�bn−i B�X�i−1

i! = B�X�

( n∑i=2

A�i��0�bn−i B�X�i−2

i!)≡ 0 mod bn�

Again gcd�B� b� = 1 gives us

n∑i=2

A�i��0�bn−i B�X�i−2

i! ≡ 0 mod bn (**)

Putting X = 0 in (**), we get A′′�0�2! ≡ 0 �mod b2�. That is, coefficient of X2 in A�X�

is divisible by b2. Proceeding like this, one gets that for all k ≥ 1, coefficient of Xk

in A�X� is divisible by bk. �

Now we prove a lemma about the structure of automorphisms inL�4��R� where R is a UFD. Let F = G2 � F2 �G1 � F1 ∈ L�4��R�� Fi = �X� Y + Ai�X�

ai�,

Gi = �X + Bi�Y�

bi� Y�� Ai�X�� Bi�X� ∈ R�X�� ai� bi ∈ R\0 and Ai�0� = Bi�0� = 0 for i =

1� 2� gcd�Ai� ai� = gcd�Bi� bi� = 1.

Lemma 2.2. We use the same notations as above. Then A2�X� = C�b1X� and B1�Y� =D�a2Y� for some C�X��D�X� ∈ R�X� and gcd�a2� b1� = 1.

Proof.

F = G2 � F2 �G1 � F1

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2958 KUTTYKRISHNAN

=

X +

B1

(Y + A1�X�

a1

)b1

+

B2

Y + A1�X�

a1+

A2

X+ B1

(Y+ A1�X�

a1

)b1

a2

b2�

Y + A1�X�

a1

+A2

(X + B1

(Y+ A1�X�

a1

)b1

)a2

(1)

Putting X = 0 in the second coordinate of F , we get A2�B1�Y�

b1� ∈ R�Y�. Then

applying Lemma 2.1, we get A2�X� = C�b1X�. Similarly, putting Y = 0 in the firstcoordinate of F−1, we get B1�Y� = D�a2Y�. Since A2�X� = C�b1X�, we know thatgcd�C�b1X�� a2� = 1 ⇒ gcd�a2� b1� = 1. �

2.1. Proof of the Main Theorem

Following useful lemma was proved in [2] when R = k�t� and was proved whenR is a UFD and a �-algebra in [5].

Lemma 2.3. Let F = F−11 �G � F1 ∈ L�3��R�, where F1 = �X� Y + A1�X�

a��G = �X +

g�Y�� Y�� A1�X� ∈ R�X�� g�Y� ∈ K�Y�� a ∈ R. Also, let gcd�A1� a� = 1 and A1�0� = 0.Then g�Y� = D�aY� for D�Y� ∈ R�Y� and a D�Y�.

Let F1 = �X� Y + A�X�

a��G1 = �X + B�Y�

b� Y�, and F = G−1

1 � F−11 �G1 � F1, where

A�X� ∈ R�X�� B�Y� ∈ R�Y�� A�0� = B�0� = 0. We may assume that gcd�A�X�� a� =gcd�B�Y�� b� = 1. By Lemma 2.2, we also know that gcd�a� b� = 1� A�X� =C�bX�� B�Y� = D�aY� for some C�X� ∈ R�X��D�Y� ∈ R�Y�� with C�0� = D�0� = 0.

F =(X + D�aY + C�bX��−D�aY + C�bX�− C�bX +D�aY + C�bX����

b�

Y + C�bX�− C�bX +D�aY + C�bX���

a

)� (2)

Claim 2.1. a D�Y�.

Let S = �1� b� b2� � � � and Rb = S−1R. Clearly, G1 ∈ GA2�Rb�, and so G1 � F ∈GA2�Rb�.

G1 � F = G1 �G−11 � F−1

1 �G1 � F1

= F−11 �G1 � F1

=(X + D�aY + C�bX��

b� Y + C�bX�− C�bX +D�aY + C�bX���

a

)∈ L�3��Rb��

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LENGTH FOUR POLYNOMIAL AUTOMORPHISMS 2959

So by Lemma 2.3 we have that a D�Y�

bin Rb. Since Rb is a UFD for b = 0 and

gcd�a� b� = 1, this implies that a D�Y�. Hence the claim.Notice that by the claim a divides all the terms in the first coordinate of F

in (2) except X. So we have

F = �X + aP�X� Y�� Y +Q�X� Y��

where

P�X� Y� = D�aY + C�bX��−D�aY + C�bX�− C�bX +D�aY + C�bX����

aband

Q�X� Y� = C�bX�− C�bX +D�aY + C�bX���

a�

Let E = �X + aW� Y�W� and L = �X� Y −Q�X� 0��W − P�X� 0��; then

F ∼st �X + aP�X� Y�� Y +Q�X� Y��W�

∼ F̃ = �X + aP�X� Y�� Y +Q�X� Y��W + P�X� Y��

∼ E � F̃ � E−1 = �X� Y +Q�X + aW� Y��W + P�X + aW� Y���

∼ F 1 = L � E−1 � F̃ � E= �X� Y +Q�X + aW� Y�−Q�X� 0��W + P�X + aW� Y�− P�X� 0��

=(X� Y + C�bX + abW�− C�bX + abW +D�aY + C�bX + abW���

a

− C�bX�− C�bX +D�C�bX���

a�W + D�aY + C�bX + abW��−D�C�bX��

ab

− D�aY + C�bX + abW�+ C�bX + abW +D�aY + C�bX + abW����

ab

+ D�C�bX�+ C�bX +D�C�bX����

ab

)�

We can compute that F 1 = G12 � F 1

2 �G11 � F 1

1 , where

F 11 =

(X� Y + C�b�X + aW��− C�bX�

a�W

)�

G11 =

(X� Y�W + D�aY + C�bX��−D�C�bX��

ab

)�

F 12 =

(X� Y − C�b�X + aW�+D�C�bX���− C�bX +D�C�bX���

a�W

)

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2960 KUTTYKRISHNAN

and

G12 =

(X� Y�W − D�aY + C�bX�− C�bX +D�C�bX����

ab

+ D�C�bX�− C�bX +D�C�bX����

ab

)�

Since F 11 ∈ EA2�R�X��, F

1 is tamely equivalent to F 1 � �F 11 �

−1 ∈ L�3��R�X�� and hencestably tame by Theorem 1.3. Thus we get that F is stably tame.

2.2. An Intriguing Example

Using the notations above we give an example of a length four automorphismwhich is not of the type discussed in the main theorem. That is, F ∈ L�4��R�,but there are no F1�G1 ∈ EAn�K� and F1 =

(X� Y + A�X�

a

)�G1 =

(X + B�Y�

b� Y

)� for

some A�X�� B�X� ∈ R�X�� a� b ∈ R\0� A�0� = B�0� = 0 for gcd�A� a� = gcd�B� b� = 1so that F = G−1

1 � F−11 �G1 � F1. Further, in this example, a = b. However, this

automorphism is stably tame!

Example 2.1. Let R be a domain and t ∈ R\�0�

F1 =(X� Y + X2

t

)F2 = �X� Y + �t − 1�X�

G1 = �X + �t + 1�Y� Y�

G2 =(X − Y 2

t� Y

)� and

F = G2 � F2 �G1 � F1

= �X + �t + 1�Y + 3X2 − t3Y 2 − tX2 − tX4 − 2t2XY

+ 2tXY − 2t2X2Y − 2tX3 + 2X3� t2Y + �t − 1�X + tX2��

Then F ∈ L�4��R�.

Claim 2.2. There are no F1�G1 ∈ EAn�K� and F1 = �X� Y + A�X�

a��G1 = �X +

B�Y�

b� Y�� for some A�X�� B�X� ∈ R�X�� a� b ∈ R\0� A�0� = B�0� = 0 for gcd�A� a� =

gcd�B� b� = 1, so that F = G−11 � F−1

1 �G1 � F1.

Suppose the claim was not true. Then by Claim 2.1, a divides �t + 1�Y + 3X2 −t3Y 2 − tX2 − tX4 − 2t2XY + 2tXY − 2t2X2Y − 2tX3 + 2X3, which implies that a is aunit in R. Similarly, by looking at the second coordinate of F−1, one can concludethat b is also a unit in R. (This can be shown by following the proof of Claim 2.1and showing that the second coordinate of F−1 is of the form Y + bP�X� Y� for someP ∈ R�X� Y�.) This implies that F ∈ EAn�R�. However, using Algorithm 1.1, we getthat F � T2�R�, a contradiction. Hence the claim.

We will now show that F is stably tame.

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LENGTH FOUR POLYNOMIAL AUTOMORPHISMS 2961

Let

P�X� Y� = �t+ 1�Y + 3X2 − t3Y 2 − tX2 − tX4 − 2t2XY + 2tXY − 2t2X2Y − 2tX3 + 2X3

Q�X� Y� = t2Y + �t − 1�X + tX2�

and

Q̃�X� Y� = X + tY + X2�

We extend F to F̃ = �F� Z� ∈ GA3�R� and define the following elementaryautomorphisms of R�X� Y� Z�:

= �X� Y� Z + Q̃�X��

� = �X� Y − tZ� Z�

� = �X − tZ� Y� Z�

Also, let

= �−Y�X� Z��

Then

� � � F̃ � = �X + tZ�X + P�X� Y�� Z + Q̃�X� Y�� and

F̃ 1 = � � � F̃ � � �= �X�X − tZ + P�X − tZ� Y�� Z + Q̃�X − tZ� Y��

= (X�X − tZ + �t + 1�Y − t3Y 2 + 3�X − tZ�2 − t�X − tZ�2

− t�X − tZ�4 − 2t2�X − tZ�Y + 2t�X − tZ�Y − 2t2�X − tZ�2Y

− 2t�X − tZ�3 + 2�X − tZ�3� Z + �X − tZ�+ tY + �X − tZ�2)

= (X� Y + tY + X − tZ − t3Y 2 + 3�X − tZ�2 − t�X − tZ�2 − t�X − tZ�4

− 2t2�X − tZ�Y + 2t�X − tZ�Y − 2t2�X − tZ�2Y

− 2t�X − tZ�3 + 2�X − tZ�3� Z + �X − tZ�+ tY + �X − tZ�2)�

Then

F̃ 1 = �X� Y − tZ + P1�X� Y� Z�� Z +Q1�X� Y� Z���

where

P1�X� Y� Z� = P�X − tZ� Y�− Y + X and Q1�X� Y� Z� = Q̃�X − tZ� Y��

Notice that F ∼st F̃1. Let � = �X� Y − P1�X� 0� 0�� Z −Q1�X� 0� 0��.

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2962 KUTTYKRISHNAN

Then F̃ 1 ∼ � � F̃ 1. Clearly, the following automorphisms are in EA3�R�:

F̃1 =(X� Y + �X − tZ�+ �X − tZ�2 − X − X2

t� Z

)

G̃1 = �X� Y� Z + tY�

H̃1 =(X� Y + �tZ +Q�X� 0��+ �tZ +Q�X� 0�2�−Q�X� 0�−Q�X� 0�2

t� Z

)�

Then we have that � � F̃ 1 = H̃1 � G̃1 � F̃1. So � � F̃ 1 ∈ T3�R�. Hence F is stablytame.

REFERENCES

[1] Berson, J., van den Essen, A., Wright, D. Stable tameness of two-dimensionalpolynomial automorphisms over a regular ring. Preprint, arXiv: 0707.3151v10.

[2] Drensky, V., Yu, J.-T. (2001). Tame and wild coordinates of K�z��x� y�. Trans. Amer.Math. Soc. 353:519–537.

[3] Edo, E. (2005). Totally stably tame variables. J. Algebra 287:15–31.[4] Jung, H. W. E. (1942). Über ganze birationale Transformationen der Ebene. J. Reine

Angew. Math. 184:161–174.[5] Kuttykrishnan, S. (2009). Some stably tame polynomial automorphisms. J. Pure Appl.

Algebra 213(1):127–135.[6] Nagata, M. (1972). On automorphism Group of k�x� y�. Kinokuniya Book-Store Co.

Ltd., Tokyo, 1972. Department of Mathematics, Kyoto University, Lectures inMathematics, No. 5.

[7] Smith, M. K. (1989). Stably tame automorphisms. J. Pure Appl. Algebra 58:209–212.[8] Shestakov, I. P., Umirbaev, U. U. (2003). The Nagata automorphism is wild. Proc.

Natl. Acad. Sci. USA 100:12561–12563 (electronic).[9] van den Essen, A. (2000). Polynomial Automorphisms and the Jacobian Conjecture.

Volume 190 of Progress in Mathematics. Basel: Birkhäuser Verlag.[10] van der Kulk, W. (1953). On polynomial rings in two variables. Nieuw Arch. Wiskunde

(3) 1:33–41.[11] Wright, D. (1976). The amalgamated free product structure of GL2�K�X1� � � � � Xn��.

Bull. Amer. Math. Soc. 82:724–726.

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