American Mathematical Society

The tame and the wild automorphisms of polynomial rings in three variables

By Ivan P. Shestakov and Ualbai U. Umirbaev

Abstract

A characterization of tame automorphisms of the algebra upper A equals upper F left-bracket x 1 comma x 2 comma x 3 right-bracket of polynomials in three variables over a field upper F of characteristic 0 is obtained. In particular, it is proved that the well-known Nagata automorphism is wild. It is also proved that the tame and the wild automorphisms of upper A are algorithmically recognizable.

1. Introduction

Let upper C equals upper F left-bracket x 1 comma x 2 comma period period period comma x Subscript n Baseline right-bracket be the polynomial ring in the variables x 1 comma x 2 comma period period period comma x Subscript n Baseline over a field upper F , and let upper A u t upper C be the group of automorphisms of upper C as an algebra over upper F . An automorphism tau element-of upper A u t upper C is called elementary if it has a form

tau colon left-parenthesis x 1 comma period period period comma x Subscript i minus 1 Baseline comma x Subscript i Baseline comma x Subscript i plus 1 Baseline comma period period period comma x Subscript n Baseline right-parenthesis right-arrow from bar left-parenthesis x 1 comma period period period comma x Subscript i minus 1 Baseline comma alpha x Subscript i Baseline plus f comma x Subscript i plus 1 Baseline comma period period period comma x Subscript n Baseline right-parenthesis comma

where 0 not-equals alpha element-of upper F comma f element-of upper F left-bracket x 1 comma period period period comma x Subscript i minus 1 Baseline comma x Subscript i plus 1 Baseline comma period period period comma x Subscript n Baseline right-bracket . The subgroup of upper A u t upper C generated by all the elementary automorphisms is called the tame subgroup, and the elements from this subgroup are called tame automorphisms of upper C . Non-tame automorphisms of the algebra upper C are called wild.

It is well known Reference6, Reference9, Reference10, Reference11 that the automorphisms of polynomial rings and free associative algebras in two variables are tame. At present, a few new proofs of these results have been found (see Reference5, Reference8). However, in the case of three or more variables the similar question was open and known as “The generation gap problem” Reference2, Reference3 or “Tame generators problem” Reference8. The general belief was that the answer is negative, and there were several candidate counterexamples (see Reference5, Reference8, Reference12, Reference7, Reference19). The best known of them is the following automorphism sigma element-of upper A u t left-parenthesis upper F left-bracket x comma y comma z right-bracket right-parenthesis , constructed by Nagata in 1972 (see Reference12):

StartLayout 1st Row 1st Column sigma left-parenthesis x right-parenthesis 2nd Column equals 3rd Column x plus left-parenthesis x squared minus y z right-parenthesis z comma 2nd Row 1st Column sigma left-parenthesis y right-parenthesis 2nd Column equals 3rd Column y plus 2 left-parenthesis x squared minus y z right-parenthesis x plus left-parenthesis x squared minus y z right-parenthesis squared z comma 3rd Row 1st Column sigma left-parenthesis z right-parenthesis 2nd Column equals 3rd Column z period EndLayout

Observe that the Nagata automorphism is stably tame Reference17; that is, it becomes tame after adding new variables.

The purpose of the present work is to give a negative answer to the above question. Our main result states that the tame automorphisms of the polynomial ring upper A equals upper F left-bracket x comma y comma z right-bracket over a field upper F of characteristic 0 are algorithmically recognizable. In particular, the Nagata automorphism sigma is wild.

The approach we use is different from the traditional ones. The novelty consists of the imbedding of the polynomial ring upper A into the free Poisson algebra (or the algebra of universal Poisson brackets) on the same set of generators and of the systematical use of brackets as an additional tool.

The crucial role in the proof is played by the description of the structure of subalgebras generated by so-called asterisk -reduced pairs of polynomials, given in Reference16. More precisely, a lower estimate for degrees of elements of these subalgebras is essentially used in most of the proofs.

We follow the so-called “method of simple automorphisms”, which was first developed in Reference1 for a characterization of tame automorphisms of two-generated free Leibniz algebras. Note that this method permits us also to establish directly the result of Reference4 concerning wild automorphisms of two-generated free matrix algebras, without using the results of Reference6, Reference9, Reference10, Reference11. In fact, the first attempt to apply this method for a characterization of tame automorphisms of polynomial rings and free associative algebras in three variables was done by C. K. Gupta and U. U. Umirbaev in 1999. At that time, some results were obtained modulo a certain conjecture, which eventually proved not to be true for polynomial rings (see Example 1, Section 3). Really, the structure of tame automorphisms turns out to be much more complicated.

Observe that no analogues of the results of Reference16 are known for free associative algebras and for polynomial rings of positive characteristic, and the question on the existence of wild automorphisms is still open for these algebras.

The paper is organized as follows. In Section 2, some results are given, mainly from Reference16, which are necessary in the sequel. Some instruments for further calculations are also created here. In Section 3, elementary reductions and reductions of types I–IV are defined and characterized for automorphisms of the algebra upper A , and simple automorphisms of upper A are defined. The main part of the work, Section 4, is devoted to the proof of Theorem 1, which states that every tame automorphism of the algebra upper A is simple. The main results are formulated and proved in Section 5 as corollaries of Theorem 1.

2. Structure of two-generated subalgebras

Let upper F be an arbitrary field of characteristic 0 , and let upper A equals upper F left-bracket x 1 comma x 2 comma x 3 right-bracket be the ring of polynomials in the variables x 1 comma x 2 comma x 3 over upper F . Following Reference16, we will identify upper A with a certain subspace of the free Poisson algebra upper P equals upper P upper L mathematical left-angle x 1 comma x 2 comma x 3 mathematical right-angle .

Recall that a vector space upper B over a field upper F , endowed with two bilinear operations x dot y (a multiplication) and left-bracket x comma y right-bracket (a Poisson bracket), is called a Poisson algebra if upper B is a commutative associative algebra under x dot y , upper B is a Lie algebra under left-bracket x comma y right-bracket , and upper B satisfies the Leibniz identity

StartLayout 1st Row with Label left-parenthesis 1 right-parenthesis EndLabel left-bracket x dot y comma z right-bracket equals left-bracket x comma z right-bracket dot y plus x dot left-bracket y comma z right-bracket period EndLayout

An important class of Poisson algebras is given by the following construction. Let upper L be a Lie algebra with a linear basis l 1 comma l 2 comma period period period comma l Subscript k Baseline comma period period period . Denote by upper P left-parenthesis upper L right-parenthesis the ring of polynomials on the variables l 1 comma l 2 comma period period period comma l Subscript k Baseline comma period period period . The operation left-bracket x comma y right-bracket of the algebra upper L can be uniquely extended to a Poisson bracket left-bracket x comma y right-bracket on the algebra upper P left-parenthesis upper L right-parenthesis by means of formula (Equation1), and upper P left-parenthesis upper L right-parenthesis becomes a Poisson algebra Reference15.

Now let upper L be a free Lie algebra with free generators x 1 comma x 2 comma period period period comma x Subscript n Baseline . Then upper P left-parenthesis upper L right-parenthesis is a free Poisson algebra Reference15 with the free generators x 1 comma x 2 comma period period period comma x Subscript n Baseline . We will denote this algebra by upper P upper L mathematical left-angle x 1 comma x 2 comma period period period comma x Subscript n Baseline mathematical right-angle . If we choose a homogeneous basis

x 1 comma x 2 comma period period period comma x Subscript n Baseline comma left-bracket x 1 comma x 2 right-bracket comma period period period comma left-bracket x 1 comma x Subscript n Baseline right-bracket comma period period period comma left-bracket x Subscript n minus 1 Baseline comma x Subscript n Baseline right-bracket comma left-bracket left-bracket x 1 comma x 2 right-bracket comma x 3 right-bracket period period period

of the algebra upper L with nondecreasing degrees, then upper P upper L mathematical left-angle x 1 comma x 2 comma period period period comma x Subscript n Baseline mathematical right-angle , as a vector space, coincides with the ring of polynomials on these elements. The space upper P upper L mathematical left-angle x 1 comma x 2 comma period period period comma x Subscript n Baseline mathematical right-angle is graded by degrees on x Subscript i , and for every element f element-of upper P upper L mathematical left-angle x 1 comma x 2 comma period period period comma x Subscript n Baseline mathematical right-angle , the highest homogeneous part f overbar and the degree degree f can be defined in an ordinary way. Note that

ModifyingAbove f g With bar equals f overbar g overbar comma degree left-parenthesis f g right-parenthesis equals degree f plus degree g comma degree left-bracket f comma g right-bracket less-than-or-equal-to degree f plus degree g period

In the sequel, we will identify the ring of polynomials upper A equals upper F left-bracket x 1 comma x 2 comma x 3 right-bracket with the subspace of the algebra upper P upper L mathematical left-angle x 1 comma x 2 comma x 3 mathematical right-angle generated by elements

x 1 Superscript r 1 Baseline x 2 Superscript r 2 Baseline x 3 Superscript r 3 Baseline comma r Subscript i Baseline greater-than-or-equal-to 0 comma 1 less-than-or-equal-to i less-than-or-equal-to 3 period

Note that if f comma g element-of upper A , then

StartLayout 1st Row left-bracket f comma g right-bracket equals gamma 12 left-bracket x 1 comma x 2 right-bracket plus gamma 23 left-bracket x 2 comma x 3 right-bracket plus gamma 13 left-bracket x 1 comma x 3 right-bracket comma 2nd Row gamma Subscript i j Baseline equals StartFraction partial-differential f Over partial-differential x Subscript i Baseline EndFraction StartFraction partial-differential g Over partial-differential x Subscript j Baseline EndFraction minus StartFraction partial-differential g Over partial-differential x Subscript i Baseline EndFraction StartFraction partial-differential f Over partial-differential x Subscript j Baseline EndFraction comma 1 less-than-or-equal-to i less-than j less-than-or-equal-to 3 period EndLayout

If f 1 comma f 2 comma period period period comma f Subscript k Baseline element-of upper A , then by mathematical left-angle f 1 comma f 2 comma period period period comma f Subscript k Baseline mathematical right-angle we denote the subalgebra of the algebra upper A generated by these elements.

The following lemma is proved in Reference16.

Lemma 1

Let f comma g comma h element-of upper A . Then the following statements are true:

1 right-parenthesis left-bracket f comma g right-bracket equals 0 iff f comma g are algebraically dependent.

2 right-parenthesis Suppose that f comma g comma h not-an-element-of upper F and m equals degree left-bracket f comma g right-bracket plus degree h , n equals degree left-bracket g comma h right-bracket plus degree f , k equals degree left-bracket h comma f right-bracket plus degree g . Then m less-than-or-equal-to max left-parenthesis n comma k right-parenthesis . If n not-equals k , then m equals max left-parenthesis n comma k right-parenthesis .

The next two simple statements are well known (see Reference5):

F1) If a comma b are nonzero homogeneous algebraically dependent elements of upper A , then there exists an element z element-of upper A such that a equals alpha z Superscript n , b equals beta z Superscript m , alpha comma beta element-of upper F . In addition, the subalgebra mathematical left-angle a comma b mathematical right-angle is one-generated iff m vertical-bar n or n vertical-bar m .

F2) Let f comma g element-of upper A and f overbar comma g overbar are algebraically independent. If h element-of mathematical left-angle f comma g mathematical right-angle , then h overbar element-of mathematical left-angle f overbar comma g overbar mathematical right-angle .

Recall that a pair of elements f comma g of the algebra upper A is called reduced (see Reference18), if f overbar not-an-element-of mathematical left-angle g overbar mathematical right-angle , g overbar not-an-element-of mathematical left-angle f overbar mathematical right-angle . A reduced pair of algebraically independent elements f comma g element-of upper A is called asterisk -reduced (see Reference16), if f overbar comma g overbar are algebraically dependent.

Let f comma g be a asterisk -reduced pair of elements of upper A and n equals degree f less-than m equals degree g . Put p equals StartFraction n Over left-parenthesis n comma m right-parenthesis EndFraction , s equals StartFraction m Over left-parenthesis n comma m right-parenthesis EndFraction ,

upper N equals upper N left-parenthesis f comma g right-parenthesis equals StartFraction m n Over left-parenthesis m comma n right-parenthesis EndFraction minus m minus n plus degree left-bracket f comma g right-bracket equals m p minus m minus n plus degree left-bracket f comma g right-bracket comma

where left-parenthesis n comma m right-parenthesis is the greatest common divisor of n comma m . Note that left-parenthesis p comma s right-parenthesis equals 1 , and by F1) there exists an element a element-of upper A such that f overbar equals beta a Superscript p , g overbar equals gamma a Superscript s . Sometimes, we will call a asterisk -reduced pair of elements f comma g also a p -reduced pair. Let upper G left-parenthesis x comma y right-parenthesis element-of upper F left-bracket x comma y right-bracket . It was proved in Reference16 that if degree Subscript y Baseline left-parenthesis upper G left-parenthesis x comma y right-parenthesis right-parenthesis equals p q plus r , 0 less-than-or-equal-to r less-than p , then

StartLayout 1st Row with Label left-parenthesis 2 right-parenthesis EndLabel degree left-parenthesis upper G left-parenthesis f comma g right-parenthesis right-parenthesis greater-than-or-equal-to q upper N plus m r comma EndLayout

and if degree Subscript x Baseline left-parenthesis upper G left-parenthesis x comma y right-parenthesis right-parenthesis equals s q 1 plus r 1 , 0 less-than-or-equal-to r 1 less-than s , then

StartLayout 1st Row with Label left-parenthesis 3 right-parenthesis EndLabel degree left-parenthesis upper G left-parenthesis f comma g right-parenthesis right-parenthesis greater-than-or-equal-to q 1 upper N plus n r 1 period EndLayout

It will be convenient for us to collect several evident properties of the asterisk -reduced pair f comma g in the following lemma.

Lemma 2

Under the above notation,

i) p greater-than-or-equal-to 2 ;

ii) upper N equals upper N left-parenthesis f comma g right-parenthesis greater-than degree left-bracket f comma g right-bracket ;

iii) if p greater-than 2 , then upper N greater-than m ;

iv) if p equals 2 , then upper N greater-than StartFraction n Over 2 EndFraction .

The properties i right-parenthesis minus i i i right-parenthesis are evident. As for i v right-parenthesis , let d equals left-parenthesis n comma m right-parenthesis ; then n equals 2 d comma m equals s d , and upper N equals s d minus 2 d plus degree left-bracket f comma g right-bracket greater-than left-parenthesis s minus 2 right-parenthesis d greater-than-or-equal-to d equals StartFraction n Over 2 EndFraction .

The statement of the following lemma is easily proved.

Lemma 3

The elements of type f Superscript i Baseline g Superscript j , where j less-than p , have different degrees for different values of i comma j .

Inequality (Equation2) and Lemma 3 imply

Corollary 1

Let upper G left-parenthesis x comma y right-parenthesis element-of upper F left-bracket x comma y right-bracket comma h equals upper G left-parenthesis f comma g right-parenthesis . Consider the following conditions:

left-parenthesis i right-parenthesis

degree h less-than upper N left-parenthesis f comma g right-parenthesis ;

left-parenthesis i i right-parenthesis

degree Subscript y Baseline left-parenthesis upper G left-parenthesis x comma y right-parenthesis right-parenthesis less-than p ;

left-parenthesis i i i right-parenthesis

h equals sigma-summation Underscript i comma j Endscripts alpha Subscript i j Baseline f Superscript i Baseline g Superscript j , where alpha Subscript i j Baseline element-of upper F and i n plus j m less-than-or-equal-to degree h for all i comma j ;

left-parenthesis i v right-parenthesis

h overbar element-of mathematical left-angle f overbar comma g overbar mathematical right-angle .

Then left-parenthesis i right-parenthesis right double arrow left-parenthesis i i right-parenthesis right double arrow left-parenthesis i i i right-parenthesis right double arrow left-parenthesis i v right-parenthesis .

Suppose that p greater-than-or-equal-to 3 or degree left-bracket f comma g right-bracket greater-than n . Then obviously upper N left-parenthesis f comma g right-parenthesis greater-than m , and in the conditions of Corollary 1 we have h overbar element-of mathematical left-angle f overbar comma g overbar mathematical right-angle or degree h greater-than m equals max left-parenthesis degree f comma degree g right-parenthesis . Note that the most complicated case in the investigation of tame automorphisms of upper A is represented by asterisk -reduced pairs f comma g with the condition upper N left-parenthesis f comma g right-parenthesis less-than-or-equal-to m , that is, by 2 -reduced pairs f comma g for which degree left-bracket f comma g right-bracket less-than-or-equal-to n .

Lemma 4

There exists a polynomial w left-parenthesis x comma y right-parenthesis element-of upper F left-bracket x comma y right-bracket of the type

StartLayout 1st Row w left-parenthesis x comma y right-parenthesis equals y Superscript p Baseline minus alpha x Superscript s Baseline minus sigma-summation alpha Subscript i j Baseline x Superscript i Baseline y Superscript j Baseline comma n i plus m j less-than m p comma EndLayout

which satisfies the following conditions:

1 right-parenthesis degree w left-parenthesis f comma g right-parenthesis less-than p m ;

2 right-parenthesis ModifyingAbove w left-parenthesis f comma g right-parenthesis With bar not-an-element-of mathematical left-angle f overbar comma g overbar mathematical right-angle .

Proof.

By F1), there exists a homogeneous element a element-of upper A such that f overbar equals beta a Superscript p , g overbar equals gamma a Superscript s . Then there exists alpha element-of upper F such that g overbar Superscript p Baseline equals alpha f overbar Superscript s , and the elements of the type f overbar Superscript i Baseline g overbar Superscript j , j less-than p , form a basis of the subalgebra mathematical left-angle f overbar comma g overbar mathematical right-angle . Putting h equals g Superscript p Baseline minus alpha f Superscript s , we have degree h less-than m p . If h overbar element-of mathematical left-angle f overbar comma g overbar mathematical right-angle , then h overbar equals alpha Subscript i j Baseline f overbar Superscript i Baseline g overbar Superscript j , where n i plus m j less-than m p . Change the element h to h minus alpha Subscript i j Baseline f Superscript i Baseline g Superscript j . Then degree left-parenthesis h minus alpha Subscript i j Baseline f Superscript i Baseline g Superscript j Baseline right-parenthesis less-than degree h . After several reductions of this type, we get an element

StartLayout 1st Row h equals g Superscript p Baseline minus alpha f Superscript s Baseline minus sigma-summation alpha Subscript i j Baseline f Superscript i Baseline g Superscript j Baseline comma n i plus m j less-than m p comma EndLayout

for which h overbar not-an-element-of mathematical left-angle f overbar comma g overbar mathematical right-angle . Since f comma g are algebraically independent, the equality h equals w left-parenthesis f comma g right-parenthesis defines uniquely a polynomial w left-parenthesis x comma y right-parenthesis that satisfies the conditions of the lemma.

A polynomial w left-parenthesis x comma y right-parenthesis satisfying the conditions of Lemma 4 we will call a derivative polynomial of the asterisk -reduced pair f comma g . Note that a derivative polynomial w left-parenthesis x comma y right-parenthesis is not uniquely defined in the general case. But the coefficient alpha in the conditions of Lemma 3 is uniquely defined by the equality g overbar Superscript p Baseline equals alpha f overbar Superscript s .

Lemma 5

Let w left-parenthesis x comma y right-parenthesis be a derivative polynomial of a asterisk -reduced pair f comma g . Then the following statements are true:

1 right-parenthesis degree w left-parenthesis f comma g right-parenthesis is uniquely defined;

2 right-parenthesis if f overbar , ModifyingAbove w left-parenthesis f comma g right-parenthesis With bar are algebraically dependent, then ModifyingAbove w left-parenthesis f comma g right-parenthesis With bar is uniquely defined;

3 right-parenthesis degree left-parenthesis w left-parenthesis f comma g right-parenthesis right-parenthesis greater-than-or-equal-to upper N left-parenthesis f comma g right-parenthesis ;

4 right-parenthesis if degree left-parenthesis w left-parenthesis f comma g right-parenthesis right-parenthesis less-than degree g equals m , then w left-parenthesis x comma y right-parenthesis is defined uniquely up to a summand q left-parenthesis x right-parenthesis , where n dot degree left-parenthesis q left-parenthesis x right-parenthesis right-parenthesis less-than-or-equal-to degree left-parenthesis w left-parenthesis f comma g right-parenthesis right-parenthesis ;

5 right-parenthesis if degree left-parenthesis w left-parenthesis f comma g right-parenthesis right-parenthesis less-than degree f , then w left-parenthesis x comma y right-parenthesis is defined uniquely up to a scalar summand from upper F .

Proof.

Let w 1 left-parenthesis x comma y right-parenthesis be another derivative polynomial of the pair f comma g . Since the coefficient alpha is uniquely defined in the conditions of Lemma 4, we have

StartLayout 1st Row h left-parenthesis x comma y right-parenthesis equals w left-parenthesis x comma y right-parenthesis minus w 1 left-parenthesis x comma y right-parenthesis equals sigma-summation gamma Subscript i j Baseline x Superscript i Baseline y Superscript j Baseline comma EndLayout

where n i plus m j less-than m p . Now, if degree left-parenthesis w left-parenthesis f comma g right-parenthesis right-parenthesis greater-than degree left-parenthesis w 1 left-parenthesis f comma g right-parenthesis right-parenthesis , then by Lemma 3 we get

StartLayout 1st Row ModifyingAbove h left-parenthesis f comma g right-parenthesis With bar equals ModifyingAbove w left-parenthesis f comma g right-parenthesis With bar element-of mathematical left-angle f overbar comma g overbar mathematical right-angle comma EndLayout

which contradicts the definition of w left-parenthesis x comma y right-parenthesis .

Suppose that ModifyingAbove w left-parenthesis f comma g right-parenthesis With bar not-equals ModifyingAbove w 1 left-parenthesis f comma g right-parenthesis With bar . Then ModifyingAbove h left-parenthesis f comma g right-parenthesis With bar equals ModifyingAbove w left-parenthesis f comma g right-parenthesis With bar minus ModifyingAbove w 1 left-parenthesis f comma g right-parenthesis With bar . Since ModifyingAbove h left-parenthesis f comma g right-parenthesis With bar element-of mathematical left-angle f overbar comma g overbar mathematical right-angle , the elements f overbar comma ModifyingAbove h left-parenthesis f comma g right-parenthesis With bar are algebraically dependent. Now, if f overbar , ModifyingAbove w left-parenthesis f comma g right-parenthesis With bar are algebraically dependent, then ModifyingAbove w left-parenthesis f comma g right-parenthesis With bar comma ModifyingAbove h left-parenthesis f comma g right-parenthesis With bar are algebraically dependent too. Furthermore, since degree left-parenthesis w left-parenthesis f comma g right-parenthesis right-parenthesis equals degree left-parenthesis h left-parenthesis f comma g right-parenthesis right-parenthesis , the elements ModifyingAbove w left-parenthesis f comma g right-parenthesis With bar comma ModifyingAbove h left-parenthesis f comma g right-parenthesis With bar are linearly dependent, and thus ModifyingAbove w left-parenthesis f comma g right-parenthesis With bar element-of mathematical left-angle f overbar comma g overbar mathematical right-angle . This again contradicts the definition of w left-parenthesis x comma y right-parenthesis .

Note that degree left-parenthesis h left-parenthesis f comma g right-parenthesis right-parenthesis less-than-or-equal-to degree left-parenthesis w left-parenthesis f comma g right-parenthesis right-parenthesis . Since ModifyingAbove w left-parenthesis f comma g right-parenthesis With bar not-an-element-of mathematical left-angle f overbar comma g overbar mathematical right-angle , Corollary 1 yields 3). If degree left-parenthesis w left-parenthesis f comma g right-parenthesis right-parenthesis less-than degree g , then by Lemma 3 we get h left-parenthesis x comma y right-parenthesis equals q left-parenthesis x right-parenthesis , degree left-parenthesis q left-parenthesis f right-parenthesis right-parenthesis equals degree left-parenthesis h left-parenthesis f comma g right-parenthesis right-parenthesis less-than-or-equal-to degree left-parenthesis w left-parenthesis f comma g right-parenthesis right-parenthesis . This proves statements 4), 5) of the lemma.

Observe that in view of 3) and Lemma 2.iii), conditions 4), 5) of Lemma 5 may take place only for 2 -reduced pairs.

Lemma 6

Let w left-parenthesis x comma y right-parenthesis be a derivative polynomial of the pair f comma g and u equals w left-parenthesis f comma g right-parenthesis . Then the highest homogeneous parts of the elements of the type

StartLayout 1st Row f Superscript i Baseline g Superscript j Baseline u Superscript t Baseline comma j less-than p comma 0 less-than-or-equal-to t less-than-or-equal-to 1 comma EndLayout

are linearly independent.

Proof.

Assuming the contrary, we get by Lemma 3 the equality of the form

StartLayout 1st Row f overbar Superscript i Baseline g overbar Superscript j Baseline u overbar equals beta f overbar Superscript i 1 Baseline g overbar Superscript j 1 Baseline comma j comma j 1 less-than p period EndLayout

If i less-than-or-equal-to i 1 , j less-than-or-equal-to j 1 , then this equality implies u overbar element-of mathematical left-angle f overbar comma g overbar mathematical right-angle , which is impossible by the definition of w left-parenthesis x comma y right-parenthesis . Assume that j less-than-or-equal-to j 1 , i greater-than i 1 . Then

StartLayout 1st Row f overbar Superscript i minus i 1 Baseline u overbar equals beta g overbar Superscript j 1 minus j Baseline period EndLayout

Since u overbar not-an-element-of mathematical left-angle f overbar comma g overbar mathematical right-angle , by Corollary 1 we have

StartLayout 1st Row degree u greater-than-or-equal-to upper N left-parenthesis f comma g right-parenthesis equals p m minus m minus n plus degree left-bracket f comma g right-bracket period EndLayout

Therefore,

StartLayout 1st Row 1st Column degree left-parenthesis g overbar Superscript j 1 minus j Baseline right-parenthesis equals m left-parenthesis j 1 minus j right-parenthesis 2nd Column greater-than-or-equal-to 3rd Column n left-parenthesis i minus i 1 right-parenthesis plus p m minus m minus n plus degree left-bracket f comma g right-bracket 2nd Row 1st Column Blank 2nd Column greater-than-or-equal-to 3rd Column left-parenthesis p minus 1 right-parenthesis m plus degree left-bracket f comma g right-bracket greater-than left-parenthesis p minus 1 right-parenthesis m period EndLayout

This contradicts the inequality j 1 minus j less-than-or-equal-to p minus 1 . If i less-than-or-equal-to i 1 , j greater-than j 1 , then

StartLayout 1st Row g overbar Superscript j minus j 1 Baseline u overbar equals beta f overbar Superscript i 1 minus i Baseline period EndLayout

Since j minus j 1 less-than-or-equal-to p minus 1 and g overbar Superscript p Baseline equals alpha f overbar Superscript s , we may assume that i 1 minus i less-than s ; otherwise u overbar element-of mathematical left-angle f overbar comma g overbar mathematical right-angle . Thus,

StartLayout 1st Row 1st Column degree left-parenthesis f overbar Superscript i 1 minus i Baseline right-parenthesis 2nd Column equals 3rd Column n left-parenthesis i 1 minus i right-parenthesis greater-than-or-equal-to m left-parenthesis j minus j 1 right-parenthesis plus p m minus m minus n plus degree left-bracket f comma g right-bracket 2nd Row 1st Column Blank 2nd Column greater-than-or-equal-to 3rd Column p m minus n plus degree left-bracket f comma g right-bracket greater-than p m minus n equals n s minus n equals n left-parenthesis s minus 1 right-parenthesis comma EndLayout

which is impossible.

Lemma 7

Let w left-parenthesis x comma y right-parenthesis be a derivative polynomial of the pair f comma g , and upper T left-parenthesis x comma y right-parenthesis element-of upper F left-bracket x comma y right-bracket , degree Subscript y Baseline left-parenthesis upper T left-parenthesis x comma y right-parenthesis right-parenthesis less-than 2 p . Then, the following statements are true:

1 right-parenthesis

upper T left-parenthesis x comma y right-parenthesis can be uniquely presented in the form StartLayout 1st Row upper T left-parenthesis x comma y right-parenthesis equals w left-parenthesis x comma y right-parenthesis q left-parenthesis x comma y right-parenthesis plus s left-parenthesis x comma y right-parenthesis comma EndLayout

where degree Subscript y Baseline left-parenthesis q left-parenthesis x comma y right-parenthesis right-parenthesis comma degree Subscript y Baseline left-parenthesis s left-parenthesis x comma y right-parenthesis right-parenthesis less-than p ;

2 right-parenthesis

if degree left-parenthesis upper T left-parenthesis f comma g right-parenthesis right-parenthesis less-than-or-equal-to t , then StartLayout 1st Row degree left-parenthesis w left-parenthesis f comma g right-parenthesis right-parenthesis plus degree left-parenthesis q left-parenthesis f comma g right-parenthesis right-parenthesis less-than-or-equal-to t comma degree left-parenthesis s left-parenthesis f comma g right-parenthesis right-parenthesis less-than-or-equal-to t semicolon EndLayout

3 right-parenthesis

if ModifyingAbove upper T left-parenthesis f comma g right-parenthesis With bar not-an-element-of mathematical left-angle f overbar comma g overbar mathematical right-angle , then q left-parenthesis x comma y right-parenthesis not-equals 0 and StartLayout 1st Row degree left-parenthesis StartFraction partial-differential upper T Over partial-differential x EndFraction left-parenthesis f comma g right-parenthesis right-parenthesis equals degree left-parenthesis q left-parenthesis f comma g right-parenthesis right-parenthesis plus n left-parenthesis s minus 1 right-parenthesis comma 2nd Row degree left-parenthesis StartFraction partial-differential upper T Over partial-differential y EndFraction left-parenthesis f comma g right-parenthesis right-parenthesis equals degree left-parenthesis q left-parenthesis f comma g right-parenthesis right-parenthesis plus m left-parenthesis p minus 1 right-parenthesis period EndLayout

4 right-parenthesis

if ModifyingAbove upper T left-parenthesis f comma g right-parenthesis With bar not-an-element-of mathematical left-angle f overbar comma g overbar mathematical right-angle and degree left-parenthesis upper T left-parenthesis f comma g right-parenthesis right-parenthesis less-than min left-brace n plus upper N comma m p right-brace , then upper T left-parenthesis x comma y right-parenthesis equals lamda w 1 left-parenthesis f comma g right-parenthesis comma 0 not-equals lamda element-of upper F comma where w 1 left-parenthesis x comma y right-parenthesis is also a derivative polynomial of the pair f comma g .

Proof.

The first statement of the lemma follows from the division algorithm in the ring left-parenthesis upper F left-bracket x right-bracket right-parenthesis left-bracket y right-bracket ; one may divide upper T left-parenthesis x comma y right-parenthesis by w left-parenthesis x comma y right-parenthesis since the last polynomial is monic. Furthermore, the right part of the equality upper T left-parenthesis f comma g right-parenthesis equals w left-parenthesis f comma g right-parenthesis q left-parenthesis f comma g right-parenthesis plus s left-parenthesis f comma g right-parenthesis is a linear combination of elements indicated in Lemma 6; therefore, by this lemma, only the elements of degree less than or equal to degree left-parenthesis upper T left-parenthesis f comma g right-parenthesis right-parenthesis may appear in this combination. This proves 2).

If ModifyingAbove upper T left-parenthesis f comma g right-parenthesis With bar not-an-element-of mathematical left-angle f overbar comma g overbar mathematical right-angle , then by Corollary 1 we have degree Subscript y Baseline upper T left-parenthesis x comma y right-parenthesis greater-than-or-equal-to p and hence q left-parenthesis x comma y right-parenthesis not-equals 0 . By Corollary 1 again, ModifyingAbove s left-parenthesis f comma g right-parenthesis With bar element-of mathematical left-angle f overbar comma g overbar mathematical right-angle ; hence ModifyingAbove upper T left-parenthesis f comma g right-parenthesis With bar not-equals ModifyingAbove s left-parenthesis f comma g right-parenthesis With bar . Consequently, by Lemma 6,

StartLayout 1st Row degree left-parenthesis upper T left-parenthesis f comma g right-parenthesis right-parenthesis equals degree left-parenthesis w left-parenthesis f comma g right-parenthesis q left-parenthesis f comma g right-parenthesis right-parenthesis greater-than-or-equal-to degree left-parenthesis s left-parenthesis f comma g right-parenthesis right-parenthesis period EndLayout

It follows from the definition of w left-parenthesis x comma y right-parenthesis in Lemma 4 that

StartLayout 1st Row degree left-parenthesis StartFraction partial-differential w Over partial-differential x EndFraction left-parenthesis f comma g right-parenthesis right-parenthesis equals n left-parenthesis s minus 1 right-parenthesis comma degree left-parenthesis StartFraction partial-differential w Over partial-differential y EndFraction left-parenthesis f comma g right-parenthesis right-parenthesis equals m left-parenthesis p minus 1 right-parenthesis period EndLayout

Furthermore,

StartLayout 1st Row StartFraction partial-differential upper T Over partial-differential x EndFraction left-parenthesis x comma y right-parenthesis equals StartFraction partial-differential w Over partial-differential x EndFraction left-parenthesis x comma y right-parenthesis q left-parenthesis x comma y right-parenthesis plus w left-parenthesis x comma y right-parenthesis StartFraction partial-differential q Over partial-differential x EndFraction left-parenthesis x comma y right-parenthesis plus StartFraction partial-differential s Over partial-differential x EndFraction left-parenthesis x comma y right-parenthesis period EndLayout

Easy calculations give

StartLayout 1st Row 1st Column degree left-parenthesis w left-parenthesis f comma g right-parenthesis StartFraction partial-differential q Over partial-differential x EndFraction left-parenthesis f comma g right-parenthesis right-parenthesis 2nd Column less-than-or-equal-to 3rd Column degree left-parenthesis w left-parenthesis f comma g right-parenthesis right-parenthesis plus degree left-parenthesis q left-parenthesis f comma g right-parenthesis right-parenthesis minus degree f comma 2nd Row 1st Column degree left-parenthesis StartFraction partial-differential s Over partial-differential x EndFraction left-parenthesis f comma g right-parenthesis right-parenthesis 2nd Column less-than-or-equal-to 3rd Column degree left-parenthesis upper T left-parenthesis f comma g right-parenthesis right-parenthesis minus degree f 3rd Row 1st Column Blank 2nd Column equals 3rd Column degree left-parenthesis w left-parenthesis f comma g right-parenthesis right-parenthesis plus degree left-parenthesis q left-parenthesis f comma g right-parenthesis right-parenthesis minus degree f comma 4th Row 1st Column degree left-parenthesis StartFraction partial-differential w Over partial-differential x EndFraction left-parenthesis f comma g right-parenthesis q left-parenthesis f comma g right-parenthesis right-parenthesis 2nd Column equals 3rd Column degree left-parenthesis q left-parenthesis f comma g right-parenthesis right-parenthesis plus n left-parenthesis s minus 1 right-parenthesis equals degree left-parenthesis q left-parenthesis f comma g right-parenthesis right-parenthesis plus m p minus n 5th Row 1st Column Blank 2nd Column equals 3rd Column degree left-parenthesis w left-parenthesis f comma g right-parenthesis right-parenthesis plus degree left-parenthesis q left-parenthesis f comma g right-parenthesis right-parenthesis minus n plus left-parenthesis m p minus degree left-parenthesis w left-parenthesis f comma g right-parenthesis right-parenthesis right-parenthesis 6th Row 1st Column Blank 2nd Column greater-than 3rd Column degree left-parenthesis w left-parenthesis f comma g right-parenthesis right-parenthesis plus degree left-parenthesis q left-parenthesis f comma g right-parenthesis right-parenthesis minus n period EndLayout

Therefore,

StartLayout 1st Row degree left-parenthesis StartFraction partial-differential upper T Over partial-differential x EndFraction left-parenthesis f comma g right-parenthesis right-parenthesis equals degree left-parenthesis q left-parenthesis f comma g right-parenthesis right-parenthesis plus n left-parenthesis s minus 1 right-parenthesis period EndLayout

Similar calculations give the value of degree left-parenthesis StartFraction partial-differential upper T Over partial-differential y EndFraction left-parenthesis f comma g right-parenthesis right-parenthesis .

To prove 4) we note first that by Lemma 5.3), degree left-parenthesis w left-parenthesis f comma g right-parenthesis right-parenthesis greater-than-or-equal-to upper N . Hence by statement 2) of this lemma, degree left-parenthesis q left-parenthesis f comma g right-parenthesis right-parenthesis less-than n and degree left-parenthesis s left-parenthesis f comma g right-parenthesis right-parenthesis less-than m p . By Corollary 1, ModifyingAbove q left-parenthesis f comma g right-parenthesis With bar element-of mathematical left-angle f overbar comma g overbar mathematical right-angle . Hence 0 not-equals q left-parenthesis x comma y right-parenthesis equals lamda element-of upper F . Now it is easy to see that the polynomial w 1 left-parenthesis x comma y right-parenthesis equals lamda Superscript negative 1 Baseline upper T left-parenthesis x comma y right-parenthesis equals w left-parenthesis x comma y right-parenthesis plus lamda Superscript negative 1 Baseline s left-parenthesis x comma y right-parenthesis is a derivative polynomial of the pair f comma g .

We give two corollaries that will be useful for references.

Corollary 2

If h element-of mathematical left-angle f comma g mathematical right-angle minus upper F and degree h less-than n , then h equals lamda w left-parenthesis f comma g right-parenthesis comma 0 not-equals lamda element-of upper F comma where w left-parenthesis x comma y right-parenthesis is a derivative polynomial of the pair f comma g .

Proof.

Put h equals upper T left-parenthesis f comma g right-parenthesis and let degree Subscript y Baseline upper T left-parenthesis x comma y right-parenthesis equals p q plus r comma r less-than p . If q equals 0 , then Corollary 1 gives h overbar element-of mathematical left-angle f overbar comma g overbar mathematical right-angle and degree h greater-than-or-equal-to n or h element-of upper F , a contradiction. Hence q greater-than 0 . Inequality (Equation2) gives degree h greater-than-or-equal-to q upper N plus m r . Consequently, r equals 0 . If q greater-than 1 , then by Lemma 2, degree h greater-than-or-equal-to 2 upper N greater-than n . Therefore, q equals 1 and degree Subscript y Baseline left-parenthesis upper T left-parenthesis x comma y right-parenthesis right-parenthesis equals p . Then Lemma 7.4) proves the corollary.

Corollary 3

If w left-parenthesis x comma y right-parenthesis is a derivative polynomial of the pair f comma g , then

StartLayout 1st Row degree left-parenthesis StartFraction partial-differential w Over partial-differential x EndFraction left-parenthesis f comma g right-parenthesis right-parenthesis equals n left-parenthesis s minus 1 right-parenthesis comma 2nd Row degree left-parenthesis StartFraction partial-differential w Over partial-differential y EndFraction left-parenthesis f comma g right-parenthesis right-parenthesis equals m left-parenthesis p minus 1 right-parenthesis period EndLayout

3. Reductions and simple automorphisms

A triple theta equals left-parenthesis f 1 comma f 2 comma f 3 right-parenthesis (or simply left-parenthesis f 1 comma f 2 comma f 3 right-parenthesis ) of elements of the algebra upper A below will always denote the automorphism theta of upper A such that theta left-parenthesis x Subscript i Baseline right-parenthesis equals f Subscript i , 1 less-than-or-equal-to i less-than-or-equal-to 3 . The number degree theta equals degree f 1 plus degree f 2 plus degree f 3 will be called a degree of the automorphism theta .

Recall that an elementary transformation of the triple left-parenthesis f 1 comma f 2 comma f 3 right-parenthesis is, by definition, a transformation that changes only one element f Subscript i to an element of the form alpha f Subscript i plus g , where 0 not-equals alpha element-of upper F comma g element-of mathematical left-angle StartSet f Subscript j Baseline vertical-bar j not-equals i EndSet mathematical right-angle . The notation

StartLayout 1st Row left-parenthesis f 1 comma f 2 comma f 3 right-parenthesis right-arrow left-parenthesis g 1 comma g 2 comma g 3 right-parenthesis EndLayout

means that the triple left-parenthesis g 1 comma g 2 comma g 3 right-parenthesis is obtained from left-parenthesis f 1 comma f 2 comma f 3 right-parenthesis by a single elementary transformation. Observe that we do not assume that degree left-parenthesis g 1 comma g 2 comma g 3 right-parenthesis should be smaller than degree theta . An automorphism left-parenthesis f 1 comma f 2 comma f 3 right-parenthesis is called tame if there exists a sequence of elementary transformations of the form

StartLayout 1st Row left-parenthesis x 1 comma x 2 comma x 3 right-parenthesis equals left-parenthesis f 1 Superscript left-parenthesis 0 right-parenthesis Baseline comma f 2 Superscript left-parenthesis 0 right-parenthesis Baseline comma f 3 Superscript left-parenthesis 0 right-parenthesis Baseline right-parenthesis right-arrow left-parenthesis f 1 Superscript left-parenthesis 1 right-parenthesis Baseline comma f 2 Superscript left-parenthesis 1 right-parenthesis Baseline comma f 3 Superscript left-parenthesis 1 right-parenthesis Baseline right-parenthesis right-arrow period period period 2nd Row left-parenthesis f 1 Superscript left-parenthesis n right-parenthesis Baseline comma f 2 Superscript left-parenthesis n right-parenthesis Baseline comma f 3 Superscript left-parenthesis n right-parenthesis Baseline right-parenthesis right-arrow left-parenthesis f 1 comma f 2 comma f 3 right-parenthesis period EndLayout

The element f 1 of the automorphism theta equals left-parenthesis f 1 comma f 2 comma f 3 right-parenthesis is called reducible, if there exists g element-of mathematical left-angle f 2 comma f 3 mathematical right-angle such that f overbar Subscript 1 Baseline equals g overbar ; otherwise it is called unreducible. Put f prime 1 equals alpha left-parenthesis f 1 minus g right-parenthesis , where 0 not-equals alpha element-of upper F ; then degree f prime 1 less-than degree f 1 and degree left-parenthesis f prime 1 comma f 2 comma f 3 right-parenthesis less-than degree theta . In this case we will say also that f 1 is reduced in theta by the automorphism left-parenthesis f prime 1 comma f 2 comma f 3 right-parenthesis . If one of the elements f 1 comma f 2 comma f 3 of theta is reducible, then we will say that theta admits an elementary reduction or simply that theta is elementary reducible.

Lemma 8

The elementary reducibility of automorphisms of the algebra upper A is algorithmically recognizable.Footnote1 In formulation of algorithmic results, we always assume that the ground field upper F is constructive.

Proof.

Let theta equals left-parenthesis f 1 comma f 2 comma f 3 right-parenthesis be an arbitrary automorphism of upper A . We will recognize the reducibility of f 3 . If f overbar Subscript 1 Baseline comma f overbar Subscript 2 Baseline are algebraically independent, then f 3 is reducible if and only if f overbar Subscript 3 Baseline element-of mathematical left-angle f overbar Subscript 1 Baseline comma f overbar Subscript 2 Baseline mathematical right-angle . Since f overbar Subscript 1 Baseline comma f overbar Subscript 2 Baseline are homogeneous, this question can be solved trivially, even without a reference to the solubility of the occurrence problem Reference13, Reference14. If f overbar Subscript 2 Baseline element-of mathematical left-angle f overbar Subscript 1 Baseline mathematical right-angle and f overbar Subscript 2 Baseline equals alpha f overbar Subscript 1 Superscript k , then the element f 3 is reducible in theta if and only if it is reducible in the automorphism left-parenthesis f 1 comma f 2 minus alpha f 1 Superscript k Baseline comma f 3 right-parenthesis . Since degree left-parenthesis f 1 comma f 2 minus alpha f 1 Superscript k Baseline comma f 3 right-parenthesis less-than degree theta , the statement of the lemma in this case can be proved by induction on degree theta .

Let now f 1 comma f 2 be a asterisk -reduced pair and degree f 1 less-than degree f 2 . Assume that there exists a polynomial upper G left-parenthesis x comma y right-parenthesis element-of upper F left-bracket x comma y right-bracket such that f overbar Subscript 3 Baseline equals ModifyingAbove upper G left-parenthesis f 1 comma f 2 right-parenthesis With bar . Inequalities (Equation2), (Equation3) gives a bound k for the numbers degree Subscript x Baseline left-parenthesis upper G left-parenthesis x comma y right-parenthesis right-parenthesis comma degree Subscript y Baseline left-parenthesis upper G left-parenthesis x comma y right-parenthesis right-parenthesis . Then upper G left-parenthesis f 1 comma f 2 right-parenthesis is in the space generated by the elements f 1 Superscript i Baseline f 2 Superscript j , where i comma j less-than-or-equal-to k . The highest homogeneous parts of elements of this space can be described by triangulation.

Now we give an example of a tame automorphism, which does not admit an elementary reduction.

Example 1. Put

StartLayout 1st Row h 1 equals x 1 comma h 2 equals x 2 plus x 1 squared comma h 3 equals x 3 plus 2 x 1 x 2 plus x 1 cubed comma 2nd Row g 1 equals 6 h 1 plus 6 h 2 h 3 plus h 3 cubed comma g 2 equals 4 h 2 plus h 3 squared comma g 3 equals h 3 period EndLayout

It is easy to show that left-parenthesis h 1 comma h 2 comma h 3 right-parenthesis and left-parenthesis g 1 comma g 2 comma g 3 right-parenthesis are tame automorphisms of the algebra upper A . Note that degree g 1 equals 9 , degree g 2 equals 6 , degree g 3 equals 3 and g 1 comma g 2 form a 2 -reduced pair. A direct calculation shows that the element

StartLayout 1st Row f equals g 1 squared minus g 2 cubed EndLayout

has degree 8. Hence degree f less-than degree g 1 , and f overbar not-an-element-of mathematical left-angle g overbar Subscript 1 Baseline comma g overbar Subscript 2 Baseline mathematical right-angle .

Now we define a tame automorphism left-parenthesis f 1 comma f 2 comma f 3 right-parenthesis by putting

StartLayout 1st Row f 1 equals g 1 plus g 3 plus f comma f 2 equals g 2 comma f 3 equals g 3 plus f period EndLayout

We have degree f 1 equals 9 , degree f 2 equals 6 , degree f 3 equals 8 and degree left-bracket f Subscript i Baseline comma f Subscript j Baseline right-bracket greater-than 9 , 1 less-than-or-equal-to i less-than j less-than-or-equal-to 3 . Then, using inequality (Equation2), it is easy to check that the automorphism left-parenthesis f 1 comma f 2 comma f 3 right-parenthesis does not admit an elementary reduction.

Proposition 1

Let theta equals left-parenthesis f 1 comma f 2 comma f 3 right-parenthesis be an automorphism of upper A such that degree f 1 equals 2 n , degree f 2 equals n s , s greater-than-or-equal-to 3 is an odd number, 2 n less-than degree f 3 less-than-or-equal-to n s , f overbar Subscript 3 Baseline not-an-element-of mathematical left-angle f overbar Subscript 1 Baseline comma f overbar Subscript 2 Baseline mathematical right-angle . Suppose that there exists 0 not-equals alpha element-of upper F such that the elements g 1 equals f 1 , g 2 equals f 2 minus alpha f 3 satisfy the conditions:

i right-parenthesis

g 1 comma g 2 is a 2 -reduced pair and degree g 1 equals 2 n , degree g 2 equals n s ;

i i right-parenthesis

the element f 3 of the automorphism left-parenthesis g 1 comma g 2 comma f 3 right-parenthesis is reduced by an automorphism left-parenthesis g 1 comma g 2 comma g 3 right-parenthesis with the condition degree left-bracket g 1 comma g 3 right-bracket less-than n s plus degree left-bracket g 1 comma g 2 right-bracket .

Then, the following statements are true:

1 right-parenthesis

ModifyingAbove left-bracket f 1 comma f 2 right-bracket With bar equals alpha ModifyingAbove left-bracket f 1 comma f 3 right-bracket With bar ;

2 right-parenthesis

degree left-bracket f Subscript i Baseline comma f Subscript j Baseline right-bracket greater-than n s , where 1 less-than-or-equal-to i less-than j less-than-or-equal-to 3 ;

3 right-parenthesis

if f element-of mathematical left-angle f Subscript i Baseline comma f Subscript j Baseline mathematical right-angle , where 1 less-than-or-equal-to i less-than j less-than-or-equal-to 3 , then either f overbar element-of mathematical left-angle f overbar Subscript i Baseline comma f overbar Subscript j Baseline mathematical right-angle or degree f greater-than n s ;

4 right-parenthesis

degree f 1 plus degree f 3 greater-than n s .

Proof.

We have

StartLayout 1st Row with Label left-parenthesis 4 right-parenthesis EndLabel g 3 equals sigma f 3 plus upper G left-parenthesis g 1 comma g 2 right-parenthesis comma degree g 3 less-than degree f 3 comma EndLayout

where 0 not-equals sigma element-of upper F , upper G left-parenthesis x comma y right-parenthesis element-of upper F left-bracket x comma y right-bracket . Hence

StartLayout 1st Row with Label left-parenthesis 5 right-parenthesis EndLabel ModifyingAbove upper G left-parenthesis g 1 comma g 2 right-parenthesis With bar equals minus sigma f overbar Subscript 3 Baseline period EndLayout

If degree f 3 equals n s , then f overbar Subscript 2 Baseline comma f overbar Subscript 3 Baseline are linearly independent, since f overbar Subscript 3 Baseline not-an-element-of mathematical left-angle f overbar Subscript 1 Baseline comma f overbar Subscript 2 Baseline mathematical right-angle . Therefore f overbar Subscript 2 Baseline comma f overbar Subscript 3 Baseline comma g overbar Subscript 2 Baseline are mutually linearly independent and f overbar Subscript 3 Baseline not-an-element-of mathematical left-angle f overbar Subscript 1 Baseline comma g overbar Subscript 2 Baseline mathematical right-angle equals mathematical left-angle g overbar Subscript 1 Baseline comma g overbar Subscript 2 Baseline mathematical right-angle . If degree f 3 less-than n s , then f overbar Subscript 2 Baseline equals g overbar Subscript 2 and again we have f overbar Subscript 3 Baseline not-an-element-of mathematical left-angle g overbar Subscript 1 Baseline comma g overbar Subscript 2 Baseline mathematical right-angle . Put degree Subscript y Baseline left-parenthesis upper G left-parenthesis x comma y right-parenthesis right-parenthesis equals k equals 2 q plus r , 0 less-than-or-equal-to r less-than-or-equal-to 1 . The condition f overbar Subscript 3 Baseline not-an-element-of mathematical left-angle g overbar Subscript 1 Baseline comma g overbar Subscript 2 Baseline mathematical right-angle implies, by Corollary 1 and (Equation5), that q greater-than-or-equal-to 1 . Then inequality (Equation2) gives r equals 0 and

StartLayout 1st Row n s greater-than-or-equal-to degree left-parenthesis upper G left-parenthesis g 1 comma g 2 right-parenthesis right-parenthesis equals degree f 3 greater-than-or-equal-to q left-parenthesis n s minus 2 n plus degree left-bracket g 1 comma g 2 right-bracket right-parenthesis period EndLayout

It is easy to deduce from here that if s greater-than 3 , then q equals 1 , k equals 2 , and if s equals 3 , then q equals 1 comma 2 , k equals 2 comma 4 . Besides, these inequalities imply degree left-bracket g 1 comma g 2 right-bracket less-than-or-equal-to 2 n and statement 4) of the proposition.

Applying (Equation1), we get from (Equation4),

StartLayout 1st Row left-bracket g 1 comma g 3 right-bracket equals sigma left-bracket g 1 comma f 3 right-bracket plus left-bracket g 1 comma g 2 right-bracket StartFraction partial-differential upper G Over partial-differential y EndFraction left-parenthesis g 1 comma g 2 right-parenthesis period EndLayout

Since degree Subscript y Baseline left-parenthesis StartFraction partial-differential upper G Over partial-differential y EndFraction right-parenthesis equals k minus 1 is an odd number, inequality (Equation2) gives

StartLayout 1st Row degree left-parenthesis left-bracket g 1 comma g 2 right-bracket StartFraction partial-differential upper G Over partial-differential y EndFraction left-parenthesis g 1 comma g 2 right-parenthesis right-parenthesis greater-than-or-equal-to degree left-bracket g 1 comma g 2 right-bracket plus n s period EndLayout

Consequently, by condition ii),

StartLayout 1st Row degree left-bracket f 1 comma f 3 right-bracket equals degree left-bracket g 1 comma f 3 right-bracket greater-than-or-equal-to degree left-bracket g 1 comma g 2 right-bracket plus n s period EndLayout

Since degree left-bracket g 1 comma g 2 right-bracket less-than-or-equal-to 2 n and alpha not-equals 0 , the equality

StartLayout 1st Row left-bracket g 1 comma g 2 right-bracket equals left-bracket f 1 comma f 2 right-bracket minus alpha left-bracket f 1 comma f 3 right-bracket EndLayout

gives statement 1) of the proposition and

StartLayout 1st Row degree left-bracket f 1 comma f 2 right-bracket equals degree left-bracket f 1 comma f 3 right-bracket greater-than n s period EndLayout

If degree f 3 equals n s , then, as was remarked earlier, f overbar Subscript 2 Baseline comma f overbar Subscript 3 Baseline are algebraically independent, and so (see Reference16)

StartLayout 1st Row degree left-bracket f 2 comma f 3 right-bracket equals degree f 2 plus degree f 3 greater-than n s period EndLayout

If degree f 3 less-than n s , then we have

StartLayout 1st Row degree left-bracket f 1 comma f 2 right-bracket plus degree f 3 less-than degree left-bracket f 1 comma f 3 right-bracket plus degree f 2 period EndLayout

By Lemma 1,

StartLayout 1st Row degree left-bracket f 2 comma f 3 right-bracket plus degree f 1 equals degree left-bracket f 1 comma f 3 right-bracket plus degree f 2 semicolon EndLayout

hence

StartLayout 1st Row degree left-bracket f 2 comma f 3 right-bracket equals degree left-bracket f 1 comma f 3 right-bracket plus n left-parenthesis s minus 2 right-parenthesis greater-than n s period EndLayout

Thus statement 2) of the proposition is proved.

To prove 3), it suffices by F2) to consider only the case when f overbar Subscript i Baseline comma f overbar Subscript j Baseline are algebraically dependent. It is easily seen that f 1 comma f 2 and f 1 comma f 3 are asterisk -reduced pairs. Suppose that f overbar Subscript 2 Baseline element-of mathematical left-angle f overbar Subscript 3 Baseline mathematical right-angle . If degree f 2 equals degree f 3 then f overbar Subscript 3 Baseline element-of mathematical left-angle f overbar Subscript 2 Baseline mathematical right-angle , which contradicts the condition of the proposition. Otherwise degree f 2 greater-than-or-equal-to 2 degree f 3 greater-than degree f 1 plus degree f 3 , which contradicts 4). Consequently, the pair f Subscript i Baseline comma f Subscript j Baseline is asterisk -reduced for every i not-equals j and Corollary 1 by Lemma 2.ii) implies 3).

Definition 1

If an automorphism theta equals left-parenthesis f 1 comma f 2 comma f 3 right-parenthesis satisfies the conditions of proposition 1 , then we will say that theta admits a reduction of type I, and the automorphism left-parenthesis g 1 comma g 2 comma g 3 right-parenthesis will be called a reduction of type I of the automorphism theta , with an active element f 3 .

The automorphism from Example 1 admits a reduction of type I.

Proposition 2

Let theta equals left-parenthesis f 1 comma f 2 comma f 3 right-parenthesis be an automorphism of upper A such that degree f 1 equals 2 n , degree f 2 equals 3 n , StartFraction 3 n Over 2 EndFraction less-than degree f 3 less-than-or-equal-to 2 n , and f overbar Subscript 1 Baseline comma f overbar Subscript 3 Baseline are linearly independent. Suppose that there exist alpha comma beta element-of upper F , where left-parenthesis alpha comma beta right-parenthesis not-equals left-parenthesis 0 comma 0 right-parenthesis , such that the elements g 1 equals f 1 minus alpha f 3 , g 2 equals f 2 minus beta f 3 satisfy the conditions:

i right-parenthesis

g 1 comma g 2 is a 2 -reduced pair and degree g 1 equals 2 n , degree g 2 equals 3 n ;

i i right-parenthesis

the element f 3 of the automorphism left-parenthesis g 1 comma g 2 comma f 3 right-parenthesis is reduced by an automorphism left-parenthesis g 1 comma g 2 comma g 3 right-parenthesis with the condition degree left-bracket g 1 comma g 3 right-bracket less-than 3 n plus degree left-bracket g 1 comma g 2 right-bracket .

Then, the following statements are true:

1 right-parenthesis

alpha element-of upper F is the solution of the equation ModifyingAbove left-bracket f 1 comma f 2 right-bracket With bar equals alpha ModifyingAbove left-bracket f 3 comma f 2 right-bracket With bar , or alpha equals 0 if it has no solution;

2 right-parenthesis

beta element-of upper F is the solution of the equation ModifyingAbove left-bracket g 1 comma f 2 right-bracket With bar equals beta ModifyingAbove left-bracket g 1 comma f 3 right-bracket With bar , or beta equals 0 if it has no solution;

3 right-parenthesis

degree left-bracket f Subscript i Baseline comma f Subscript j Baseline right-bracket greater-than 3 n , where 1 less-than-or-equal-to i less-than j less-than-or-equal-to 3 ;

4 right-parenthesis

if f element-of mathematical left-angle f Subscript i Baseline comma f Subscript j Baseline mathematical right-angle , where 1 less-than-or-equal-to i less-than j less-than-or-equal-to 3 , then either f overbar element-of mathematical left-angle f overbar Subscript i Baseline comma f overbar Subscript j Baseline mathematical right-angle or degree f greater-than 3 n .

Proof.

Consider equalities (Equation4), (Equation5). If degree f 3 less-than 2 n , then obviously ModifyingAbove upper G left-parenthesis g 1 comma g 2 right-parenthesis With bar not-an-element-of mathematical left-angle g overbar Subscript 1 Baseline comma g overbar Subscript 2 Baseline mathematical right-angle . If degree f 3 equals 2 n , then by the condition of the proposition, f overbar Subscript 1 Baseline comma f overbar Subscript 3 Baseline are linearly independent. Therefore, either alpha equals 0 and g 1 equals f 1 , or alpha not-equals 0 and g overbar Subscript 1 Baseline comma f overbar Subscript 1 Baseline comma f overbar Subscript 3 Baseline are mutually linearly independent. In any case, ModifyingAbove upper G left-parenthesis g 1 comma g 2 right-parenthesis With bar not-an-element-of mathematical left-angle g overbar Subscript 1 Baseline comma g overbar Subscript 2 Baseline mathematical right-angle . Since degree f 3 less-than-or-equal-to 2 n , then, as in the proof of Proposition 1, inequality (Equation2) gives that degree Subscript y Baseline left-parenthesis upper G left-parenthesis x comma y right-parenthesis right-parenthesis equals 2 , degree left-bracket g 1 comma g 2 right-bracket less-than-or-equal-to n . Consequently, Lemma 7.4) gives that upper G left-parenthesis x comma y right-parenthesis is a derivative polynomial (up to a nonzero scalar factor) of the pair g 1 comma g 2 , and by Corollary 3,

StartLayout 1st Row degree left-parenthesis StartFraction partial-differential upper G Over partial-differential y EndFraction left-parenthesis g 1 comma g 2 right-parenthesis right-parenthesis equals 3 n period EndLayout

From here, as in the proof of Proposition 1, we get

StartLayout 1st Row with Label left-parenthesis 6 right-parenthesis EndLabel degree left-bracket f 1 comma f 3 right-bracket equals degree left-bracket g 1 comma f 3 right-bracket equals degree left-bracket g 1 comma g 2 right-bracket plus 3 n period EndLayout

Consider the triple left-parenthesis g 1 comma g 2 comma f 3 right-parenthesis . By (Equation6) and Lemma 1,

StartLayout 1st Row degree left-bracket g 2 comma f 3 right-bracket plus degree g 1 equals degree left-bracket g 1 comma f 3 right-bracket plus degree g 2 comma EndLayout

which yields

StartLayout 1st Row with Label left-parenthesis 7 right-parenthesis EndLabel degree left-bracket f 2 comma f 3 right-bracket equals degree left-bracket g 2 comma f 3 right-bracket equals degree left-bracket g 1 comma f 3 right-bracket plus n period EndLayout

Furthermore,

StartLayout 1st Row left-bracket f 1 comma f 2 right-bracket equals left-bracket g 1 plus alpha f 3 comma g 2 plus beta f 3 right-bracket equals left-bracket g 1 comma g 2 right-bracket plus beta left-bracket g 1 comma f 3 right-bracket plus alpha left-bracket f 3 comma g 2 right-bracket period EndLayout

Since degree left-bracket g 1 comma g 2 right-bracket less-than-or-equal-to n , this implies, by (Equation6) and (Equation7), that ModifyingAbove left-bracket f 1 comma f 2 right-bracket With bar equals alpha ModifyingAbove left-bracket f 3 comma g 2 right-bracket With bar equals alpha ModifyingAbove left-bracket f 3 comma f 2 right-bracket With bar if alpha not-equals 0 , and ModifyingAbove left-bracket f 1 comma f 2 right-bracket With bar equals beta ModifyingAbove left-bracket g 1 comma f 3 right-bracket With bar if alpha equals 0 . Hence degree left-bracket f 1 comma f 2 right-bracket greater-than 3 n , and if alpha equals 0 , then left-bracket f 1 comma f 2 right-bracket , left-bracket f 3 comma f 2 right-bracket have different degrees. We have also

StartLayout 1st Row left-bracket g 1 comma f 2 right-bracket equals left-bracket g 1 comma g 2 plus beta f 3 right-bracket equals left-bracket g 1 comma g 2 right-bracket plus beta left-bracket g 1 comma f 3 right-bracket period EndLayout

Hence either beta not-equals 0 and ModifyingAbove left-bracket g 1 comma f 2 right-bracket With bar equals beta ModifyingAbove left-bracket g 1 comma f 3 right-bracket With bar , or the elements left-bracket g 1 comma f 2 right-bracket equals left-bracket g 1 comma g 2 right-bracket and left-bracket g 1 comma f 3 right-bracket have different degrees. This proves the statements 1), 2), 3) of the proposition. Finally, as in the proof of Proposition 1, Corollary 1 and Lemma 2.ii) give 4).

Definition 2

If an automorphism theta equals left-parenthesis f 1 comma f 2 comma f 3 right-parenthesis satisfies the conditions of proposition 2 , then we will say that theta admits a reduction of type II, and the automorphism left-parenthesis g 1 comma g 2 comma g 3 right-parenthesis will be called a reduction of type II of the automorphism theta , with an active element f 3 .

Proposition 3

Let theta equals left-parenthesis f 1 comma f 2 comma f 3 right-parenthesis be an automorphism of upper A such that degree f 1 equals 2 n , and either degree f 2 equals 3 n , n less-than degree f 3 less-than-or-equal-to StartFraction 3 n Over 2 EndFraction , or StartFraction 5 n Over 2 EndFraction less-than degree f 2 less-than-or-equal-to 3 n , degree f 3 equals StartFraction 3 n Over 2 EndFraction . Suppose that there exist alpha comma beta comma gamma element-of upper F such that the elements g 1 equals f 1 minus beta f 3 , g 2 equals f 2 minus gamma f 3 minus alpha f 3 squared satisfy the conditions:

i right-parenthesis

g 1 comma g 2 is a 2 -reduced pair and degree g 1 equals 2 n , degree g 2 equals 3 n ;

i i right-parenthesis

there exists an element g 3 of the form StartLayout 1st Row g 3 equals sigma f 3 plus g comma EndLayout

where 0 not-equals sigma element-of upper F , g element-of mathematical left-angle g 1 comma g 2 mathematical right-angle minus upper F , such that degree g 3 less-than-or-equal-to StartFraction 3 n Over 2 EndFraction , degree left-bracket g 1 comma g 3 right-bracket less-than 3 n plus degree left-bracket g 1 comma g 2 right-bracket .

Then, the following statements are true:

1 right-parenthesis

alpha element-of upper F is the solution of the equation ModifyingAbove left-bracket f 1 comma f 2 right-bracket With bar equals 2 alpha ModifyingAbove left-bracket f 1 comma f 3 right-bracket With bar f overbar Subscript 3 , or alpha equals 0 if it has no solution;

2 right-parenthesis

beta element-of upper F is the solution of the equation ModifyingAbove left-bracket f 2 minus alpha f 3 squared comma f 1 right-bracket With bar equals beta ModifyingAbove left-bracket f 2 comma f 3 right-bracket With bar , or beta equals 0 if it has no solution;

3 right-parenthesis

gamma element-of upper F is the solution of the equation ModifyingAbove left-bracket g 1 comma f 2 minus alpha f 3 squared right-bracket With bar equals gamma ModifyingAbove left-bracket g 1 comma f 3 right-bracket With bar , or gamma equals 0 if it has no solution;

4 right-parenthesis

degree left-bracket f 1 comma f 3 right-bracket comma degree left-bracket f 2 comma f 3 right-bracket greater-than 3 n ;

5 right-parenthesis

if left-parenthesis alpha comma beta comma gamma right-parenthesis not-equals left-parenthesis 0 comma 0 comma 0 right-parenthesis , then degree left-bracket f 1 comma f 2 right-bracket greater-than 3 n ; otherwise, degree left-bracket f 1 comma f 2 right-bracket equals degree left-bracket g 1 comma g 2 right-bracket less-than-or-equal-to StartFraction n Over 2 EndFraction ;

6 right-parenthesis

if g overbar Subscript 2 Baseline equals minus alpha f overbar Subscript 3 Superscript 2 , then StartFraction 5 n Over 2 EndFraction plus degree left-bracket g 1 comma g 2 right-bracket less-than-or-equal-to degree f 2 less-than 3 n ; otherwise, degree f 2 equals 3 n .

Proof.

Since degree f 3 comma degree g 3 less-than-or-equal-to StartFraction 3 n Over 2 EndFraction , condition i i right-parenthesis yields that degree g less-than-or-equal-to StartFraction 3 n Over 2 EndFraction . Then, by Corollary 2, g equals lamda w left-parenthesis g 1 comma g 2 right-parenthesis , where w left-parenthesis x comma y right-parenthesis is a derivative polynomial of the g 1 comma g 2 . Inequality (Equation2) gives also degree left-bracket g 1 comma g 2 right-bracket less-than-or-equal-to StartFraction n Over 2 EndFraction . As in the proof of Propositions 1, 2, we obtain also (Equation6), (Equation7), which yields 4). Besides, we have

StartLayout 1st Row left-bracket f 1 comma f 2 right-bracket equals left-bracket g 1 comma g 2 right-bracket plus gamma left-bracket g 1 comma f 3 right-bracket plus beta left-bracket f 3 comma g 2 right-bracket plus 2 alpha left-bracket g 1 comma f 3 right-bracket f 3 period EndLayout

Since degree f 3 greater-than n , this equality yields statements 1), 5) of the proposition. If g overbar Subscript 2 Baseline equals minus alpha f overbar Subscript 3 Superscript 2 , then alpha not-equals 0 , degree f 3 equals StartFraction 3 n Over 2 EndFraction , degree f 2 less-than 3 n , and

StartLayout 1st Row degree left-bracket f 1 comma f 2 right-bracket equals degree left-bracket g 1 comma f 3 right-bracket plus degree f 3 equals degree left-bracket g 1 comma g 2 right-bracket plus 3 n plus StartFraction 3 n Over 2 EndFraction period EndLayout

Consequently, degree f 2 greater-than-or-equal-to degree left-bracket g 1 comma g 2 right-bracket plus StartFraction 5 n Over 2 EndFraction , which proves 6). We have also

StartLayout 1st Row 1st Column left-bracket f 2 minus alpha f 3 squared comma f 1 right-bracket 2nd Column equals 3rd Column left-bracket g 2 comma g 1 right-bracket plus gamma left-bracket f 3 comma g 1 right-bracket plus beta left-bracket g 2 comma f 3 right-bracket comma 2nd Row 1st Column left-bracket g 1 comma f 2 minus alpha f 3 squared right-bracket 2nd Column equals 3rd Column left-bracket g 1 comma g 2 right-bracket plus gamma left-bracket g 1 comma f 3 right-bracket period EndLayout

These equalities imply statements 2), 3) of the proposition.

Definition 3

If an automorphism theta equals left-parenthesis f 1 comma f 2 comma f 3 right-parenthesis satisfies the conditions of Proposition 3 , and left-parenthesis alpha comma beta comma gamma right-parenthesis not-equals left-parenthesis 0 comma 0 comma 0 right-parenthesis , degree g 3 less-than n plus degree left-bracket g 1 comma g 2 right-bracket , then we will say that theta admits a reduction of type III, and the automorphism left-parenthesis g 1 comma g 2 comma g 3 right-parenthesis will be called a reduction of type III of the automorphism theta , with an active element f 3 .

Corollary 4

In the conditions of Proposition 3 , if the automorphism theta equals left-parenthesis f 1 comma f 2 comma f 3 right-parenthesis admits a reduction of type III, then degree left-parenthesis g 1 comma g 2 comma g 3 right-parenthesis less-than degree theta .

Proof.

By Definition 3, we have

StartLayout 1st Row degree left-parenthesis g 1 comma g 2 comma g 3 right-parenthesis less-than 6 n plus degree left-bracket g 1 comma g 2 right-bracket period EndLayout

Hence it is sufficient to prove that

StartLayout 1st Row with Label left-parenthesis 8 right-parenthesis EndLabel degree theta greater-than-or-equal-to 6 n plus degree left-bracket g 1 comma g 2 right-bracket period EndLayout

It follows from (Equation6) that degree f 3 greater-than-or-equal-to n plus degree left-bracket g 1 comma g 2 right-bracket . If g overbar Subscript 2 Baseline not-equals minus alpha f overbar Subscript 3 Superscript 2 , then Proposition 3.6) gives degree f 2 equals 3 n , which proves (Equation8). If g overbar Subscript 2 Baseline equals minus alpha f overbar Subscript 3 Superscript 2 , then, as shown above, degree f 3 equals StartFraction 3 n Over 2 EndFraction