# 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 of polynomials in three variables over a field of characteristic 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 are algorithmically recognizable.

## 1. Introduction

Let be the polynomial ring in the variables over a field , and let be the group of automorphisms of as an algebra over . An automorphism is called elementary if it has a form

where . The subgroup of generated by all the elementary automorphisms is called the tame subgroup, and the elements from this subgroup are called tame automorphisms of . Non-tame automorphisms of the algebra 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 , constructed by Nagata in 1972 (see Reference12):

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 over a field of characteristic are algorithmically recognizable. In particular, the Nagata automorphism is wild.

The approach we use is different from the traditional ones. The novelty consists of the imbedding of the polynomial ring 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 -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 , and simple automorphisms of 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 is simple. The main results are formulated and proved in Section 5 as corollaries of Theorem 1.

## 2. Structure of two-generated subalgebras

Let be an arbitrary field of characteristic , and let be the ring of polynomials in the variables over . Following Reference16, we will identify with a certain subspace of the free Poisson algebra .

Recall that a vector space over a field , endowed with two bilinear operations (a multiplication) and (a Poisson bracket), is called a Poisson algebra if is a commutative associative algebra under , is a Lie algebra under , and satisfies the Leibniz identity

An important class of Poisson algebras is given by the following construction. Let be a Lie algebra with a linear basis . Denote by the ring of polynomials on the variables . The operation of the algebra can be uniquely extended to a Poisson bracket on the algebra by means of formula (Equation1), and becomes a Poisson algebra Reference15.

Now let be a free Lie algebra with free generators . Then is a free Poisson algebra Reference15 with the free generators . We will denote this algebra by . If we choose a homogeneous basis

of the algebra with nondecreasing degrees, then , as a vector space, coincides with the ring of polynomials on these elements. The space is graded by degrees on , and for every element , the highest homogeneous part and the degree can be defined in an ordinary way. Note that

In the sequel, we will identify the ring of polynomials with the subspace of the algebra generated by elements

Note that if , then

If , then by we denote the subalgebra of the algebra generated by these elements.

The following lemma is proved in Reference16.

### Lemma 1

Let . Then the following statements are true:

iff are algebraically dependent.

Suppose that and , , . Then . If , then .

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

F1) If are nonzero homogeneous algebraically dependent elements of , then there exists an element such that , , . In addition, the subalgebra is one-generated iff or .

F2) Let and are algebraically independent. If , then .

Recall that a pair of elements of the algebra is called reduced (see Reference18), if , . A reduced pair of algebraically independent elements is called -reduced (see Reference16), if are algebraically dependent.

Let be a -reduced pair of elements of and . Put , ,

where is the greatest common divisor of . Note that , and by F1) there exists an element such that , . Sometimes, we will call a -reduced pair of elements also a -reduced pair. Let . It was proved in Reference16 that if , , then

and if , , then

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

### Lemma 2

Under the above notation,

i) ;

ii) ;

iii) if , then ;

iv) if , then .

The properties are evident. As for , let ; then , and .

The statement of the following lemma is easily proved.

### Lemma 3

The elements of type , where , have different degrees for different values of .

Inequality (Equation2) and Lemma 3 imply

### Corollary 1

Let . Consider the following conditions:

;

;

, where and for all ;

.

Then .

Suppose that or . Then obviously , and in the conditions of Corollary 1 we have or . Note that the most complicated case in the investigation of tame automorphisms of is represented by -reduced pairs with the condition , that is, by -reduced pairs for which .

### Lemma 4

There exists a polynomial of the type

which satisfies the following conditions:

;

.

### Proof.

By F1), there exists a homogeneous element such that , . Then there exists such that , and the elements of the type , , form a basis of the subalgebra . Putting , we have . If , then , where . Change the element to . Then . After several reductions of this type, we get an element

for which . Since are algebraically independent, the equality defines uniquely a polynomial that satisfies the conditions of the lemma.

A polynomial satisfying the conditions of Lemma 4 we will call a derivative polynomial of the -reduced pair . Note that a derivative polynomial is not uniquely defined in the general case. But the coefficient in the conditions of Lemma 3 is uniquely defined by the equality .

### Lemma 5

Let be a derivative polynomial of a -reduced pair . Then the following statements are true:

is uniquely defined;

if , are algebraically dependent, then is uniquely defined;

;

if , then is defined uniquely up to a summand , where ;

if , then is defined uniquely up to a scalar summand from .

### Proof.

Let be another derivative polynomial of the pair . Since the coefficient is uniquely defined in the conditions of Lemma 4, we have

where . Now, if , then by Lemma 3 we get

which contradicts the definition of .

Suppose that . Then . Since , the elements are algebraically dependent. Now, if , are algebraically dependent, then are algebraically dependent too. Furthermore, since , the elements are linearly dependent, and thus . This again contradicts the definition of .

Note that . Since , Corollary 1 yields 3). If , then by Lemma 3 we get , . 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 -reduced pairs.

### Lemma 6

Let be a derivative polynomial of the pair and . Then the highest homogeneous parts of the elements of the type

are linearly independent.

### Proof.

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

If , , then this equality implies , which is impossible by the definition of . Assume that , . Then

Since , by Corollary 1 we have

Therefore,

This contradicts the inequality . If , , then

Since and , we may assume that ; otherwise . Thus,

which is impossible.

### Lemma 7

Let be a derivative polynomial of the pair , and , . Then, the following statements are true:

can be uniquely presented in the form

where ;

if , then

if , then and

if and , then where is also a derivative polynomial of the pair .

### Proof.

The first statement of the lemma follows from the division algorithm in the ring ; one may divide by since the last polynomial is monic. Furthermore, the right part of the equality is a linear combination of elements indicated in Lemma 6; therefore, by this lemma, only the elements of degree less than or equal to may appear in this combination. This proves 2).

If , then by Corollary 1 we have and hence . By Corollary 1 again, ; hence . Consequently, by Lemma 6,

It follows from the definition of in Lemma 4 that

Furthermore,

Easy calculations give

Therefore,

Similar calculations give the value of .

To prove 4) we note first that by Lemma 5.3), . Hence by statement 2) of this lemma, and . By Corollary 1, . Hence . Now it is easy to see that the polynomial is a derivative polynomial of the pair .

We give two corollaries that will be useful for references.

### Corollary 2

If and , then where is a derivative polynomial of the pair .

### Proof.

Put and let . If , then Corollary 1 gives and or , a contradiction. Hence . Inequality (Equation2) gives . Consequently, . If , then by Lemma 2, . Therefore, and . Then Lemma 7.4) proves the corollary.

### Corollary 3

If is a derivative polynomial of the pair , then

## 3. Reductions and simple automorphisms

A triple (or simply ) of elements of the algebra below will always denote the automorphism of such that , . The number will be called a degree of the automorphism .

Recall that an elementary transformation of the triple is, by definition, a transformation that changes only one element to an element of the form , where . The notation

means that the triple is obtained from by a single elementary transformation. Observe that we do not assume that should be smaller than . An automorphism is called tame if there exists a sequence of elementary transformations of the form

The element of the automorphism is called reducible, if there exists such that ; otherwise it is called unreducible. Put , where ; then and . In this case we will say also that is reduced in by the automorphism . If one of the elements of is reducible, then we will say that admits an elementary reduction or simply that is elementary reducible.

### Lemma 8

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

### Proof.

Let be an arbitrary automorphism of . We will recognize the reducibility of . If are algebraically independent, then is reducible if and only if . Since are homogeneous, this question can be solved trivially, even without a reference to the solubility of the occurrence problem Reference13, Reference14. If and , then the element is reducible in if and only if it is reducible in the automorphism . Since , the statement of the lemma in this case can be proved by induction on .

Let now be a -reduced pair and . Assume that there exists a polynomial such that . Inequalities (Equation2), (Equation3) gives a bound for the numbers . Then is in the space generated by the elements , where . 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

It is easy to show that and are tame automorphisms of the algebra . Note that , , and form a -reduced pair. A direct calculation shows that the element

has degree 8. Hence , and .

Now we define a tame automorphism by putting

We have , , and , . Then, using inequality (Equation2), it is easy to check that the automorphism does not admit an elementary reduction.

### Proposition 1

Let be an automorphism of such that , , is an odd number, , . Suppose that there exists such that the elements , satisfy the conditions:

is a -reduced pair and , ;

the element of the automorphism is reduced by an automorphism with the condition .

Then, the following statements are true:

;

, where ;

if , where , then either or ;

.

### Proof.

We have

where , . Hence

If , then are linearly independent, since . Therefore are mutually linearly independent and . If , then and again we have . Put , . The condition implies, by Corollary 1 and (Equation5), that . Then inequality (Equation2) gives and

It is easy to deduce from here that if , then , , and if , then , . Besides, these inequalities imply and statement 4) of the proposition.

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

Since is an odd number, inequality (Equation2) gives

Consequently, by condition ii),

Since and , the equality

gives statement 1) of the proposition and

If , then, as was remarked earlier, are algebraically independent, and so (see Reference16)

If , then we have

By Lemma 1,

hence

Thus statement 2) of the proposition is proved.

To prove 3), it suffices by F2) to consider only the case when are algebraically dependent. It is easily seen that and are -reduced pairs. Suppose that . If then , which contradicts the condition of the proposition. Otherwise , which contradicts 4). Consequently, the pair is -reduced for every and Corollary 1 by Lemma 2.ii) implies 3).

### Definition 1

If an automorphism satisfies the conditions of proposition , then we will say that admits a reduction of type I, and the automorphism will be called a reduction of type I of the automorphism , with an active element .

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

### Proposition 2

Let be an automorphism of such that , , , and are linearly independent. Suppose that there exist , where , such that the elements , satisfy the conditions:

is a -reduced pair and , ;

the element of the automorphism is reduced by an automorphism with the condition .

Then, the following statements are true:

is the solution of the equation , or if it has no solution;

is the solution of the equation , or if it has no solution;

, where ;

if , where , then either or .

### Proof.

Consider equalities (Equation4), (Equation5). If , then obviously . If , then by the condition of the proposition, are linearly independent. Therefore, either and , or and are mutually linearly independent. In any case, . Since , then, as in the proof of Proposition 1, inequality (Equation2) gives that , . Consequently, Lemma 7.4) gives that is a derivative polynomial (up to a nonzero scalar factor) of the pair , and by Corollary 3,

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

Consider the triple . By (Equation6) and Lemma 1,

which yields

Furthermore,

Since , this implies, by (Equation6) and (Equation7), that if , and if . Hence , and if , then , have different degrees. We have also

Hence either and , or the elements and 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 satisfies the conditions of proposition , then we will say that admits a reduction of type II, and the automorphism will be called a reduction of type II of the automorphism , with an active element .

### Proposition 3

Let be an automorphism of such that , and either , , or , . Suppose that there exist such that the elements , satisfy the conditions:

is a -reduced pair and , ;

there exists an element of the form

where , , such that , .

Then, the following statements are true:

is the solution of the equation , or if it has no solution;

is the solution of the equation , or if it has no solution;

is the solution of the equation , or if it has no solution;

;

if , then ; otherwise, ;

if , then ; otherwise, .

### Proof.

Since , condition yields that . Then, by Corollary 2, , where is a derivative polynomial of the . Inequality (Equation2) gives also . As in the proof of Propositions 1, 2, we obtain also (Equation6), (Equation7), which yields 4). Besides, we have

Since , this equality yields statements 1), 5) of the proposition. If , then , , , and

Consequently, , which proves 6). We have also

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

### Definition 3

If an automorphism satisfies the conditions of Proposition , and , , then we will say that admits a reduction of type III, and the automorphism will be called a reduction of type III of the automorphism , with an active element .

### Corollary 4

In the conditions of Proposition , if the automorphism admits a reduction of type III, then .

### Proof.

By Definition 3, we have

Hence it is sufficient to prove that

It follows from (Equation6) that . If , then Proposition 3.6) gives , which proves (Equation8). If , then, as shown above,