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 Reference 6, Reference 9, Reference 10, Reference 11 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 Reference 5, Reference 8). However, in the case of three or more variables the similar question was open and known as “The generation gap problem” Reference 2, Reference 3 or “Tame generators problem” Reference 8. The general belief was that the answer is negative, and there were several candidate counterexamples (see Reference 5, Reference 8, Reference 12, Reference 7, Reference 19). The best known of them is the following automorphism , constructed by Nagata in 1972 (see Reference 12):

Observe that the Nagata automorphism is stably tame Reference 17; 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 Reference 16. 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 Reference 1 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 Reference 4 concerning wild automorphisms of two-generated free matrix algebras, without using the results of Reference 6, Reference 9, Reference 10, Reference 11. 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 Reference 16 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 Reference 16, 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 Reference 16, 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 (Equation 1), and becomes a Poisson algebra Reference 15.

Now let be a free Lie algebra with free generators . Then is a free Poisson algebra Reference 15 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 Reference 16.

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 Reference 5):

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 Reference 18), if , . A reduced pair of algebraically independent elements is called -reduced (see Reference 16), 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 Reference 16 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 (Equation 2) 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 (Equation 2) 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

1

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 Reference 13, Reference 14. 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 (Equation 2), (Equation 3) 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 (Equation 2), 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 (Equation 5), that . Then inequality (Equation 2) 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 (Equation 1), we get from (Equation 4),

Since is an odd number, inequality (Equation 2) 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 Reference 16)

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 1, 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 (Equation 4), (Equation 5). 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 (Equation 2) 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 (Equation 6) and Lemma 1,

which yields

Furthermore,

Since , this implies, by (Equation 6) and (Equation 7), 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 2, 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 (Equation 2) gives also . As in the proof of Propositions 1, 2, we obtain also (Equation 6), (Equation 7), 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 3, 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 3, 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 (Equation 6) that . If , then Proposition 3.6) gives , which proves (Equation 8). If , then, as shown above, and by Proposition 3.6), , which also gives (Equation 8).

Definition 4.

If an automorphism satisfies the conditions of Proposition 3, and there exists such that , then we will say that admits a reduction of type IV, and the automorphism will be called a reduction of type IV of the automorphism , with an active element . In this case we will also call the automorphism a predreduction of type IV of .

Corollary 5.

If an automorphism satisfies all the conditions of Proposition 3 and Definition 4, then .

Proof.

Since , then and . Consequently, . Inequality (Equation 8) completes the proof.

Reductions of types I–IV, together with elementary reductions, permit us to introduce an auxiliary notion of a simple automorphism.

Definition 5.

By induction on degree, we will define simple automorphisms of the algebra  as follows.

All the automorphisms of degree are simple.

Suppose that the simple automorphisms of degree are already defined.

An automorphism of degree is called simple if there exists a simple automorphism of degree that is either an elementary reduction or a reduction of type I–IV of .

Evidently, any simple automorphism is tame. Our principal goal is to prove the converse statement, that every tame automorphism is simple. We will do it in the next section.

Remark 1.

If is a simple automorphism, then the automorphisms and , where , , are simple as well.

Really, if is an elementary reduction or a reduction of type I–IV of , then is a reduction of the same type for . It is also clear that we can always choose such that becomes a corresponding reduction for .

For convenience of terminology, we introduce also

Definition 6.

An element of the automorphism is called simple reducible if it is reduced by a simple automorphism.

4. A characterization of tame automorphisms

This section is devoted to the proof of our main result.

Theorem 1.

Every tame automorphism of the algebra is simple.

The plan of the proof.

Assume that the statement of the theorem is not true. Then there exist tame automorphisms , of such that is simple, is not simple, and

In the set of all pairs of automorphisms with this property we choose and fix a pair with the minimal .

In order to obtain a contradiction, it is enough to prove that is simple. The proof will consist of analysis of the cases, when admits an elementary reduction or a reduction of type I–IV to a simple automorphism of lower degree. If admits a reduction of type I–IV, then it will be convenient for us to fix the reduction of and consider one of the following variants for :

Here, the restriction on is imposed in order to exclude the trivial case when is an elementary reduction of . In the case when admits an elementary reduction, we will assume that has form (Equation 11). The proof of the theorem will be completed by Lemmas 917 and by Propositions 4, 5.

The following evident statement is formulated for convenience of references.

Lemma 9.

Let be a simple automorphism of such that . If , then is simple too.

The proof follows immediately from the minimality condition for .

Corollary 6.

Suppose that there exists a sequence of automorphisms

of the algebra such that for , and is a simple automorphism. Then is also simple.

Lemma 10.

If admits a reduction of type I, then is simple.

Proof.

We adopt all the conditions and notation of Proposition 1 and by the definition of reduction we have that is simple. Without loss of generality, we can also put

If is of form (Equation 9), then by Proposition 1.3) we have . Since , this implies . Hence is simple by Remark 1.

Assume that has form (Equation 10). By statements 3) and 4) of Proposition 1, we have

Consider . Then again and ; hence , and it is easy to see that . Thus we have

Since , where is odd, . Furthermore, can be less than only if . But ; hence . Put

Then , and . If , then (Equation 12) implies that is a reduction of type I of . If , then and the element is reduced in by . It remains to note that, by Lemma 9, is a simple automorphism.

Now consider the case when has form (Equation 11). Proposition 1 gives, as before,

and is possible only if . Consider 3 cases:

1) , 2) , 3) .

In case 1) we put

Since , the equality is impossible, . By Lemma 9, the automorphism is simple again. Since

is a reduction of type I of .

In case 2) we put

A direct calculation gives

where

Since , then is a reduction of type I of .

In case 3) we have

Therefore, , and it is easy to check that is reduced in by . By Remark 1, is simple. Thus, by Lemma 9, is simple too.

Lemma 11.

If admits a reduction of type II, then is simple.

Proof.

We adopt all the conditions and notation of Proposition 2, as well as equality (Equation 12). If has form (Equation 9), then Proposition 2.4) and the condition give . By Remark 1 we may assume that . So

If , then it is easily checked that is a reduction of type II of . Otherwise, the element is reduced in the automorphism by .

If has form (Equation 10), then by Proposition 2.4) (and Remark 1) we get

Furthermore,

If , then is a reduction of type II of . Otherwise, reduces the element of .

Assume that has form (Equation 11). Proposition 2 and Remark 1 give

and is possible only if . Since are linearly independent, .

If , a direct calculation gives

Since , it is easy to check that is a reduction of type II of . By Corollary 6, the automorphism is simple.

If , then

If , then the element is reduced in by . Otherwise, it is easily checked that is a reduction of type II of with an active element .

Lemma 12.

If admits a reduction of type III or IV, then is simple.

Proof.

We adopt the conditions and notation of Proposition 3. Assume that has form (Equation 9). Let us show that in this case .

Evidently, it suffices to prove that, for any with , holds. Note that . By F2) we may assume, without loss of generality, that are algebraically dependent. If , then the pair is -reduced, and the statement holds by Proposition 3.4), Corollary 1, and Lemma 2.ii). Otherwise , where . Since , then , i.e., . Consider . Then

Since , we have, as in the proof of Proposition 3, that , which yields . Note that . If are algebraically independent, then by F2) we get . Otherwise, form a -reduced pair, and since , we have again by Corollary 1 and Lemma 2.ii). But , and so .

Thus , and so

Since is preserved in the structure of , it is easily checked that if admits a reduction of type IV, then is a predreduction of type IV of . Suppose that admits a reduction of type III. If , then is a reduction of type III of . Otherwise, since (see the proof of Corollary 4), the element is reduced in by .

Suppose that has form (Equation 10). Then, by Proposition 3.4) and Corollary 1 we have . Note that and . So

and is possible only if . Therefore,

Hence, if , then , i.e., , . By Proposition 3,

Since

then

Consequently, . Now it is easy to show, that if admits a reduction of type IV, then is a predreduction of type IV of . Suppose now that admits a reduction of type III. If , then is a reduction of type III of . Otherwise, the element is reduced in by .

Assume that has form (Equation 11). If , then Proposition 3 yields . If , then admits a reduction of type IV. We have

where . Since , we have

where and . If , then is a predreduction of type IV of . Otherwise, by Remark 1 we can take and the element is reduced in by .

It remains now to consider the principal case, when admits an elementary reduction. It follows from Lemma 9 that if is simple and , then every reducible element of is simple reducible. But one should carefully distinguish these notions if .

Lemma 13.

Let be a simple automorphism and . If is a simple reducible element of , then every elementary transformation of changing only gives a simple automorphism.

Proof.

Assume that the element is reduced in by a simple automorphism . Then and , where , . Put . Then

Hence and Lemma 9 completes the proof.

In the sequel we will assume that has form (Equation 11) and admits an elementary reduction. Then Lemma 13 gives

Corollary 7.

If is a simple reducible element of , then is a simple automorphism.

In the remainder of this section we will assume that is not a simple reducible element of . Then either or are simple reducible. For definiteness, we assume that is reduced in by a simple automorphism , where

Lemma 14.

The automorphism is simple if one of the following conditions is satisfied:

;

;

does not depend on ;

are algebraically independent.

Proof.

If , then by Lemma 13 we can choose an element satisfying (Equation 13) such that . Then . Consequently, there exists a sequence of elementary transformations of the form

Since , (by (Equation 13)), it follows from Corollary 6 that the automorphism is simple.

If and , then we put . There exists a sequence

By Corollary 6, the automorphism is simple, and consequently, is a simple reducible element of , which is impossible.

If does not depend on , then sequence (Equation 14) proves the simplicity of .

Assume that are algebraically independent. Then by (Equation 13) we obtain that . By 1), we can assume that , i.e., depends on . Then . Observe that are algebraically independent; otherwise, would be algebraically dependent. Therefore, . By 3), we can also assume that contains . Then (Equation 11) gives , and so . Thus

We have the equalities

which induce the following sequence of elementary transformations:

Since is simple, by Remark 1, is simple too. By Lemma 9, sequence (Equation 15) gives simple reducibility of in , a contradiction.

Thus, by Lemma 14, we can suppose that are algebraically dependent and . We will consider separately the 3 cases:

1) is a -reduced pair, ;

2) is a -reduced pair, ;

3) , .

First, we will prove two propositions.

Proposition 4.

Let be a simple automorphism satisfying the following conditions:

;

is a -reduced pair and , ;

are linearly independent, , and is not a simple reducible element of .

Then one of the following statements is satisfied:

;

admits a reduction of type IV with an active element ;

, and there exists such that .

Proof.

Assume that the proposition is not true, and let be a counterexample of minimal degree. Then we have

Hence , and by Propositions 1, 2, 3, the automorphism does not admit a reduction of types I–III. By its choice, neither admits a reduction of type IV. So, by the definition of a simple automorphism, admits an elementary reduction. By , either or is a simple reducible element of . Since are algebraically dependent, this implies that are mutually algebraically dependent. It follows from and (Equation 16) that is a -reduced pair.

Case 1. .

Assume that is a simple reducible element of . The inequalities imposed on make impossible the inclusion ; hence Corollary 1 yields

Thus . If , then , is an odd number, which contradicts (Equation 16). Hence . Putting , we have , , , , . Applying again Corollary 1, we get

i.e., , which is impossible.

If is a simple reducible element of , then the pair is -reduced, and Corollary 1 yields

Thus and , is an odd number. Since , we have and , a contradiction.

Case 2. .

Since are algebraically dependent, they would be linearly dependent if , which contradicts . Therefore, . If is a simple reducible element of , then Corollary 1 gives

Therefore, . If , then , is an odd number. Hence , and the inequality yields , which is impossible. If , then , , . Thus and , which is also impossible.

If is a simple reducible element of , then it follows from (Equation 17) that , , is an odd number. Since , then , which is impossible. So case 2 is done.

Now we can assume that . Then (Equation 16) gives , and consequently,

Then, by Lemma 1, we have

i.e.,

Case 3. .

By putting , we have , , . Put also (recall that and are algebraically dependent). We will first show that is a simple reducible element of . Assume that is a simple reducible element of which is reduced by a simple automorphism . Then , and by Lemma 9 the automorphism is also simple. The sequence

proves that is a simple reducible element of .

Put

where is an unreducible element of . Let , . Since , inequality (Equation 2) yields that either , , or , .

Consider first the case . Note that is a 3-reduced pair and , where is a derivative polynomial of the pair . By Lemma 7, the polynomial can be presented in the form

where , . By Lemma 6.3),

Since , Lemma 7.2) gives . The polynomials satisfy condition (ii) of Corollary 1. Hence they satisfy also (iii), and we get . Therefore,

Consequently, by Corollary 3,

By (Equation 1) and (Equation 20),

This yields, by (Equation 16) and (Equation 19),

Therefore, . Since , this yields . Now, applying F2) and Corollary 1, it is easy to show that are unreducible elements of . Since , by Propositions 1, 2, 3, we conclude that does not admit reductions of types I–IV. This contradicts the simplicity of .

Hence , . By Corollary 1 ( (ii) implies (iii) ), we have

Then and by (Equation 19), (Equation 22), we conclude that

Now we will assume that and show that this case is impossible. If is a simple reducible element of , then form a -reduced pair, and it follows easily from (Equation 2) that this pair is -reduced. But it is impossible since . Furthermore, if is a simple reducible element of , then form a -reduced pair, that is, . Put

where is an unreducible element of the simple automorphism .

Observe that satisfies all the conditions of the present proposition. By (Equation 2) and (Equation 23), we get . Lemma 7.4) and Corollary 3 give

Since

by (Equation 23) we have

Therefore,

These inequalities show that does not satisfy the conclusions of the proposition. Since , this contradicts our choice of .

We have thus shown that does not admit elementary reductions. As for reductions of type I, the only possibility for to have it is when is an active element of reduction and . Then there should exist such that the element of the automorphism is reducible. Recall that . Set . Then and we can replace by in our previous arguments. Then, as above, the condition for to be a reducible element of gives a contradiction.

Furthermore, the comparison of the degrees of the components of shows that does not admit a reduction of type II. Thus admits a reduction of type III or IV, with an active element . Let be a reduction of type III or a predreduction of type IV of ,

Observe that, by definition, . We have

By Proposition 3, applied to the triple , , . Furthermore, by (Equation 16) and (Equation 19), . Comparing the degrees of the left and right parts in the last equation gives

Thus , , , where . Since is unreducible in , . But then

which contradicts Corollary 4.

Therefore, the automorphism is not a reduction of type III of . Assume that it is a predreduction of type IV of ; then it is easy to see that in this case also admits a reduction of type IV with an active element , which is impossible.

We have thus shown that the inequality is impossible. Hence . Since , this implies . Furthermore, it follows from (Equation 23) that

i.e., . Thus the automorphism satisfies statement 3) of the present proposition, which is impossible.

Case 4. .

If is a simple reducible element of , then (Equation 17) gives that . Applying once more (Equation 17) and (Equation 19), we get

i.e., , a contradiction.

Therefore, is a simple reducible element of . By (Equation 18), .

Consider first the case . Then , , . Hence , and since , we have . By putting , we have , , . Then (Equation 18) gives .

Consider equality (Equation 20) and put again . Since , inequality (Equation 2) yields that either or . In the first case, by Corollary 1, , which is impossible. Thus and by Lemma 7.4) is a derivative polynomial (up to a nonzero scalar multiplier) of the pair . Then Corollary 3 gives

From (Equation 21), (Equation 22), taking into account (Equation 16) and (Equation 19), we get

Hence , and Corollary 1 yields that are unreducible elements of the automorphism . Since , it is easy to check that does not admit reductions of types I–IV, which contradicts the simplicity of .

Now we consider the case . By putting , we get , , is an odd number. Then and (Equation 16) implies . Hence and , , . Consider again equality (Equation 20) and put , . Assume that . Then (Equation 2) yields , i.e., . By Lemma 7,

where is a derivative polynomial of the 2-reduced pair , and , . By Lemma 5.3),

Now Lemma 7.2) gives , and then Lemma 3 yields . Thus, by Lemma 7.3), we get

Now (Equation 21) and (Equation 22) yield, by means of (Equation 16) and (Equation 19),

Hence , and the elements of are unreducible. Since , it is easy to check that does not admit reductions of type I–IV, a contradiction.

Thus , . Inequality (Equation 2) gives and

Hence, by means of (Equation 19), (Equation 22),

In particular,

Assume that . Since is an unreducible element of , the elements are linearly independent, if . Note that and do not compose -reduced pairs. Consequently, the elements of are unreducible. In fact, assume that there exists such that . Since , the elements are algebraically dependent and the pair is -reduced. It follows easily from (2) that this pair should be 2-reduced, a contradiction. Similarly, is unreducible. Furthermore, it follows from (Equation 25) that . Hence, due to Definitions 14 and Propositions 13, does not admit reductions of types I–IV. This contradicts the simplicity of .

Therefore, and satisfies all the conditions of the present proposition. Since , then, by the choice of should satisfy the conclusion of the proposition. It follows from (Equation 24), (Equation 25) that does not satisfy statements 1), 3); hence admits a reduction of type IV with an active element . In this case, and we have in (Equation 25), which implies . By statement 5) of Proposition 3, has a predreduction of the form , where

By Corollary 2 we have , where is a derivative polynomial of the pair . Then . Now and , where . If we substitute instead of in (Equation 20) (with instead of ), we will have again , , since the equality is impossible. Then inequalities (Equation 24) and (Equation 25) hold also for . In particular,

which contradicts the previous inequality.

Proposition 5.

Let be a simple automorphism satisfying the following conditions:

;

is a -reduced pair and , , where is an odd number;

, , and is not a simple reducible element of .

Then .

Proof.

Assuming the contrary, we have

Hence . It is now easy to check that does not admit reductions of type I–IV. Then is elementary reducible, and either or is a simple reducible element of . In particular, are mutually algebraically dependent. It follows from (Equation 26) that , . Consequently, and are -reduced pairs as well.

If is a simple reducible element of , then by Corollary 1 we get

Observe that by Lemma 2.i) we have . If , inequalities (Equation 26), (Equation 27) give . Hence , which contradicts condition of the proposition.

If , then , , where is an odd number. From (Equation 26) we get , which is impossible.

Therefore, . By putting , we have , , , where , . Inequalities (Equation 26) and (Equation 27) yield

Hence , i.e., , , . Applying (Equation 26), (Equation 27) once more, we get

Let again , where and is unreducible in . Since and , inequality (Equation 2) yields . By Lemma 7.4), up to a scalar multiplier, is equal to a derivative polynomial of the pair . Hence by Corollary 3,

Since

Lemma 1 implies that

Therefore, . Now equalities (Equation 21), (Equation 22), with and permuted, give

Hence , which implies easily by (Equation 2) that the elements are unreducible in . Since , it is easily checked that does not admit reductions of types I–IV. Thus, is not a simple automorphism.

Now suppose that is a simple reducible element of . Then Corollary 1 gives

Therefore, and , where is an odd number. Then and (Equation 26) gives , i.e., . Since , this yields , . By putting , we have again , , . By (Equation 26), we have also . Suppose that

where is an unreducible element of . Since , by Corollary 2 we have , where is a derivative polynomial of the pair . Therefore, by Corollary 3,

Since

a comparison of degrees gives

Thus,

It remains to note that satisfies all the conditions of Proposition 4 and does not satisfy conclusions 1), 3) of this proposition. If had admitted a reduction of type IV with an active element , then by the definition there would exist an element such that , i.e., . Then by Corollary 2 we would have again , which contradicts the equality .

Lemma 15.

If is a -reduced pair and , then is a simple automorphism.

Proof.

We consider first the case when . By (Equation 11), we have , . By Lemma 14, we may assume that and . Since , then by F2), are algebraically dependent. By Corollary 1, . Hence and by Lemma 2.iii), the pair is -reduced, that is, , , where is an odd number, and

Since is not a simple reducible element of (see Corollary 7), satisfies the conditions of one of Propositions 4, 5. Since , statements 2), 3) of Proposition 4 are not fulfilled for . Therefore,

which contradicts the previous inequality.

Suppose now that . If and are linearly dependent, then the element in (Equation 13) can be chosen as . Then sequence (Equation 15) gives the simplicity of .

Thus we can assume that . Then using Lemma 14.i) we can furthermore assume that . Since , by Corollary 1 , and by Lemma 2.iii) the elements form a -reduced pair. Put , , where is an odd number. The inequality gives also . We can also assume that is an unreducible element of . Then satisfies the conditions of one of Propositions 4, 5.

Let , where . Then (Equation 2) implies that if , and if . Consequently, , and (Equation 2) yields

By (Equation 13),

Assume first that . Then (Equation 28) yields

Therefore, and . Consequently, either the elements are algebraically independent or form a -reduced pair. Since , we have by Lemma 2.ii) and Corollary 1,

where only if . Since is a -reduced pair, then is also a -reduced pair. By Lemma 9, is a simple automorphism. We have

If , then it is easily checked that is a reduction of of types I–III, with an active element . Note that the type of the reduction depends on the degree of :

1) if , then , , and admits a reduction of type I;

2) if , then (otherwise is not reducible in ), and admits a reduction of type II;

3) if , then and admits a reduction of type III.

The case follows from Lemma 14.

It remains to consider the cases when satisfies one of the conclusions 2), 3) of Proposition 4. If satisfies 3), then (Equation 28) again gives (Equation 29). Besides, in this case and . Consequently, . By Lemma 14 we can put . Then the automorphism gives a reduction of type I or II of with an active element .

Assume finally that admits a reduction of type IV with an active element . Since , the scalars in Definition 4 are by Proposition 3.5). Consequently, the reduction of type IV of has the form , where

Since , then . Note that, by Definition 4, the automorphism satisfies statement 3) of Proposition 4.

If , then we may take , and this case may be reduced to the previous one.

Assume that . Then we write

If , then admits the reduction of type IV , with an active element , , and the predreduction . This case was considered in Lemma 12. If , then and the automorphism is simple by Remark 1. Consequently, is a simple reducible element of , a contradiction.

Lemma 16.

If is a -reduced pair and , then is a simple automorphism.

Proof.

Consider first the case when . Since by (Equation 11), in this case . Then the inclusion implies by F2) that the elements are algebraically dependent. By Lemma 14, we may assume that form a -reduced pair. Since , inequality (Equation 2) gives that , , . Besides, if , then Corollary 1 implies . Thus, we may assume that

Observe that satisfies all the conditions of Proposition 4; hence it should satisfy one of conlusions 2), 3) of this proposition. The case when admits a reduction of type IV was considered in Lemma 12. Therefore, we may assume that , and

Observe that . By Lemma 13, the automorphism is simple. If , then gives a reduction of type IV of .

Consider now the case when . If , then by Lemma 14 we may assume that are linearly independent. Suppose . If are linearly dependent, sequence (Equation 15) proves that is simple. Suppose that , . Since and , under the assumption that , inequality (Equation 2) yields , , and

Therefore, . Since is a simple reducible element of , it follows from Corollary 1 that and by Lemma 2.iii) the pair is 2-reduced. Without loss of generality, we can assume in (Equation 13) to be unreducible in ; then satisfies the conditions of one of Propositions 4, 5. The rest of the proof can now be fulfilled similarly to that of Lemma 15.

Lemma 17.

If and , then is a simple automorphism.

Proof.

Put , , . Then . Consider the simple automorphism defined by (Equation 13). Since , Lemma 9 implies that is also a simple automorphism. Therefore, by Lemma 9 again, the sequence

proves that is a simple reducible element of . Interchanging the elements , by Lemmas 14, 15, 16, we can restrict ourselves to the case when , . Thus, we can take , . By Lemma 14, we can assume that is a -reduced pair, with . Then . Hence by Corollary 1, , which evidently implies that .

This finishes the proof of Theorem 1.

5. The main results

Theorem 1 implies immediately

Theorem 2.

Let be a tame automorphism of the ring of polynomials over a field of characteristic . If , then admits either an elementary reduction or a reduction of types I–IV.

Corollary 8.

Under the conditions of Theorem 2, if , then admits an elementary reduction.

Proof.

In fact, it is easy to see that in this case does not admit reductions of types I–IV.

Now we will consider the Nagata automorphism (see Reference 12) of the polynomial ring , where

Corollary 9.

The Nagata automorphism of the polynomial ring over a field of characteristic is wild.

Proof.

Note that , , are mutually algebraically independent, and none of the elements is contained in the subalgebra generated by the other two elements. Consequently, the automorphism does not admit an elementary reduction. By Corollary 8, is wild.

In Reference 7, some examples of wild automorphisms of the algebra were constructed. The next corollary shows that all those automorphisms are also wild as automorphisms of the algebra .

Corollary 10.

Let be a field of characteristic . An automorphism of the -algebra is tame if and only if the automorphism of the -algebra is tame.

The proof follows easily from Corollary 8.

Theorem 3.

The tame and the wild automorphisms of the algebra of polynomials in three variables over a constructive field of characteristic are algorithmically recognizable.

Proof.

By induction on degree, it suffices to recognize, for every automorphism with , whether admits either an elementary reduction or a reduction of types I–IV. By Lemma 8, the elementary reducibility of is algorithmically recognizable.

Suppose that admits a reduction of types I–IV. If we assume that is an active element of the reduction and , then the roles of the elements of are defined uniquely. It follows from Propositions 13 that the coefficients and the elements satisfying the conditions of these propositions are uniquely defined and can be found effectively. Put .

If the element of is unreducible, then it remains to check whether admits or not a reduction of type IV with an active element . By Corollary 2, the element satisfying the condition of Proposition 3 will be defined effectively by the conditions

where is a derivative polynomial of the pair . Now it is easy to check the existence of such that .

If is a reducible element of , then by Lemma 8, we can effectively find an unreducible reduction of . If

then is a reduction of types I, II, III of . Otherwise it remains to check the validity of statements 2), 3) of Proposition 4 for . It was already shown above that it can be done effectively. Then, as in the proof of Lemmas 15, 16, one can define which type of reduction admits.

Note that a reduction of type I consists of two elementary transformations, reductions of types II, III consist of three elementary transformations, and a reduction of type IV in general case consists of four elementary transformations. Then it follows from Theorem 2 that the degree of any tame nonlinear automorphism of can be reduced by at most four elementary transformations. In this context, the following question seems very interesting.

Problem 1.

Construct examples of tame automorphisms of the algebra that admit reductions of types II–IV.

In other words, do there exist tame automorphisms of whose degrees cannot be reduced by two (or even by three) elementary transformations? The automorphism from Example 1 admits a reduction of type I and so its degree can be reduced by two (but not by one!) elementary transformations.

Acknowledgments

The second author is grateful to the Institute of Mathematics and Statistics of the University of Sao Paulo for its warm hospitality during his visit from October 9, 2000 until August 31, 2001 when the main part of this work was done.

We are grateful to L. Makar-Limanov, A. A. Mikhalev, V. Shpilrain, and J.-T. Yu for thoroughly going over the details of the proofs during an informal seminar at the University of Hong Kong in June 2002. We also thank A. van den Essen, D. Wright, and E. Zelmanov for numerous helpful comments.

Mathematical Fragments

Equation (1)
Lemma 1.

Let . Then the following statements are true:

iff are algebraically dependent.

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

Equation (2)
Equation (3)
Lemma 2.

Under the above notation,

i) ;

ii) ;

iii) if , then ;

iv) if , then .

Lemma 3.

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

Corollary 1.

Let . Consider the following conditions:

;

;

, where and for all ;

.

Then .

Lemma 4.

There exists a polynomial of the type

which satisfies the following conditions:

;

.

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 .

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.

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 .

Corollary 2.

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

Corollary 3.

If is a derivative polynomial of the pair , then

Lemma 8.

The elementary reducibility of automorphisms of the algebra is algorithmically recognizable.⁠Footnote1

1

In formulation of algorithmic results, we always assume that the ground field is constructive.

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 ;

.

Equation (4)
Equation (5)
Definition 1.

If an automorphism satisfies the conditions of proposition 1, 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 .

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 .

Equation (6)
Equation (7)
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, .

Definition 3.

If an automorphism satisfies the conditions of Proposition 3, 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 3, if the automorphism admits a reduction of type III, then .

Equation (8)
Definition 4.

If an automorphism satisfies the conditions of Proposition 3, and there exists such that , then we will say that admits a reduction of type IV, and the automorphism will be called a reduction of type IV of the automorphism , with an active element . In this case we will also call the automorphism a predreduction of type IV of .

Remark 1.

If is a simple automorphism, then the automorphisms and , where , , are simple as well.

Theorem 1.

Every tame automorphism of the algebra is simple.

Equation (9)
Equation (10)
Equation (11)
Lemma 9.

Let be a simple automorphism of such that . If , then is simple too.

Corollary 6.

Suppose that there exists a sequence of automorphisms

of the algebra such that for , and is a simple automorphism. Then is also simple.

Equation (12)
Lemma 12.

If admits a reduction of type III or IV, then is simple.

Lemma 13.

Let be a simple automorphism and . If is a simple reducible element of , then every elementary transformation of changing only gives a simple automorphism.

Corollary 7.

If is a simple reducible element of , then is a simple automorphism.

Equation (13)
Lemma 14.

The automorphism is simple if one of the following conditions is satisfied:

;

;

does not depend on ;

are algebraically independent.

Equation (14)
Equation (15)
Proposition 4.

Let be a simple automorphism satisfying the following conditions:

;

is a -reduced pair and , ;

are linearly independent, , and is not a simple reducible element of .

Then one of the following statements is satisfied:

;

admits a reduction of type IV with an active element ;

, and there exists such that .

Equation (16)
Equation (17)
Equation (18)
Equation (19)
Equation (20)
Equations (21), (22)
Equation (23)
Equation (24)
Equation (25)
Proposition 5.

Let be a simple automorphism satisfying the following conditions:

;

is a -reduced pair and , , where is an odd number;

, , and is not a simple reducible element of .

Then .

Equation (26)
Equation (27)
Lemma 15.

If is a -reduced pair and , then is a simple automorphism.

Equation (28)
Equation (29)
Lemma 16.

If is a -reduced pair and , then is a simple automorphism.

Lemma 17.

If and , then is a simple automorphism.

Theorem 2.

Let be a tame automorphism of the ring of polynomials over a field of characteristic . If , then admits either an elementary reduction or a reduction of types I–IV.

Corollary 8.

Under the conditions of Theorem 2, if , then admits an elementary reduction.

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Article Information

MSC 2000
Primary: 13F20 (Polynomial rings and ideals; rings of integer-valued polynomials), 13P10 (Polynomial ideals, Gröbner bases), 14H37 (Automorphisms)
Secondary: 14R10 (Affine spaces), 14R15 (Jacobian problem)
Keywords
  • Rings of polynomials
  • automorphisms
  • subalgebras
Author Information
Ivan P. Shestakov
Instituto de Matemática e Estatística, Universidade de São Paulo, Caixa Postal 66281, São Paulo - SP, 05311–970, Brazil; Sobolev Institute of Mathematics, Novosibirsk, 630090, Russia
shestak@ime.usp.br
MathSciNet
Ualbai U. Umirbaev
Department of Mathematics, Eurasian National University, Astana, 473021, Kazakhstan
umirbaev@yahoo.com
Additional Notes

The first author was supported by CNPq.

The second author was supported by the FAPESP Proc.00/06832-8.

Journal Information
Journal of the American Mathematical Society, Volume 17, Issue 1, ISSN 1088-6834, published by the American Mathematical Society, Providence, Rhode Island.
Publication History
This article was received on and published on .
Copyright Information
Copyright 2003 American Mathematical Society
Article References
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  • DOI 10.1090/S0894-0347-03-00440-5
  • MathSciNet Review: 2015334
  • Show rawAMSref \bib{2015334}{article}{ author={Shestakov, Ivan}, author={Umirbaev, Ualbai}, title={The tame and the wild automorphisms of polynomial rings in three variables}, journal={J. Amer. Math. Soc.}, volume={17}, number={1}, date={2004-01}, pages={197-227}, issn={0894-0347}, review={2015334}, doi={10.1090/S0894-0347-03-00440-5}, }

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