Tame and wild automorphisms of polynomial algebras and free associative algebras

U. U. Umirbaev Eurasian National University, Astana, 473021, Kazakhstan e-mail: umirbaev@yahoo.com

Abstract

Let F be an arbitrary field of characteristic 0. We describe a set of defining relations of the group of tame automorphisms of the polynomial algebra F [x, y, z]. We also prove that the well-known Anick automorphism (see [3, p. 343]) of the free associative algebra F < x, y, z > is wild.

Mathematics Subject Classification (2000): Primary 14H37, 14R10, 16W20; Secondary 14R15, 14J50. Key words: automorphism, polynomial algebra, free associative algebra.

1 Introduction

Let B_{n} = F < y_{1}, y_{2}, …, y_{n} > be the free associative algebra with free generators y_{1}, y_{2}, …, y_{n} over a field F, and let Aut B_{n} be the automorphism group of B_{n}. Let ϕ = (f_{1}, f_{2}, …, f_{n}) denote an automorphism ϕ of B_{n} such that ϕ(y_{i}) = f_{i}, 1 ≤ i ≤ n. An automorphism

σ(i, α, f) = (y_{1}, …, y_{i−1}, αy_{i} + f, y_{i+1}, …, y_{n}),

where 0 ≠ α ∈ F, f ∈ F < y_{1}, …, y_{i−1}, y_{i+1}, …, y_{n} >, is called elementary. The subgroup T Aut B_{n} of Aut B_{n} generated by all elementary automorphisms is called the tame subgroup, and the elements of this subgroup are called the tame automorphisms of B_{n}. Non-tame automorphisms of B_{n} are called wild.

In 1968 P.Cohn [1] proved that the automorphisms of free Lie algebras with a finite set of generators are tame. Later this result was generalized for the free algebras of Schreier varieties [15]. Remind that the varieties of all nonassociative algebras[14], commutative and anticommutative algebras [25], Lie algebras [25, 31] are Schreier. Another examples of Schreier varieties can be found in [17, 21, 26].

It is well known [5, 12, 13, 16] that the automorphisms of polynomial algebras and free associative algebras in two variables are tame. It was recently proved in [28, 22, 23, 24] that the well-known Nagata automorphism (see [19])

σ = (x + (x^{2} − yz)z, y + 2(x^{2} − yz)x + (x^{2} − yz)^{2}z, z)

of the polynomial algebra F [x, y, z] over a field F of characteristic 0 is wild. Moreover, it was shown [29] that the first two coordinates of the Nagata automorphism cannot be a coordinate of any tame automorphism of the algebra F [x, y, z].

In the case of free associative algebras the question about the existence of wild automorphisms was open and was formulated in [8, Problem 1.74] as a problem of P.Cohn. The best known candidate to be nontame was the Anick automorphism (see [3, p. 343])

δ = (x + z(xz − zy), y + (xz − zy)z, z)

of the algebra F < x, y, z >. It is well known [20] that the Nagata and the Anick automorphisms are stably tame.

In combinatorial group theory (see, for example [30]) there are several well-known descriptions of the group of automorphisms of free groups by generators and defining relations. It is well known (see, for example [3]) that the groups of automorphisms of polynomial algebras and free associative algebras in two variables are isomorphic and have a nice representation as a free product of groups. Similar results are well known for two dimensional Cremona groups [11, 33]. Main part of the investigations of the automorphism groups in the case of free linear algebras consist in finding a system of generators.

In this paper we describe a set of defining relations of the group of tame automorphisms T Aut F [x, y, z] of the polynomial algebra F [x, y, z] over a field F of characteristic 0. Later we use this result to prove that the Anick automorphism δ of the free associative algebra F < x, y, z > is wild.

The paper is organized as follows. Sections 2–4 are devoted to studying the group T Aut F [x, y, z]. In Section 2 we give some preliminaries from [23, 24]. It was proved in [24] that every tame automorphism of the polynomial algebra F [x, y, z] with degree > 3 admits either an elementary reduction or a reduction of types I–IV. In this paper we use some extension of the definition of automorphisms admiting a reduction of type IV in the sence of [24]. Furthermore, in Section 3 we define essential reductions of the tame automorphisms and study the uniqueness of these reductions. In particular, it is proved that every tame automorphism admits only one type of the essential reductions. A set of defining relations of the group of tame automorphisms T Aut F [x, y, z] is given in Section 4.

Recall that an associative algebra with the identity

[x, y][z, t] = 0

is called metabelian. In [27] there were introduced and investigated the Jacobian matrices of endomorphisms of free metabelian associative algebras. In Section 5 we investigate the Jacobian matrices of tame automorphisms of the free metabelian algebra in three variables. It is well known [2] that the matrix

( \begin{matrix}\frac{∂δ(x)}{∂x} & \frac{∂δ(y)}{∂x} \\ \frac{∂δ(x)}{∂y} & \frac{∂δ(y)}{∂y} \end{matrix} ) = ( \begin{matrix}1 + z′ ⊗ z & 1 ⊗ z^{2} \\ −(z′ ⊗ 1)^{2} & 1 − z′ ⊗ z \end{matrix} )

is not a product of elementary matrices over the polynomial algebra F [z′ ⊗1, 1⊗z]. From this fact we deduce that the Anick automorphism δ is wild.

2 Reductions of tame automorphisms

Let F be an arbitrary field of characteristic 0, and let A = F [x_{1}, x_{2}, x_{3}] be the polynomial algebra in the variables x_{1}, x_{2}, x_{3} over F. As in [23], we will identify the algebra A with the corresponding subspace of the free Poisson algebra P = P L⟨x_{1}, x_{2}, x_{3}⟩. The highest homogeneous part \overline{f} and the degree deg f can be defined in an ordinary way. Note that

\overline{fg} = \overline{f}\overline{g}, deg(fg) = deg f + deg g, deg[f, g] ≤ deg f + deg g.

If f_{1}, f_{2}, …, f_{k} ∈ A, then we denote by ⟨f_{1}, f_{2}, …, f_{k}⟩ the subalgebra of A generated by these elements.

We will often use the terminology and results of [24]. Therefore, let us make an agreement that Corollary 2A, Lemma 4A, Proposition 5A, and Theorem 7A mean Corollary 2, Lemma 4, Proposition 5, and Theorem 7 of [24], respectively.

Let us remind now the more necessary definitions and statements from [23, 24].

Lemma 2.1 [23] Let f, g, h ∈ A. Then the following statements are true:

1) [f, g] = 0 if and only if f, g are algebraically dependent.

2) Suppose that f, g, h ∉ F and m = deg[f, g] + deg h, n = deg[g, h] + deg f, k = deg[h, f] + deg g. Then m ≤ max(n, k). If n ≠ k, then m = max(n, k).

The following statements are well known (see [3]):

F1) If a, b are homogeneous algebraically dependent elements of the algebra A, then there exists an element z ∈ A such that a = αz^{n}, b = βz^{m}, α, β ∈ F. The subalgebra ⟨a, b⟩ is single generated if and only if m|n or n|m.

F2) Let f, g ∈ A be such that \overline{f}, \overline{g} are algebraically independent. If h ∈ ⟨f, g⟩, then \overline{h} ∈ ⟨\overline{f}, \overline{g}⟩.

A pair of elements f, g of the algebra A is called reduced if \overline{f} ∉ ⟨\overline{g}⟩ and \overline{g} ∉ ⟨\overline{f}⟩. A reduced pair of algebraically independent elements f, g ∈ A is called ∗-reduced if \overline{f} and \overline{g} are algebraically dependent.

Consider a ∗-reduced pair of elements f, g of the algebra A and let n = deg f < m = deg g. Put p = \frac{n}{(n,m)}, s = \frac{m}{(n,m)}, and

N = N(f, g) = \frac{mn}{(m, n)} − m − n + deg[f, g] = mp − m − n + deg[f, g],

where (n, m) is the greatest common divisor of n, m. Note that (p, s) = 1, and since \overline{f}, \overline{g} are algebraically dependent, there exists an element a ∈ A such that \overline{f} = βa^{p}, \overline{g} = γa^{s}. Sometimes we will call a ∗-reduced pair of elements f, g also a p-reduced pair. Assume that G(x, y) ∈ F [x, y]. It was proved in [23] that if deg_{y}(G(x, y)) = pq + r, 0 ≤ r < p, then

deg(G(f, g)) ≥ qN + mr,  (1)

and if deg_{x}(G(x, y)) = sq_{1} + r_{1}, 0 ≤ r_{1} < s, then

deg(G(f, g)) ≥ q_{1}N + nr_{1}.  (2)

Corollary 2.1 [24] Assume that G(x, y) ∈ F [x, y] and h = G(f, g). Consider the following conditions:

(i) deg h < N(f, g);

(ii) deg_{y}(G(x, y)) < p;

(iii) h = ∑_{i,j} α_{ij}f^{i}g^{j}, where α_{ij} ∈ F and in + jm ≤ deg h for all i, j;

(iv) \overline{h} ∈ ⟨\overline{f}, \overline{g}⟩.

Then (i) ⇒ (ii) ⇒ (iii) ⇒ (iv).

Lemma 2.2 [24] There exists a polynomial w(x, y) ∈ F [x, y] of the type

w(x, y) = y^{p} − αx^{s} − ∑ α_{ij}x^{i}y^{j}, ni + mj < mp,

which satisfies the following conditions:

1) deg w(f, g) < pm;

2) \overline{w(f, g)} ∉ ⟨\overline{f}, \overline{g}⟩.

A polynomial w(x, y) satisfying the conditions of Lemma 2.2 is called a derivative polynomial of the ∗-reduced pair f, g.

Corollary 2.2 [24] If h ∈ ⟨f, g⟩\F and deg h < n, then h = λw(f, g), 0 ≠ λ ∈ F, where w(x, y) is a derivative polynomial of the pair f, g.

Corollary 2.3 [24] If w(x, y) is a derivative polynomial of the pair f, g, then

deg(\frac{∂w}{∂x}(f, g)) = n(s − 1),

deg(\frac{∂w}{∂y}(f, g)) = m(p − 1).

Let θ = (f_{1}, f_{2}, f_{3}) be an arbitrary automorphism of the algebra A. The number

deg θ = deg f_{1} + deg f_{2} + deg f_{3}

is called the degree of θ.

Recall that an elementary transformation of a triple (f_{1}, f_{2}, f_{3}) is, by definition, a transformation that changes only one element f_{i} to an element of the form αf_{i} + g, where 0 ≠ α ∈ F and g ∈ ⟨{f_{j}|j ≠ i}⟩. The notation

θ → τ

means that the triple τ is obtained from θ by a single elementary transformation. An automorphism θ is called elementarily reducible or admits an elementary reduction if there exists τ ∈ Aut A such that θ → τ and deg τ < deg θ. The element f_{i} of the automorphism θ which was changed in τ to an element of less degree is called reducible and we will say also that f_{i} is reduced in θ by the automorphism τ.

Definition 2.1 [24] Let θ = (f_{1}, f_{2}, f_{3}) be an automorphism of A such that deg f_{1} = 2n, deg f_{2} = ns, s ≥ 3 is an odd number, 2n < deg f_{3} ≤ ns, \overline{f}_{3} ∉ ⟨\overline{f}_{1}, \overline{f}_{2}⟩. Suppose that there exists 0 ≠ β ∈ F such that the elements g_{1} = f_{1}, g_{2} = f_{2} − βf_{3} satisfy the conditions:

i) g_{1}, g_{2} is a 2-reduced pair and deg g_{1} = deg f_{1}, deg g_{2} = deg f_{2};

ii) the element f_{3} of the automorphism (g_{1}, g_{2}, f_{3}) is reduced by an automorphism (g_{1}, g_{2}, g_{3}) with the condition deg[g_{1}, g_{3}] < deg g_{2} + deg[g_{1}, g_{2}].

Then we will say that θ admits a reduction of type I and the automorphism (g_{1}, g_{2}, g_{3}) will be called a reduction of θ of type I with an active element f_{3}.

Definition 2.2 [24] Let θ = (f_{1}, f_{2}, f_{3}) be an automorphism of A such that deg f_{1} = 2n, deg f_{2} = 3n, \frac{3n}{2} < deg f_{3} ≤ 2n, and let \overline{f}_{1}, \overline{f}_{3} be linearly independent. Suppose that there exist α, β ∈ F, where (α, β) ≠ 0, such that the elements g_{1} = f_{1} − αf_{3}, g_{2} = f_{2} − βf_{3} satisfy the conditions i),ii) of Definition 2.1. Then we will say that θ admits a reduction of type II and the automorphism (g_{1}, g_{2}, g_{3}) will be called a reduction of θ of type II with an active element f_{3}.

Definition 2.3 [24] Let θ = (f_{1}, f_{2}, f_{3}) be an automorphism of A such that deg f_{1} = 2n, and either deg f_{2} = 3n, n < deg f_{3} ≤ \frac{3n}{2}, or \frac{5n}{2} < deg f_{2} ≤ 3n, deg f_{3} = \frac{3n}{2}. Suppose that there exist α, β_{1}, β_{2} ∈ F, where (α, β_{1}, β_{2}) ≠ 0, such that the elements g_{1} = f_{1} − αf_{3}, g_{2} = f_{2} − β_{1}f_{3} − β_{2}f_{3}^{2} satisfy the conditions:

i) g_{1}, g_{2} is a 2-reduced pair and deg g_{1} = 2n, deg g_{2} = 3n;

ii) the element f_{3} of the automorphism (g_{1}, g_{2}, f_{3}) is reduced by an automorphism (g_{1}, g_{2}, g_{3}) with the condition deg g_{3} < n + deg[g_{1}, g_{2}].

Then we will say that θ admits a reduction of type III and the automorphism (g_{1}, g_{2}, g_{3}) will be called a reduction of θ of type III with an active element f_{3}.

The next remark can be extracted from the proofs of Propositions 1A, 2A, and 3A.

Remark 2.1 i) deg [g_{1}, g_{2}] ≤ 2n in the conditions of Definition 2.1;

ii) deg [g_{1}, g_{2}] ≤ n in the conditions of Definition 2.2;

iii) deg [g_{1}, g_{2}] ≤ \frac{n}{2} in the conditions of Definition 2.3.

Lemma 2.3 Let θ = (f_{1}, f_{2}, f_{3}) satisfy the conditions of Definition 2.3. Then the following statements are true:

i) if a ∈< f_{2}, f_{3} > and deg a ≤ \frac{5n}{2}, then a ∈< f_{3} >;

ii) if a ∈< f_{1}, f_{3} > and deg a ≤ \frac{7n}{2}, then \overline{a} ∈< \overline{f_{1}}, \overline{f_{3}} >;

iii) if a ∈< f_{1}, f_{2} > and deg a < 2n, then a = δ ∈ F.

We omit the proof of this lemma, since later in Lemma 2.4 we consider analogous statements for a more complicated case.

The next definition is some extension of the definition of automorphisms admiting a reduction of type IV in the sence of [24].

Definition 2.4 Let θ = (f_{1}, f_{2}, f_{3}) be an automorphism of A such that deg f_{3} ≤ \frac{3n}{2}, deg θ ≤ \frac{13n}{2}. Suppose that there exist α_{1}, α_{2}, β_{1}, β_{2}, β_{3}, β_{4} ∈ F such that the elements g_{1} = f_{1} − α_{1}f_{3} − α_{2}f_{3}^{2}, g_{2} = f_{2} − β_{1}f_{3} − β_{2}f_{3}^{2} − β_{3}f_{1}f_{3} − β_{4}f_{3}^{3} satisfy the conditions:

i) g_{1}, g_{2} is a 2-reduced pair and deg g_{1} = 2n, deg g_{2} = 3n;

ii) there exists an element g_{3} of the form

g_{3} = f_{3} − γw(g_{1}, g_{2}), 0 ≠ γ ∈ F,

where w(x, y) is a derivative polynomial of the 2-reduced pair g_{1}, g_{2}, such that

a) deg g_{3} = \frac{3n}{2} and deg[g_{1}, g_{3}] < 3n + deg[g_{1}, g_{2}];

b) there exists 0 ≠ μ ∈ F such that deg(g_{2} − μg_{3}^{2}) ≤ 2n.

Then we will say that θ admits a reduction of type IV and the automorphism (g_{1}, g_{2} − μg_{3}^{2}, g_{3}) will be called a reduction of θ of type IV with an active element f_{3}. Proposition 2.1 Let θ = (f_{1}, f_{2}, f_{3}) be an automorphism of A satisfying the conditions of Definition 2.4. Then the following statements are true:

1) deg [g_{1}, g_{2}] ≤ \frac{n}{2}, deg f_{3} ≥ n + deg [g_{1}, g_{2}];

2) 2n ≤ deg f_{1} < \frac{5n}{2}, \frac{5n}{2} + deg [g_{1}, g_{2}] ≤ deg f_{2} < \frac{7n}{2};

3) deg [f_{1}, f_{3}] = deg [g_{1}, f_{3}] > 3n, deg [f_{2}, f_{3}] > 3n;

4) deg f_{2} + deg f_{3} > 4n;

5) if (β_{3}, β_{4}) = 0, then either deg f_{2} = 3n or deg f_{3} = \frac{3n}{2};

6) the coefficients α_{1}, α_{2}, β_{1}, β_{2}, β_{3}, β_{4} are uniquely defined;

7) if (α_{1}, α_{2}, β_{1}, β_{2}, β_{3}, β_{4}) ≠ 0, then deg [f_{1}, f_{2}] > 3n. Proof. Since deg g_{3}, deg f_{3} ≤ \frac{3n}{2}, and γ ≠ 0, Definition 2.4 gives deg w(g_{1}, g_{2}) ≤ \frac{3n}{2}. By the definition of a derivative polynomial, applying Corollary 2.1 we get

\frac{3n}{2} ≥ deg w(g_{1}, g_{2} ≥ N(g_{1}, g_{2}) = n + deg [g_{1}, g_{2}].

Consequently, deg [g_{1}, g_{2}] ≤ \frac{n}{2}. According to Corollary 2.3 we have deg \frac{∂w}{∂y}(g_{1}, g_{2}) = 3n. By Definition 2.4, we have also

deg [g_{1}, g_{3}] < 3n + deg [g_{1}, g_{2}] = deg [g_{1}, g_{2}]\frac{∂w}{∂y}(g_{1}, g_{2}).

Since

[f_{1}, f_{3}] = [g_{1}, f_{3}] = [g_{1}, g_{3}] + [g_{1}, g_{2}]\frac{∂w}{∂y}(g_{1}, g_{2}),

comparing the degrees of elements here we find that

deg [f_{1}, f_{3}] = deg [g_{1}, f_{3}] = 3n + deg [g_{1}, g_{2}].

Since deg g_{1} = 2n, it follows also that deg f_{3} ≥ n + deg [g_{1}, g_{2}]. Consequently, deg f_{3} < deg g_{1} = 2n < deg f_{3}^{2}, which gives deg f_{1} ≥ 2n.

Obviously,

deg [g_{1}, g_{2}] + deg f_{3} < deg [g_{1}, f_{3}] + deg g_{2}.

Then by the statement 2) of Lemma 2.1 we obtain

deg [g_{2}, f_{3}] + deg g_{1} = deg [g_{1}, f_{3}] + deg g_{2},

i.e.

deg [g_{2}, f_{3}] = 4n + deg [g_{1}, g_{2}].

Next we have

deg [g_{2}, f_{3}] < 4n + 2 deg [g_{1}, g_{2}] ≤ deg [f_{1}, f_{3}]f_{3}.

Since

[f_{2}, f_{3}] = [g_{2}, f_{3}] + β_{3}[f_{1}, f_{3}]f_{3},

comparing the degrees of elements we get

deg [f_{2}, f_{3}] ≥ 4n + deg [g_{1}, g_{2}].

Then

deg f_{2} ≥ 4n + deg [g_{1}, g_{2}] − deg f_{3} ≥ \frac{5n}{2} + deg [g_{1}, g_{2}].

Assume that (β_{3}, β_{4}) ≠ 0. Then it is easy to check that deg (β_{3}f_{1} + β_{4}f_{3}^{2})f_{3} > 3n. Since deg f_{3} < deg f_{3}^{2} ≤ deg g_{2} = 3n, we have deg f_{2} > 3n. Now if (β_{3}, β_{4}) = 0, then we have deg f_{2} ≤ 3n. And the inequality deg f_{2} < 3n is possible only if deg f_{3} = \frac{3n}{2}. It remains to note that the inequality deg f_{2} + deg f_{3} > 4n is fulfilled in both cases. Since deg θ ≤ \frac{13n}{2}, this gives also deg f_{1} < \frac{5n}{2}. Note that the inequality deg f_{1} + deg f_{3} > 3n implies that deg f_{2} < \frac{7n}{2}.

We have

f_{1} = g_{1} + α_{1}f_{3} + α_{2}f_{3}^{2},
f_{2} = g_{2} + β_{1}f_{3} + β_{2}f_{3}^{2} + β_{3}f_{1}f_{3} + β_{4}f_{3}^{3}
= g_{2} + β_{1}f_{3} + (β_{2} + α_{1}β_{3})f_{3}^{2} + β_{3}g_{1}f_{3} + (β_{4} + α_{2}β_{3})f_{3}^{3}.

Consequently,

[f_{1}, f_{2}] = [g_{1}, g_{2}] + β_{1}[g_{1}, f_{3}] + α_{1}[f_{3}, g_{2}] + (2β_{2} + α_{1}β_{3})[g_{1}, f_{3}]f_{3}
+β_{3}g_{1}[g_{1}, f_{3}] + 2α_{2}f_{3}[f_{3}, g_{2}] + (3β_{4} + α_{2}β_{3})[g_{1}, f_{3}]f_{3}^{2}.

Since

deg [g_{1}, g_{2}] < deg [g_{1}, f_{3}] < deg [f_{3}, g_{2}] < deg [g_{1}, f_{3}]f_{3}
< deg g_{1}[g_{1}, f_{3}] < deg [f_{3}, g_{2}]f_{3} < deg [g_{1}, f_{3}]f_{3}^{2},

as in the proofs of Propositions 1A, 2A, and 3A we can easily deduce the statements 6),

7) of Proposition 2.1. □

Lemma 2.4 Let θ = (f_{1}, f_{2}, f_{3}) satisfy the conditions of Definition 2.4. Then the following statements are true:

i) if a ∈< f_{2}, f_{3} > and deg a ≤ \frac{5n}{2}, then a ∈< f_{3} >;

ii) if a ∈< f_{1}, f_{3} > and deg a ≤ \frac{7n}{2}, then \overline{a} ∈< \overline{g_{1}}, \overline{f_{3}} >;

iii) if a ∈< f_{1}, f_{2} > and deg a < 2n, then we have a = δ ∈ F if (α_{1}, α_{2}, β_{1}, β_{2}, β_{3}, β_{4}) ≠ 0, while a = δw(g_{1}, g_{2}) + λ otherwise, where w(x, y) is a derivative polynomial of the 2-reduced pair g_{1}, g_{2}.

Proof. Assume that a ∈< f_{2}, f_{3} >. The subalgebra < f_{2}, f_{3} > does not change if we replace f_{2} by g_{2} + β_{3}g_{1}f_{3}. So, we can assume that f_{2} = g_{2} + β_{3}g_{1}f_{3}. If \overline{f_{2}} and \overline{f_{3}} are algebraically independent, then by F2) we have \overline{a} ∈< \overline{f_{2}}, \overline{f_{3}} >. By Proposition 2.1, we have deg f_{2} > \frac{5n}{2}. Consequently, a ∈< f_{3} >. Suppose that \overline{f_{2}} and \overline{f_{3}} are algebraically dependent. If β_{3} ≠ 0, then f_{2}, f_{3} is a ∗-reduced pair. If β_{3} = 0, then either f_{2}, f_{3} is a ∗-reduced pair, or there exists μ ∈ F such that \overline{g_{2}} = μ\overline{f_{3}^{2}}. In the last case we can assume that f_{2} = g_{2} − μf_{3}^{2}. Then again f_{2}, f_{3} is a ∗-reduced pair. Since

N(f_{2}, f_{3}) > deg [f_{2}, f_{3}] > 4n > \frac{5n}{2} ≥ deg a,

the inequality deg a ≥ N(f_{2}, f_{3}) leads to a contradiction. Consequently, deg a < N(f_{2}, f_{3}). Since deg f_{2} > \frac{5n}{2} ≥ deg a, by Corollary 2.1 we get again a ∈< f_{3} >.

Suppose that a ∈< f_{1}, f_{3} >. As above we can assume that f_{1} = g_{1}. If \overline{g_{1}} and \overline{f_{3}} are algebraically dependent, then g_{1}, f_{3} is a p-reduced pair. Consider the case when

\frac{5n}{2} ≥ deg a ≥ N(g_{1}, f_{3}) = (p − 1) deg g_{1} − deg f_{3} + deg[g_{1}, f_{3}].

If p ≥ 3, then with regard to Proposition 2.1 we have N(g_{1}, f_{3}) > 5n. If p = 2, then this is possible only if

deg g_{1} = 2n = 3k, deg f_{3} = 2k ≤ \frac{3n}{2}.

Consequently, repeately applying Proposition 2.1 gives

N(g_{1}, f_{3}) = k + deg[g_{1}, f_{3}] > \frac{2n}{3} + 3n = \frac{11n}{3} > \frac{7n}{2}.

Thus deg a < N(g_{1}, f_{3}). Then Corollary 2.1 gives \overline{a} ∈< \overline{g_{1}}, \overline{f_{3}} >.

Now suppose that a ∈< f_{1}, f_{2} >. If (α_{1}, α_{2}, β_{1}, β_{2}, β_{3}, β_{4}) ≠ 0, then by Proposition 2.1 we have deg [f_{1}, f_{2}] > 3n. From here it is easy to deduce that \overline{a} ∈< \overline{f_{1}}, \overline{f_{2}} > and a = δ ∈ F if deg a < 2n. Otherwise Corollary 2.2 gives the statement iii) of the lemma. □

It is easy to deduce from Proposition 2.1 that if (α_{2}, β_{3}, β_{4}) = 0, then Definition 2.4 gives exactly the automorphisms admiting a reduction of type IV in the sence of [24].

Theorem 2.1 [24] Let θ = (f_{1}, f_{2}, f_{3}) be a tame automorphism of the polynomial algebra A = F [x_{1}, x_{2}, x_{3}] over a field F of characteristic 0. If deg θ > 3, then θ admits either an elementary reduction or a reduction of one of the types I–IV.

Further we need the following proposition.

Proposition 2.2 Let (f_{1}, f_{2}, f_{3}) be a tame automorphism of the algebra A satisfying the following conditions:

i) f_{1}, f_{2} is a 2-reduced pair and deg f_{1} = 2n, deg f_{2} = sn, where s ≥ 3 is an odd number;

ii) deg f_{3} = m < sn and f_{3} is an irreducible element of the automorphism (f_{1}, f_{2}, f_{3}).

Then one of the following statements is true:

1) m < n(s − 2) + deg[f_{1}, f_{2}];

2) (f_{1}, f_{2}, f_{3}) admits a reduction of type IV with the active element f_{3}, and the coefficients α_{1}, α_{2}, β_{1}, β_{2}, β_{3}, β_{4} of Definition 2.4 are equal to 0;

3) deg[f_{1}, f_{3}] < 3n + deg[f_{1}, f_{2}] and there exists μ ∈ F such that deg(f_{2} − μf_{3}^{2}) ≤ 2n.

Proof. The conditions of this proposition coincide with the conditions of Propositions 4A and 5A if we take into account the fact that the tame automorphisms of the algebra A are already simple and every elementary reducible element is simple reducible in the sence of [24]. It is easy to check that the proofs of Propositions 4A and 5A give also the proof of Proposition 2.2. □

3 On the uniqueness of reductions

Every tame automorphism θ has a sequence of elementary transformations

(x_{1}, x_{2}, x_{3}) = θ_{0} → θ_{1} → … → θ_{k−1} → θ_{k} = θ.  (3)

Put d = max{deg θ_{i} | 0 ≤ i ≤ k}. The number d will be called the width of the sequence (3). The minimal width of all sequences of the type (3) for θ will be called the width of the automorphism θ.

Let θ ∈ T Aut A, deg θ > 3. An automorphism ϕ will be called an essential reduction of θ if there exists a sequence of elementary transformations

ϕ → θ_{0} → θ_{1} → … → θ_{t−1} → θ_{t} = θ  (4)

such that d(ϕ) < d(θ) and deg (θ_{i}) ≤ d(θ), where 0 ≤ i ≤ t.

The minimal number t for which there exists an essential reduction ϕ of θ of the type (4) will be called the height of θ, and the corresponding sequence (4) will be called a minimal essential reduction of θ. If t = 0, then we will say that θ admits an essential elementary reduction.

Later we will see that elementary reductions of automorphisms admiting a reduction of types III, IV are not essential.

If deg θ = 3, then we put t = ∞. Introduce a lexicographic order on the set of all pairs (d, t), where d is the width and t is the height of some tame automorphism, by putting (d_{1}, t_{1}) < (d_{2}, t_{2}) if either d_{1} < d_{2} or d_{1} = d_{2}, t_{1} < t_{2}.

Lemma 3.1 If (4) is a minimal essential reduction of θ, then deg θ_{0} = d(θ) and

(d(ϕ), t(ϕ)) < (d(θ_{0}), t(θ_{0})) < (d(θ_{1}), t(θ_{1})) < … < (d(θ_{t}), t(θ_{t})) = (d(θ), t(θ)).

Proof. If deg θ_{0} < d(θ), then (4) gives d(θ_{0}) < d(θ). In this case instead of ϕ we can take θ_{0}, which contradicts the minimality of (4). In addition, by the minimality of (4), ϕ is also a minimal essential reduction of each θ_{i}, 0 ≤ i ≤ k. Consequently,

deg θ_{0} = d(θ_{i}), 0 ≤ i ≤ k, t(θ_{i+1}) = t(θ_{i}) + 1, 0 ≤ i ≤ t − 1,

which gives the statement of the lemma. □

Let d be the width and t be the height of θ. The sequence (3) will be called a minimal representation of the automorphism θ if

θ_{k−t−1} → θ_{k−t} → θ_{k−t+1} → … → θ_{k} = θ

is a minimal essential reduction of θ and deg θ_{i} < d(θ), 0 ≤ i ≤ k − t − 2.

So, every minimal essential reduction of a tame automorphism can be continued to a minimal representation.

Proposition 3.1 Let θ ∈ T Aut A, deg θ > 3, let d be the width and t be the height of θ, and let (4) be a minimal essential reduction of θ. Then one of the following statements is true:

a) If θ admits a reduction of type I, then d = deg θ, t = 1, and in the conditions of Definition 2.1 we have

θ_{t−1} = (f_{1}, ηg_{2} + κg_{1} + ν, f_{3}), η ≠ 0.  (5)

b) If θ admits a reduction of type II, then d = deg θ and in the conditions of Definition 2.2 the automorphism θ_{t−1} and the height t will be calculated by the following way:

1) if αβ ≠ 0, then t = 2 and θ_{t−1} can be written down simultaneously in the form (5) and in the form

θ_{t−1} = (ρg_{1} + σ, f_{2}, f_{3}), ρ ≠ 0;  (6)

2) if α = 0, then t = 1 and θ_{t−1} has the form (5);

3) if β = 0, then t = 1 and θ_{t−1} has the form (6).

c) If θ admits a reduction of type III, then in the conditions of Definition 2.3 we have d = 5n + deg f_{3} and the automorphism θ_{t−1} and the height t will be calculated by the following way:

1) if α ≠ 0 and (β_{1}, β_{2}) ≠ 0, then t = 2 and θ_{t−1} can be written down simultaneously in the form (5) and (6);

2) if α = 0, then t = 1 and θ_{t−1} has the form (5);

3) if (β_{1}, β_{2}) = 0, then t = 1 and θ_{t−1} has the form (6).

d) If θ admits a reduction of type IV, then in the conditions of Definition 2.4 we have d = \frac{13n}{2} and the automorphism θ_{t−1} and the height t will be calculated by the following way:

1) if (α_{1}, α_{2}) ≠ 0 and (β_{1}, β_{2}, β_{3}, β_{4}) ≠ 0, then t = 3 and θ_{t−1} can be written down simultaneously in the form (5) and (6);

2) if (α_{1}, α_{2}) = 0 and (β_{1}, β_{2}, β_{3}, β_{4}) ≠ 0, then t = 2 and θ_{t−1} has the form (5);

3) if (α_{1}, α_{2}) ≠ 0 and (β_{1}, β_{2}, β_{3}, β_{4}) = 0, then t = 2 and θ_{t−1} has the form (6);

4) if (α_{1}, α_{2}) = 0 and (β_{1}, β_{2}, β_{3}, β_{4}) = 0, then t = 1 and θ_{t−1} = (g_{1}, g_{2}, ρg_{3} + σ).

e) In the other cases θ admits an essential elementary reduction, i.e. d = deg θ and t = 0.

Proof. Let θ ∈ T Aut A and let d be the width and t be the height of θ. Denote by (d_{1}, t_{1}) the value of (d, t) indicated in Proposition 3.1. We first show that

(d, t) ≤ (d_{1}, t_{1})  (7)

by induction on deg θ. So, we can assume that the inequality (7) is true for automorphisms with less degree.

Consider the case d). In this case θ admits a reduction of type IV, and we adopt all the conditions and notation of Definition 2.4. Consider the sequence of elementary transformations

ϕ = (g_{1}, g_{2} − μg_{3}^{2}, g_{3}) ⟶ (g_{1}, g_{2}, g_{3}) ⟶ (g_{1}, g_{2}, f_{3}) ⟶ (f_{1}, g_{2}, f_{3}) ⟶ (f_{1}, f_{2}, f_{3}). (8)

Since deg ϕ < deg θ, it follows that the inequality (7) is valid for ϕ.

We will show that

d(ϕ) < deg θ.  (9)

First of all we will give a standart argument that deduces (7) from (9). In the sequence (8) the automorphism (g_{1}, g_{2}, g_{3}) has the maximal degree and deg (g_{1}, g_{2}, g_{3}) = \frac{13n}{2} = d_{1}.

If (9) is fulfilled, then we get d ≤ d_{1}. If d < d_{1}, then (7) is fulfilled. Suppose that d = d_{1}. Then the sequence (8) gives that ϕ is an essential reduction of θ. Consequently, t ≤ 3. We have f_{1} = g_{1} if (α_{1}, α_{2}) = 0, and f_{2} = g_{2} if (β_{1}, β_{2}, β_{3}, β_{4}) = 0. In these cases, excluding (f_{1}, g_{2}, f_{3}) from (8), we get t ≤ 2. If (α_{1}, α_{2}, β_{1}, β_{2}, β_{3}, β_{4}) = 0, then excluding also (g_{1}, g_{2}, f_{3}) from (8), we get t ≤ 1.

If ϕ satisfies one of the conditions a), b), and e) of the proposition, then by (7) we have d(ϕ) ≤ deg (ϕ), which gives (9).

We have

deg g_{1} = 2n, deg (g_{2} − μg_{3}^{2}) ≤ 2n, deg g_{3} = \frac{3n}{2}.  (10)

Suppose that ϕ admits a reduction of types III, IV. According to (10), the element g_{1} has the highest degree among the components of ϕ. Therefore g_{2} − μg_{3}^{2} or g_{3} can be an active element of the reduction. If g_{3} is an active element of the reduction, then there exists m such that

m < deg g_{3} ≤ \frac{3m}{2}, 2m ≤ deg (g_{2} − μg_{3}^{2}) < \frac{5m}{2}, \frac{5m}{2} < deg g_{1} < \frac{7m}{2}.

Taking account of (10), from here we deduce the inequalties

2m ≤ 2n, \frac{3n}{2} ≤ \frac{3m}{2}, \frac{5m}{2} < deg g_{1} = 2m,

which are in a contradiction.

If g_{2} − μg_{3}^{2} is an active element of the reduction of ϕ, then there exists m such that

m < deg (g_{2} − μg_{3}^{2}) ≤ \frac{3m}{2}, 2m ≤ deg g_{3} < \frac{5m}{2}, \frac{5m}{2} < deg g_{1} < \frac{7m}{2}.

Consequently, 2m ≤ \frac{3n}{2}, i.e. m ≤ \frac{3n}{4}. By (7), we have d(ϕ) ≤ \frac{13m}{2} ≤ \frac{39n}{8} < 5n. From Proposition 2.1 we have deg θ > 6n, i.e. the inequality (9) is fulfilled.

The case when θ satisfies one of the conditions a), b), and c), can be considered similarly.

Consider the case e). If θ does not admit a reduction of types I–IV, then according to Theorem 2.1 θ admits an elementary reduction. Let ϕ be an elementary reduction of θ. If ϕ satisfies one of the conditions a), b), and e), then by (7) we have d(ϕ) ≤ deg ϕ < deg θ, i.e. the inequality (9) is fulfilled.

Assume that ϕ admits a reduction of type IV. Temporarily we assume that ϕ = (f_{1}, f_{2}, f_{3}) and that (f_{1}, f_{2}, f_{3}) satisfies the conditions of Definition 2.4. According to (7), we have d(ϕ) ≤ \frac{13n}{2}. Put also θ = (q_{1}, q_{2}, q_{3}). If deg θ > \frac{13n}{2}, then (9) is fulfilled. Therefore we may assume that

deg θ ≤ \frac{13n}{2}.  (11)

We first consider the case in which

q_{1} = ρf_{1} + a, q_{2} = f_{2}, q_{3} = f_{3}, a ∈< f_{2}, f_{3} > .  (12)

Changing ρg_{1} to g_{1}, we may assume that ρ = 1. Proposition 2.1 gives deg f_{2}+deg f_{3} > 6n and deg f_{1} < \frac{5n}{2}. Then from (11) we obtain deg g_{1} < \frac{5n}{2}. Consequently, deg a < \frac{5n}{2}. By Lemma 2.4 we have a ∈< f_{3} > and

a = α′_{0} + α′_{1}f_{3} + α′_{2}f_{3}^{2}.

Consequently,

q_{1} = g_{1} + α′_{0} + (α_{1} + α′_{1})f_{3} + (α_{2} + α′_{2})f_{3}^{2}.

Changing g_{1} + α′_{0} to g_{1}, we may assume that α′_{0} = 0. Then θ admits a reduction of type IV, which contradicts the condition e).

Let

q_{1} = f_{1}, q_{2} = ηf_{2} + a, q_{3} = f_{3}, a ∈< f_{1}, f_{3} > .  (13)

As above we can take η = 1. By Proposition 2.1 we have deg f_{1} + deg f_{3} > 3n and \frac{5n}{2} < deg f_{2} < \frac{7n}{2}. Applying (11) we find that deg q_{2} < \frac{7n}{2}, i.e. deg a < \frac{7n}{2}. By Lemma 2.4 we obtain \overline{a} ∈< \overline{g_{1}}, \overline{f_{3}} >. Now it is easy to deduce that

a = γ_{0} + γ_{1}g_{1} + β′_{1}f_{3} + β′_{2}f_{3}^{2} + β′_{3}f_{1}f_{3} + β′_{4}f_{3}^{3}.

Consequently,

q_{2} = g_{2} + γ_{0} + γ_{1}g_{1} + (β_{1} + β′_{1})f_{3} + (β_{2} + β′_{2})f_{3}^{2} + (β_{3} + β′_{3})f_{1}f_{3} + (β_{4} + β′_{4})f_{3}^{3}.

Changing g_{2}+γ_{0}+γ_{1}g_{1} to g_{2}, we may assume that γ_{0} = γ_{1} = 0. Then θ admits a reduction of type IV.

Now we consider the case in which

q_{1} = f_{1}, q_{2} = f_{2}, q_{3} = ρf_{3} + a, a ∈< f_{1}, f_{2} > .  (14)

Changing ρg_{3} to g_{3}, we may assume that ρ = 1. Proposition 2.1 gives deg f_{1}+deg f_{2} > \frac{9n}{2}. By (11), from here we get deg q_{3} < 2n, i.e. deg a < 2n. If (α_{1}, α_{2}, β_{1}, β_{2}, β_{3}, β_{4}) ≠ 0, then according to Lemma 2.4 we obtain a = δ ∈ F, i.e. θ admits a reduction of type IV. If (α_{1}, α_{2}, β_{1}, β_{2}, β_{3}, β_{4}) = 0, then f_{1} = g_{1}, f_{2} = g_{2}, and by Lemma 2.4 we have

a = δw(g_{1}, g_{2}) + σ.

Then

q_{3} = g_{3} + (γ + δ)w(g_{1}, g_{2}) + σ.

Changing g_{3} + σ to g_{3}, we can take σ = 0. If γ + δ ≠ 0, then θ admits a reduction of type IV. If γ + δ = 0, then θ = (g_{1}, g_{2}, g_{3}) and instead of ϕ we can take

ϕ = (g_{1}, g_{2} − μq_{3}^{2}, q_{3}).

Recall that, considering the case when θ admits a reduction of type IV, we have simultaneously proved that

d(ϕ) < \frac{13n}{2}.

Since deg (θ) = \frac{13n}{2}, the inequality (9) is also fulfilled.

If ϕ admits a reduction of type III, then as above we can assume that ϕ = (f_{1}, f_{2}, f_{3}) and that (f_{1}, f_{2}, f_{3}) satisfies the conditions of Definition 2.3. This case can be settled by analogy to the case when ϕ admits a reduction of type IV. We only note that instead of (11) in this case we have a stronger inequality

deg θ ≤ 5n + deg f_{3}.

The inequality (7) is thus proved if θ satisfies one of the conditions d), e). The case when θ satisfies one of the conditions a), b), and c) can be considered similarly. Therefore we will assume that the inequality (7) is proved.

Now we begin a full proof of Proposition 3.1. Assume that the statement of the proposition is not true. Let θ be an automorphism with minimal (d, t), which does not satisfy the statement of the proposition. Note that if θ satisfies the condition e), then we have nothing to prove.

Restrict ourselves to the case when θ admits a reduction of type IV. Let θ satisfy the conditions of Definition 2.4. Put θ_{t−1} = (q_{1}, q_{2}, q_{3}). Since

(d(θ_{t−1}), t(θ_{t−1})) < (d, t)  (15)

by Lemma 3.1, it follows that Proposition 3.1 is valid for θ_{t−1}. By (7) we have d = d(θ) ≤ \frac{13n}{2}. Then

d(θ_{t−1}) ≤ \frac{13n}{2}.  (16)

Suppose that (q_{1}, q_{2}, q_{3}) has the form (12). Repeating the same arguments which were given after (12), we can assume that

q_{1} = g_{1} + (α_{1} + α′_{1})f_{3} + (α_{2} + α′_{2})f_{3}^{2},

i.e. θ_{t−1} admits a reduction of type IV.

Let (β_{1}, β_{2}, β_{3}, β_{4}) ≠ 0. If (α_{1} + α′_{1}, α_{2} + α′_{2}) ≠ 0, then according to the statement d) of Proposition 3.1 we have d(θ_{t−1}) = \frac{13n}{2} and t(θ_{t−1}) = 3. Since deg θ ≤ \frac{13n}{2}, from here we get d = \frac{13n}{2} and t = 4, which contradicts (7). Consequently, (α_{1} + α′_{1}, α_{2} + α′_{2}) = 0. In this case Proposition 3.1 gives d(θ_{t−1}) = \frac{13n}{2} and t(θ_{t−1}) = 2. Hence, d = \frac{13n}{2} and t = 3. If (α_{1}, α_{2}) ≠ 0, then Proposition 3.1 is valid. If (α_{1}, α_{2}) = 0, then the inequality (7) is not fulfilled. It means that θ_{t−1} is obtained from θ by another elementary transformation!

Now let (β_{1}, β_{2}, β_{3}, β_{4}) = 0. As above we have (α_{1} + α′_{1}, α_{2} + α′_{2}) = 0. Then d(θ_{t−1}) = \frac{13n}{2} and t(θ_{t−1}) = 1. Consequently, d = \frac{13n}{2} and t = 2. If (α_{1}, α_{2}) ≠ 0, then Proposition 3.1 is valid, and if (α_{1}, α_{2}) = 0, then the inequality (7) is not fulfilled.

The above discussion is standart and the other cases can be examined similarly. □

Corollary 3.1 Reductions of types I–IV are essential minimal reductions.

Corollary 3.2 If θ ∈ T Aut A admits a reduction of types I–IV, then the type of this reduction is uniquely defined.

Proof. Let θ = (f_{1}, f_{2}, f_{3}) admit a reduction of types I–IV. By Proposition 3.1, an active element f_{3} of this reduction is uniquely characterized as a component of θ which does not change at a minimal reduction before appearing an automorphism (g_{1}, g_{2}, f_{3}) with the property deg [g_{1}, g_{2}] ≤ deg f_{3}. The last inequality is a generalization of the estimaties of the degree of [g_{1}, g_{2}] in Remark 2.1 and in Proposition 2.1. In addition, if deg f_{1} ≤ deg f_{2}, then the roles of elements in Definitions 2.1–2.4 are uniquely defined.

Now reductions of types I, II, and III can be easiely distinguished among themselves by the degree of deg f_{3}.

Assume that θ admits a reduction of type IV and satisfies the conditions of Definition 2.4. If α_{2} ≠ 0, then \overline{f_{1}} = α_{2}\overline{f_{3}}^{2}. Consequently, f_{1} is an elementarily reducible element of θ. Note that if θ admits a reduction of types I–III, then θ does not admit such an elementary reduction. If α_{2} = 0, then deg f_{1} = 2n and deg f_{3} ≤ \frac{3n}{2}. Consequently, θ does not admit a reduction of type I or II. If θ admits a reduction of type III, then the elements g_{1}, g_{2} are uniquely defined for both reductions. We have g_{3} = f_{3} − γw(g_{1}, g_{2}). Since deg [g_{1}, g_{3}] < 3n+deg [g_{1}, g_{2}] and deg [g_{1}, w(g_{1}, g_{2})] = 3n+deg [g_{1}, g_{2}], it follows that γ is also uniquely defined from the equality \overline{[g_{1}, f_{3}]} = γ\overline{[g_{1}, w(g_{1}, g_{2})]}. Now if deg g_{3} < \frac{3n}{2}, then θ admits a reduction of type III, and if deg g_{3} = \frac{3n}{2}, then θ admits a reduction of type IV. So, θ does not admit reductions of types III, IV simultaneously. □

As we can see from the proof of Corollary 3.2, not only the type of reduction of θ but also the elements g_{1}, g_{2} are uniquely defined. The element g_{3} in Definitions 2.3, 2.4 is uniquely defined up to a summand from the field F, since it is with this exactness that the element w(g_{1}, g_{2}) is defined. In Definition 2.1 instead of g_{3} we can take any element of the form ρg_{3} + F (g_{1}), where 0 ≠ ρ ∈ F and deg F (g_{1}) ≤ deg g_{3}. Furthermore, in Definition 2.2 instead of g_{3} we can take any element of the form ρg_{3} +γ, where 0 ≠ ρ ∈ F.

Proposition 3.1 and Corollary 3.2 immediately give the next stronger form of Theorem 2.1.

Theorem 3.1 Let θ ∈ T Aut A, deg θ > 3. Then θ admits only one type of the following reductions: an essential elementary reduction or a reduction of types I–IV.

Besides, from the proof of Proposition 3.1 we can extract the next corollary.

Corollary 3.3 Let θ ∈ T Aut A, deg θ > 3. If θ admits a reduction of types I–IV, then every elementary reduction of θ again admits a reduction of the same type.

In fact, the elementary reductions considered in this corollary are not essential.

Note that if θ admits a reduction of types I, II, then θ does not admit any elementary reduction. If θ admits a reduction of type III, then in the conditions of Definition 2.3 θ admits an elementary reduction if and only if \overline{f_{2}} = ς\overline{f_{3}^{2}}, deg f_{3} = \frac{3n}{2}, and deg f_{2} = 3n.

Assume that θ admits a reduction of type IV. If (α_{2}, β_{3}, β_{4}) = 0 in the conditions of Definition 2.4, then θ also admits an elementary reduction if and only if \overline{f_{2}} = ς\overline{f_{3}^{2}}, deg f_{3} = \frac{3n}{2}, and deg f_{2} = 3n. If (α_{2}, β_{3}, β_{4}) ≠ 0, then θ admits obvious elementary reductions.

4 Defining relations of the group of tame automorphisms

Let A_{n} = F [x_{1}, x_{2}, …, x_{n}] be the polynomial algebra over a field F with the set of free generators X = {x_{1}, x_{2}, …, x_{n}}. Then the group T Aut A_{n} is generated by all elementary automorphisms

σ(i, α, f) = (x_{1}, …, x_{i−1}, αx_{i} + f, x_{i+1}, …, x_{n}),  (17)

where 0 ≠ α ∈ F, f ∈< X \ {x_{i}} >.

Our aim in this section is to describe defining relations of the group T Aut A_{n} with respect to the generators (17). It is easy to check that

σ(i, α, f)σ(i, β, g) = σ(i, αβ, βf + g).  (18)

Note that from here we obtain trivial relations

σ(i, 1, 0) = id, 1 ≤ i ≤ n.

If i ≠ j and f ∈< X \ {x_{i}, x_{j}} >, then we have also

σ(i, α, f)^{−1}σ(j, β, g)σ(i, α, f) = σ(j, β, σ(i, α, f)^{−1}(g)).  (19)

Consequently, if i ≠ j and f, g ∈< X \ {x_{i}, x_{j}} >, then the automorphisms σ(i, α, f), σ(j, β, g) commute.

For every pair k, s, where 1 ≤ k ≠ s ≤ n, we define a tame automorphism (ks) by putting

(ks) = σ(s, −1, x_{k})σ(k, 1, −x_{s})σ(s, 1, x_{k}).

Note that the automorphism (ks) of the algebra A_{n} changes only the positions of the variables x_{k} and x_{s}. Now it is easy to see that

σ(i, α, f)^{(ks)} = σ(j, α, (ks)(f)),  (20)

where j = (ks)(x_{i}).

Lemma 4.1 From the relations (18)–(20) it follows that the subgroup generated by all automorphisms (ks), where 1 ≤ k ≠ s ≤ n, is isomorphic to the symmetric group S_{n}.

Proof. By (18) and (19), we have

(ks)^{2} = σ(s, −1, x_{k})σ(k, 1, −x_{s})σ(s, 1, x_{k})σ(s, −1, x_{k})σ(k, 1, −x_{s})σ(s, 1, x_{k})
= σ(s, −1, x_{k})σ(k, 1, −x_{s})σ(s, −1, 0)σ(k, 1, −x_{s})σ(s, 1, x_{k})
= σ(s, −1, x_{k})σ(s, −1, 0)σ(k, 1, −x_{s})^{σ(s,−1,0)}σ(k, 1, −x_{s})σ(s, 1, x_{k})
= σ(s, 1, −x_{k})σ(k, 1, x_{s})σ(k, 1, −x_{s})σ(s, 1, x_{k}) = σ(s, 1, −x_{k})σ(s, 1, x_{k}) = id.

Then (20) gives

(ks)^{(sk)} = σ(s, −1, x_{k})^{(sk)}σ(k, 1, −x_{s})^{(sk)}σ(s, 1, x_{k})^{(sk)}
= σ(k, −1, x_{s})σ(s, 1, −x_{k})σ(k, 1, x_{s}) = (sk),

i.e. (ks) = (sk). Now it is not difficult to deduce from (18)–(20) that

[(ij), (ks)] = id, (ik)^{(is)} = (ks),

where i, j, k, s are distinct. It is immediate that the given relations imply the defining relations of the group S_{n} with respect to the system of generators (i i + 1), where 1 ≤ i ≤ n + 1, which are indicated in [4]. □

By Lemma 4.1, elements of the symmetric group S_{n} can be identified with automorphisms of the algebra A_{n} if we put

π = (x_{π(1)}, x_{π(2)}, …, x_{π(n)})

for every π ∈ S_{n}. We note that (20) can be rewritten as

σ(i, α, f)^{π} = σ(π^{−1}(i), α, π^{−1}(f)).

It is well known that the group of affine automorphisms GA_{n}(F) of the algebra A_{n} is generated by all affine elementary automorphisms.

Lemma 4.2 The relations (18)–(20) for elementary affine automorphisms are defining relations of the group GA_{n}(F).

Proof. Let φ be a product of elementary affine automorphisms. Suppose that φ = id. By (18), (19), we can represent φ in the form

φ = σ(1, 1, α_{1})σ(2, 1, α_{2}) … σ(n, 1, α_{n})φ′,

where φ′ is a product of elementary linear automorphisms. Obviously, α_{1} = α_{2} = … = α_{n} = 0. Therefore we can assume that φ is a product of elementary linear automorphisms. By (18), (19), we can easily represent φ in the form

φ = σ(1, α_{1}, 0)σ(2, α_{2}, 0) … σ(n, α_{n}, 0)φ′,

where φ′ is a product of elementary automorphisms of the type σ(i, 1, f). By (18)–(20), we have

σ(k, α, 0) = σ(s, α, 0)^{(ks)}
= σ(s, −1, x_{k})σ(k, 1, −x_{s})σ(s, 1, x_{k})σ(s, α, 0)σ(s, −1, x_{k})σ(k, 1, −x_{s})σ(s, 1, x_{k})
= σ(s, −1, x_{k})σ(k, 1, −x_{s})σ(s, −α, 0)σ(s, 1, (1 − α)x_{k})σ(k, 1, −x_{s})σ(s, 1, x_{k})
= σ(s, −1, x_{k})σ(s, −α, 0)σ(k, 1, α^{−1}x_{s})σ(s, 1, (1 − α)x_{k})σ(k, 1, −x_{s})σ(s, 1, x_{k})
= σ(s, α, 0)σ(s, 1, −αx_{k})σ(k, 1, α^{−1}x_{s})σ(s, 1, (1 − α)x_{k})σ(k, 1, −x_{s})σ(s, 1, x_{k}).

By this relation, we can represent φ in the form

φ = σ(n, α_{n}, 0)φ′,

where φ′ is a product of elementary linear automorphisms of the form σ(i, 1, f). Hence α_{n} = 1. Note that σ(i, 1, f) can be represented as a product of automorphisms

X_{ij}(λ) = σ(j, 1, λx_{i}), λ ∈ F, i ≠ j.  (21)

Thus, we can assume that φ is a product of automorphisms of the form (21).

Let G be the subgroup of T Aut A_{n} generated by all automorphisms of the form (21). We define a map

J : G ⟶ SL_{n}(F),

where J(ψ) is the Jacobian matrix of ψ ∈ G. It is easy to check that

J(X_{ij}(λ)) = E_{ij}(λ),

and that J is an isomorphism of the groups.

Now it is sufficient to prove that every relation of the group SL_{n}(F) is a corollary of (18)–(20). Obviously, (18)–(19) cover the Steinberg relations (see, for example [18]). Besides, according to [18], we need to check the relation

{u, v} = id, 0 ≠ u, v ∈ F,

where

{u, v} = h_{ij}(uv)h_{ij}(u)^{−1}h_{ij}(v)^{−1},
h_{ij}(u) = w_{ij}(u)w_{ij}(−1),
w_{ij}(u) = X_{ij}(u)X_{ji}(−u^{−1})X_{ij}(u).

Applying (18)–(20) we have

w_{ij}(u) = σ(j, 1, ux_{i})σ(i, 1, −u^{−1}x_{j})σ(j, 1, ux_{i})
= σ(j, 1, ux_{i})σ(i, u, 0)σ(i, 1, −x_{j})σ(i, u^{−1}, 0)σ(j, 1, ux_{i})
= σ(i, u, 0)σ(j, 1, ux_{i})^{σ(i,u,0)}σ(i, 1, −x_{j})σ(j, 1, ux_{i})^{σ(i,u,0)}σ(i, u^{−1}, 0)
= σ(i, u, 0)σ(j, 1, x_{i})σ(i, 1, −x_{j})σ(j, 1, x_{i})σ(i, u^{−1}, 0)
= σ(i, u, 0)σ(j, −1, 0)σ(j, −1, x_{i})σ(i, 1, −x_{j})σ(j, 1, x_{i})σ(i, u^{−1}, 0)
= σ(i, u, 0)σ(j, −1, 0)(ij)σ(i, u^{−1}, 0) = (ij)σ(i, u, 0)^{(ij)}σ(j, −1, 0)^{(ij)}σ(i, u^{−1}, 0)
= (ij)σ(j, u, 0)σ(i, −1, 0)σ(i, u^{−1}, 0) = (ij)σ(j, u, 0)σ(i, −u^{−1}, 0).

Consequently,

h_{ij}(u) = w_{ij}(u)w_{ij}(−1) = (ij)σ(j, u, 0)σ(i, −u^{−1}, 0)(ij)σ(j, −1, 0)σ(i, 1, 0)
= σ(j, u, 0)^{(ij)}σ(i, −u^{−1}, 0)^{(ij)}σ(j, −1, 0)
= σ(i, u, 0)σ(j, −u^{−1}, 0)σ(j, −1, 0) = σ(i, u, 0)σ(j, u^{−1}, 0).

Hence

{u, v} = h_{ij}(uv)h_{ij}(u)^{−1}h_{ij}(v)^{−1}
= σ(i, uv, 0)σ(j, (uv)^{−1}, 0)σ(i, u, 0)σ(j, u^{−1}, 0)σ(i, v, 0)σ(j, v^{−1}, 0) = id.

Thus we can say that every relation of the group SL_{n}(F) follows from (18)–(20). □ Assume that

θ = ϕ_{1}ϕ_{2} … ϕ_{r} ∈ T Aut A_{n},  (22)

where ϕ_{i}, 1 ≤ i ≤ r, are elementary automorphisms. Put

ψ_{i} = ϕ_{1}ϕ_{2} … ϕ_{i}, 0 ≤ i ≤ r.

In particular, we have

ψ_{r} = θ, ψ_{0} = id.

To (22) corresponds the sequence of elementary transformations

id = ψ_{0} →^{ϕ_{1}} ψ_{1} →^{ϕ_{2}} ψ_{2} →^{ϕ_{3}} … →^{ϕ_{r}} ψ_{r} = θ.  (23)

Note that the representations (22) and (23) of the automorphism θ are equivalent. If (23) is a minimal representation of θ, then the representation (22) will be also called a minimal representation of θ.

Theorem 4.1 Let F be a field of characteristic 0 and A = F [x_{1}, x_{2}, x_{3}]. Then the relations (18), (19), (20) are defining relations of the group T Aut A with respect to the generators (17).

Plan of the proof. Assume that

φ_{1}φ_{2} … φ_{k} = id = (x_{1}, x_{2}, x_{3}),  (24)

where φ_{i}, 1 ≤ i ≤ k, are elementary automorphisms. Put

θ_{i} = φ_{1}φ_{2} … φ_{i}, 0 ≤ i ≤ k.

In particular, we have θ_{0} = θ_{k} = (x_{1}, x_{2}, x_{3}). To (24) corresponds the sequence of elementary transformations

id = θ_{0} →^{φ_{1}} θ_{1} →^{φ_{2}} … →^{φ_{k}} θ_{k} = id.  (25)

Denote by d = max{deg θ_{i}|0 ≤ i ≤ k} the width of the sequence (25). Let i_{1} be the minimal number and i_{2} be the maximal number which satisfy the equations deg θ_{i_{1}} = d and deg θ_{i_{2}} = d. Put q = i_{2} −i_{1}. The pair (d, q) will be called the exponent of the relation (24).

To prove the theorem, we show that (24) follows from (18)–(20). Assume that our theorem is not true. Call a relation of the form (24) trivial if it follows from (18)–(20). We choose a nontrivial relation (24) with the minimal exponent (d, q) with respect to the lexicographical order. To arrive at a contradiction, we show that (24) is also trivial.

If d = 3, then Lemma 4.2 gives the triviality of the relation (24). Therefore we can assume that d > 3.

Our plan is to change the product (24) using (18)–(20) and to obtain a new sequence (25) with the exponent strictly less than (d, q). The proof of the theorem will be completed by Lemmas 4.3–4.10. □

Denote by t = [\frac{q}{2}] the integral part of \frac{q}{2}. Put also

ϕ = θ_{i_{1}+t−1}, θ = θ_{i_{1}+t}, τ = θ_{i_{1}+t+1}.

Then we have

ϕ ⟶^{φ_{i_{1}+t}} θ ⟶^{φ_{i_{1}+t+1}} τ.  (26)

Lemma 4.3 The following statements are true:

1) d is the width of θ, t is the height of θ, and

θ = φ_{1}φ_{2} … φ_{i_{1}+t}  (27)

is a minimal representation of θ.

2) If q is an even number, then

θ = φ_{k}^{−1}φ_{k−1}^{−1} … φ_{i_{1}+t+1}^{−1}

is also a minimal representation of θ.

3) If q is an odd number, then (d(τ), t(τ)) = (d, t) and

τ = φ_{k}^{−1}φ_{k−1}^{−1} … φ_{i_{1}+t+2}^{−1}

is a minimal representation of τ. Moreover, in (24) the product (27) can be replaced by an arbitrary minimal representation of θ.

Proof. Assume that (d(θ), t(θ)) < (d, t) and let (22) be a minimal representation of θ. Then (24) is a consequence of the equalities

φ_{1}φ_{2} … φ_{i_{1}+t}ϕ_{r}^{−1} … ϕ_{2}^{−1}ϕ_{1}^{−1} = id,  (28)

ϕ_{1}ϕ_{2} … ϕ_{r}φ_{i_{1}+t+1} … φ_{k−1}φ_{k} = id.  (29)

To (28) corresponds the sequence of elementary transformations

(x_{1}, x_{2}, x_{3}) → θ_{1} → … → θ_{i_{1}+t} = θ = ψ_{r} → ψ_{r−1} … → ψ_{1} → (x_{1}, x_{2}, x_{3}),

and to (29) corresponds

(x_{1}, x_{2}, x_{3}) → ψ_{1} → … → ψ_{r} = θ = θ_{i_{1}+t} → θ_{i_{1}+t+1} → … → θ_{k−1} → (x_{1}, x_{2}, x_{3}).

Since (d(θ), t(θ)) < (d, t), it follows that (28) and (29) have the exponents strictly less than (d, q). This gives the first statement of the lemma.

It is obvious that

φ_{k}^{−1}φ_{k−1}^{−1} … φ_{1}^{−1} = id

has the same exponent (d, q). Applying the first statement of the lemma to this relation, we get the second statement of the lemma as well as the minimality of the representation of τ if q is an odd number. If q is an odd number, then (28) has the exponent strictly less than (d, q), and (29) has the exponent (d, q). Consequently, (24) and (29) are equivalent modulo (18)–(20). Thus θ can be changed by an arbitrary minimal representation in (29). □

Lemma 4.4 If t ≥ 1, then the relation (24) is trivial.

Proof. Our aim is to change the product φ_{i_{1}+t}φ_{i_{1}+t+1} by (18)–(20) and to get a new sequence (25) with the exponent strictly less than (d, q), i.e. to show that (24) is trivial.

Since t ≥ 1, according to Proposition 3.1 θ admits a reduction of types I–IV. We restrict ourselves only to the case when θ admits a reduction of type IV. The other cases can be considered similarly.

Let θ = (f_{1}, f_{2}, f_{3}) satisfy the conditions of Definition 2.4. According to Lemma 4.3, the representation (27) of θ is minimal. Then ϕ can be calculated by using Proposition 3.1.

Case I: q is an even, q = 2t. According to Lemma 4.3 τ can also be calculated by using Proposition 3.1.

Assume that t = 2. By Proposition 3.1 the automorphisms ϕ and τ have the same form, i.e. either

ϕ = (f_{1}, η_{1}g_{2} + κ_{1}g_{1} + ν_{1}, f_{3}), τ = (f_{1}, η_{2}g_{2} + κ_{2}g_{1} + ν_{2}, f_{3}),

or

ϕ = (ρ_{1}g_{1} + σ_{1}, f_{2}, f_{3}), τ = (ρ_{2}g_{1} + σ_{2}, f_{2}, f_{3}).

By (18), in both cases φ_{i_{1}+t}φ_{i_{1}+t+1} can be replaced by an elementary automorphism. Obviously, (24) will then be replaced by a relation with the exponent strictly less than (d, q).

Assume that t = 3. According to Proposition 3.1, the automorphisms ϕ, τ have the forms (5), (6). If ϕ, τ have the same form (5) (or (6)), then as above, by (18), we obtain the triviality of (24). Suppose that ϕ has the form (5), and τ has the form (6). It is immediate that

φ_{i_{1}+t} = σ(2, η^{−1}, F (x_{1}, x_{3})), φ_{i_{1}+t+1} = σ(1, ρ, G(x_{3})).

By (19) we get

φ_{i_{1}+t}φ_{i_{1}+t+1} = σ(1, ρ, G(x_{3}))σ(2, η^{−1}, F_{1}(x_{1}, x_{3})).

We replace φ_{i_{1}+t}φ_{i_{1}+t+1} in (24) according to this equality,. Then θ in (25) can be changed to

θ′ = (ρg_{1} + α, ηg_{2} + κg_{1} + ν, f_{3}).

Note that deg θ′ = \frac{13n}{2} and after such replacement the exponent (d, q) of the sequence (25) does not change. As before θ′ admits a reduction of type IV. But according to Proposition 3.1, in this case we have t(θ′) = 1. This contradicts Lemma 4.3, since t(θ′) < t.

If t = 1, then according to Proposition 3.1 we have

ϕ = (g_{1}, g_{2}, ρ_{1}g_{3} + α_{1}), τ = (g_{1}, g_{2}, ρ_{2}g_{3} + α_{2}).

By (18), we obtain the triviality of (24) as above.

Case II: q is odd, q = 2t+1. If (α_{1}, α_{2}) = 0 and (β_{1}, β_{2}, β_{3}, β_{4}) = 0 in the conditions of Definition 2.4, then θ = (g_{1}, g_{2}, f_{3}). Moreover, according to Proposition 3.1, we have t = 1 and

ϕ = (g_{1}, g_{2}, ρg_{3} + α).

Assume that

τ = (λf_{1} + a, f_{2}, f_{3}), a ∈< f_{2}, f_{3} > .  (30)

By Lemma 4.3 we have (d(τ), t(τ)) = (d, t). Since d = d(θ) = \frac{13n}{2}, from here we get deg τ ≤ \frac{13n}{2}. With regard to the inequalities of Proposition 2.1, we have deg (λf_{1}+a) < \frac{5n}{2} and deg a < \frac{5n}{2}. Applying Lemma 2.4 gives also

a = α′_{0} + α′_{1}f_{3} + α′_{2}f_{3}^{2}.

Consequently,

λf_{1} + a = λg_{1} + α′_{0} + α′_{1}f_{3} + α′_{2}f_{3}^{2}.

If (α′_{1}, α′_{2}) ≠ 0, then τ admits a reduction of type IV. Moreover, t(τ) > 1, which contradicts the equality (d(τ), t(τ)) = (d, t). Consequently,

τ = (λg_{1} + α′_{0}, g_{2}, f_{3}).

We have

φ_{i_{1}+t} = σ(3, ρ^{−1}, F (x_{1}, x_{2})), φ_{i_{1}+t+1} = σ(1, λ, α′_{0}).

According to (19), we obtain

φ_{i_{1}+t}φ_{i_{1}+t+1} = σ(1, λ, α′_{0})σ(3, ρ^{−1}, F_{1}(x_{1}, x_{2})).

After the corresponding replacement, instead of θ we have

θ′ = (λg_{1} + α′_{0}, g_{2}, ρg_{3} + α).

Note that θ′ admits an essential elementary reduction, i.e. t(θ′) = 0.

Now assume that

τ = (f_{1}, ηf_{2} + a, f_{3}), a ∈< f_{1}, f_{3} > .  (31)

Using the same arguments as above we get

τ = (g_{1}, ηg_{2} + κg_{1} + ν, f_{3}).

We have

φ_{i_{1}+t} = σ(3, ρ^{−1}, F (x_{1}, x_{2})), φ_{i_{1}+t+1} = σ(2, η, κx_{1} + ν).

The relation (19) gives

φ_{i_{1}+t}φ_{i_{1}+t+1} = σ(2, η, κx_{1} + ν)σ(3, ρ^{−1}, F_{1}(x_{1}, x_{2})).

After such replacement, instead of θ we have

θ′ = (g_{1}, ηg_{2} + κg_{1} + ν, ρg_{3} + α),

and this automorphism also admits an essential elementary reduction.

If the elementary reduction θ → τ replaces the element f_{3} of the automorphism θ, then applying (18) also gives the triviality of (24).

We now consider the case when (α_{1}, α_{2}) = 0 and (β_{1}, β_{2}, β_{3}, β_{4}) ≠ 0. Then according to Proposition 3.1 we have t = 2 and ϕ has the form (5), i.e.

ϕ = (g_{1}, ηg_{2} + κg_{1} + ν, f_{3}).

If τ has the form (31), then (18) gives the triviality of (24). Assume that τ has the form (30). Then by the same discussion as above we get

τ = (ρg_{1} + α, f_{2}, f_{3}).

We have

φ_{i_{1}+t} = σ(2, η^{−1}, F (x_{1}, x_{2})), φ_{i_{1}+t+1} = σ(1, ρ, α).

By (19), we get

φ_{i_{1}+t}φ_{i_{1}+t+1} = σ(1, ρ, α)σ(3, η^{−1}, F_{1}(x_{1}, x_{2})).

After the corresponding replacement, instead of θ we obtain

θ′ = (ρg_{1} + α, ηg_{2} + κg_{1} + ν, f_{3}).

Proposition 3.1 gives t(θ′) = 1 < t; a contradiction.

Assume that

τ = (f_{1}, f_{2}, ρf_{3} + a), a ∈< f_{1}, f_{2} > .  (32)

By Proposition 2.1 we have deg f_{1} + deg f_{2} > \frac{9n}{2}. Since d(θ) = \frac{13n}{2} ≥ deg τ, we have deg (ρf_{3} + a) < 2n and deg a < 2n. Lemma 2.4 gives a = α ∈ F. Then φ_{i_{1}+t+1} = σ(3, ρ, α). Hence

φ_{i_{1}+t}φ_{i_{1}+t+1} = σ(3, ρ, α)σ(2, η^{−1}, F_{1}(x_{1}, x_{2})).

After this replacement, θ will be changed to

θ′ = (g_{1}, ηg_{2} + κg_{1} + ν, ρf_{3} + a).

Proposition 3.1 gives t(θ′) = 1, which also leads to a contradiction.

The case when (α_{1}, α_{2}) ≠ 0 and (β_{1}, β_{2}, β_{3}, β_{4}) = 0 can be considered analogously. Assume that (α_{1}, α_{2}) ≠ 0 and (β_{1}, β_{2}, β_{3}, β_{4}) ≠ 0. According to Proposition 3.1, we

have t = 3. Now we use the statement 3) of Lemma 4.3 about the arbitrariness of the minimal representation of θ in (24). If τ has the form (30), then we can assume that ϕ has the form (6), and if τ has the form (31), then we can assume that ϕ has the form (5). By (18), in both cases we can reduce the exponent of (24). If τ has the form (32), then using the same arguments we get

τ = (f_{1}, f_{2}, ρf_{3} + α).

Assume that ϕ = (λg_{1} + β, f_{2}, f_{3}). Applying (19) to φ_{i_{1}+t}φ_{i_{1}+t+1} changes θ to

θ′ = (g_{1}, g_{2}, ρf_{3} + a).

Then we have t(θ′) = 2, which also leads to a contradiction. □

Now we begin to consider the most complicated case when t = 0, i.e. q = 0, 1. Put θ = (f_{1}, f_{2}, f_{3}). According to Lemma 4.3, t = 0 is the height of θ, i.e. θ admits an essential elementary reduction. Moreover, ϕ is an essential elementary reduction of θ. If q = 0, then τ is also an esssential elementary reduction of θ. If q = 1, then we have (d(τ), t(τ)) = (d, t). Consequently, deg τ ≤ deg θ = d. Thus we can assume that

τ = (f_{1}, f_{2}, f), f = αf_{3} + a, a = a(f_{1}, f_{2}) ∈ ⟨f_{1}, f_{2}⟩, deg a ≤ deg f_{3}.  (33)

Lemma 4.5 If ϕ reduces the element f_{3} of θ, then the relation (24) is trivial.

Proof. Applying (18) we can replace φ_{i_{1}+t}φ_{i_{1}+t+1} by an elementary automorphism. Obviously, this replacement also reduces the exponent of (25). □

By Lemma 4.5, we can assume that ϕ reduces one of the elements f_{1}, f_{2} of θ. Without loss of generality, later on we consider that ϕ reduces the element f_{2} of θ, i.e. ϕ = (f_{1}, g_{2}, f_{3}), deg g_{2} < deg f_{2}.

Lemma 4.6 If ϕ′ reduces the element f_{2} of θ, then in (26) the automorphism ϕ can be replaced by ϕ′.

Proof. According to (18), in this case the elementary transformation ϕ → θ can be changed to ϕ → ϕ′ → θ. Since deg ϕ′ < deg θ = d, the exponent (d, q) of the sequence (25) does not change after this replacement. But in the new sequence (25) we have ϕ′ instead of ϕ. □

Taking this lemma into account, we can assume that

ϕ = (f_{1}, g_{2}, f_{3}), g_{2} = f_{2} − b, b ∈ ⟨f_{1}, f_{3}⟩, deg g_{2} < deg f_{2}.  (34)

Lemma 4.7 If one of the following conditions is fulfilled, then (24) is trivial: 1) \overline{f_{2}} ∈ ⟨\overline{f_{1}}⟩;

2) \overline{f_{3}} ∈ ⟨\overline{f_{1}}⟩;

3) a does not depend on f_{2};

4) \overline{f_{2}} = β\overline{f_{3}} + γ\overline{f_{1}^{k}};

5) \overline{f_{1}}, \overline{f_{3}} are algebraically independent.

Proof. Assume that \overline{f_{2}} ∈ ⟨\overline{f_{1}}⟩ and \overline{f_{2}} = β\overline{f_{1}}^{l}. According to Lemma 4.6, we can suppose that g_{2} = f_{2} − βf_{1}^{l}. Then

φ_{i_{1}+t} = σ(2, 1, βx_{1}^{l}), φ_{i_{1}+t+1} = σ(3, α, a(x_{1}, x_{2})).

By (19), we have

φ_{i_{1}+t}φ_{i_{1}+t+1} = σ(3, α, a_{1}(x_{1}, x_{2}))σ(2, 1, βx_{1}^{l}).

After the corresponding replacement in (24), θ will be replaced by θ′ = (f_{1}, g_{2}, f) in (25). Since deg f_{1} + deg g_{2} + deg f < d, the exponent of (24) is decreased.

Assume that \overline{f_{3}} ∈ ⟨\overline{f_{1}}⟩ and \overline{f_{3}} = T (\overline{f_{1}}). Put g_{3} = f_{3} − T (f_{1}). According to (19), we have

φ_{i_{1}+t} = σ(2, 1, b(x_{1}, x_{3})) = σ(3, 1, −T (x_{1}))σ(2, 1, b_{1}(x_{1}, x_{3}))σ(3, 1, T (x_{1})).

After the corresponding replacement in (24), the elementary transformation ϕ → θ will be replaced by the sequence elementary transformations

ϕ = (f_{1}, g_{2}, f_{3}) → (f_{1}, g_{2}, g_{3}) → (f_{1}, f_{2}, g_{3}) → (f_{1}, f_{2}, f_{3}) = θ.

Since d(ϕ), deg (f_{1}, g_{2}, g_{3}), deg (f_{1}, f_{2}, g_{3}) < d = deg θ, the new sequence (25) has the same exponent (d, q). However, instead of ϕ we have (f_{1}, f_{2}, g_{3}). By Lemma 4.5, we obtain the triviality of (24).

Assume that a does not depend on f_{2}. By (19) we have

φ_{i_{1}+t}φ_{i_{1}+t+1} = σ(2, 1, b(x_{1}, x_{3}))σ(3, α, a(x_{1})) = σ(3, α, a(x_{1}))σ(2, 1, b_{1}(x_{1}, x_{3})).

After the corresponding replacement in (24), instead of θ we obtain θ′ = (f_{1}, g_{2}, f). Since deg (f_{1}, g_{2}, f) < d, this replacement also decreases the exponent of (24).

Consider the case when \overline{f_{2}} = β\overline{f_{3}} + γ\overline{f_{1}^{k}}. By the statement 1) we can assume that β ≠ 0. By Lemma 4.6 we can also assume that b = βf_{3} + γf_{1}^{k}. Consequently,

g_{2} = f_{2} − βf_{3} − γf_{1}^{k},
f_{2} = g_{2} + βf_{3} + γf_{1}^{k},
f_{3} = −\frac{1}{β} g_{2} + \frac{1}{β} f_{2} − \frac{γ}{β} f_{1}^{k}.

These equalities justify the sequence of elementary transformations

(f_{1}, f_{3}, g_{2}) → (f_{1}, f_{2}, g_{2}) → (f_{1}, f_{2}, f_{3}) = θ.

We have

φ_{i_{1}+t} = σ(2, 1, βx_{3} + γx_{1}^{k})).

Applying (18) and (19) we get

φ_{i_{1}+t} = σ(2, 1, βx_{3})σ(2, 1, γx_{1}^{k})

= σ(2, 1, βx_{3})σ(3, −β, x_{2})σ(3, −\frac{1}{β} , \frac{1}{β} x_{2})σ(2, 1, γx_{1}^{k})

= σ(2, 1, βx_{3})σ(3, −β, x_{2})σ(2, 1, γx_{1}^{k})σ(3, −\frac{1}{β} , \frac{1}{β} x_{2})^{σ(2,1,γx_{1}^{k})}

= σ(2, 1, βx_{3})σ(3, −β, x_{2})σ(2, 1, γx_{1}^{k})σ(3, −\frac{1}{β} , \frac{1}{β} (x_{2} − γx_{1}^{k}))

= σ(2, 1, βx_{3})σ(3, −β, x_{2})σ(2, \frac{1}{β} , −\frac{1}{β} x_{3})σ(2, β, x_{3} + γx_{1}^{k})σ(3, −\frac{1}{β} , \frac{1}{β} (x_{2} − γx_{1}^{k}))).

Since the transposition (23) ∈ S_{3} has a linear representation

(23) = σ(2, 1, βx_{3})σ(3, −β, x_{2})σ(2, \frac{1}{β} , −\frac{1}{β} x_{3}),

we obtain

φ_{i_{1}+t} = (23)σ(2, β, x_{3} + γx_{1}^{k}))σ(3, −\frac{1}{β} , \frac{1}{β} (x_{2} − γx_{1}^{k})).

Then

θ = (23)φ_{1}^{(23)}φ_{2}^{(23)} … φ_{i_{1}+t−1}^{(23)}σ(2, β, x_{3} + γx_{1}^{k}))σ(3, −\frac{1}{β} , \frac{1}{β} (x_{2} − γx_{1}^{k})),  (35)

where φ_{i}^{(23)} are elementary automorphisms, according to (20). To (35) corresponds the sequence of elementary transformations

(x_{1}, x_{2}, x_{3}) ↦ (x_{1}, x_{3}, x_{2}) → θ′_{1} → θ′_{2} → … → θ′_{i_{1}+t−1}
= (f_{1}, f_{3}, g_{2}) → (f_{1}, f_{2}, g_{2}) → (f_{1}, f_{2}, f_{3}) = θ,

where θ′_{i} is obtained from θ_{i} only by the permutation of the second and the third components, and (x_{1}, x_{2}, x_{3}) ↦ (x_{1}, x_{3}, x_{2}) denotes a composition of elementary linear transformations corresponding to the definition of the automorphism (23) ∈ S_{3}.

If in (24) we replace θ by (35), then the exponent of (25) remains the same. But instead of ϕ we have (f_{1}, f_{2}, g_{2}), and Lemma 4.5 gives the triviality of (24).

We now consider the case when \overline{f_{1}}, \overline{f_{3}} are algebraically independent. Then \overline{f_{2}} = \overline{b} ∈ ⟨\overline{f_{1}}, \overline{f_{3}}⟩. By 1) we can assume that \overline{f_{2}} ∉ ⟨\overline{f_{1}}⟩, i.e. \overline{f_{2}} depends on \overline{f_{3}}. Consequently, deg f_{3} ≤ deg f_{2}. If \overline{f_{1}}, \overline{f_{2}} are algebraically dependent, then it follows that \overline{f_{1}}, \overline{f_{3}} are algebraically dependent. Consequently, \overline{f_{1}}, \overline{f_{2}} are algebraically independent. Then \overline{a} ∈ ⟨\overline{f_{1}}, \overline{f_{2}}⟩. By 3) we can assume that a necessarily contains f_{2}. Then deg f_{2} ≤ deg a ≤ deg f_{3}, i.e. deg f_{2} = deg f_{3}. Hence

\overline{b} = \overline{f_{2}} = β\overline{f_{3}} + γ\overline{f_{1}^{k}}.

From the statement 4) of the lemma we obtain that (24) is trivial. □

So, by Lemma 4.7, we can assume that \overline{f_{1}}, \overline{f_{3}} are algebraically dependent and that \overline{f_{3}} ∉ ⟨\overline{f_{1}}⟩. It remains to consider the following three cases separately:

1) f_{1}, f_{3} is a ∗-reduced pair and deg f_{1} < deg f_{3};

2) f_{1}, f_{3} is a ∗-reduced pair and deg f_{3} < deg f_{1};

3) \overline{f_{1}} ∈ ⟨\overline{f_{3}}⟩ and deg f_{1} > deg f_{3}.

Lemma 4.8 If f_{1}, f_{3} is a ∗-reduced pair and deg f_{1} < deg f_{3}, then (24) is trivial.

Proof. We first consider the case when deg f_{2} > deg f_{3}. If \overline{f_{1}}, \overline{f_{2}} are algebraically independent, then \overline{a} ∈< \overline{f_{1}}, \overline{f_{2}} >. Since deg a ≤ deg f_{3} < deg f_{2}, we have a ∈ ⟨f_{1}⟩. Lemma 4.7 gives the triviality of (24).

Suppose that \overline{f_{1}}, \overline{f_{2}} are algebraically dependent. By Lemma 4.7, we can assume that \overline{f_{2}} ∉ ⟨\overline{f_{1}}⟩. Since deg f_{1} < deg f_{3} < deg f_{2}, therefore f_{1}, f_{2} is a ∗-reduced pair.

Assume that deg a < N = N(f_{1}, f_{2}). Since deg a ≤ deg f_{3} < deg f_{2}, Corollary 2.1 gives a ∈< f_{1} >. By Lemma 4.7, the relation (24) is trivial.

Therefore we can assume that deg a ≥ N. Then deg f_{2} > N = N(f_{1}, f_{2}). By the definition of N = N(f_{1}, f_{2}), this is possible only if f_{1}, f_{2} is a 2-reduced pair. Consequently, deg f_{1} = 2n, deg f_{2} = sn, where s ≥ 3 is an odd number, and moreover

N = n(s − 2) + deg [f_{1}, f_{2}] ≤ deg a ≤ deg f_{3}.

Let (f_{1}, f_{2}, g_{3}) be an elementary reduction of θ such that g_{3} is an irreducible element of (f_{1}, f_{2}, g_{3}). If f_{3} is an irreducible element of θ, then we put g_{3} = f_{3}. Then (f_{1}, f_{2}, g_{3}) satisfies the conditions of Proposition 2.2. Assume that the statement 1) of Proposition 2.2 is valid for (f_{1}, f_{2}, g_{3}), i.e.

deg g_{3} < n(s − 2) + deg [f_{1}, f_{2}] = N.

Since deg f_{3} ≥ N, it means that

f_{3} = g_{3} + c(f_{1}, f_{2}), \overline{c(f_{1}, f_{2})} = \overline{f_{3}}.

By Lemma 4.7 we can assume that \overline{c(f_{1}, f_{2})} = \overline{f_{3}} ∉< f_{1} >. Since deg c(f_{1}, f_{2}) = deg f_{3} < deg f_{2}, we have \overline{c(f_{1}, f_{2})} ∉< \overline{f_{1}}, \overline{f_{2}} >. It is easy to deduce from (1) that deg_{y} c(x, y) ∈ \{2, 4\}. Then deg_{y} \frac{∂c}{∂y}(x, y) ∈ \{1, 3\}, and (1) gives \frac{∂c}{∂y}(f_{1}, f_{2}) ≥ deg f_{2}. We have

deg [f_{1}, g_{3}] ≤ deg f_{1} + deg g_{3} < ns + deg [f_{1}, f_{2}] = deg f_{2} + deg [f_{1}, f_{2}].

Then

[f_{1}, f_{3}] = [f_{1}, g_{3}] + [f_{1}, f_{2}]\frac{∂c}{∂y} (f_{1}, f_{2})

gives that

deg [f_{1}, f_{3}] = deg [f_{1}, f_{2}]\frac{∂c}{∂y} (f_{1}, f_{2}) ≥ deg [f_{1}, f_{2}] + deg f_{2}.

Consequently, N(f_{1}, f_{3}) > deg f_{2}. By (34) we have deg b = deg f_{2} < N(f_{1}, f_{3}). Corollary 2.1 gives \overline{f_{2}} ∈< \overline{f_{1}}, \overline{f_{3}} >. Since deg f_{1}f_{3}, deg f_{3}^{2} > deg f_{2}, we obtain \overline{f_{2}} ∈< \overline{f_{1}} >; a contradiction.

We now assume that one of the statements 2), 3) of Proposition 2.2 is valid for (f_{1}, f_{2}, g_{3}). Combining both the cases we can say that

f_{3} = h_{3} + r(f_{1}, f_{2}), deg f_{2} = 3n, deg (f_{2} − βh_{3}^{2}) ≤ 2n, deg [f_{1}, h_{3}] < n + deg [f_{1}, f_{2}].

Since deg f_{3} < 2n and deg h_{3} = \frac{3n}{2}, it follows that deg r(f_{1}, f_{2}) < 2n. By Corollary 2.2, we have r(f_{1}, f_{2}) = λw(f_{1}, f_{2}) + μ, where w(x, y) is a derivative polynomail of the pair f_{1}, f_{2}. If λ ≠ 0, then an immediate calculation gives deg [f_{1}, f_{3}] > deg f_{2}. This leads to a contradiction, as above. Then λ = 0, and we can assume that f_{3} = h_{3}. Combining (33) and Corollary 2.2 we also conclude that

a = λw(f_{1}, f_{2}) + μ, f = αf_{3} + λw(f_{1}, f_{2}) + μ.

If λ ≠ 0, then τ admits a reduction of type IV. Otherwise a = μ ∈ F and we can apply Lemma 4.7.

Now, suppose that deg f_{2} ≤ deg f_{3}. If \overline{f_{2}} ∈< \overline{f_{1}}, \overline{f_{3}} >, then \overline{f_{2}} = β\overline{f_{3}} + γ\overline{f_{1}^{k}}. Hence Lemma 4.7 gives the triviality of (24).

Let \overline{b} = \overline{f_{2}} ∉< \overline{f_{1}}, \overline{f_{3}} >. Since deg b = deg f_{2} ≤ deg f_{3}, applying Corollary 2.1 gives deg f_{3} ≥ N(f_{1}, f_{3}). By the definition of N(f_{1}, f_{3}), we see that f_{1}, f_{3} is a 2-reduced pair, i.e. deg f_{1} = 2n and deg f_{3} = ns, where s ≥ 3 is an odd number. According to Lemma 4.6, we can assume that g_{2} is an irreducible element of ϕ = (f_{1}, g_{2}, f_{3}). Then ϕ satisfies the conditions of Proposition 2.2.

If deg_{y}(b(x, y)) = k, then (1) gives k = 2 if s > 3, and k = 2, 4 if s = 3. Consequently, deg_{y}(\frac{∂b}{∂y}(x, y)) = k − 1 ∈ \{1, 3\}, and (1) gives also deg(\frac{∂b}{∂y}(f_{1}, f_{3})) ≥ ns. We have

[f_{1}, f_{2}] = [f_{1}, g_{2}] + [f_{1}, f_{3}]\frac{∂b}{∂y} (f_{1}, f_{3}).  (36)

Assume that the statement 1) of Proposition 2.2 is valid for ϕ, i.e. deg g_{2} < n(s − 2) + deg[f_{1}, f_{3}]. Comparing the degrees of elements in (36) we obtain

deg[f_{1}, f_{2}] ≥ deg[f_{1}, f_{3}] + ns.  (37)

Consequently, deg f_{2} ≥ n(s − 2) + deg[f_{1}, f_{3}] > n and \overline{f_{1}} ∉ ⟨\overline{f_{2}}⟩. Then either \overline{f_{1}} and \overline{f_{2}} are algebraically independent or f_{1}, f_{2} is a ∗-reduced pair. Note that N(f_{1}, f_{2}) > deg [f_{1}, f_{2}] > ns ≥ deg a. By Corollary 2.1, in both cases we have \overline{a} ∈ ⟨\overline{f_{1}}, \overline{f_{2}}⟩. Since deg f_{2}^{3} > deg f_{3} ≥ deg a and deg f_{1}f_{2} > deg f_{3}, it follows that

a = βf_{2}^{2} + γf_{2} + G(f_{1}), deg(G(f_{1})) ≤ deg f_{3};

moreover, we can have β ≠ 0 only if deg f_{3} ≥ 2 deg f_{2}. Note that the equality deg(G(f_{1})) = deg f_{3} is impossible. Consequently, deg(G(f_{1})) < deg f_{3}. Since f_{1}, f_{3} is a 2-reduced pair, it follows that f_{1}, αf_{3} + G(f_{1}) is also a 2-reduced pair. We have

τ = (f_{1}, f_{2}, αf_{3} + βf_{2}^{2} + γf_{2} + G(f_{1})).

If (β, γ) ≠ (0, 0), then it is easy to check that (f_{1}, g_{2}, αf_{3} + G(f_{1})) is a reduction of types I–III of the automorphism τ with the active element f_{2}. Proposition 3.1 gives d(τ) = n(s + 2) + deg f_{2} = d = deg θ and t(τ) ≥ 1. This contradicts the inequality (d(τ), t(τ)) ≤ (d, t). If β = γ = 0, then we apply Lemma 4.7.

We now consider the case when the statement 3) of Proposition 2.2 is valid for ϕ. Then (36) again gives (37). Besides, in this case s = 3 and deg f_{2} > deg g_{2} = \frac{3n}{2}. Consequently, the same argument as above gives a = βf_{2} + γf_{1} + δ. By Lemma 4.7, we can assume that β ≠ 0. Then (f_{1}, g_{2}, αf_{3} + γf_{1} + δ) is a reduction of type I or II of τ with the active element f_{2}. The value of (d(τ), t(τ)), which can be calculated by Proposition 3.1, again contradicts the inequality (d(τ), t(τ)) ≤ (d, t).

At last we consider the case when the statement 2) of Proposition 3.1 is valid for ϕ. Let (f_{1}, h_{2}, f_{3} − δh_{2}^{2}) be a reduction of type IV of ϕ, where

g_{2} = h_{2} + g, g ∈ ⟨f_{1}, f_{3}⟩ \ F, deg(f_{3} − δh_{2}^{2}) ≤ 2n.

Since deg g_{2} ≤ deg h_{2} = \frac{3n}{2}, we have deg g ≤ \frac{3n}{2}. By Definition 2.4, the statement 3) of Proposition 2.2 is valid for (f_{1}, h_{2}, f_{3}).

If deg f_{2} > \frac{3n}{2}, then by Lemma 4.6 we can assume that ϕ = (f_{1}, h_{2}, f_{3}) and this case reduces to the preceding case.

Assume that deg f_{2} ≤ \frac{3n}{2}. By Proposition 3.1, we have d(ϕ) = \frac{13n}{2}, t(ϕ) = 1. Since d = deg θ ≤ \frac{13n}{2}, this contradicts the inequality (d(ϕ), t(ϕ)) < (d, t). □

Lemma 4.9 If f_{1}, f_{3} is a ∗-reduced pair and deg f_{3} < deg f_{1}, then (24) is trivial.

Proof. We first consider the case when deg f_{1} < deg f_{2}. Since deg a ≤ deg f_{3} < deg f_{3} < deg f_{1}, applying Lemma 4.7 we conclude that \overline{a} ∉ ⟨\overline{f_{1}}, \overline{f_{2}}⟩. Consequently, \overline{f_{1}}, \overline{f_{2}} are algebraically dependent. By Lemma 4.7, we assume that \overline{f_{2}} ∉ ⟨\overline{f_{1}}⟩. Then f_{1}, f_{2} is a ∗-reduced pair. Corollary 2.1 gives also deg a ≥ N(f_{1}, f_{2}). Combining these inequalities and the definition of N(f_{1}, f_{2}) we conclude that deg f_{1} = 2n and deg f_{2} = 3n. Consider an elementary reduction (f_{1}, f_{2}, g_{3}) of θ with an irreducible g_{3} (assume that g_{3} = f_{3} if f_{3} is an irreducible element of θ). Then (f_{1}, f_{2}, g_{3}) satisfies the conditions of Proposition 2.2. By the same discussions related to the automorphism (f_{1}, f_{2}, g_{3}) as in the proof of Lemma 4.8, we obtain the triviality of (24).

If deg f_{1} = deg f_{2}, then by Lemma 4.7 we may assume that \overline{f_{1}}, \overline{f_{2}} are linearly independent. Then \overline{f_{1}}, \overline{f_{2}} are algebraically independent. Consequently, a ∈ F.

It remains to consider the case when deg f_{2} < deg f_{1}. Assume that \overline{f_{2}} ∈ ⟨\overline{f_{3}}⟩. By Lemma 4.7 we can assume that \overline{f_{2}}, \overline{f_{3}} are linearly independent. Then deg f_{2} = t · deg f_{3}, t ≥ 2. In this case \overline{f_{1}} ∉ ⟨\overline{f_{2}}⟩, since f_{1}, f_{3} is a ∗-reduced pair. Consequently, f_{1}, f_{2} is also a ∗-reduced pair. Since deg a ≤ deg f_{3} < min{deg f_{1}, deg f_{2}}, applying Corollary 2.1 and Lemma 4.7 we obtain deg f_{2} = 2n, deg f_{1} = 3n, and

deg f_{3} ≥ deg a ≥ N(f_{1}, f_{2}) = n + deg [f_{1}, f_{2}] > \frac{2n}{t} = deg f_{3},

which gives a contradiction.

Therefore we can assume that \overline{f_{2}} ∉ ⟨\overline{f_{3}}⟩, i.e. \overline{b} = \overline{f_{2}} ∉ ⟨\overline{f_{1}}, \overline{f_{3}}⟩. Then Corollary 2.1 gives N(f_{1}, f_{3}) ≤ deg f_{2} < deg f_{1}. By the definition of N(f_{1}, f_{3}) this is possible only if f_{1}, f_{3} is a 2-reduced pair, and deg f_{3} = 2n, deg f_{1} = ns, where s ≥ 3 is an odd number. By Lemma 4.6 we can assume that g_{2} is an irreducible element of ϕ = (f_{1}, g_{2}, f_{3}). Then ϕ satisfies the conditions of Proposition 2.2. Applying the same part of discussions related to the automorphism ϕ as in the proof of Lemma 4.8, we obtain the triviality of (24). □

Lemma 4.10 If \overline{f_{1}} ∈ ⟨\overline{f_{3}}⟩ and deg f_{1} > deg f_{3}, then the relation (24) is trivial.

Proof. Assume that n = deg f_{3} and \overline{f_{1}} = β\overline{f_{3}}^{k}, k ≥ 2. Then deg f_{1} = nk. By (19) we have

φ_{i_{1}+t} = σ(2, 1, b(x_{1}, x_{2})) = σ(1, 1, −βx_{3}^{k})σ(2, 1, b(x_{1}, x_{2}))σ(1, 1, βx_{3}^{k}).

After such replacement, instead of ϕ → θ we obtain

ϕ → (f_{1} − βf_{3}^{k}, g_{2}, f_{3}) → (f_{1} − βf_{3}^{k}, f_{2}, f_{3}) → θ.

Since deg (f_{1} − βf_{3}^{k}, g_{2}, f_{3}), deg (f_{1} − βf_{3}^{k}, f_{2}, f_{3}) < d = deg θ, the new sequence (25) has the same exponent. Then instead of ϕ we can take (f_{1} − βf_{3}^{k}, f_{2}, f_{3}), i.e. we can assume that ϕ reduces the element f_{1} of θ. Interchanging the elements f_{1}, f_{2} and applying Lemmas 4.7, 4.8, and 4.9 we can restrict ourselves to the case when \overline{f_{2}} ∈ ⟨\overline{f_{3}}⟩, deg f_{2} > deg f_{3}. Then deg f_{2} = nr, r ≥ 2. By Lemma 4.7 we may assume that f_{1}, f_{2} is a ∗-reduced pair, i.e. deg f_{1} = mp, deg f_{2} = ms, (p, s) = 1, m ≥ n. Hence N(f_{1}, f_{2}) > m(ps − p − s) > n ≥ deg a. Corollary 2.1 gives \overline{a} ∈ ⟨\overline{f_{1}}, \overline{f_{2}}⟩, i.e. a ∈ F. □

This finishes the proof of Theorem 4.1.

5 Free metabelian associative algebras

Let us remind that the variety of associative algebras defined by the identity

[x, y][z, t] = 0

is called metabelian [27]. We denote by C the free metabelian algebra (with unit 1) with free generators z_{1}, z_{2}, z_{3}. Then the generators (17) of the group of tame automorphisms T Aut C satisfy the relations (18)–(20). However these relations may fail to be a full system of defining relations of the group T Aut C. Consider the homomorphism

τ : C ⟶ A = F [x_{1}, x_{2}, x_{3}], τ(z_{i}) = x_{i}, 1 ≤ i ≤ 3.

If θ = (f_{1}, f_{2}, f_{3}) ∈ Aut C, then obviously

τ^{∗}(θ) = (τ(f_{1}), τ(f_{2}), τ(f_{3})) ∈ Aut A.

It is easy to check that the map

τ^{∗} : Aut C ⟶ Aut A

is a homomorphism of the groups. We denote by T the restriction of the homomorphism τ^{∗} to T Aut C, i.e.

T : T Aut C ⟶ T Aut A, T (θ) = τ^{∗}(θ), θ ∈ T Aut C.

Note that the kernel I = Ker τ of the homomorphism τ is the commutator ideal of C. Obviously, any automorphism of the form

σ(i, α, f), α = 1, f ∈ I,  (38)

is contained in Ker T.

Theorem 5.1 Let F be a field of characteristic 0. Then the kernel of the homomorphism

T : T Aut C ⟶ T Aut A,

as a normal subgroup of the group T Aut C, is generated by all automorphisms of the form (38).

Proof. Let N be a normal subgroup of T Aut C generated by all elements of the form (38). Since N ⊆ Ker T, the homomorphism T induces the group homomorphism

μ : G = T Aut C/N ⟶ T Aut A.

We show that μ is an isomorphism. Every element f ∈ C has a unique representation in the form f = f_{0} + f_{1}, where f_{0} is a linear combination of the words z_{1}^{i}z_{2}^{j}z_{3}^{k} and f_{1} ∈ I. We have

σ(i, α, f) = σ(i, α, f_{0})σ(i, 1, f_{1}).

Since σ(i, 1, f_{1}) ∈ N, the group G is generated by all elements of the form

σ(i, α, f_{0})N, f ∈ C.  (39)

Consequently, μ establishes a one to one correspondence between the set of generators (39) of the group G and the set of generators (17) of the group T Aut A. Of course, the relations (18)–(20) are also true for elementary automorphisms of C. They induce the same relations for elements (39). By Theorem 4.1, there exists a homomorphism

ν : T Aut A ⟶ G

which is innverse to μ on the set of generators. Consequently, μ is an isomorphism. □

It was proved in [27] that every automorphism of the algebra A can be lifted to an automorphism of the algebra C. Thus, the algebra C has wild automorphisms. Now our aim is to find another set of wild automorphisms of the algebra C which can be lifted to automorphisms of the free associative algebra B = F < y_{1}, y_{2}, y_{3} >.

The Fox derivatives in free associative algebras can be found in [9]. We recall the definition of Fox derivatives of elements of the algebra C, following [27].

Let C′ be the algebra antiisomorphic to C, and A′ be the algebra antiisomorphic to A. We will denote the corresponding antiisomorphisms by ′ : C → C′, ′ : A → A′. Put also U(C) = C′ ⊗_{F} C, U(A) = A′ ⊗_{F} A. Given homomorphisms

φ : C → C, φ : A → A, φ : C → A,

we will denote by the same symbol the corresponding homomorphims

φ : U(C) → U(C), φ : U(A) → U(A), φ : U(C) → U(A),

defined as φ(f′ ⊗ g) = φ(f)′ ⊗ φ(g). If φ is an endomorphism of the algebra C, then we will denote the induced endomorphism of the algebra A by \overline{φ}.

A linear map

\frac{∂}{∂z_{i}} : C → U(A), 1 ≤ i ≤ 3,

is uniquely defined by the rules

\frac{∂z_{j}}{∂z_{i}} = δ_{ij}, \frac{∂fg}{∂z_{i}} = \frac{∂f}{∂z_{i}} τ(1 ⊗ g) + \frac{∂g}{∂z_{i}} τ(f′ ⊗ 1),

where δ_{ij} is Kronecker symbol. For every f ∈ C we put

∂(f) = (\frac{∂f}{∂z_{1}} , \frac{∂f}{∂z_{2}} , \frac{∂f}{∂z_{3}} )^{t},

where t means transposition. If φ is an endomorphism of C, then the matrix

J(φ) = (∂(φ(z_{1})), ∂(φ(z_{2})), ∂(φ(z_{3})))

is called the Jacobian matrix of φ. For arbitrary endomorphisms φ, ψ we have

J(φψ) = J(φ)\overline{φ}(J(ψ)).  (40)

Consequently, J(φ) is invertible over U(A) if φ is an automorphism. The converse statement is also true [27].

The ideal I has a right U(C)-module structure with respect to the action

m · (f′ ⊗ g) = fmg, m ∈ I, f, g ∈ C.

Since I^{2} = 0, this action induces a right U(A)-module structure on I. If f ∈ I, then

f = [z_{1}, z_{2}] · f_{12} + [z_{1}, z_{3}] · f_{13} + [z_{2}, z_{3}] · f_{23}, f_{ij} ∈ U(A_{n}).  (41)

Put Δ(x_{i}) = x′_{i} ⊗ 1 − 1 ⊗ x_{i}, 1 ≤ i ≤ 3. It is easy to check that if f has the form (41), then

\frac{∂f}{∂z_{3}} = Δ(x_{1})f_{13} + Δ(x_{2})f_{23}.  (42)

We denote by e_{i} the vector-row of length 3 whose ith component is 1 and the others are 0, 1 ≤ i ≤ 3.

Lemma 5.1 Let φ be an automorphism of the algebra C of the type (38) and let ψ ∈ Aut C. Then

J(ψφψ^{−1}) = E + u^{t} · v,

where u = (u_{1}, u_{2}, u_{3}), v = (v_{1}, v_{2}, v_{3}), u_{i}, v_{i} ∈ U(A_{n}).

Proof. Assume that φ = σ(i, 1, f), where f ∈ I. Then J(φ) = E + ∂(f) · e_{i}. We have

J(ψφψ^{−1}) = J(ψ)\overline{ψ}(J(φ)\overline{φ}(J(ψ))) = J(ψ)\overline{ψ}(J(φ))\overline{ψ}(J(ψ))
= J(ψ)\overline{ψ}(J(φ))J(ψ)^{−1} = E + J(ψ)\overline{ψ}(∂(f)) · \overline{ψ}(e_{i})J(ψ)^{−1}.

Put u^{t} = J(ψ)\overline{ψ}(∂(f)), v = \overline{ψ}(e_{i})J(ψ)^{−1}. □

For an arbitrary automorphism φ of C we put

J_{2}(φ) = ( \begin{matrix}\frac{∂φ(z_{1})}{∂z_{1}} & \frac{∂φ(z_{2})}{∂z_{1}} \\ \frac{∂φ(z_{1})}{∂z_{2}} & \frac{∂φ(z_{2})}{∂z_{2}} \end{matrix} ) .

We now consider the homomorphism

η : A ⟶ F [x_{3}], η(x_{1}) = η(x_{2}) = 0, η(x_{3}) = x_{3}.

Lemma 5.2 Let φ ∈ Aut C and \overline{φ} = id. Then

η(J(φ)) = ( \begin{matrix}η(J_{2}(φ)) & ∗ \\ 0 & 1 \end{matrix} )

Proof. We have φ = (z_{1} + f_{1}, z_{2} + f_{2}, z_{3} + f_{3}), where f_{i} ∈ I. From (42) it follows that η(\frac{∂f_{i}}{∂z_{3}} ) = 0. □

Denote the standart matrix units by e_{ij}. A matrix E_{ij}(a) = E + ae_{ij}, where i ≠ j, a ∈ U(A), is called an elementary matrix over U(A). We denote the subgroup of SL_{2}(U(A)) generated by all elementary matrices by E_{2}(U(A)). We also denote by diag(α, β) a diagonal matrix with the diagonal entries α, β.

For every 0 ≠ λ ∈ U(A) we will denote by U_{λ}(A) the subring of the ring of quotients Q(U(A)) generated by U(A) ∪ {λ^{−1}}.

Now we put S_{λ} = E_{2}(U_{λ}(A)) · diag(λ, 1). It is easy to check that

E_{12}(a) · diag(λ, 1) = diag(λ, 1) · E_{12}(\frac{a}{λ}), E_{21}(a) · diag(λ, 1) = diag(λ, 1) · E_{21}(λa),

diag(λ, 1) = E_{12}(1) · E_{21}(−1) · E_{12}(1) · diag(1, λ) · E_{12}(−1) · E_{21}(1) · E_{12}(−1).

From here we get

S_{λ} = E_{2}(U_{λ}(A)) · diag(1, λ) = diag(λ, 1) · E_{2}(U_{λ}(A)) = diag(1, λ) · E_{2}(U_{λ}(A)). (43)

Lemma 5.3 Let φ be an automorphism of C of the form (38) and let ψ ∈ T Aut C. Then

λ = det(J_{2}(ψφψ^{−1})) ≠ 0, J_{2}(ψφψ^{−1}) ∈ S_{λ}.

Proof. Since \overline{ψφψ^{−1}} = id, it follows that det(J(ψφψ^{−1})) = 1. By Lemma 5.2, we have η(λ) = 1, i.e. λ ≠ 0.

Assume that φ = σ(i, 1, f), where f ∈ I. Then

J(φ) = E + ∂(f) · e_{i}.

If i = 3, then J_{2}(φ) = E. Since f does not contain z_{i}, the ith component of ∂(f) is 0. Consequently, if i ≤ 2, then the matrix J_{2}(φ) is elementary. Since det(J_{2}(φ)) = 1, the statement of the lemma is valid for φ.

We have

σ(i, α, g) = (1i)σ(1, α, (1i)(g))(1i)

if i > 1. Since

(1i) = σ(1, −1, 0)σ(1, 1, z_{i})σ(i, 1, −z_{1})σ(1, 1, z_{i}),

the relations (18), (19) give

σ(i, α, g) = σ(1, 1, z_{i})σ(i, 1, z_{1})σ(1, α, (1 − α)z_{i} − (1i)(g))σ(i, 1, −z_{1})σ(1, 1, z_{i}).

Therefore we can assume that ψ is a product of the automorphisms

σ(1, α, g), X_{ij}(β) = σ(j, 1, βx_{i}), 1 < j, i ≠ j.  (44)

Consider ψ = ψ_{1}ψ_{2} … ψ_{k}, where ψ_{i}, 1 ≤ i ≤ k, are automorphisms of the form (44). We now begin an induction on k. Above we have checked that the lemma is true if k = 0.

Assume that the statement of the lemma is valid for

φ_{1} = ψ_{2} … ψ_{k}φ(ψ_{2} … ψ_{k})^{−1},

i.e. J_{2}(φ_{1}) ∈ S_{λ}, where λ = det(J_{2}(φ_{1})). By Lemma 5.3, we have

J(φ_{1}) = E + u^{t} · v,

where u = (u_{1}, u_{2}, u_{3}), v = (v_{1}, v_{2}, v_{3}). Then

J_{2}(φ_{1}) = E + (u_{1}, u_{2})^{t} · (v_{1}, v_{2}) ∈ S_{λ}.

Consequently, \overline{ψ}_{1}(J_{2}(φ_{1})) ∈ S_{μ}, where μ = \overline{ψ}_{1}(λ) = det(\overline{ψ}_{1}(J_{2}(φ_{1}))).

Put

\overline{ψ}_{1}(u) = (b_{1}, b_{2}, b_{3}) = b, \overline{ψ}_{1}(v) = (c_{1}, c_{2}, c_{3}) = c,

P = \overline{ψ}_{1}(J_{2}(φ_{1})), Q = J_{2}(ψ_{1}φ_{1}ψ_{1}^{−1}).

P = E + (b_{1}, b_{2})^{t} · (c_{1}, c_{2}).

Since \overline{φ}_{1} = id, it follows that

J(ψ_{1}φ_{1}ψ_{1}^{−1}) = J(ψ_{1})\overline{ψ}_{1}(J(φ_{1}))J(ψ_{1})^{−1} = E + J(ψ_{1})b^{t} · cJ(ψ_{1})^{−1}.

We first consider the case when ψ_{1} = σ(1, α, g). We have

∂(ψ_{1}(z_{1})) = (α, a_{2}, a_{3}), J(ψ_{1}) = (∂(ψ_{1}(z_{1})), e_{2}^{t}, e_{3}^{t}).

An immediate calculation gives

Q = E + (αb_{1}, a_{2}b_{1} + b_{2})^{t} · (α^{−1}(c_{1} − c_{2}a_{2}), c_{2}) + (αb_{1}, a_{2}b_{1} + b_{2})^{t} · (−α^{−1}c_{3}a_{3}, 0).

Hence

Q_{1} = E + (αb_{1}, a_{2}b_{1} + b_{2})^{t} · (α^{−1}(c_{1} − c_{2}a_{2}), c_{2})
= diag(α, 1) · E_{21}(a_{2}) · P · E_{21}(−a_{2}) · diag(α^{−1}, 1).

It follows from (43) that Q_{1} ∈ S_{μ}. By (43), we can also assume that Q_{1} = P_{1}D, where P_{1} ∈ E_{2}(U_{μ}(A)), D = diag(1, μ). Then

Q = P_{1}(D + P_{1}^{−1}(αb_{1}, a_{2}b_{1} + b_{2})^{t} · (−α^{−1}c_{3}a_{3}, 0)).

We have

D + P_{1}^{−1}(αb_{1}, a_{2}b_{1} + b_{2})^{t} · (−α^{−1}c_{3}a_{3}, 0) = ( \begin{matrix}λ_{1} & 0 \\ λ_{2} & μ \end{matrix} ) .

It follows from (43) that

D + P_{1}^{−1}(αb_{1}, a_{2}b_{1} + b_{2})^{t} · (−α^{−1}c_{3}a_{3}, 0) ∈ S_{ν},

where ν = λ_{1}μ = det(Q). Since μ|ν, we have U_{μ}(A) ⊆ U_{ν}(A) and E_{2}(U_{μ}(A)) ⊆ E_{2}(U_{ν}(A)). Consequently, Q ∈ S_{ν}.

We now assume that ψ_{1} = X_{ij}(β), j > 1, i ≠ j. We have J(ψ_{1}) = E_{ij}(β). Suppose that i = 1. If j = 2, then

Q = E + (b_{1} + βb_{2}, b_{2})^{t} · (c_{1}, c_{2} − βc_{1}) = E_{12}(β) · P · E_{12}(−β) ∈ S_{μ}.

If j = 3, then

Q = E + (b_{1} + βb_{3}, b_{2})^{t} · (c_{1}, c_{2}) = P + (βb_{3}, 0)^{t} · (c_{1}, c_{2}).

By (43), we can assume that P = DP_{1}, where P_{1} ∈ E_{2}(U_{μ}(A)), D = diag(1, μ). Then

Q = (D + (βb_{3}, 0)^{t} · (c_{1}, c_{2})P^{−1})P ∈ S_{ν}, μ|ν, ν = det(Q).

Suppose that i = 2. Then j = 3 and

Q = E + (b_{1}, b_{2} + βb_{3})^{t} · (c_{1}, c_{2}) = P + (0, βb_{3})^{t} · (c_{1}, c_{2}).

By (43), in this case we can assume that P = DP_{1}, where P_{1} ∈ E_{2}(U_{μ}(A)), D = diag(μ, 1). Hence

Q = (D + (0, βb_{3})^{t} · (c_{1}, c_{2})P_{1}^{−1})P_{1} ∈ S_{ν}, μ|ν, ν = det(Q).

Now suppose that i = 3. Then j = 2 and

Q = E + (b_{1}, b_{2})^{t} · (c_{1}, c_{2} − βc_{3}) = P + (b_{1}, b_{2})^{t} · (0, −βc_{3}).

Put P = P_{1}D, where P_{1} ∈ E_{2}(U_{μ}(A)), D = diag(μ, 1). Then we have

Q = P_{1}(D + P_{1}^{−1}(b_{1}, b_{2})^{t} · (0, −βc_{3})) ∈ S_{ν}, μ|ν, ν = det(Q).

Thus J_{2}(ψφψ^{−1}) = Q ∈ S_{ν}. □

Lemma 5.4 Let φ be an automorphism of C of the form (38) and let ψ ∈ T Aut C. Then

η(J_{2}(ψφψ^{−1})) ∈ E_{2}(U(F [x_{3}])).

Proof. Assume that det(J_{2}(ψφψ^{−1})) = λ. By Lemma 5.2, we have η(λ) = 1. Consequently, the homomorphism η may be extended to the homomorphism of the algebra U_{λ}(A) to U(F [x_{3}]). Moreover, we have η(S_{λ}) ⊆ E_{2}(U(F [x_{3}])). Now Lemma 5.3 gives our statement. □

Theorem 5.2 Assume that φ ∈ T Aut C and \overline{φ} = id. Then η(J_{2}(φ)) ∈ E_{2}(U(F [x_{3}])).

Proof. According to Theorem 5.1, we have

φ = θ_{1}θ_{2} … θ_{s}, θ_{i} = ψ_{i}φ_{i}ψ_{i}^{−1},

where φ_{i} is an automorphism of the form (38) and ψ_{i} is a tame automorphism. Since \overline{θ}_{i} = id, applying (40) we have

J(φ) = J(θ_{1})J(θ_{2}) … J(θ_{s}).

Lemma 5.2 implies that

η(J_{2}(φ)) = η(J_{2}(θ_{1}))η(J_{2}(θ_{2})) … η(J_{2}(θ_{s})).

By Lemma 5.4, we have η(J_{2}(θ_{i})) ∈ E_{2}(U(F [x_{3}])). Hence η(J_{2}(φ)) ∈ E_{2}(U(F [x_{3}])). □

Corollary 5.1 The automorphism

φ = (z_{1} + z_{3}(z_{1}z_{3} − z_{3}z_{2}), z_{2} + (z_{1}z_{3} − z_{3}z_{2})z_{3}, z_{3})

of the algebra C over a field F of characteristic 0 is wild.

Proof. We have

η(J_{2}(φ)) = ( \begin{matrix}1 + z′_{3} ⊗ z_{3} & 1 ⊗ z_{3}^{2} \\ −(z′_{3} ⊗ 1)^{2} & 1 − z′_{3} ⊗ z_{3} \end{matrix} ) .

It is well known [2] that η(J_{2}(φ)) ∉ E_{2}(F [z′_{3} ⊗ 1, 1 ⊗ z_{3}]). Put φσ = ψ, where

σ = σ(1, 1, −z_{2})σ(2, 1, −z_{1}z_{3}^{2})σ(1, 1, z_{2}).

We can immediately check that

\overline{ψ} = id, η(J_{2}(ψ)) = η(J_{2}(φ))η(\overline{φ}J_{2}(σ))), η(\overline{φ}J_{2}(σ))) ∈ E_{2}(F [z′_{3} ⊗ 1, 1 ⊗ z_{3}]).

Consequently, η(J_{2}(ψ)) ∉ E_{2}(F [z′_{3} ⊗1, 1⊗z_{3}]). Then Theorem 5.2 gives that ψ ∉ T Aut C. Therefore φ ∉ T Aut C, i.e. φ is wild. □

Thus the Anick automorphism

δ = (x + z(xz − zy), y + (xz − zy)z, z)

of the free associative algebra F < x, y, z > induces a wild automorphism of the algebra C.

Theorem 5.3 The Anick automorphism δ of the free associative algebra F < x, y, z > over a field F of characteristic 0 is wild.

Combining the results of [32, 6] and [24], we conclude that an automorphism φ = (f, g, z) of the polynomial algebra F [x, y, z] is tame if and only if

J_{2}(φ) ∈ GL_{2}(F)E_{2}(F [x, y, z]).

Theorem 5.4 Let φ = (f, g, z) be an automorphism of the free associative algebra F < x, y, z > over a field F of characteristic 0 such that deg_{x,y} f, deg_{x,y} g ≤ 1. Then φ is tame if and only if

J_{2}(φ) ∈ GL_{2}(F)E_{2}(F [z′ ⊗ 1, 1 ⊗ z]).

Proof. Assume that the matrix J_{2}(φ) satisfies the condition of the theorem. Without loss of generality we can suppose that

J_{2}(φ) ∈ E_{2}(F [x, y, z]).

Since deg_{x,y} f, deg_{x,y} g ≤ 1, to each elementary matrix of E_{2}(F [x, y, z]) corresponds an elementary transformation of the first two coordinates of φ. Applying elementary transformations corresponding to a decomposition of J_{2}(φ), we can get (x + a(z), y + b(z), z) from φ. Consequently, φ is tame.

Now, assume that φ is tame. Let φ̃ be an automorphism of the free metabelian associative algebra induced by φ. Since deg_{x,y} f, deg_{x,y} g ≤ 1, it is easy to check that

J_{2}(φ) = ν(J_{2}(φ̃)).

If J_{2}(φ) ∉ E_{2}(F [x, y, z]), then as in the proof of Theorem 5.3 we can obtain a contradiction. □

Automorphisms of the form considered in Theorem 5.4 were investigated in [7].

Now we formulate some problems related to the relations (18)–(20) and discuss why it is impotant to study them.

Let us denote by G(A_{n}) the group defined by the system of generators (17) and the system of relations (18)–(20).

Problem 5.1 Find a canonical form of elements of the group G(A_{n}). In particular, is the word problem solvable in this group?

This problem is also interesting for the group G(B_{n}), where B_{n} is a free associative algebra in n variables, which can be defined similarly G(A_{n}). Note that the group G(A_{2}) ≅ G(B_{2}) ≅ Aut A_{2} ≅ Aut B_{2} has a nice representation (see [3]). By Theorem 4.1, if n = 3 and F is a field of characteristic 0, then the group G(A_{3}) has the solvable word problem. But we have no canonical form of elements of G(A_{3}).

Let L_{n} be a free Lie algebra (or a free algebra of some Schreier variety of algebras) with n free generators. Certainly, we can define the group G(L_{n}) also. Applying only general ideas of the proof of Theorem 4.1 it is not difficult to prove that G(L_{n}) ≅ Aut L_{n}.

Now consider the homomorphism

s_{n} : G(A_{n}) ⟶ Aut A_{n}

which sends the generators (17) to the corresponding automorphisms of A_{n}.

Problem 5.2 Find the kernel of s_{n}.

We know that the kernel of s_{2} is trivial. By Theorem 4.1 the kernel of s_{3} is also trivial if char(F) = 0.

A solution of Problem 5.2 gives a description of the group T Aut A_{n} by generators and relations. If n ≥ 4, then it is possible also that Aut A_{n} = T Aut A_{n}.

Acknowledgments

I’m grateful to N. Dairbekov, M. Jibladze, L. Makar-Limanov, J. -T. Yu, and M. Zaidenberg for very helpful discussions and comments. I am also grateful to Max-Planck Institute für Mathematik for hospitality and exellent working conditions.

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