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

Abstract

A characterization of tame automorphisms of the algebra $A=F[x_1,x_2,x_3]$ of polynomials in three variables over a field $F$ of characteristic $0$ is obtained. In particular, it is proved that the well-known Nagata automorphism is wild. It is also proved that the tame and the wild automorphisms of $A$ are algorithmically recognizable.

1. Introduction

Let $C=F[x_1,x_2,\ldots ,x_n]$ be the polynomial ring in the variables $x_1,x_2,\ldots ,x_n$ over a field $F$, and let $Aut\,C$ be the group of automorphisms of $C$ as an algebra over $F$. An automorphism $\tau \in Aut\,C$ is called elementary if it has a form

$$\begin{equation*} \tau : (x_1,\ldots ,x_{i-1},x_i,x_{i+1},\ldots ,x_n)\mapsto (x_1,\ldots ,x_{i-1}, \alpha x_i+f, x_{i+1},\ldots ,x_n), \end{equation*}$$

where $0\neq \alpha \in F,\ f\in F[x_1,\ldots ,x_{i-1},x_{i+1},\ldots ,x_n]$. The subgroup of $Aut\,C$ generated by all the elementary automorphisms is called the tame subgroup, and the elements from this subgroup are called tame automorphisms of $C$. Non-tame automorphisms of the algebra $C$ are called wild.

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

$$\begin{eqnarray*} \sigma (x)&=&x+(x^2-yz)z,\\ \sigma (y)&=&y+2(x^2-yz)x+(x^2-yz)^2z,\\ \sigma (z)&=&z. \end{eqnarray*}$$

Observe that the Nagata automorphism is stably tame 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 $A=F[x,y,z]$ over a field $F$ of characteristic $0$ are algorithmically recognizable. In particular, the Nagata automorphism $\sigma$ is wild.

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

The crucial role in the proof is played by the description of the structure of subalgebras generated by so-called $\ast$-reduced pairs of polynomials, given in 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 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 4 concerning wild automorphisms of two-generated free matrix algebras, without using the results of 6, 9, 10, 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 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 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 $A$, and simple automorphisms of $A$ are defined. The main part of the work, Section 4, is devoted to the proof of Theorem 1, which states that every tame automorphism of the algebra $A$ is simple. The main results are formulated and proved in Section 5 as corollaries of Theorem 1.

2. Structure of two-generated subalgebras

Let $F$ be an arbitrary field of characteristic $0$, and let $A=F[x_1,x_2,x_3]$ be the ring of polynomials in the variables $x_1,x_2,x_3$ over $F$. Following 16, we will identify $A$ with a certain subspace of the free Poisson algebra $P=PL\langle x_1,x_2,x_3\rangle$.

Recall that a vector space $B$ over a field $F$, endowed with two bilinear operations $x\cdot y$ (a multiplication) and $[x,y]$ (a Poisson bracket), is called a Poisson algebra if $B$ is a commutative associative algebra under $x\cdot y$, $B$ is a Lie algebra under $[x,y]$, and $B$ satisfies the Leibniz identity

$$\begin{eqnarray} [x\cdot y,z]=[x,z]\cdot y+x\cdot [y,z]. {\tag{1}} \end{eqnarray}$$

An important class of Poisson algebras is given by the following construction. Let $L$ be a Lie algebra with a linear basis $l_1, l_2, \ldots , l_k, \ldots$. Denote by $P(L)$ the ring of polynomials on the variables $l_1, l_2, \ldots , l_k, \ldots$. The operation $[x,y]$ of the algebra $L$ can be uniquely extended to a Poisson bracket $[x,y]$ on the algebra $P(L)$ by means of formula (1), and $P(L)$ becomes a Poisson algebra 15.

Now let $L$ be a free Lie algebra with free generators $x_1,x_2,\ldots ,x_n$. Then $P(L)$ is a free Poisson algebra 15 with the free generators $x_1,x_2,\ldots ,x_n$. We will denote this algebra by $PL\langle x_1,x_2,\ldots ,x_n\rangle$. If we choose a homogeneous basis

$$\begin{equation*} x_1,x_2,\ldots ,x_n,\,[x_1,x_2],\ldots ,[x_1,x_n],\ldots ,[x_{n-1},x_n],\,[[x_1,x_2],x_3]\ldots \end{equation*}$$

of the algebra $L$ with nondecreasing degrees, then $PL\langle x_1,x_2,\ldots ,x_n\rangle$, as a vector space, coincides with the ring of polynomials on these elements. The space $PL\langle x_1,x_2,\ldots ,x_n\rangle$ is graded by degrees on $x_i$, and for every element $f \in$ $PL\langle x_1,x_2,\ldots ,x_n\rangle$, the highest homogeneous part $\bar{f}$ and the degree $\deg f$ can be defined in an ordinary way. Note that

$$\begin{equation*} \overline{fg}=\bar{f} \bar{g},\quad \deg (fg)=\deg f+\deg g,\quad \deg [f,g]\leq \deg f+\deg g. \end{equation*}$$

In the sequel, we will identify the ring of polynomials $A=F[x_1,x_2,x_3]$ with the subspace of the algebra $PL\langle x_1,x_2,x_3\rangle$ generated by elements

$$\begin{equation*} x_1^{r_1}x_2^{r_2}x_3^{r_3},\quad r_i\geq 0,\ 1\leq i \leq 3. \end{equation*}$$

Note that if $f,g\in A$, then

$$\begin{eqnarray*} [f,g]=\gamma _{12}[x_1,x_2]+\gamma _{23}[x_2,x_3]+\gamma _{13}[x_1,x_3],\\ \gamma _{ij}=\frac{\partial f}{\partial x_i}\frac{\partial g}{\partial x_j}- \frac{\partial g}{\partial x_i}\frac{\partial f}{\partial x_j}, \ \ \ 1 \leq i<j\leq 3. \end{eqnarray*}$$

If $f_1,f_2,\ldots ,f_k\in A$, then by $\langle f_1,f_2,\ldots ,f_k\rangle$ we denote the subalgebra of the algebra $A$ generated by these elements.

The following lemma is proved in 16.

Lemma 1.

Let $f,g,h\in A$. Then the following statements are true:

$1)$ $[f,g]=0$ iff $f,g$ are algebraically dependent.

$2)$ Suppose that $f,g,h \notin F$ and $m=\deg [f,g]+\deg h$, $n=\deg [g,h]+\deg f$, $k=\deg [h,f]+\deg g$. Then $m\leq \max (n,k)$. If $n\neq k$, then $m=\max (n,k)$.

The next two simple statements are well known (see 5):

F1) If $a,b$ are nonzero homogeneous algebraically dependent elements of $A$, then there exists an element $z\in A$ such that $a=\alpha z^n$, $b=\beta z^m$, $\alpha ,\beta \in F$. In addition, the subalgebra $\langle a,b \rangle$ is one-generated iff $m|n$ or $n|m$.

F2) Let $f,g\in A$ and $\bar{f},\bar{g}$ are algebraically independent. If $h\in \langle f,g \rangle$, then $\bar{h}\in \langle \bar{f},\bar{g}\rangle$.

Recall that a pair of elements $f,g$ of the algebra $A$ is called reduced (see 18), if $\bar{f}\notin \langle \bar{g}\rangle$, $\bar{g}\notin \langle \bar{f}\rangle$. A reduced pair of algebraically independent elements $f,g\in A$ is called $\ast$-reduced (see 16), if $\bar{f},\bar{g}$ are algebraically dependent.

Let $f,g$ be a $\ast$-reduced pair of elements of $A$ and $n=\deg f<m=\deg g$. Put $p=\frac{n}{(n,m)}$, $s=\frac{m}{(n,m)}$,

$$\begin{equation*} N=N(f,g)=\frac{mn}{(m,n)}-m-n+\deg [f,g]=mp-m-n+\deg [f,g], \end{equation*}$$

where $(n,m)$ is the greatest common divisor of $n,m$. Note that $(p,s)=1$, and by F1) there exists an element $a\in A$ such that $\bar{f}=\beta a^p$, $\bar{g}=\gamma a^s$. Sometimes, we will call a $\ast$-reduced pair of elements $f,g$ also a $p$-reduced pair. Let $G(x,y)\in F[x,y]$. It was proved in 16 that if $\deg _y(G(x,y))=pq+r$, $0\leq r<p$, then

$$\begin{eqnarray} \deg (G(f,g))\geq qN+mr, {\tag{2}} \end{eqnarray}$$

and if $\deg _x(G(x,y))=sq_1+r_1$, $0\leq r_1<s$, then

$$\begin{eqnarray} \deg (G(f,g))\geq q_1N+nr_1. {\tag{3}} \end{eqnarray}$$

It will be convenient for us to collect several evident properties of the $\ast$-reduced pair $f,g$ in the following lemma.

Lemma 2.

Under the above notation,

i) $p \geq 2$;

ii) $N=N(f,g)>\deg [f,g]$;

iii) if $p>2$, then $N>m$;

iv) if $p=2$, then $N>\frac{n}{2}$.

The properties $i) - iii)$ are evident. As for $iv)$, let $d=(n,m)$; then $n=2d,\ m=sd$, and $N=sd-2d+\deg [f,g]> (s-2)d\geq d=\frac{n}{2}$.

The statement of the following lemma is easily proved.

Lemma 3.

The elements of type $f^ig^j$, where $j<p$, have different degrees for different values of $i,j$.

Inequality (2) and Lemma 3 imply

Corollary 1.

Let $G(x,y)\in F[x,y],\ h=G(f,g)$. Consider the following conditions:

$(i)$

$\deg h<N(f,g)$;

$(ii)$

$\deg _y(G(x,y))<p$;

$(iii)$

$h=\sum _{i,j}\alpha _{ij}f^ig^j$, where $\alpha _{ij}\in F$ and $in+jm\leq \deg h$ for all $i,j$;

$(iv)$

$\bar{h}\in \langle \bar{f},\bar{g}\rangle$.

Then $(i)\Rightarrow (ii)\Rightarrow (iii)\Rightarrow (iv)$.

Suppose that $p\geq 3$ or $\deg [f,g]>n$. Then obviously $N(f,g)>m$, and in the conditions of Corollary 1 we have $\bar{h}\in \langle \bar{f},\bar{g}\rangle$ or $\deg h>m=\max (\deg f,\deg g)$. Note that the most complicated case in the investigation of tame automorphisms of $A$ is represented by $\ast$-reduced pairs $f,g$ with the condition $N(f,g)\leq m$, that is, by $2$-reduced pairs $f,g$ for which $\deg [f,g]\leq n$.

Lemma 4.

There exists a polynomial $w(x,y)\in F[x,y]$ of the type

$$\begin{eqnarray*} w(x,y)=y^p-\alpha x^s-\sum \alpha _{ij}x^iy^j, \ ni+mj<mp, \end{eqnarray*}$$

which satisfies the following conditions:

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

$2)$ $\overline{w(f,g)}\notin \langle \bar{f},\bar{g}\rangle$.

Proof.

By F1), there exists a homogeneous element $a\in A$ such that $\bar{f}=\beta a^p$, $\bar{g}=\gamma a^s$. Then there exists $\alpha \in F$ such that $\bar{g}^p=\alpha \bar{f}^s$, and the elements of the type $\bar{f}^i\bar{g}^j$, $j<p$, form a basis of the subalgebra $\langle \bar{f},\bar{g}\rangle$. Putting $h=g^p-\alpha f^s$, we have $\deg h<mp$. If $\bar{h}\in \langle \bar{f},\bar{g}\rangle$, then $\bar{h}=\alpha _{ij}\bar{f}^i\bar{g}^j$, where $ni+mj<mp$. Change the element $h$ to $h-\alpha _{ij}f^ig^j$. Then $\deg (h-\alpha _{ij}f^ig^j)<\deg h$. After several reductions of this type, we get an element

$$\begin{eqnarray*} h=g^p-\alpha f^s-\sum \alpha _{ij}f^ig^j,\ ni+mj<mp, \end{eqnarray*}$$

for which $\bar{h}\notin \langle \bar{f},\bar{g}\rangle$. Since $f,g$ are algebraically independent, the equality $h=w(f,g)$ defines uniquely a polynomial $w(x,y)$ that satisfies the conditions of the lemma.

■

A polynomial $w(x,y)$ satisfying the conditions of Lemma 4 we will call a derivative polynomial of the $\ast$-reduced pair $f,g$. Note that a derivative polynomial $w(x,y)$ is not uniquely defined in the general case. But the coefficient $\alpha$ in the conditions of Lemma 3 is uniquely defined by the equality $\bar{g}^p=\alpha \bar{f}^s$.

Lemma 5.

Let $w(x,y)$ be a derivative polynomial of a $\ast$-reduced pair $f,g$. Then the following statements are true:

$1)$ $\deg w(f,g)$ is uniquely defined;

$2)$ if $\bar{f}$, $\overline{w(f,g)}$ are algebraically dependent, then $\overline{w(f,g)}$ is uniquely defined;

$3)$ $\deg (w(f,g))\geq N(f,g)$;

$4)$ if $\deg (w(f,g))<\deg g =m$, then $w(x,y)$ is defined uniquely up to a summand $q(x)$, where $n\cdot \deg (q(x))\leq \deg (w(f,g))$;

$5)$ if $\deg (w(f,g))<\deg f$, then $w(x,y)$ is defined uniquely up to a scalar summand from $F$.

Proof.

Let $w_1(x,y)$ be another derivative polynomial of the pair $f,g$. Since the coefficient $\alpha$ is uniquely defined in the conditions of Lemma 4, we have

$$\begin{eqnarray*} h(x,y)=w(x,y)-w_1(x,y)=\sum \gamma _{ij}x^iy^j, \end{eqnarray*}$$

where $ni+mj<mp$. Now, if $\deg (w(f,g))>\deg (w_1(f,g))$, then by Lemma 3 we get

$$\begin{eqnarray*} \overline{h(f,g)}=\overline{w(f,g)}\in \langle \bar{f},\bar{g}\rangle , \end{eqnarray*}$$

which contradicts the definition of $w(x,y)$.

Suppose that $\overline{w(f,g)}\neq \overline{w_1(f,g)}$. Then $\overline{h(f,g)}= \overline{w(f,g)}-\overline{w_1(f,g)}$. Since $\overline{h(f,g)}\in \langle \bar{f},\bar{g}\rangle$, the elements $\bar{f},\ \overline{h(f,g)}$ are algebraically dependent. Now, if $\bar{f}$, $\overline{w(f,g)}$ are algebraically dependent, then $\overline{w(f,g)}, \ \overline{h(f,g)}$ are algebraically dependent too. Furthermore, since $\deg (w(f,g))=\deg (h(f,g))$, the elements $\overline{w(f,g)},\ \overline{h(f,g)}$ are linearly dependent, and thus $\overline{w(f,g)}\in \langle \bar{f},\bar{g}\rangle$. This again contradicts the definition of $w(x,y)$.

Note that $\deg (h(f,g))\leq \deg (w(f,g))$. Since $\overline{w(f,g)}\!\not \in \!\langle \bar{f},\bar{g}\rangle$, Corollary 1 yields 3). If $\deg (w(f,g))<\deg g$, then by Lemma 3 we get $h(x,y)=q(x)$, $\deg (q(f))=\deg (h(f,g))\leq \deg (w(f,g))$. This proves statements 4), 5) of the lemma.

■

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

Lemma 6.

Let $w(x,y)$ be a derivative polynomial of the pair $f,g$ and $u=w(f,g)$. Then the highest homogeneous parts of the elements of the type

$$\begin{eqnarray*} f^ig^ju^t,\ \ j<p,\ \ 0\leq t\leq 1, \end{eqnarray*}$$

are linearly independent.

Proof.

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

$$\begin{eqnarray*} \bar{f}^i\bar{g}^j\bar{u}=\beta \bar{f}^{i_1}\bar{g}^{j_1},\ \ \ j,j_1<p. \end{eqnarray*}$$

If $i\leq i_1$, $j\leq j_1$, then this equality implies $\bar{u}\in \langle \bar{f},\bar{g}\rangle$, which is impossible by the definition of $w(x,y)$. Assume that $j\leq j_1$, $i>i_1$. Then

$$\begin{eqnarray*} \bar{f}^{i-i_1}\bar{u}=\beta \bar{g}^{j_1-j}. \end{eqnarray*}$$

Since $\bar{u}\notin \langle \bar{f},\bar{g}\rangle$, by Corollary 1 we have

$$\begin{eqnarray*} \deg u\geq N(f,g)=pm-m-n+\deg [f,g]. \end{eqnarray*}$$

Therefore,

$$\begin{eqnarray*} \deg (\bar{g}^{j_1-j})=m(j_1-j)&\geq & n(i-i_1)+pm-m-n+\deg [f,g]\\ &\geq & (p-1)m+\deg [f,g]>(p-1)m. \end{eqnarray*}$$

This contradicts the inequality $j_1-j\leq p-1$. If $i\leq i_1$, $j>j_1$, then

$$\begin{eqnarray*} \bar{g}^{j-j_1}\bar{u}=\beta \bar{f}^{i_1-i}. \end{eqnarray*}$$

Since $j-j_1\leq p-1$ and $\bar{g}^p=\alpha \bar{f}^s$, we may assume that $i_1-i<s$; otherwise $\bar{u}\in \langle \bar{f},\bar{g}\rangle$. Thus,

$$\begin{eqnarray*} \deg (\bar{f}^{i_1-i})&=&n(i_1-i)\geq m(j-j_1)+pm-m-n+\deg [f,g]\\ &\geq & pm-n+\deg [f,g]>pm-n=ns-n=n(s-1), \end{eqnarray*}$$

which is impossible.

■

Lemma 7.

Let $w(x,y)$ be a derivative polynomial of the pair $f,g$, and $T(x,y)\in F[x,y]$, $\deg _y(T(x,y))<2p$. Then, the following statements are true:

$1)$

$T(x,y)$ can be uniquely presented in the form$$\begin{eqnarray*} T(x,y)=w(x,y) q(x,y)+s(x,y), \end{eqnarray*}$$

where $\deg _y(q(x,y)),\ \deg _y(s(x,y))<p$;

$2)$

if $\deg (T(f,g))\leq t$, then$$\begin{eqnarray*} \deg (w(f,g))+\deg (q(f,g))\leq t, \ \deg (s(f,g))\leq t; \end{eqnarray*}$$

$3)$

if $\overline{T(f,g)}\notin \langle \bar{f},\bar{g}\rangle$, then $q(x,y)\neq 0$ and$$\begin{eqnarray*} \deg \left( \frac{\partial T}{\partial x}(f,g) \right)=\deg (q(f,g))+n(s-1), \\ \deg \left( \frac{\partial T}{\partial y}(f,g) \right)=\deg (q(f,g))+m(p-1). \end{eqnarray*}$$

$4)$

if $\overline{T(f,g)}\notin \langle \bar{f},\bar{g}\rangle$ and $\deg (T(f,g))<\min \{n+N,mp\}$, then $T(x,y)=\lambda w_1(f,g), 0\neq \lambda \in F,$ where $w_1(x,y)$ is also a derivative polynomial of the pair $f,g$.

Proof.

The first statement of the lemma follows from the division algorithm in the ring $(F[x])[y]$; one may divide $T(x,y)$ by $w(x,y)$ since the last polynomial is monic. Furthermore, the right part of the equality $T(f,g)=w(f,g)q(f,g)+s(f,g)$ is a linear combination of elements indicated in Lemma 6; therefore, by this lemma, only the elements of degree less than or equal to $\deg (T(f,g))$ may appear in this combination. This proves 2).

If $\overline{T(f,g)}\notin \langle \bar{f},\bar{g}\rangle$, then by Corollary 1 we have $\deg _y{T(x,y)}\geq p$ and hence $q(x,y)\neq 0$. By Corollary 1 again, $\overline{s(f,g)}\in \langle \bar{f},\bar{g}\rangle$; hence $\overline{T(f,g)}\neq \overline{s(f,g)}$. Consequently, by Lemma 6,

$$\begin{eqnarray*} \deg (T(f,g))=\deg (w(f,g) q(f,g))\geq \deg (s(f,g)). \end{eqnarray*}$$

It follows from the definition of $w(x,y)$ in Lemma 4 that

$$\begin{eqnarray*} \deg \left(\frac{\partial w}{\partial x}(f,g)\right) =n(s-1),\ \ \deg \left(\frac{\partial w}{\partial y}(f,g)\right) =m(p-1). \end{eqnarray*}$$

Furthermore,

$$\begin{eqnarray*} \frac{\partial T}{\partial x}(x,y)=\frac{\partial w}{\partial x}(x,y) q(x,y)+ w(x,y) \frac{\partial q}{\partial x}(x,y)+\frac{\partial s}{\partial x}(x,y). \end{eqnarray*}$$

Easy calculations give

$$\begin{eqnarray*} \deg (w(f,g) \frac{\partial q}{\partial x}(f,g))&\leq & \deg (w(f,g))+\deg (q(f,g))-\deg f, \\ \deg (\frac{\partial s}{\partial x}(f,g))&\leq & \deg (T(f,g))-\deg f\\ & = &\deg (w(f,g))+\deg (q(f,g))-\deg f, \\ \deg (\frac{\partial w}{\partial x}(f,g) q(f,g))&=&\deg (q(f,g))+n(s-1)= \deg (q(f,g))+mp-n \\ &=&\deg (w(f,g))+\deg (q(f,g))-n+(mp-\deg (w(f,g))) \\ &>&\deg (w(f,g))+\deg (q(f,g))-n. \end{eqnarray*}$$

Therefore,

$$\begin{eqnarray*} \deg \left(\frac{\partial T}{\partial x}(f,g)\right) =\deg (q(f,g))+n(s-1). \end{eqnarray*}$$

Similar calculations give the value of $\deg (\frac{\partial T}{\partial y}(f,g))$.

To prove 4) we note first that by Lemma 5.3), $\deg (w(f,g))\geq N$. Hence by statement 2) of this lemma, $\deg (q(f,g))<n$ and $\deg (s(f,g))<mp$. By Corollary 1, $\overline{q(f,g)}\in \langle \bar{f},\bar{g}\rangle$. Hence $0\neq q(x,y)=\lambda \in F$. Now it is easy to see that the polynomial $w_1(x,y)=\lambda ^{-1}T(x,y)=w(x,y)+\lambda ^{-1}s(x,y)$ is a derivative polynomial of the pair $f,g$.

■

We give two corollaries that will be useful for references.

Corollary 2.

If $h\in \langle f,g\rangle \setminus F$ and $\deg h<n$, then $h=\lambda w(f,g),\ 0\neq \lambda \in F,$ where $w(x,y)$ is a derivative polynomial of the pair $f,g$.

Proof.

Put $h=T(f,g)$ and let $\deg _yT(x,y)=pq+r,\ r<p$. If $q=0$, then Corollary 1 gives $\bar{h}\in \langle \bar{f},\bar{g}\rangle$ and $\deg h\geq n$ or $h\in F$, a contradiction. Hence $q>0$. Inequality (2) gives $\deg h\geq qN+mr$. Consequently, $r=0$. If $q>1$, then by Lemma 2, $\deg h\geq 2N>n$. Therefore, $q=1$ and $\deg _y(T(x,y))=p$. Then Lemma 7.4) proves the corollary.

■

Corollary 3.

If $w(x,y)$ is a derivative polynomial of the pair $f,g$, then

$$\begin{eqnarray*} \deg \left(\frac{\partial w}{\partial x}(f,g)\right)=n(s-1), \\ \deg \left(\frac{\partial w}{\partial y}(f,g)\right)=m(p-1). \end{eqnarray*}$$

3. Reductions and simple automorphisms

A triple $\theta =(f_1,f_2,f_3)$ (or simply $(f_1,f_2,f_3)$) of elements of the algebra $A$ below will always denote the automorphism $\theta$ of $A$ such that $\theta (x_i)=f_i$, $1\leq i\leq 3$. The number $\deg \theta =\deg f_1+\deg f_2+\deg f_3$ will be called a degree of the automorphism $\theta$.

Recall that an elementary transformation of the 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 $\alpha f_i+g$, where $0\neq \alpha \in F,\ g\in \langle \{f_j|j\neq i\}\rangle$. The notation

$$\begin{eqnarray*} (f_1,f_2,f_3)\rightarrow (g_1,g_2,g_3) \end{eqnarray*}$$

means that the triple $(g_1,g_2,g_3)$ is obtained from $(f_1,f_2,f_3)$ by a single elementary transformation. Observe that we do not assume that $\deg \,(g_1,g_2,g_3)$ should be smaller than $\deg \theta$. An automorphism $(f_1,f_2,f_3)$ is called tame if there exists a sequence of elementary transformations of the form

$$\begin{multline*} (x_1,x_2,x_3)=(f_1^{(0)},f_2^{(0)},f_3^{(0)})\rightarrow (f_1^{(1)},f_2^{(1)},f_3^{(1)})\rightarrow \ldots \\ \rightarrow (f_1^{(n)},f_2^{(n)},f_3^{(n)})=(f_1,f_2,f_3). \end{multline*}$$

The element $f_1$ of the automorphism $\theta =(f_1,f_2,f_3)$ is called reducible, if there exists $g\in \langle f_2,f_3 \rangle$ such that $\bar{f}_1=\bar{g}$; otherwise it is called unreducible. Put $f_1'=\alpha (f_1-g)$, where $0\neq \alpha \in F$; then $\deg f_1'<\deg f_1$ and $\deg (f_1',f_2,f_3)<\deg \theta$. In this case we will say also that $f_1$ is reduced in $\theta$ by the automorphism $(f_1',f_2,f_3)$. If one of the elements $f_1,f_2,f_3$ of $\theta$ is reducible, then we will say that $\theta$ admits an elementary reduction or simply that $\theta$ is elementary reducible.

Lemma 8.

The elementary reducibility of automorphisms of the algebra $A$ is algorithmically recognizable.⁠¹

¹

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

✖

Proof.

Let $\theta =(f_1,f_2,f_3)$ be an arbitrary automorphism of $A$. We will recognize the reducibility of $f_3$. If $\bar{f}_1,\bar{f}_2$ are algebraically independent, then $f_3$ is reducible if and only if $\bar{f}_3\in \langle \bar{f}_1,\bar{f}_2\rangle$. Since $\bar{f}_1, \bar{f}_2$ are homogeneous, this question can be solved trivially, even without a reference to the solubility of the occurrence problem 13, 14. If $\bar{f}_2\in \langle \bar{f}_1\rangle$ and $\bar{f}_2=\alpha \bar{f}_1^k$, then the element $f_3$ is reducible in $\theta$ if and only if it is reducible in the automorphism $(f_1,f_2-\alpha f_1^k,f_3)$. Since $\deg (f_1,f_2-\alpha f_1^k,f_3)<\deg \theta$, the statement of the lemma in this case can be proved by induction on $\deg \theta$.

Let now $f_1,f_2$ be a $\ast$-reduced pair and $\deg f_1<\deg f_2$. Assume that there exists a polynomial $G(x,y)\in F[x,y]$ such that $\bar{f}_3=\overline{G(f_1,f_2)}$. Inequalities (2), (3) gives a bound $k$ for the numbers $\deg _x(G(x,y)),\deg _y(G(x,y))$. Then $G(f_1,f_2)$ is in the space generated by the elements $f_1^if_2^j$, where $i,j\leq k$. The highest homogeneous parts of elements of this space can be described by triangulation.

■

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

Example 1. Put

$$\begin{eqnarray*} h_1=x_1,\ \ h_2=x_2+x_1^2,\ \ h_3=x_3+2x_1x_2+x_1^3, \\ g_1=6h_1+6h_2h_3+h_3^3,\ \ g_2=4h_2+h_3^2,\ \ g_3=h_3. \end{eqnarray*}$$

It is easy to show that $(h_1,h_2,h_3)$ and $(g_1,g_2,g_3)$ are tame automorphisms of the algebra $A$. Note that $\deg g_1=9$, $\deg g_2=6$, $\deg g_3=3$ and $g_1,g_2$ form a $2$-reduced pair. A direct calculation shows that the element

$$\begin{eqnarray*} f=g_1^2-g_2^3 \end{eqnarray*}$$

has degree 8. Hence $\deg f <\deg g_1$, and $\bar{f}\notin \langle \bar{g}_1,\bar{g}_2\rangle$.

Now we define a tame automorphism $(f_1,f_2,f_3)$ by putting

$$\begin{eqnarray*} f_1=g_1+g_3+f,\ \ f_2=g_2,\ \ f_3=g_3+f. \end{eqnarray*}$$

We have $\deg f_1=9$, $\deg f_2=6$, $\deg f_3=8$ and $\deg [f_i,f_j]>9$, $1\leq i<j\leq 3$. Then, using inequality (2), it is easy to check that the automorphism $(f_1,f_2,f_3)$ does not admit an elementary reduction.

Proposition 1.

Let $\theta =(f_1,f_2,f_3)$ be an automorphism of $A$ such that $\deg f_1=2n$, $\deg f_2=ns$, $s\geq 3$ is an odd number, $2n<\deg f_3\leq ns$, $\bar{f}_3\notin \langle \bar{f}_1,\bar{f}_2\rangle$. Suppose that there exists $0\neq \alpha \in F$ such that the elements $g_1=f_1$, $g_2=f_2-\alpha f_3$ satisfy the conditions:

$i)$

$g_1,g_2$ is a $2$-reduced pair and $\deg g_1=2n$, $\deg g_2=ns$;

$ii)$

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 $\deg [g_1,g_3]<ns+\deg [g_1,g_2]$.

Then, the following statements are true:

$1)$

$\overline{[f_1,f_2]}=\alpha \overline{[f_1,f_3]}$;

$2)$

$\deg [f_i,f_j]>ns$, where $1\leq i<j\leq 3$;

$3)$

if $f\in \langle f_i,f_j\rangle$, where $1\leq i<j\leq 3$, then either $\bar{f}\in \langle \bar{f}_i,\bar{f}_j\rangle$ or $\deg f >ns$;

$4)$

$\deg f_1+\deg f_3>ns$.

Proof.

We have

$$\begin{eqnarray} g_3=\sigma f_3+G(g_1,g_2),\ \ \deg g_3<\deg f_3 , {\tag{4}} \end{eqnarray}$$

where $0\neq \sigma \in F$, $G(x,y)\in F[x,y]$. Hence

$$\begin{eqnarray} \overline{G(g_1,g_2)}=-\sigma \bar{f}_3. {\tag{5}} \end{eqnarray}$$

If $\deg f_3 =ns$, then $\bar{f}_2,\bar{f}_3$ are linearly independent, since $\bar{f}_3\notin \langle \bar{f}_1,\bar{f}_2\rangle$. Therefore $\bar{f}_2,\bar{f}_3,\bar{g}_2$ are mutually linearly independent and $\bar{f}_3\notin \langle \bar{f}_1,\bar{g}_2\rangle = \langle \bar{g}_1,\bar{g}_2\rangle$. If $\deg f_3 <ns$, then $\bar{f}_2=\bar{g}_2$ and again we have $\bar{f}_3\notin \langle \bar{g}_1,\bar{g}_2\rangle$. Put $\deg _y(G(x,y))=k=2q+r$, $0\leq r\leq 1$. The condition $\bar{f}_3\notin \langle \bar{g}_1,\bar{g}_2\rangle$ implies, by Corollary 1 and (5), that $q\geq 1$. Then inequality (2) gives $r=0$ and

$$\begin{eqnarray*} ns\geq \deg (G(g_1,g_2))= \deg f_3\geq q(ns-2n+\deg [g_1,g_2]). \end{eqnarray*}$$

It is easy to deduce from here that if $s>3$, then $q=1$, $k=2$, and if $s=3$, then $q=1,2$, $k=2,4$. Besides, these inequalities imply $\deg [g_1,g_2]\leq 2n$ and statement 4) of the proposition.

Applying (1), we get from (4),

$$\begin{eqnarray*} [g_1,g_3]=\sigma [g_1,f_3]+[g_1,g_2]\frac{\partial G}{\partial y}(g_1,g_2). \end{eqnarray*}$$

Since $\deg _y(\frac{\partial G}{\partial y})=k-1$ is an odd number, inequality (2) gives

$$\begin{eqnarray*} \deg ([g_1,g_2]\frac{\partial G}{\partial y}(g_1,g_2))\geq \deg [g_1,g_2]+ns. \end{eqnarray*}$$

Consequently, by condition ii),

$$\begin{eqnarray*} \deg [f_1,f_3]=\deg [g_1,f_3]\geq \deg [g_1,g_2]+ns. \end{eqnarray*}$$

Since $\deg [g_1,g_2]\leq 2n$ and $\alpha \neq 0$, the equality

$$\begin{eqnarray*} [g_1,g_2]=[f_1,f_2]-\alpha [f_1,f_3] \end{eqnarray*}$$

gives statement 1) of the proposition and

$$\begin{eqnarray*} \deg [f_1,f_2]=\deg [f_1,f_3]>ns. \end{eqnarray*}$$

If $\deg f_3 =ns$, then, as was remarked earlier, $\bar{f}_2,\bar{f}_3$ are algebraically independent, and so (see 16)

$$\begin{eqnarray*} \deg [f_2,f_3]=\deg f_2+\deg f_3 >ns. \end{eqnarray*}$$

If $\deg f_3 <ns$, then we have

$$\begin{eqnarray*} \deg [f_1,f_2]+\deg f_3 <\deg [f_1,f_3]+\deg f_2. \end{eqnarray*}$$

By Lemma 1,

$$\begin{eqnarray*} \deg [f_2,f_3]+\deg f_1=\deg [f_1,f_3]+\deg f_2; \end{eqnarray*}$$

hence

$$\begin{eqnarray*} \deg [f_2,f_3]=\deg [f_1,f_3]+n(s-2)>ns. \end{eqnarray*}$$

Thus statement 2) of the proposition is proved.

To prove 3), it suffices by F2) to consider only the case when $\bar{f}_i, \bar{f}_j$ are algebraically dependent. It is easily seen that $f_1,f_2$ and $f_1,f_3$ are $\ast$-reduced pairs. Suppose that $\bar{f}_2\in \langle \bar{f}_3\rangle$. If $\deg f_2=\deg f_3$ then $\bar{f}_3\in \langle \bar{f}_2\rangle$, which contradicts the condition of the proposition. Otherwise $\deg f_2\geq 2\deg f_3>\deg f_1+\deg f_3$, which contradicts 4). Consequently, the pair $f_i, f_j$ is $\ast$-reduced for every $i\neq j$ and Corollary 1 by Lemma 2.ii) implies 3).

■

Definition 1.

If an automorphism $\theta =(f_1,f_2,f_3)$ satisfies the conditions of proposition 1, then we will say that $\theta$ admits a reduction of type I, and the automorphism $(g_1,g_2,g_3)$ will be called a reduction of type I of the automorphism $\theta$, with an active element $f_3$.

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

Proposition 2.

Let $\theta =(f_1,f_2,f_3)$ be an automorphism of $A$ such that $\deg f_1=2n$, $\deg f_2=3n$, $\frac{3n}{2}<\deg f_3 \leq 2n$, and $\bar{f}_1,\bar{f}_3$ are linearly independent. Suppose that there exist $\alpha ,\beta \in F$, where $(\alpha ,\beta )\neq (0,0)$, such that the elements $g_1=f_1-\alpha f_3$, $g_2=f_2-\beta f_3$ 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 $\deg [g_1,g_3]<3n+\deg [g_1,g_2]$.

Then, the following statements are true:

$1)$

$\alpha \in F$ is the solution of the equation $\overline{[f_1,f_2]}=\alpha \overline{[f_3,f_2]}$, or $\alpha =0$ if it has no solution;

$2)$

$\beta \in F$ is the solution of the equation $\overline{[g_1,f_2]}=\beta \overline{[g_1,f_3]}$, or $\beta =0$ if it has no solution;

$3)$

$\deg [f_i,f_j]>3n$, where $1\leq i<j\leq 3$;

$4)$

if $f\in \langle f_i,f_j\rangle$, where $1\leq i<j\leq 3$, then either $\bar{f}\in \langle \bar{f}_i,\bar{f}_j\rangle$ or $\deg f >3n$.

Proof.

Consider equalities (4), (5). If $\deg f_3 <2n$, then obviously $\overline{G(g_1,g_2)}\notin \langle \bar{g}_1,\bar{g}_2\rangle$. If $\deg f_3 =2n$, then by the condition of the proposition, $\bar{f}_1,\bar{f}_3$ are linearly independent. Therefore, either $\alpha =0$ and $g_1=f_1$, or $\alpha \neq 0$ and $\bar{g}_1,\bar{f}_1,\bar{f}_3$ are mutually linearly independent. In any case, $\overline{G(g_1,g_2)}\notin \langle \bar{g}_1,\bar{g}_2\rangle$. Since $\deg f_3 \leq 2n$, then, as in the proof of Proposition 1, inequality (2) gives that $\deg _y(G(x,y))=\nobreakspace 2$, $\deg [g_1,g_2]\leq n$. Consequently, Lemma 7.4) gives that $G(x,y)$ is a derivative polynomial (up to a nonzero scalar factor) of the pair $g_1,g_2$, and by Corollary 3,

$$\begin{eqnarray*} \deg \left(\frac{\partial G}{\partial y}(g_1,g_2)\right)=3n. \end{eqnarray*}$$

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

$$\begin{eqnarray} \deg [f_1,f_3]=\deg [g_1,f_3]=\deg [g_1,g_2]+3n. {\tag{6}} \end{eqnarray}$$

Consider the triple $(g_1,g_2,f_3)$. By (6) and Lemma 1,

$$\begin{eqnarray*} \deg [g_2,f_3]+\deg g_1=\deg [g_1,f_3]+\deg g_2, \end{eqnarray*}$$

which yields

$$\begin{eqnarray} \deg [f_2,f_3]=\deg [g_2,f_3]=\deg [g_1,f_3]+n. {\tag{7}} \end{eqnarray}$$

Furthermore,

$$\begin{eqnarray*} [f_1,f_2]=[g_1+\alpha f_3,g_2+\beta f_3]=[g_1,g_2]+\beta [g_1,f_3]+\alpha [f_3,g_2]. \end{eqnarray*}$$

Since $\deg [g_1,g_2]\leq n$, this implies, by (6) and (7), that $\overline{[f_1,f_2]}=\alpha \overline{[f_3,g_2]}=\alpha \overline{[f_3,f_2]}$ if $\alpha \neq 0$, and $\overline{[f_1,f_2]}=\beta \overline{[g_1,f_3]}$ if $\alpha =0$. Hence $\deg [f_1,f_2]>3n$, and if $\alpha =0$, then $[f_1,f_2]$, $[f_3,f_2]$ have different degrees. We have also

$$\begin{eqnarray*} [g_1,f_2]=[g_1,g_2+\beta f_3]=[g_1,g_2]+\beta [g_1,f_3]. \end{eqnarray*}$$

Hence either $\beta \neq 0$ and $\overline{[g_1,f_2]}=\beta \overline{[g_1,f_3]}$, or the elements $[g_1,f_2]=[g_1,g_2]$ and $[g_1,f_3]$ have different degrees. This proves the statements 1), 2), 3) of the proposition. Finally, as in the proof of Proposition 1, Corollary 1 and Lemma 2.ii) give 4).

■

Definition 2.

If an automorphism $\theta =(f_1,f_2,f_3)$ satisfies the conditions of proposition 2, then we will say that $\theta$ admits a reduction of type II, and the automorphism $(g_1,g_2,g_3)$ will be called a reduction of type II of the automorphism $\theta$, with an active element $f_3$.

Proposition 3.

Let $\theta =(f_1,f_2,f_3)$ be an automorphism of $A$ such that $\deg f_1=2n$, and either $\deg f_2=3n$, $n<\deg f_3 \leq \frac{3n}{2}$, or $\frac{5n}{2}<\deg f_2\leq 3n$, $\deg f_3 =\frac{3n}{2}$. Suppose that there exist $\alpha ,\beta ,\gamma \in F$ such that the elements $g_1=f_1-\beta f_3$, $g_2=f_2-\gamma f_3-\alpha f_3^2$ satisfy the conditions:

$i)$

$g_1,g_2$ is a $2$-reduced pair and $\deg g_1=2n$, $\deg g_2=3n$;

$ii)$

there exists an element $g_3$ of the form$$\begin{eqnarray*} g_3=\sigma f_3+g, \end{eqnarray*}$$

where $0\neq \sigma \in F$, $g\in \langle g_1,g_2\rangle \setminus F$, such that $\deg g_3\leq \frac{3n}{2}$, $\deg [g_1,g_3]<3n+\deg [g_1,g_2]$.

Then, the following statements are true:

$1)$

$\alpha \in F$ is the solution of the equation $\overline{[f_1,f_2]}=2\alpha \overline{[f_1,f_3]}\bar{f}_3$, or $\alpha =0$ if it has no solution;

$2)$

$\beta \in F$ is the solution of the equation $\overline{[f_2-\alpha f_3^2,f_1]}=\beta \overline{[f_2,f_3]}$, or $\beta =0$ if it has no solution;

$3)$

$\gamma \in F$ is the solution of the equation $\overline{[g_1,f_2-\alpha f_3^2]}=\gamma \overline{[g_1,f_3]}$, or $\gamma =0$ if it has no solution;

$4)$

$\deg [f_1,f_3],\ \deg [f_2,f_3]>3n$;

$5)$

if $(\alpha ,\beta ,\gamma )\neq (0,0,0)$, then $\deg [f_1,f_2]>3n$; otherwise, $\deg [f_1,f_2]=$ $\deg [g_1,g_2]\leq \frac{n}{2}$;

$6)$

if $\bar{g}_2=-\alpha \bar{f}_3^2$, then $\frac{5n}{2}+\deg [g_1,g_2]\leq \deg f_2<3n$; otherwise, $\deg f_2=3n$.

Proof.

Since $\deg f_3, \deg g_3\leq \frac{3n}{2}$, condition $ii)$ yields that $\deg g\leq \frac{3n}{2}$. Then, by Corollary 2, $g=\lambda w(g_1,g_2)$, where $w(x,y)$ is a derivative polynomial of the $g_1,g_2$. Inequality (2) gives also $\deg [g_1,g_2]\leq \frac{n}{2}$. As in the proof of Propositions 1, 2, we obtain also (6), (7), which yields 4). Besides, we have

$$\begin{eqnarray*} [f_1,f_2]=[g_1,g_2]+\gamma [g_1,f_3]+\beta [f_3,g_2]+2\alpha [g_1,f_3]f_3. \end{eqnarray*}$$

Since $\deg f_3 >n$, this equality yields statements 1), 5) of the proposition. If $\bar{g}_2=-\alpha \bar{f}_3^2$, then $\alpha \neq 0$, $\deg f_3=\frac{3n}{2}$, $\deg f_2<3n$, and

$$\begin{eqnarray*} \deg [f_1,f_2]=\deg [g_1,f_3]+\deg f_3 =\deg [g_1,g_2]+3n+\frac{3n}{2}. \end{eqnarray*}$$

Consequently, $\deg f_2\geq \deg [g_1,g_2]+\frac{5n}{2}$, which proves 6). We have also

$$\begin{eqnarray*} [f_2-\alpha f_3^2,f_1]&=&[g_2,g_1]+\gamma [f_3,g_1]+\beta [g_2,f_3],\\ \ [g_1,f_2-\alpha f_3^2]&=&[g_1,g_2]+\gamma [g_1,f_3]. \end{eqnarray*}$$

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

■

Definition 3.

If an automorphism $\theta =(f_1,f_2,f_3)$ satisfies the conditions of Proposition 3, and $(\alpha ,\beta ,\gamma )\neq (0,0,0)$, $\deg g_3<n+\deg [g_1,g_2]$, then we will say that $\theta$ admits a reduction of type III, and the automorphism $(g_1,g_2,g_3)$ will be called a reduction of type III of the automorphism $\theta$, with an active element $f_3$.

Corollary 4.

In the conditions of Proposition 3, if the automorphism $\theta =$ $(f_1,f_2,f_3)$ admits a reduction of type III, then $\deg (g_1,g_2,g_3)<\deg \theta$.

Proof.

By Definition 3, we have

$$\begin{eqnarray*} \deg (g_1,g_2,g_3)<6n+\deg [g_1,g_2]. \end{eqnarray*}$$

Hence it is sufficient to prove that

$$\begin{eqnarray} \deg \theta \geq 6n+\deg [g_1,g_2]. {\tag{8}} \end{eqnarray}$$

It follows from (6) that $\deg f_3 \geq n+\deg [g_1,g_2]$. If $\bar{g}_2\neq -\alpha \bar{f}_3^2$, then Proposition 3.6) gives $\deg f_2=3n$, which proves (8). If $\bar{g}_2= -\alpha \bar{f}_3^2$, then, as shown above, $\deg f_3 =\frac{3n}{2}$ and by Proposition 3.6), $\deg f_2\geq \frac{5n}{2}+\deg [g_1,g_2]$, which also gives (8).

■

Definition 4.

If an automorphism $\theta =(f_1,f_2,f_3)$ satisfies the conditions of Proposition 3, and there exists $0\neq \mu \in F$ such that $\deg (g_2-\mu g_3^2)\leq 2n$, then we will say that $\theta$ admits a reduction of type IV, and the automorphism $(g_1,g_2-\mu g_3^2,g_3)$ will be called a reduction of type IV of the automorphism $\theta$, with an active element $f_3$. In this case we will also call the automorphism $(g_1,g_2,g_3)$ a predreduction of type IV of $\theta$.

Corollary 5.

If an automorphism $\theta =(f_1,f_2,f_3)$ satisfies all the conditions of Proposition 3 and Definition 4, then $\deg (g_1,g_2-\mu g_3^2,g_3)<\deg \theta$.

Proof.

Since $\deg (g_2-\mu g_3^2)\leq 2n$, then $\bar{g}_2=\mu \bar{g_3}^2$ and $\deg g_3=\frac{3n}{2}$. Consequently, $\deg (g_1,g_2-\mu g_3^2,g_3)\leq \frac{11n}{2}$. Inequality (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 $A$ as follows.

$1)$ All the automorphisms of degree $3$ are simple.

$2)$ Suppose that the simple automorphisms of degree $<n$ are already defined.

$3)$ An automorphism $\theta$ of degree $n>3$ is called simple if there exists a simple automorphism of degree $<n$ that is either an elementary reduction or a reduction of type I–IV of $\theta$.

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 $(f_1,f_2,f_3)$ is a simple automorphism, then the automorphisms $(f_2,f_1,f_3)$ and $(\alpha f_1+\beta ,f_2,f_3)$, where $\alpha ,\beta \in F$, $\alpha \neq 0$, are simple as well.

Really, if $(g_1,g_2,g_3)$ is an elementary reduction or a reduction of type I–IV of $(f_1,f_2,f_3)$, then $(g_2,g_1,g_3)$ is a reduction of the same type for $(f_2,f_1,f_3)$. It is also clear that we can always choose $\alpha _i,\beta _i\in F$ such that $(\alpha _1g_1+\beta _1,\alpha _2g_2+\beta _2,\alpha _3g_3+\beta _3)$ becomes a corresponding reduction for $(\alpha f_1+\beta ,f_2,f_3)$.

For convenience of terminology, we introduce also

Definition 6.

An element $f_i,\ i=1,2,3,$ of the automorphism $\theta =(f_1,f_2,f_3)$ 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 $A$ is simple.

The plan of the proof.

Assume that the statement of the theorem is not true. Then there exist tame automorphisms $\theta =(f_1,f_2,f_3)$, $\tau$ of $A$ such that $\theta$ is simple, $\tau$ is not simple, and

$$\begin{eqnarray*} \theta =(f_1,f_2,f_3)\rightarrow \tau . \end{eqnarray*}$$

In the set of all pairs of automorphisms with this property we choose and fix a pair $\theta , \tau$ with the minimal $\deg \theta$.

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

$$\begin{eqnarray} \tau =(f,f_2,f_3),\ \ f=f_1+a,\ \ a\in \langle f_2,f_3 \rangle ,\ \ \deg a\leq \deg f_1, {\tag{9}} \end{eqnarray}$$

$$\begin{eqnarray} \tau =(f_1,f,f_3),\ \ f=f_2+a,\ \ a\in \langle f_1,f_3 \rangle ,\ \ \deg a\leq \deg f_2, {\tag{10}} \end{eqnarray}$$

$$\begin{eqnarray} \tau =(f_1,f_2,f),\ \ f=f_3+a,\ \ a\in \langle f_1,f_2 \rangle ,\ \ \deg a\leq \deg f_3 . {\tag{11}} \end{eqnarray}$$

Here, the restriction on $\deg a$ is imposed in order to exclude the trivial case when $\theta$ is an elementary reduction of $\tau$. In the case when $\theta$ admits an elementary reduction, we will assume that $\tau$ has form (11). The proof of the theorem will be completed by Lemmas 9–17 and by Propositions 4, 5.

■

The following evident statement is formulated for convenience of references.

Lemma 9.

Let $\phi$ be a simple automorphism of $A$ such that $\deg \phi <\deg \theta$. If $\phi \rightarrow \psi$, then $\psi$ is simple too.

The proof follows immediately from the minimality condition for $\deg \theta$.

Corollary 6.

Suppose that there exists a sequence of automorphisms

$$\begin{eqnarray*} \phi _0\rightarrow \phi _1\rightarrow \ldots \rightarrow \phi _{k-1} \rightarrow \phi _k \end{eqnarray*}$$

of the algebra $A$ such that $\deg \phi _i<\deg \theta$ for $0\leq i\leq k-1$, and $\phi _0$ is a simple automorphism. Then $\phi _k$ is also simple.

Lemma 10.

If $\theta$ admits a reduction of type I, then $\tau$ is simple.

Proof.

We adopt all the conditions and notation of Proposition 1 and by the definition of reduction we have that $(g_1,g_2,g_3)$ is simple. Without loss of generality, we can also put

$$\begin{eqnarray} g_3=f_3+g,\ \ g\in \langle g_1,g_2\rangle ,\ \deg g_3<\deg f_3. {\tag{12}} \end{eqnarray}$$

If $\tau$ is of form (9), then by Proposition 1.3) we have $\bar{a}\in \langle \bar{f}_2,\bar{f}_3\rangle$. Since $\deg a\leq \deg f_1<\min \{\deg f_2,\deg f_3 \}$, this implies $a\in F$. Hence $\tau$ is simple by Remark 1.

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

$$\begin{eqnarray*} \bar{a}=\beta \bar{f}_3+\gamma (\bar{f}_1)^k,\ \beta ,\gamma \in F,\ 2nk\leq \deg f_2. \end{eqnarray*}$$

Consider $a_1=a-\beta f_3-\gamma (f_1)^k$. Then again $a_1\in \langle f_1,f_3\rangle$ and $\deg a_1<\deg a$; hence $\bar{a}_1\in \langle \bar{f}_1\rangle$, and it is easy to see that $a_1\in \langle f_1\rangle$. Thus we have

$$\begin{eqnarray*} a=\beta f_3+T(f_1),\ \ \deg (T(f_1))\leq \deg f_2,\ \ f=f_2+\beta f_3+T(f_1). \end{eqnarray*}$$

Since $\deg f_1=2n,\ \deg f_2=sn$, where $s$ is odd, $\deg (T(f_1))<\deg f_2$. Furthermore, $\deg f$ can be less than $\deg f_2$ only if $\bar{f}_2=-\beta \bar{f}_3$. But $\bar{f}_3\notin \langle \bar{f}_1,\bar{f}_2\rangle$; hence $\deg f =\deg f_2$. Put

$$\begin{eqnarray*} g_2'=f-(\alpha +\beta )f_3=(f_2-\alpha f_3)+T(f_1)=g_2+T(g_1). \end{eqnarray*}$$

Then $\overline{g_2'}=\bar{g}_2$, and $g\in \langle g_1,g_2\rangle =\langle g_1,g_2'\rangle$. If $\alpha +\beta \neq 0$, then (12) implies that $(g_1,g_2',g_3)$ is a reduction of type I of $\tau$. If $\alpha +\beta =0$, then $\tau =(g_1,g_2',f_3)$ and the element $f_3$ is reduced in $\tau$ by $(g_1,g_2',g_3)$. It remains to note that, by Lemma 9, $(g_1,g_2',g_3)$ is a simple automorphism.

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

$$\begin{eqnarray*} a=\beta f_2+T(f_1),\ \ \deg (T(f_1))\leq \deg f_3 ,\ \ f=f_3+\beta f_2+T(f_1), \end{eqnarray*}$$

and $\beta \neq 0$ is possible only if $\deg f_3 =\deg f_2$. Consider 3 cases:

1) $\beta =0$, 2) $\beta (1+\alpha \beta )\neq 0$, 3) $1+\alpha \beta =0$.

In case 1) we put

$$\begin{eqnarray*} g_2'=f_2-\alpha f=g_2-\alpha T(g_1). \end{eqnarray*}$$

Since $\deg g_1\not |\,\deg g_2$, the equality $\deg (T(g_1))=\deg g_2$ is impossible, $\overline{g_2'}=\bar{g_2}$. By Lemma 9, the automorphism $\phi =(g_1,g_2',g_3)$ is simple again. Since

$$\begin{eqnarray*} f=f_3+T(f_1)=g_3-g+T(g_1),\ \ -g+T(g_1)\in \langle g_1,g_2\rangle =\langle g_1,g_2'\rangle , \end{eqnarray*}$$

$\phi$ is a reduction of type I of $\tau$.

In case 2) we put

$$\begin{eqnarray*} g_2'=f_2-\frac{\alpha }{1+\alpha \beta }\,f=\frac{1}{1+\alpha \beta }\,g_2-\frac{\alpha }{1+\alpha \beta }\,T(g_1). \end{eqnarray*}$$

A direct calculation gives

$$\begin{eqnarray*} f=(1+\alpha \beta )g_3+g', \end{eqnarray*}$$

where

$$\begin{eqnarray*} g'=-(1+\alpha \beta )g+\beta (1+\alpha \beta )g_2'+(1+\alpha \beta )T(g_1),\ \ g'\in \langle g_1,g_2\rangle =\langle g_1,g_2'\rangle . \end{eqnarray*}$$

Since $\overline{g_2'}=\frac{1}{1+\alpha \beta }\,\bar{g_2}$, then $(g_1,g_2',g_3)$ is a reduction of type I of $\tau$.

In case 3) we have

$$\begin{eqnarray*} f&=&\beta (f_2-\alpha f_3)+T(f_1)=\beta g_2+T(g_1),\\ f_2&=&\alpha g_3+(g_2-\alpha g), \ g_2-\alpha g \in \langle g_1,g_2\rangle = \langle g_1,f\rangle . \end{eqnarray*}$$

Therefore, $\bar{f}=\beta \bar{g_2}$, and it is easy to check that $f_2$ is reduced in $\tau =(g_1,f_2,f)$ by $(g_1,g_3,f)$. By Remark 1, $(g_1,g_3,g_2)$ is simple. Thus, by Lemma 9, $(g_1,g_3,f)$ is simple too.

■

Lemma 11.

If $\theta$ admits a reduction of type II, then $\tau$ is simple.

Proof.

We adopt all the conditions and notation of Proposition 2, as well as equality (12). If $\tau$ has form (9), then Proposition 2.4) and the condition $\deg a \leq \deg f_1$ give $a=\gamma f_3+\lambda$. By Remark 1 we may assume that $\lambda =0$. So

$$\begin{eqnarray*} a=\gamma f_3,\ \ f=f_1+\gamma f_3,\\ f-(\gamma +\alpha )f_3=g_1,\ \ f_2-\beta f_3=g_2. \end{eqnarray*}$$

If $(\gamma +\alpha ,\beta )\neq (0,0)$, then it is easily checked that $(g_1,g_2,g_3)$ is a reduction of type II of $\tau$. Otherwise, the element $f_3$ is reduced in the automorphism $\tau =(g_1,g_2,f_3)$ by $(g_1,g_2,g_3)$.

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

$$\begin{eqnarray*} a=\gamma f_3+\delta f_1,\ \ f=f_2+\gamma f_3+ \delta f_1. \end{eqnarray*}$$

Furthermore,

$$\begin{eqnarray*} f-(\gamma +\beta +\delta \alpha )f_3=g_2+\delta g_1=g_2',\ \ f_1-\alpha f_3=g_1. \end{eqnarray*}$$

If $(\alpha ,\gamma +\beta +\delta \alpha )\neq (0,0)$, then $(g_1,g_2',g_3)$ is a reduction of type II of $\tau$. Otherwise, $(g_1,g_2',g_3)$ reduces the element $f_3$ of $\tau =(g_1,g_2',f_3)$.

Assume that $\tau$ has form (11). Proposition 2 and Remark 1 give

$$\begin{eqnarray*} a=\gamma f_1,\ \ f=f_3+\gamma f_1, \end{eqnarray*}$$

and $\gamma \neq 0$ is possible only if $\deg f_1=\deg f_3$. Since $\bar{f_1},\bar{f_3}$ are linearly independent, $\deg f =\deg f_3$.

If $1+\alpha \gamma \neq 0$, a direct calculation gives

$$\begin{eqnarray*} f=f_3+\gamma f_1&=&(1+\alpha \gamma )g_3+(\gamma g_1-(1+\alpha \gamma )g),\\ f_1-\frac{\alpha }{1+\alpha \gamma }\,f&=& \frac{1}{1+\alpha \gamma }\,g_1=g_1',\\ f_2-\frac{\beta }{1+\alpha \gamma }\,f&=&g_2-\frac{\beta \gamma }{1+\alpha \gamma }\,g_1=g_2'. \end{eqnarray*}$$

Since $\langle g_1,g_2\rangle =\langle g_1',g_2'\rangle$, it is easy to check that $(g_1',g_2',g_3)$ is a reduction of type II of $\tau$. By Corollary 6, the automorphism $(g_1',g_2',g_3)$ is simple.

If $1+\alpha \gamma =0$, then

$$\begin{eqnarray*} f=\gamma g_1,\ f_1=\alpha g_3+(g_1-\alpha g),\\ f_2-\frac{\beta }{\alpha }f_1=g_2-\frac{\beta }{\alpha }g_1=g_2'. \end{eqnarray*}$$

If $\beta =0$, then the element $f_1$ is reduced in $\tau =(f_1,g_2',\gamma g_1)$ by $(g_3,g_2',\gamma g_1)$. Otherwise, it is easily checked that $(g_3,g_2',\gamma g_1)$ is a reduction of type II of $\tau$ with an active element $f_1$.

■

Lemma 12.

If $\theta$ admits a reduction of type III or IV, then $\tau$ is simple.

Proof.

We adopt the conditions and notation of Proposition 3. Assume that $\tau$ has form (9). Let us show that in this case $a=\delta _1f_3$.

Evidently, it suffices to prove that, for any $b\in \langle f_2,f_3\rangle$ with $\deg b\leq \deg f_1=2n$, $\bar{b}\in \langle \bar{f}_3\rangle$ holds. Note that $\deg f_2>\frac{5n}{2}>\deg b$. By F2) we may assume, without loss of generality, that $\bar{f}_2,\bar{f}_3$ are algebraically dependent. If $\bar{f}_2\notin \langle \bar{f}_3\rangle$, then the pair $f_2,f_3$ is $\ast$-reduced, and the statement holds by Proposition 3.4), Corollary 1, and Lemma 2.ii). Otherwise $\deg f_3=\frac{3n}{2},\ \deg f_2=3n, \ \bar{f}_2=\lambda (\bar{f}_3)^2$, where $0\neq \lambda \in F$. Since $\deg g_2=3n$, then $\bar{f}_2\neq \alpha \bar{f}_3^2$, i.e., $\lambda \neq \alpha$. Consider $f_2'=f_2-\lambda f_3^2=g_2+\gamma f_3+(\alpha -\lambda )f_3^2$. Then

$$\begin{equation*} [f_1,f_2']= [g_1,g_2]+\gamma [g_1,f_3]+\beta [f_3,g_2]+2(\alpha -\lambda )[g_1,f_3]f_3. \end{equation*}$$

Since $(\alpha -\lambda )\neq 0$, we have, as in the proof of Proposition 3, that $\deg [f_1,f_2']=\deg [g_1,g_2]+\frac{9n}{2}$, which yields $\deg f_2'>\frac{5n}{2}$. Note that $\langle f_2',f_3\rangle =\langle f_2,f_3 \rangle$. If $\bar{f}_2',\bar{f}_3$ are algebraically independent, then by F2) we get $\bar{b}\in \langle \bar{f}_2',\bar{f}_3\rangle$. Otherwise, $f_2', f_3$ form a $\ast$-reduced pair, and since $\deg [f_2',f_3]=\deg [f_2,f_3]>3n$, we have again $\bar{b}\in \langle \bar{f}_2',\bar{f}_3\rangle$ by Corollary 1 and Lemma 2.ii). But $\deg f_2'>\frac{5n}{2}>\deg b$, and so $\bar{b}\in \langle \bar{f}_3\rangle$.

Thus $a=\delta _1f_3$, and so

$$\begin{eqnarray*} f=f_1+\delta _1f_3=g_1+(\beta +\delta _1)f_3,\ \ f_2=g_2+\gamma f_3+\alpha f_3^2. \end{eqnarray*}$$

Since $f_3$ is preserved in the structure of $\tau =(f,f_2,f_3)$, it is easily checked that if $\theta$ admits a reduction of type IV, then $(g_1,g_2,g_3)$ is a predreduction of type IV of $\tau$. Suppose that $\theta$ admits a reduction of type III. If $(\alpha ,\beta +\delta _1,\gamma )\neq (0,0,0)$, then $(g_1,g_2,g_3)$ is a reduction of type III of $\tau$. Otherwise, since $\deg g_3<\deg f_3$ (see the proof of Corollary 4), the element $f_3$ is reduced in $\tau =(g_1,g_2,f_3)$ by $(g_1,g_2,g_3)$.

Suppose that $\tau$ has form (10). Then, by Proposition 3.4) and Corollary 1 we have $\bar{a}\in \langle \bar{f}_1,\bar{f}_3\rangle$. Note that $\deg a\leq \deg f_2\leq 3n$ and $\deg (f_1^2),\ \deg (f_1f_3),\ \deg (f_3^3)>3n$. So

$$\begin{eqnarray*} a=\delta _1 f_3^2+\sigma _1 f_3+\mu _1 f_1, \end{eqnarray*}$$

and $\delta _1\neq 0$ is possible only if $\deg f_2\geq 2 \deg f_3$. Therefore,

$$\begin{eqnarray*} f_1=g_1+\beta f_3,\ \ f=g_2+\mu _1g_1+(\alpha +\delta _1)f_3^2+(\gamma +\sigma _1+\mu _1\beta )f_3. \end{eqnarray*}$$

Hence, if $\deg f \neq 3n$, then $\bar{g_2}+(\alpha +\delta _1)\bar{f_3}^2=0$, i.e., $\alpha +\delta _1\neq 0$, $\deg f_3 =\frac{3n}{2}$. By Proposition 3,

$$\begin{eqnarray*} \deg [f_3,g_1]=\deg [f_3,f_1]>3n. \end{eqnarray*}$$

Since

$$\begin{eqnarray*} [f,g_1]=[g_2,g_1]+2(\alpha +\delta _1)[f_3,g_1]f_3+(\gamma +\sigma _1+\mu _1\beta )[f_3,g_1], \end{eqnarray*}$$

then

$$\begin{eqnarray*} \deg [f,g_1]>3n+\frac{3n}{2}=\frac{9n}{2}. \end{eqnarray*}$$

Consequently, $\deg f >\frac{5n}{2}$. Now it is easy to show, that if $\theta$ admits a reduction of type IV, then $(g_1,g_2+\mu _1g_1,g_3)$ is a predreduction of type IV of $\tau$. Suppose now that $\theta$ admits a reduction of type III. If $(\alpha +\delta _1,\beta ,\gamma +\sigma _1+\mu _1\beta )\neq 0$, then $(g_1,g_2+\mu _1g_1,g_3)$ is a reduction of type III of $\tau$. Otherwise, the element $f_3$ is reduced in $\tau =(g_1,g_2+\mu _1g_1,f_3)$ by $(g_1,g_2+\mu _1g_1,g_3)$.

Assume that $\tau$ has form (11). If $(\alpha ,\beta ,\gamma )\neq (0,0,0)$, then Proposition 3 yields $a\in F$. If $\alpha =\beta =\gamma =0$, then $\theta$ admits a reduction of type IV. We have

$$\begin{eqnarray*} f_1=g_1,\ \ f_2=g_2,\ \ f_3=\frac{1}{\sigma }g_3-\frac{1}{\sigma }g, \end{eqnarray*}$$

where $g\in \langle g_1,g_2\rangle \setminus F$. Since $\deg a\leq \deg f_3 \leq \frac{3n}{2}$, we have

$$\begin{eqnarray*} f=\frac{1}{\sigma }g_3+c, \end{eqnarray*}$$

where $c=-\frac{1}{\sigma }g+a\in \langle g_1,g_2\rangle$ and $\deg c\leq \frac{3n}{2}$. If $c\notin F$, then $(g_1,g_2,g_3)$ is a predreduction of type IV of $\tau$. Otherwise, by Remark 1 we can take $\tau =(g_1,g_2,\frac{1}{\sigma }g_3),$ and the element $g_2$ is reduced in $\tau$ by $(g_1,g_2-\mu g_3^2,\frac{1}{\sigma }g_3)$.

■

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

Lemma 13.

Let $\phi =(g_1,g_2,g_3)$ be a simple automorphism and $\deg \phi \leq \deg \theta$. If $g_1$ is a simple reducible element of $\phi$, then every elementary transformation $\psi$ of $\phi$ changing only $g_1$ gives a simple automorphism.

Proof.

Assume that the element $g_1$ is reduced in $\phi$ by a simple automorphism $\phi '=(h_1,g_2,g_3)$. Then $\deg \phi '<\deg \theta$ and $h_1=\alpha g_1+g$, where $0\neq \alpha \in F$, $g\in \langle g_2,g_3\rangle$. Put $\psi =(\beta g_1+T(g_2,g_3),g_2,g_3)$. Then

$$\begin{eqnarray*} \beta g_1+T(g_2,g_3)=\frac{\beta }{\alpha }h_1+g',\ \ g'\in \langle g_2,g_3\rangle . \end{eqnarray*}$$

Hence $\phi '\rightarrow \psi$ and Lemma 9 completes the proof.

■

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

Corollary 7.

If $f_3$ is a simple reducible element of $\theta$, then $\tau$ is a simple automorphism.

In the remainder of this section we will assume that $f_3$ is not a simple reducible element of $\theta$. Then either $f_1$ or $f_2$ are simple reducible. For definiteness, we assume that $f_2$ is reduced in $\theta$ by a simple automorphism $\phi =(f_1,g_2,f_3)$, where

$$\begin{eqnarray} g_2=f_2+b,\ \ b\in \langle f_1,f_3 \rangle ,\ \ \deg g_2<\deg f_2. {\tag{13}} \end{eqnarray}$$

Lemma 14.

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

$1)$ $\bar{f_2}\in \langle \bar{f_1}\rangle$;

$2)$ $\bar{f_3}\in \langle \bar{f_1}\rangle$;

$3)$ $a$ does not depend on $f_2$;

$4)$ $\bar{f_1},\bar{f_3}$ are algebraically independent.

Proof.

If $\bar{f_2}\in \langle \bar{f_1}\rangle$, then by Lemma 13 we can choose an element $b$ satisfying (13) such that $b\in \langle f_1\rangle$. Then $\langle f_1,f_2 \rangle =\langle f_1,g_2\rangle$. Consequently, there exists a sequence of elementary transformations of the form

$$\begin{eqnarray} \phi =(f_1,g_2,f_3)\rightarrow (f_1,g_2,f)\rightarrow (f_1,f_2,f)=\tau . {\tag{14}} \end{eqnarray}$$

Since $\deg \phi$, $\deg (f_1,g_2,f)<\deg \theta$ (by (13)), it follows from Corollary 6 that the automorphism $\tau$ is simple.

If $\bar{f_3}\in \langle \bar{f_1}\rangle$ and $\bar{f_3}=T(\bar{f_1})$, then we put $g_3=f_3-T(f_1)$. There exists a sequence

$$\begin{eqnarray*} \phi =(f_1,g_2,f_3)\rightarrow (f_1,g_2,g_3)\rightarrow (f_1,f_2,g_3) \rightarrow (f_1,f_2,f_3)=\theta . \end{eqnarray*}$$

By Corollary 6, the automorphism $(f_1,f_2,g_3)$ is simple, and consequently, $f_3$ is a simple reducible element of $\theta$, which is impossible.

If $a$ does not depend on $f_2$, then sequence (14) proves the simplicity of $\tau$.

Assume that $\bar{f_1},\bar{f_3}$ are algebraically independent. Then by (13) we obtain that $\bar{f_2}=-\bar{b}\in \langle \bar{f_1},\bar{f_3}\rangle$. By 1), we can assume that $\bar{f_2}\notin \langle \bar{f_1}\rangle$, i.e., $\bar{f_2}$ depends on $\bar{f_3}$. Then $\deg f_3 \leq \deg f_2$. Observe that $\bar{f_1},\bar{f_2}$ are algebraically independent; otherwise, $\bar{f}_1, \bar{f}_3$ would be algebraically dependent. Therefore, $\bar{a}\in \langle \bar{f}_1,\bar{f}_2\rangle$. By 3), we can also assume that $a$ contains $f_2$. Then (11) gives $\deg f_2\leq \deg a\leq \deg f_3$, and so $\deg f_2=\deg f_3$. Thus

$$\begin{eqnarray*} b=\alpha f_3+T(f_1),\ \ \alpha \neq 0,\ \ \deg (T(f_1))\leq \deg f_2. \end{eqnarray*}$$

We have the equalities

$$\begin{eqnarray*} g_2&=&f_2+\alpha f_3+T(f_1),\\ f_2&=&g_2-\alpha f_3-T(f_1),\\ f_3&=&\frac{1}{\alpha } g_2-\frac{1}{\alpha }f_2-\frac{1}{\alpha }T(f_1), \end{eqnarray*}$$

which induce the following sequence of elementary transformations:

$$\begin{eqnarray} (f_1,f_3,g_2)\rightarrow (f_1,f_2,g_2)\rightarrow (f_1,f_2,f_3)=\theta . {\tag{15}} \end{eqnarray}$$

Since $\phi =(f_1,g_2,f_3)$ is simple, by Remark 1, $(f_1,f_3,g_2)$ is simple too. By Lemma 9, sequence (15) gives simple reducibility of $f_3$ in $\theta$, a contradiction.

■

Thus, by Lemma 14, we can suppose that $\bar{f_1},\bar{f_3}$ are algebraically dependent and $\bar{f_3}\notin \langle \bar{f_1}\rangle$. We will consider separately the 3 cases:

1) $f_1,f_3$ is a $\ast$-reduced pair, $\deg f_1<\deg f_3$;

2) $f_1,f_3$ is a $\ast$-reduced pair, $\deg f_3 <\deg f_1$;

3) $\bar{f_1}\in \langle \bar{f_3}\rangle$, $\deg f_1>\deg f_3$.

First, we will prove two propositions.

Proposition 4.

Let $\psi =(g_1,g_2,g_3)$ be a simple automorphism satisfying the following conditions:

$i)$ $\deg \psi \leq \deg \theta$;

$ii)$ $g_1,g_2$ is a $2$-reduced pair and $\deg g_1=2n$, $\deg g_2=3n$;

$iii)$ $\bar{g_1},\bar{g_3}$ are linearly independent, $\deg g_3=m<3n$, and $g_3$ is not a simple reducible element of $\psi$.

Then one of the following statements is satisfied:

$1)$ $m<n+\deg [g_1,g_2]$;

$2)$ $\psi$ admits a reduction of type IV with an active element $g_3$;

$3)$ $\deg [g_1,g_3]<3n+\deg [g_1,g_2]$, and there exists $\alpha \in F$ such that $\deg (g_2-\alpha g_3^2)\leq 2n$.

Proof.

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

$$\begin{eqnarray} n+\deg [g_1,g_2]\leq m<3n. {\tag{16}} \end{eqnarray}$$

Hence $\deg [g_1,g_2]<2n$, and by Propositions 1, 2, 3, the automorphism $\psi$ does not admit a reduction of types I–III. By its choice, neither admits $\psi$ a reduction of type IV. So, by the definition of a simple automorphism, $\psi$ admits an elementary reduction. By $iii)$, either $g_1$ or $g_2$ is a simple reducible element of $\psi$. Since $\bar{g_1},\bar{g_2}$ are algebraically dependent, this implies that $\bar{g_1},\bar{g_2},\bar{g_3}$ are mutually algebraically dependent. It follows from $iii)$ and (16) that $g_1,g_3$ is a $\ast$-reduced pair.

Case 1. $2n<m<3n$.

Assume that $g_2$ is a simple reducible element of $\psi$. The inequalities imposed on $m$ make impossible the inclusion $\bar{g_2}\in \langle \bar{g_1},\bar{g_3}\rangle$; hence Corollary 1 yields

$$\begin{eqnarray*} \deg g_2=3n\geq N(g_1,g_3)=\frac{2n}{(2n,m)} m-m-2n+\deg [g_1,g_3]. \end{eqnarray*}$$

Thus $\frac{2n}{(2n,m)}\leq 3$. If $\frac{2n}{(2n,m)}=2$, then $m=nt$, $t\geq 3$ is an odd number, which contradicts (16). Hence $\frac{2n}{(2n,m)}=3$. Putting $(2n,m)=2\rho$, we have $n=3\rho$, $\deg g_1=6\rho$, $m=2\rho t$, $t>3$, $3\not |\,t$. Applying again Corollary 1, we get

$$\begin{eqnarray*} 9\rho >6\rho t-2\rho t-6\rho , \end{eqnarray*}$$

i.e., $t<\frac{15}{4}$, which is impossible.

If $g_1$ is a simple reducible element of $\psi$, then the pair $g_2,g_3$ is $\ast$-reduced, and Corollary 1 yields

$$\begin{eqnarray} \deg g_1=2n\geq N(g_2,g_3)=\frac{m}{(3n,m)} 3n-3n-m+\deg [g_2,g_3]. {\tag{17}} \end{eqnarray}$$

Thus $\frac{m}{(3n,m)}=2$ and $3n=(3n,m)t$, $t\geq 3$ is an odd number. Since $m>2n$, we have $(3n,m)>n$ and $3n=(3n,m)t\geq (3n,m) 3>3n$, a contradiction.

Case 2. $\frac{3n}{2}<m\leq 2n$.

Since $\bar{g_1},\bar{g_3}$ are algebraically dependent, they would be linearly dependent if $m=2n$, which contradicts $iii)$. Therefore, $m<2n$. If $g_2$ is a simple reducible element of $\psi$, then Corollary 1 gives

$$\begin{eqnarray} \deg g_2=3n\geq N(g_3,g_1)=\frac{m}{(2n,m)} 2n-2n-m+\deg [g_1,g_3]. {\tag{18}} \end{eqnarray}$$

Therefore, $\frac{m}{(2n,m)}\leq 3$. If $\frac{m}{(2n,m)}=2$, then $2n=(2n,m)t$, $t\geq 3$ is an odd number. Hence $m=\frac{4n}{t}$, and the inequality $m>\frac{3n}{2}$ yields $t<\frac{8}{3}$, which is impossible. If $\frac{m}{(2n,m)}=3$, then $2n=(2n,m)t$, $t>3$, $3\not |t$. Thus $m=\frac{6n}{t}>\frac{3n}{2}$ and $t<4$, which is also impossible.

If $g_1$ is a simple reducible element of $\psi$, then it follows from (17) that $\frac{m}{(3n,m)}=2$, $3n=(3n,m)t$, $t\geq 3$ is an odd number. Since $\frac{3n}{2}<m=\frac{6n}{t}<2n$, then $3<t<4$, which is impossible. So case 2 is done.

Now we can assume that $m\leq \frac{3n}{2}$. Then (16) gives $\deg [g_1,g_2]\leq \frac{n}{2}$, and consequently,