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 Reference 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
$$\begin{eqnarray} [x\cdot y,z]=[x,z]\cdot y+x\cdot [y,z]. \cssId{for1}{\tag{1}} \end{eqnarray}$$
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 (Equation 1 ), and $P(L)$ becomes a Poisson algebra Reference 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 Reference 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 Reference 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 Reference 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 Reference 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 Reference 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 Reference 16 that if $\deg _y(G(x,y))=pq+r$ , $0\leq r<p$ , then
$$\begin{eqnarray} \deg (G(f,g))\geq qN+mr, \cssId{for2}{\tag{2}} \end{eqnarray}$$
and if $\deg _x(G(x,y))=sq_1+r_1$ , $0\leq r_1<s$ , then
$$\begin{eqnarray} \deg (G(f,g))\geq q_1N+nr_1. \cssId{for3}{\tag{3}} \end{eqnarray}$$
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 (Equation 2 ) and Lemma 3 imply
Corollary 1.
Let $G(x,y)\in F[x,y],\ h=G(f,g)$ . Consider the following conditions:
$(ii)$ $\deg _y(G(x,y))<p$ ;
$(iii)$ $h=\sum _{i,j}\alpha _{ij}f^ig^j$ , 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.
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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.
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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.
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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 $f,g$ .
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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 (Equation 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.
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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.
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 Reference 13 , Reference 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 (Equation 2 ), (Equation 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.
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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 (Equation 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
$$\begin{eqnarray} g_3=\sigma f_3+G(g_1,g_2),\ \ \deg g_3<\deg f_3 , \cssId{for5}{\tag{4}} \end{eqnarray}$$
where $0\neq \sigma \in F$ , $G(x,y)\in F[x,y]$ . Hence
$$\begin{eqnarray} \overline{G(g_1,g_2)}=-\sigma \bar{f}_3. \cssId{for6}{\tag{5}} \end{eqnarray}$$
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 (Equation 5 ), that $q\geq 1$ . Then inequality (Equation 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 (Equation 1 ), we get from (Equation 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 (Equation 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 Reference 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).
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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 (Equation 4 ), (Equation 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 (Equation 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
$$\begin{eqnarray} \deg [f_1,f_3]=\deg [g_1,f_3]=\deg [g_1,g_2]+3n. \cssId{for7}{\tag{6}} \end{eqnarray}$$
Consider the triple $(g_1,g_2,f_3)$ . By (Equation 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
$$\begin{eqnarray} \deg [f_2,f_3]=\deg [g_2,f_3]=\deg [g_1,f_3]+n. \cssId{for8}{\tag{7}} \end{eqnarray}$$
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 (Equation 6 ) and (Equation 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).
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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 (Equation 2 ) gives also $\deg [g_1,g_2]\leq \frac{n}{2}$ . As in the proof of Propositions 1 , 2 , we obtain also (Equation 6 ), (Equation 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.
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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
$$\begin{eqnarray} \deg \theta \geq 6n+\deg [g_1,g_2]. \cssId{for81}{\tag{8}} \end{eqnarray}$$
It follows from (Equation 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 (Equation 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 (Equation 8 ).
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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 (Equation 8 ) completes the proof.
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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 $\tau$ :
$$\begin{eqnarray} \tau =(f,f_2,f_3),\ \ f=f_1+a,\ \ a\in \langle f_2,f_3 \rangle ,\ \ \deg a\leq \deg f_1, \cssId{for9}{\tag{9}} \end{eqnarray}$$
$$\begin{eqnarray} \tau =(f_1,f,f_3),\ \ f=f_2+a,\ \ a\in \langle f_1,f_3 \rangle ,\ \ \deg a\leq \deg f_2, \cssId{for10}{\tag{10}} \end{eqnarray}$$
$$\begin{eqnarray} \tau =(f_1,f_2,f),\ \ f=f_3+a,\ \ a\in \langle f_1,f_2 \rangle ,\ \ \deg a\leq \deg f_3 . \cssId{for11}{\tag{11}} \end{eqnarray}$$
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 (Equation 11 ). The proof of the theorem will be completed by Lemmas 9 –17 and by Propositions 4 , 5 .
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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
$$\begin{eqnarray} g_3=f_3+g,\ \ g\in \langle g_1,g_2\rangle ,\ \deg g_3<\deg f_3. \cssId{for12}{\tag{12}} \end{eqnarray}$$
If $\tau$ is of form (Equation 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 (Equation 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 (Equation 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 (Equation 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.
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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 (Equation 12 ). If $\tau$ has form (Equation 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 (Equation 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 (Equation 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 $f_1$ .
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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 (Equation 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 (Equation 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 (Equation 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 $(g_1,g_2-\mu g_3^2,\frac{1}{\sigma }g_3)$ .
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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.
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In the sequel we will assume that $\tau$ has form (Equation 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
$$\begin{eqnarray} g_2=f_2+b,\ \ b\in \langle f_1,f_3 \rangle ,\ \ \deg g_2<\deg f_2. \cssId{for13}{\tag{13}} \end{eqnarray}$$
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 (Equation 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
$$\begin{eqnarray} \phi =(f_1,g_2,f_3)\rightarrow (f_1,g_2,f)\rightarrow (f_1,f_2,f)=\tau . \cssId{for14}{\tag{14}} \end{eqnarray}$$
Since $\deg \phi$ , $\deg (f_1,g_2,f)<\deg \theta$ (by (Equation 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 (Equation 14 ) proves the simplicity of $\tau$ .
Assume that $\bar{f_1},\bar{f_3}$ are algebraically independent. Then by (Equation 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 (Equation 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:
$$\begin{eqnarray} (f_1,f_3,g_2)\rightarrow (f_1,f_2,g_2)\rightarrow (f_1,f_2,f_3)=\theta . \cssId{for15}{\tag{15}} \end{eqnarray}$$
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 (Equation 15 ) gives simple reducibility of $f_3$ in $\theta$ , a contradiction.
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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
$$\begin{eqnarray} n+\deg [g_1,g_2]\leq m<3n. \cssId{for16}{\tag{16}} \end{eqnarray}$$
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 (Equation 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 (Equation 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
$$\begin{eqnarray} \deg g_1=2n\geq N(g_2,g_3)=\frac{m}{(3n,m)} 3n-3n-m+\deg [g_2,g_3]. \cssId{for17}{\tag{17}} \end{eqnarray}$$
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
$$\begin{eqnarray} \deg g_2=3n\geq N(g_3,g_1)=\frac{m}{(2n,m)} 2n-2n-m+\deg [g_1,g_3]. \cssId{for18}{\tag{18}} \end{eqnarray}$$
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 (Equation 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 (Equation 16 ) gives $\deg [g_1,g_2]\leq \frac{n}{2}$ , and consequently,
$$\begin{eqnarray*} \deg [g_1,g_2]+\deg g_3\leq 2n. \end{eqnarray*}$$
Then, by Lemma 1 , we have
$$\begin{eqnarray*} \deg [g_2,g_3]+\deg g_1=\deg [g_1,g_3]+\deg g_2, \end{eqnarray*}$$
i.e.,
$$\begin{eqnarray} \deg [g_2,g_3]=\deg [g_1,g_3]+n. \cssId{for19}{\tag{19}} \end{eqnarray}$$
Case 3. $m=\frac{3n}{2}$ .
By putting $n=2\rho$ , we have $\deg g_1=4\rho$ , $\deg g_2=6\rho$ , $\deg g_3=3\rho$ . Put also $\bar{g_2}=\alpha \bar{g_3}^2$ (recall that $\bar{g}_2$ and $\bar{g}_3$ are algebraically dependent). We will first show that $g_2$ is a simple reducible element of $\psi$ . Assume that $g_1$ is a simple reducible element of $\psi$ which is reduced by a simple automorphism $\psi _1=(h_1,g_2,g_3)$ . Then $\deg \psi _1<\deg \theta$ , and by Lemma 9 the automorphism $\psi _2=(h_1,g_2-\alpha g_3^2,g_3)$ is also simple. The sequence
$$\begin{eqnarray*} \psi _2\rightarrow (g_1,g_2-\alpha g_3^2,g_3)\rightarrow (g_1,g_2,g_3)=\psi \end{eqnarray*}$$
proves that $g_2$ is a simple reducible element of $\psi$ .
Put
$$\begin{eqnarray} h_2=g_2-T(g_3,g_1),\ \deg h_2 <\deg g_2, \cssId{for20}{\tag{20}} \end{eqnarray}$$
where $h_2$ is an unreducible element of $\xi =(g_1,h_2,g_3)$ . Let $\deg _y(T(x,y))=k=3q+r$ , $0\leq r<3$ . Since $\deg (T(g_3,g_1))=\deg g_2=6\rho$ , inequality (Equation 2 ) yields that either $q=1$ , $r=0$ , or $q=0$ , $r=0,1$ .
Consider first the case $q=1$ . Note that $g_3,g_1$ is a 3-reduced pair and $\deg _y(w(x,y))=3$ , where $w(x,y)$ is a derivative polynomial of the pair $g_3,g_1$ . By Lemma 7 , the polynomial $T(x,y)$ can be presented in the form
$$\begin{equation*} T(x,y)=w(x,y)q(x,y)+s(x,y), \end{equation*}$$
where $\deg _y(q(x,y))=0$ , $\deg _y(s(x,y))<3$ . By Lemma 6 .3),
$$\begin{equation*} \deg (w(g_3,g_1))\geq N(g_3,g_1)=3\cdot 4\rho -4\rho -3\rho +\deg [g_1,g_3]> 5\rho . \end{equation*}$$
Since $\deg (T(g_3,g_1))=6\rho$ , Lemma 7 .2) gives $\deg (q(g_3,g_1))<\rho , \ \deg (s(g_3,g_1))\leq 6\rho$ . The polynomials $q(x,y),\,s(x,y)$ satisfy condition (ii) of Corollary 1 . Hence they satisfy also (iii), and we get $q(x,y)=\lambda \neq 0,\ s(x,y)=\gamma x^2+\delta x+\mu y$ . Therefore,
$$\begin{equation*} T(x,y)=\lambda w(x,y)+\gamma x^2+\delta x+\mu y. \end{equation*}$$
Consequently, by Corollary 3 ,
$$\begin{eqnarray*} \deg \left(\frac{\partial T}{\partial x}(g_3,g_1)\right) =9\rho ,\ \ \deg \left(\frac{\partial T}{\partial y}(g_3,g_1)\right)=8\rho . \end{eqnarray*}$$
By (Equation 1 ) and (Equation 20 ),
$$\begin{eqnarray} [g_1,h_2]&=&[g_1,g_2]-[g_1,g_3]\frac{\partial T}{\partial x}(g_3,g_1),\cssId{for21}{\tag{21}}\\ \ [h_2,g_3]&=&[g_2,g_3]-[g_1,g_3]\frac{\partial T}{\partial y}(g_3,g_1). \cssId{for22}{\tag{22}} \end{eqnarray}$$
This yields, by (Equation 16 ) and (Equation 19 ),
$$\begin{eqnarray*} \deg [g_1,h_2]&=&\deg [g_1,g_3]+9\rho ,\\ \deg [h_2,g_3]&=&\deg [g_1,g_3]+8\rho . \end{eqnarray*}$$
Therefore, $\deg h_2 \geq \deg [g_1,g_3]+5\rho$ . Since $\deg h_2 <\deg g_2=6\rho$ , this yields $\deg [g_1,g_3]<\rho$ . Now, applying F2) and Corollary 1 , it is easy to show that $g_1,g_3$ are unreducible elements of $\xi =(g_1,h_2,g_3)$ . Since $\deg [g_1,g_3]<\rho$ , by Propositions 1 , 2 , 3 , we conclude that $\xi$ does not admit reductions of types I–IV. This contradicts the simplicity of $\xi$ .
Hence $q=0$ , $r=0,1$ . By Corollary 1 ( (ii) implies (iii) ), we have
$$\begin{eqnarray*} T(x,y)=\alpha x^2+\beta x+\gamma y. \end{eqnarray*}$$
Then $\deg (\frac{\partial T}{\partial y}(g_3,g_1))\leq 0$ and by (Equation 19 ), (Equation 22 ), we conclude that
$$\begin{eqnarray} \deg [h_2,g_3]=\deg [g_2,g_3]=\deg [g_1,g_3]+2\rho . \cssId{for23}{\tag{23}} \end{eqnarray}$$
Now we will assume that $\deg h_2 >2n$ and show that this case is impossible. If $g_3$ is a simple reducible element of $\xi$ , then $g_1,h_2$ form a $\ast$ -reduced pair, and it follows easily from (Equation 2 ) that this pair is $2$ -reduced. But it is impossible since $\deg h_2<3n$ . Furthermore, if $g_1$ is a simple reducible element of $\xi$ , then $g_3,h_2$ form a $2$ -reduced pair, that is, $\deg h_2 =\frac{9}{2}\rho$ . Put
$$\begin{eqnarray*} h_1=g_1+Q(g_3,h_2),\ \deg h_1 <\deg g_1, \end{eqnarray*}$$
where $h_1$ is an unreducible element of the simple automorphism $\xi _1=(h_1,h_2,g_3)$ .
Observe that $\xi _1$ satisfies all the conditions of the present proposition. By (Equation 2 ) and (Equation 23 ), we get $\deg _y(Q(x,y))=2$ . Lemma 7 .4) and Corollary 3 give
$$\begin{eqnarray*} \deg \left(\frac{\partial Q}{\partial y}(g_3,h_2)\right) =\deg h_2 =\frac{9}{2}\rho . \end{eqnarray*}$$
Since
$$\begin{eqnarray*} [h_1,g_3]=[g_1,g_3]+[h_2,g_3]\frac{\partial Q}{\partial y}(g_3,h_2), \end{eqnarray*}$$
by (Equation 23 ) we have
$$\begin{eqnarray*} \deg [h_1,g_3]=\deg [h_2,g_3]+\frac{9}{2}\rho . \end{eqnarray*}$$
Therefore,
$$\begin{eqnarray*} \deg h_1\geq \deg [h_2,g_3]+\frac{3}{2}\rho =\frac{7}{2}\rho +\deg [g_1,g_3]. \end{eqnarray*}$$
These inequalities show that $\xi _1$ does not satisfy the conclusions of the proposition. Since $\deg \xi _1<\deg \psi$ , this contradicts our choice of $\psi$ .
We have thus shown that $\xi$ does not admit elementary reductions. As for reductions of type I, the only possibility for $\xi$ to have it is when $g_1$ is an active element of reduction and $\deg h_2 =\frac{9}{2}\rho$ . Then there should exist $\alpha _1\in F$ such that the element $g_1$ of the automorphism $(g_1,h_2-\alpha _1 g_1,g_3)$ is reducible. Recall that $h_2=g_2-\alpha g_3^2-\beta g_3-\gamma g_1$ . Set $h_2'=h_2-\alpha _1 g_1=g_2-\alpha g_3^2-\beta g_3-(\gamma +\alpha _1) g_1$ . Then $\overline{h_2'}=\overline{h_2}$ and we can replace $h_2$ by $h_2'$ in our previous arguments. Then, as above, the condition for $g_1$ to be a reducible element of $(g_1,h'_2,g_3)$ gives a contradiction.
Furthermore, the comparison of the degrees of the components of $\xi$ shows that $\xi$ does not admit a reduction of type II. Thus $\xi$ admits a reduction of type III or IV, with an active element $g_3$ . Let $(r_1,r_2,r_3)$ be a reduction of type III or a predreduction of type IV of $\xi$ ,
$$\begin{eqnarray*} r_1=g_1-\beta _1 g_3,\ \ r_2=h_2-\alpha _1 g_3^2-\gamma _1 g_3. \end{eqnarray*}$$
Observe that, by definition, $\deg r_1=4\rho ,\ \deg r_2=6\rho$ . We have
$$\begin{eqnarray*} r_2&=&g_2-(\alpha +\alpha _1)g_3^2-(\beta +\gamma _1)g_3-\gamma g_1,\\ \ [r_1,r_2]&=&[g_1,g_2]-2(\alpha +\alpha _1)[g_1,g_3]g_3 \\ &&-\,(\beta +\gamma _1+\gamma \beta _1)[g_1,g_3]+\beta _1[g_2,g_3]. \end{eqnarray*}$$
By Proposition 3 , applied to the triple $\xi =(g_1,h_2,g_3)$ , $\deg [r_1,r_2]\leq \rho$ , $\deg [g_1,g_3]>6\rho$ . Furthermore, by (Equation 16 ) and (Equation 19 ), $\deg [g_1,g_2]\leq \rho ,\ \deg [g_1,g_3]>8\rho$ . Comparing the degrees of the left and right parts in the last equation gives
$$\begin{eqnarray*} \alpha +\alpha _1=0,\ \ \beta +\gamma _1+\gamma \beta _1=0,\ \ \beta _1=0. \end{eqnarray*}$$
Thus $r_1=g_1$ , $r_2=g_2-\gamma g_1$ , $r_3=\sigma g_3+g$ , where $g\in \langle r_1,r_2\rangle \setminus F$ . Since $g_3$ is unreducible in $\psi$ , $\deg r_3=\deg g_3=3\rho$ . But then
$$\begin{equation*} \deg r_1+\deg r_2+\deg r_3=13\rho >\deg g_1+\deg h_2+\deg g_3=\deg \xi , \end{equation*}$$
which contradicts Corollary 4 .
Therefore, the automorphism $(r_1,r_2,r_3)$ is not a reduction of type III of $\xi$ . Assume that it is a predreduction of type IV of $\xi$ ; then it is easy to see that in this case $\psi$ also admits a reduction of type IV with an active element $g_3$ , which is impossible.
We have thus shown that the inequality $\deg h_2 >2n$ is impossible. Hence $\deg h_2 \leq 2n$ . Since $h_2=g_2-\alpha g_3^2-\beta g_3-\gamma g_1$ , this implies $\deg (g_2-\alpha g_3^2)\leq 2n$ . Furthermore, it follows from (Equation 23 ) that
$$\begin{eqnarray*} \deg [g_1,g_3]+2\rho \leq \deg h_2 +\deg g_3\leq 7\rho , \end{eqnarray*}$$
i.e., $\deg [g_1,g_3]\leq 5\rho$ . Thus the automorphism $\psi$ satisfies statement 3) of the present proposition, which is impossible.
Case 4. $m<\frac{3n}{2}$ .
If $g_1$ is a simple reducible element of $\psi$ , then (Equation 17 ) gives that $\frac{m}{(3n,m)}=2$ . Applying once more (Equation 17 ) and (Equation 19 ), we get
$$\begin{eqnarray*} 2n\geq 3n-m+\deg [g_2,g_3]=4n-m+\deg [g_1,g_3], \end{eqnarray*}$$
i.e., $m>2n$ , a contradiction.
Therefore, $g_2$ is a simple reducible element of $\psi$ . By (Equation 18 ), $\frac{m}{(2n,m)}\leq 3$ .
Consider first the case $\frac{m}{(2n,m)}=3$ . Then $2n=(2n,m)t$ , $t>3$ , $3\,{\nmid }\,t$ . Hence $m=\frac{6n}{t}$ , and since $n<m<\frac{3n}{2}$ , we have $t=5$ . By putting $n=5\rho$ , we have $\deg g_1=10\rho$ , $\deg g_2=15\rho$ , $\deg g_3=6\rho$ . Then (Equation 18 ) gives $\deg [g_1,g_3]\leq \rho$ .
Consider equality (Equation 20 ) and put again $\deg _yT(x,y)=k=3q+r, \ 0\leq r<3$ . Since $\deg (T(g_3,g_1))=\deg g_2=15\rho$ , inequality (Equation 2 ) yields that either $q=0,\,r=1$ or $q=1,\,r=0$ . In the first case, by Corollary 1 , $\overline{T(g_3,g_1)}=\bar{g}_2\in \langle \bar{g}_1,\bar{g}_3\rangle$ , which is impossible. Thus $k=3$ and by Lemma 7 .4) $T(x,y)$ is a derivative polynomial (up to a nonzero scalar multiplier) of the pair $g_3,g_1$ . Then Corollary 3 gives
$$\begin{eqnarray*} \deg \left(\frac{\partial T}{\partial x}(g_3,g_1)\right) =24\rho ,\ \ \deg \left(\frac{\partial T}{\partial y}(g_3,g_1)\right)=20\rho . \end{eqnarray*}$$
From (Equation 21 ), (Equation 22 ), taking into account (Equation 16 ) and (Equation 19 ), we get
$$\begin{eqnarray*} \deg [g_1,h_2]=\deg [g_1,g_3]+24\rho ,\\ \deg [h_2,g_3]=\deg [g_1,g_3]+20\rho . \end{eqnarray*}$$
Hence $15\rho >\deg h_2 \geq 14\rho +\deg [g_1,g_3]$ , and Corollary 1 yields that $g_1,g_3$ are unreducible elements of the automorphism $\xi =(g_1,h_2,g_3)$ . Since $\deg [g_1,g_3]<\rho$ , it is easy to check that $\xi$ does not admit reductions of types I–IV, which contradicts the simplicity of $\xi$ .
Now we consider the case $\frac{m}{(2n,m)}=2$ . By putting $(2n,m)=2\rho$ , we get $m=4\rho$ , $2n=2\rho t$ , $t\geq 3$ is an odd number. Then $m=\frac{4n}{t}$ and (Equation 16 ) implies $t<4$ . Hence $t=3$ and $\deg g_1=6\rho$ , $\deg g_2=9\rho$ , $\deg g_3=4\rho$ . Consider again equality (Equation 20 ) and put $\deg _y(T(x,y))=k=2q+r$ , $0\leq r<2$ . Assume that $r=1$ . Then (Equation 2 ) yields $q=1$ , i.e., $k=3$ . By Lemma 7 ,
$$\begin{eqnarray*} T(x,y)=w(x,y)q(x,y)+s(x,y), \end{eqnarray*}$$
where $w(x,y)$ is a derivative polynomial of the 2-reduced pair $g_3,g_1$ , and $\deg _y(q(x,y))=1$ , $\deg _y(s(x,y))\leq 1$ . By Lemma 5 .3),
$$\begin{equation*} \deg (w(g_3,g_1))\geq N(g_3,g_1)=12\rho -6\rho -4\rho +\deg [g_1,g_3]>2\rho . \end{equation*}$$
Now Lemma 7 .2) gives $\deg (q(g_3,g_1))<7\rho$ , and then Lemma 3 yields $\deg (q(g_3,g_1)) =6\rho$ . Thus, by Lemma 7 .3), we get
$$\begin{eqnarray*} \deg \left(\frac{\partial T}{\partial x}(g_3,g_1)\right) =14\rho ,\ \ \deg \left(\frac{\partial T}{\partial y}(g_3,g_1)\right)=12\rho . \end{eqnarray*}$$
Now (Equation 21 ) and (Equation 22 ) yield, by means of (Equation 16 ) and (Equation 19 ),
$$\begin{eqnarray*} \deg [g_1,h_2]&=&\deg [g_1,g_3]+14\rho ,\\ \deg [h_2,g_3]&=&\deg [g_1,g_3]+12\rho . \end{eqnarray*}$$
Hence $\deg h_2 \geq 8\rho +\deg [g_1,g_3]$ , and the elements $g_1,g_3$ of $\xi =(g_1,h_2,g_3)$ are unreducible. Since $\deg [g_1,g_3]<\rho$ , it is easy to check that $\xi$ does not admit reductions of type I–IV, a contradiction.
Thus $r=0$ , $k=2q$ . Inequality (Equation 2 ) gives $1\leq q\leq 4$ and
$$\begin{eqnarray*} \deg \left(\frac{\partial T}{\partial y}(g_3,g_1)\right) \geq (q-1)(2\rho +\deg [g_3,g_1])+6\rho . \end{eqnarray*}$$
Hence, by means of (Equation 19 ), (Equation 22 ),
$$\begin{eqnarray} \deg [h_2,g_3]\geq q(2\rho +\deg [g_3,g_1])+4\rho . \cssId{for231}{\tag{24}} \end{eqnarray}$$
In particular,
$$\begin{eqnarray} \deg h_2 \geq q(2\rho +\deg [g_3,g_1]). \cssId{for24}{\tag{25}} \end{eqnarray}$$
Assume that $\deg h_2 \geq 6\rho$ . Since $h_2$ is an unreducible element of $\xi =(g_1,h_2,g_3)$ , the elements $\bar{g_1},\bar{h}_2$ are linearly independent, if $\deg h_2 =6\rho$ . Note that $g_1,h_2$ and $g_3,h_2$ do not compose $2$ -reduced pairs. Consequently, the elements $g_1,g_3$ of $\xi$ are unreducible. In fact, assume that there exists $f\in \langle h_2,g_3\rangle$ such that $\bar{g}_1=\bar{f}$ . Since $\bar{g}_1\notin \langle \bar{h}_2,\bar{g}_3\rangle$ , the elements $\bar{h}_2, \bar{g}_3$ are algebraically dependent and the pair $g_3,h_2$ is $\ast$ -reduced. It follows easily from (2) that this pair should be 2-reduced, a contradiction. Similarly, $g_3$ is unreducible. Furthermore, it follows from (Equation 25 ) that $\deg [g_3,g_1]<\deg h_2$ . Hence, due to Definitions 1 –4 and Propositions 1 –3 , $\xi$ does not admit reductions of types I–IV. This contradicts the simplicity of $\xi$ .
Therefore, $\deg h_2 <6\rho$ and $\xi$ satisfies all the conditions of the present proposition. Since $\deg \xi <\deg \psi$ , then, by the choice of $\psi ,\ \xi$ should satisfy the conclusion of the proposition. It follows from (Equation 24 ), (Equation 25 ) that $\xi$ does not satisfy statements 1), 3); hence $\xi$ admits a reduction of type IV with an active element $h_2$ . In this case, $\deg h_2\leq 3\rho$ and we have $q=1$ in (Equation 25 ), which implies $\deg [g_1,g_3]\leq \rho$ . By statement 5) of Proposition 3 , $\xi$ has a predreduction of the form $(g_1,\tilde{h}_2,g_3)$ , where
$$\begin{equation*} \tilde{h}_2=\sigma h_2+g,\ g\in \langle g_3,g_1\rangle \setminus F,\ \deg \tilde{h}_2=3\rho ,\ \deg [g_3,\tilde{h}_2]<6\rho +\deg [g_3,g_1]. \end{equation*}$$
By Corollary 2 we have $g=\lambda w(g_3,g_1)$ , where $w(x,y)$ is a derivative polynomial of the pair $g_3,g_1$ . Then $\deg _y(w(x,y))=2$ . Now $\tilde{h}_2=\sigma (g_2-T(g_3,g_1))+\lambda w(g_3,g_1)$ and $\sigma ^{-1}\tilde{h}_2=g_2-\tilde{T}(g_3,g_1)$ , where $\tilde{T}(x,y)=T(x,y)-\sigma ^{-1}\lambda w(x,y)$ . If we substitute $\sigma ^{-1}\tilde{h}_2$ instead of $h_2$ in (Equation 20 ) (with $\tilde{T}$ instead of $T$ ), we will have again $\deg _y(\tilde{T}(x,y))=2q$ , $1\leq q\leq 4$ , since the equality $\deg _y(\tilde{T}(x,y))=3$ is impossible. Then inequalities (Equation 24 ) and (Equation 25 ) hold also for $\tilde{h}_2$ . In particular,
$$\begin{equation*} \deg [g_3,\tilde{h}_2]\geq 6\rho +\deg [g_3,g_1], \end{equation*}$$
which contradicts the previous inequality.
■
Proposition 5.
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 $\deg g_1=2n$ , $\deg g_2=ns$ , where $s\geq 5$ is an odd number;
$iii)$ $\bar{g_3}\notin \langle \bar{g_1}\rangle$ , $\deg g_3=m<ns$ , and $g_3$ is not a simple reducible element of $\psi$ .
Then $m<n(s-2)+\deg [g_1,g_2]$ .
Proof.
Assuming the contrary, we have
$$\begin{eqnarray} n(s-2)+\deg [g_1,g_2]\leq m<ns. \cssId{for26}{\tag{26}} \end{eqnarray}$$
Hence $\deg [g_1,g_2]<2n$ . It is now easy to check that $\psi$ does not admit reductions of type I–IV. Then $\psi$ is elementary reducible, and either $g_1$ or $g_2$ is a simple reducible element of $\psi$ . In particular, $\bar{g_1},\bar{g_2},\bar{g_3}$ are mutually algebraically dependent. It follows from (Equation 26 ) that $\bar{g_2}\notin \langle \bar{g_1},\bar{g_3}\rangle$ , $\bar{g_1}\notin \langle \bar{g_2},\bar{g_3}\rangle$ . Consequently, $g_1,g_3$ and $g_2,g_3$ are $\ast$ -reduced pairs as well.
If $g_2$ is a simple reducible element of $\psi$ , then by Corollary 1 we get
$$\begin{eqnarray} \deg g_2=ns\geq N(g_1,g_3)=\frac{2n}{(2n,m)}m-m-2n+\deg [g_1,g_3]. \cssId{for27}{\tag{27}} \end{eqnarray}$$
Observe that by Lemma 2 .i) we have $p=\frac{2n}{(2n,m)}\geq 2$ . If $\frac{2n}{(2n,m)}\geq 4$ , inequalities (Equation 26 ), (Equation 27 ) give $ns>3m-2n>3n(s-2)-2n$ . Hence $s<4$ , which contradicts condition $ii)$ of the proposition.
If $\frac{2n}{(2n,m)}=2$ , then $n=(2n,m)$ , $m=nt$ , where $t\geq 3$ is an odd number. From (Equation 26 ) we get $s-2<t<s$ , which is impossible.
Therefore, $\frac{2n}{(2n,m)}=3$ . By putting $(2n,m)=2\rho$ , we have $n=3\rho$ , $\deg g_1=6\rho$ , $m=2\rho t$ , where $t>3$ , $3\not |\,t$ . Inequalities (Equation 26 ) and (Equation 27 ) yield
$$\begin{eqnarray*} 3(s+2)>4t,\ \ 3(s-2)<2t<3s. \end{eqnarray*}$$
Hence $t=s=5$ , i.e., $\deg g_1=6\rho$ , $\deg g_2=15\rho$ , $\deg g_3=10\rho$ . Applying (Equation 26 ), (Equation 27 ) once more, we get
$$\begin{eqnarray*} \deg [g_1,g_2]\leq \rho , \ \deg [g_1,g_3]\leq \rho . \end{eqnarray*}$$
Let again $h_2=g_2-T(g_1,g_3)$ , where $\deg h_2<\deg g_2$ and $h_2$ is unreducible in $\xi =(g_1,h_2,g_3)$ . Since $\deg (T(g_1,g_3))=\deg g_2= 15\rho$ and $\overline{T(g_1,g_3)}=\bar{g}_2\not \in \langle \bar{g}_1,\bar{g}_3\rangle$ , inequality (Equation 2 ) yields $\deg _y(T(x,y))=3$ . By Lemma 7 .4), up to a scalar multiplier, $T(x,y)$ is equal to a derivative polynomial of the pair $g_1,g_3$ . Hence by Corollary 3 ,
$$\begin{eqnarray*} \deg \left(\frac{\partial T}{\partial x}(g_1,g_3)\right)=24\rho ,\ \ \deg \left(\frac{\partial T}{\partial y}(g_1,g_3)\right)=20\rho . \end{eqnarray*}$$
Since
$$\begin{eqnarray*} \deg [g_1,g_3]+\deg g_2&\leq & 16\rho ,\\ \deg [g_1,g_2]+\deg g_3&\leq & 11\rho , \end{eqnarray*}$$
Lemma 1 implies that
$$\begin{eqnarray*} \deg [g_2,g_3]+\deg g_1\leq 16\rho . \end{eqnarray*}$$
Therefore, $\deg [g_2,g_3]\leq 10\rho$ . Now equalities (Equation 21 ), (Equation 22 ), with $g_1$ and $g_3$ permuted, give
$$\begin{eqnarray*} \deg [g_1,h_2]&=&\deg [g_1,g_3]+20\rho ,\\ \deg [h_2,g_3]&=&\deg [g_1,g_3]+24\rho . \end{eqnarray*}$$
Hence $\deg h_2 \geq 14\rho +\deg [g_1,g_3]$ , which implies easily by (Equation 2 ) that the elements $g_1,g_3$ are unreducible in $\xi =(g_1,h_2,g_3)$ . Since $\deg [g_1,g_3]\leq \rho$ , it is easily checked that $\xi$ does not admit reductions of types I–IV. Thus, $\xi$ is not a simple automorphism.
Now suppose that $g_1$ is a simple reducible element of $\psi$ . Then Corollary 1 gives
$$\begin{eqnarray*} \deg g_1=2n\geq N(g_3,g_2)=\frac{m}{(ns,m)} ns-ns-m+\deg [g_2,g_3]. \end{eqnarray*}$$
Therefore, $\frac{m}{(ns,m)}=2$ and $ns=(ns,m)t$ , where $t\geq 3$ is an odd number. Then $m=\frac{2ns}{t}$ and (Equation 26 ) gives $s-2<\frac{2s}{t}$ , i.e., $(t-2)(s-2)<4$ . Since $s\geq 5$ , this yields $t=3$ , $s=5$ . By putting $(ns,m)=5\rho$ , we have again $\deg g_1= 6\rho$ , $\deg g_2=15\rho$ , $\deg g_3=10\rho$ . By (Equation 26 ), we have also $\deg [g_1,g_2]\leq \rho$ . Suppose that
$$\begin{eqnarray*} h_1=g_1+G(g_3,g_2), \ \ \deg h_1 <\deg g_1, \end{eqnarray*}$$
where $h_1$ is an unreducible element of $\xi =(h_1,g_2,g_3)$ . Since $\deg (G(g_3,g_2))=\deg g_1=6\rho$ , by Corollary 2 we have $G(x,y)=\lambda w(x,y)$ , where $w(x,y)$ is a derivative polynomial of the pair $g_3,g_2$ . Therefore, by Corollary 3 ,
$$\begin{eqnarray*} \deg \left(\frac{\partial G}{\partial x}(g_3,g_2)\right) =2\cdot \deg g_3=20\rho . \end{eqnarray*}$$
Since
$$\begin{eqnarray*} [h_1,g_2]=[g_1,g_2]+[g_3,g_2]\frac{\partial G}{\partial x}(g_3,g_2), \end{eqnarray*}$$
a comparison of degrees gives
$$\begin{eqnarray*} \deg [h_1,g_2]=\deg [g_3,g_2]+20\rho . \end{eqnarray*}$$
Thus,
$$\begin{eqnarray*} \deg h_1 \geq 5\rho +\deg [g_3,g_2],\qquad \deg [g_3,g_2]<\rho . \end{eqnarray*}$$
It remains to note that $\xi =(h_1,g_2,g_3)$ satisfies all the conditions of Proposition 4 and does not satisfy conclusions 1), 3) of this proposition. If $\xi$ had admitted a reduction of type IV with an active element $h_1$ , then by the definition there would exist an element $g\in \langle g_3,g_2\rangle \setminus F$ such that $\deg (h_1+g)= \frac{15\rho }{2}$ , i.e., $\deg g = \frac{15\rho }{2}$ . Then by Corollary 2 we would have again $g=\alpha _1 w(g_3,g_2)$ , which contradicts the equality $\deg (w(g_3,g_2))=\deg (G(g_3,g_2))=6\rho$ .
■
Lemma 15.
If $f_1,f_3$ is a $\ast$ -reduced pair and $\deg f_1<\deg f_3$ , then $\tau$ is a simple automorphism.
Proof.
We consider first the case when $\deg f_2>\deg f_3$ . By (Equation 11 ), we have $a\in \langle f_1,f_2 \rangle$ , $\deg a\leq \deg f_3$ . By Lemma 14 , we may assume that $\bar{f_2}\notin \langle \bar{f_1}\rangle$ and $a\notin \langle f_1\rangle$ . Since $\deg a<\deg f_2$ , then by F2), $\bar{f_1},\bar{f_2}$ are algebraically dependent. By Corollary 1 , $\deg a\geq N=N(f_1,f_2)$ . Hence $\deg f_2>N$ and by Lemma 2 .iii), the pair $f_1,f_2$ is $2$ -reduced, that is, $\deg f_1=2n$ , $\deg f_2=ns$ , where $s\geq 3$ is an odd number, and
$$\begin{eqnarray*} n(s-2)+\deg [f_1,f_2]\leq \deg a\leq \deg f_3 . \end{eqnarray*}$$
Since $f_3$ is not a simple reducible element of $\theta$ (see Corollary 7 ), $\theta$ satisfies the conditions of one of Propositions 4 , 5 . Since $\deg f_3 >\deg f_1$ , statements 2), 3) of Proposition 4 are not fulfilled for $\theta$ . Therefore,
$$\begin{eqnarray*} \deg f_3 <n(s-2)+\deg [f_1,f_2], \end{eqnarray*}$$
which contradicts the previous inequality.
Suppose now that $\deg f_2\leq \deg f_3$ . If $\deg f_2=\deg f_3$ and $\bar{f_2},\bar{f_3}$ are linearly dependent, then the element $b$ in (Equation 13 ) can be chosen as $b=\alpha f_3$ . Then sequence (Equation 15 ) gives the simplicity of $\tau$ .
Thus we can assume that $\deg f_2<\deg f_3$ . Then using Lemma 14 .i) we can furthermore assume that $\bar{b}=-\bar{f_2}\notin \langle \bar{f_1},\bar{f_3}\rangle$ . Since $\deg b\leq \deg f_3$ , by Corollary 1 $N(f_1,f_3)\leq \deg f_3$ , and by Lemma 2 .iii) the elements $f_1,f_3$ form a $2$ -reduced pair. Put $\deg f_1=2n$ , $\deg f_3 =ns$ , where $s\geq 3$ is an odd number. The inequality $\deg f_3=ns\geq N(f_1,f_3)$ gives also $\deg [f_1,f_3]\leq 2n$ . We can also assume that $g_2$ is an unreducible element of $\phi =(f_1,g_2,f_3)$ . Then $\phi$ satisfies the conditions of one of Propositions 4 , 5 .
Let $b=T(f_1,f_3)$ , where $\deg _y(T(x,y))=k$ . Then (Equation 2 ) implies that $k=2$ if $s>3$ , and $k=2,4$ if $s=3$ . Consequently, $\deg _y(\frac{\partial T}{\partial y}(x,y))=k-1\in \{1,3\}$ , and (Equation 2 ) yields
$$\begin{eqnarray*} \deg \left( \frac{\partial T}{\partial y}(f_1,f_3)\right) \geq ns. \end{eqnarray*}$$
By (Equation 13 ),
$$\begin{eqnarray} [f_1,g_2]=[f_1,f_2]+[f_1,f_3]\frac{\partial T}{\partial y}(f_1,f_3). \cssId{for28}{\tag{28}} \end{eqnarray}$$
Assume first that $\deg g_2<n(s-2)+\deg [f_1,f_3]$ . Then (Equation 28 ) yields
$$\begin{eqnarray} \deg [f_1,f_2]\geq \deg [f_1,f_3]+ns. \cssId{for29}{\tag{29}} \end{eqnarray}$$
Therefore, $\deg f_2\geq n(s-2)+\deg [f_1,f_3]>n$ and $\bar{f}_1 \notin \langle \bar{f}_2\rangle$ . Consequently, either the elements $\bar{f}_1, \bar{f}_2$ are algebraically independent or $f_1, f_2$ form a $\ast$ -reduced pair. Since $\deg [f_1,f_2]>ns\geq \deg a$ , we have by Lemma 2 .ii) and Corollary 1 ,
$$\begin{eqnarray*} a=\alpha f_2^2+\gamma f_2+G(f_1),\ \ \deg (G(f_1))<\deg f_3 , \end{eqnarray*}$$
where $\alpha \neq 0$ only if $\deg f_3 \geq 2 \deg f_2$ . Since $f_1,f_3$ is a $2$ -reduced pair, then $f_1,f_3+G(f_1)$ is also a $2$ -reduced pair. By Lemma 9 , $(f_1,g_2,f_3+G(f_1))$ is a simple automorphism. We have
$$\begin{equation*} (f_1,g_2,f_3+G(f_1))\rightarrow (f_1,f_2,f_3+G(f_1))\rightarrow (f_1,f_2,f_3+\alpha f_2^2+\gamma f_2+G(f_1))=\tau . \end{equation*}$$
If $(\alpha ,\gamma )\neq (0,0)$ , then it is easily checked that $(f_1,g_2,f_3+G(f_1))$ is a reduction of $\tau$ of types I–III, with an active element $f_2$ . Note that the type of the reduction depends on the degree of $f_2$ :
1) if $2n<\deg f_2$ , then $\deg \,(f_2^2)>n(s-2)+ \deg [f_1,f_3]+2n>ns = \deg f_3$ , $\alpha =0$ , and $\tau$ admits a reduction of type I;
2) if $\frac{3n}{2}<\deg f_2\leq 2n$ , then $s=3$ (otherwise $f_2$ is not reducible in $\theta$ ), $\alpha =0$ and $\tau$ admits a reduction of type II;
3) if $\deg f_2\leq \frac{3n}{2}$ , then $s=3$ and $\tau$ admits a reduction of type III.
The case $\alpha =\gamma =0$ follows from Lemma 14 .
It remains to consider the cases when $\phi$ satisfies one of the conclusions 2), 3) of Proposition 4 . If $\phi$ satisfies 3), then (Equation 28 ) again gives (Equation 29 ). Besides, in this case $s=3$ and $\deg f_2>\deg g_2=\frac{3n}{2}$ . Consequently, $a=\beta f_2+\gamma f_1$ . By Lemma 14 we can put $\beta \neq 0$ . Then the automorphism $(f_1,g_2,f_3+\gamma f_1)$ gives a reduction of type I or II of $\tau$ with an active element $f_2$ .
Assume finally that $\phi$ admits a reduction of type IV with an active element $g_2$ . Since $\deg [f_1,f_3]\leq 2n$ , the scalars $\alpha ,\beta ,\gamma$ in Definition 4 are $0$ by Proposition 3 .5). Consequently, the reduction of type IV of $\phi$ has the form $(f_1,h_2,f_3-\delta h_2^2)$ , where
$$\begin{eqnarray*} g_2=h_2+g,\ g\in \langle f_1,f_3\rangle \setminus F, \qquad \deg (f_3-\delta h_2^2)\leq 2n. \end{eqnarray*}$$
Since $\deg g_2\leq \deg h_2 =\frac{3n}{2}$ , then $\deg g\leq \frac{3n}{2}$ . Note that, by Definition 4 , the automorphism $(f_1,h_2,f_3)$ satisfies statement 3) of Proposition 4 .
If $\deg f_2>\frac{3n}{2}$ , then we may take $\phi =(f_1,h_2,f_3)$ , and this case may be reduced to the previous one.
Assume that $\deg f_2\leq \frac{3n}{2}$ . Then we write
$$\begin{eqnarray*} f_2=g_2-b=h_2+(g-b), \quad g-b\in \langle f_1,f_3\rangle . \end{eqnarray*}$$
If $g-b\notin F$ , then $\theta$ admits the reduction of type IV $(f_1,h_2,f_3-\delta h_2^2)$ , with an active element $f_2$ , $\alpha =\beta =\gamma =0$ , and the predreduction $(f_1,h_2,f_3)$ . This case was considered in Lemma 12 . If $g-b=\alpha \in F$ , then $f_2=h_2+\alpha$ and the automorphism $(f_1,f_2,f_3-\delta h_2^2)$ is simple by Remark 1 . Consequently, $f_3$ is a simple reducible element of $\theta$ , a contradiction.
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Lemma 16.
If $f_1,f_3$ is a $\ast$ -reduced pair and $\deg f_3 <\deg f_1$ , then $\tau$ is a simple automorphism.
Proof.
Consider first the case when $\deg f_1<\deg f_2$ . Since $\deg a\leq \deg f_3$ by (Equation 11 ), in this case $\bar{a}\not \in \langle \bar{f}_1,\bar{f}_2 \rangle$ . Then the inclusion $a\in \langle f_1,f_2 \rangle$ implies by F2) that the elements $\bar{f_1},\bar{f_2}$ are algebraically dependent. By Lemma 14 , we may assume that $f_1,f_2$ form a $*$ -reduced pair. Since $\deg a<\deg f_1$ , inequality (Equation 2 ) gives that $\deg f_1=2n$ , $\deg f_2=3n$ , $\deg [f_1,f_2]\leq 2n$ . Besides, if $\deg f_3 <n+\deg [f_1,f_2]=N(f_1,f_2)$ , then Corollary 1 implies $a\in F$ . Thus, we may assume that
$$\begin{eqnarray*} \deg f_3 \geq n+\deg [f_1,f_2]. \end{eqnarray*}$$
Observe that $\theta$ satisfies all the conditions of Proposition 4 ; hence it should satisfy one of conlusions 2), 3) of this proposition. The case when $\theta$ admits a reduction of type IV was considered in Lemma 12 . Therefore, we may assume that $\deg f_3 =\frac{3n}{2}$ , and
$$\begin{eqnarray*} \deg (f_2-\alpha f_3^2)\leq 2n,\ \ \deg [f_1,f_3]<3n+\deg [f_1,f_2]. \end{eqnarray*}$$
Observe that $\deg a,\,\deg f\leq \frac{3n}{2}$ . By Lemma 13 , the automorphism $(f_1,f_2-\alpha f_3^2,f_3)$ is simple. If $a\notin F$ , then $(f_1,f_2-\alpha f_3^2,f_3)$ gives a reduction of type IV of $\tau$ .
Consider now the case when $\deg f_2\leq \deg f_1$ . If $\deg f_1=\deg f_2$ , then by Lemma 14 we may assume that $\bar{f_1},\bar{f_2}$ are linearly independent. Suppose $\bar{f_2}\in \langle \bar{f_3}\rangle$ . If $\bar{f_2},\bar{f_3}$ are linearly dependent, sequence (Equation 15 ) proves that $\tau$ is simple. Suppose that $\deg f_2=t\cdot \deg f_3$ , $t\geq 2$ . Since $\deg f_3 < \min \{\deg f_1,\deg f_2\}$ and $\deg a\leq \deg f_3$ , under the assumption that $a\notin F$ , inequality (Equation 2 ) yields $\deg f_2=2n$ , $\deg f_1=3n$ , and
$$\begin{eqnarray*} \deg f_3 \geq n+\deg [f_1,f_2]>\frac{2n}{t}=\deg f_3 . \end{eqnarray*}$$
Therefore, $\bar{f_2}\notin \langle \bar{f_1},\bar{f_3}\rangle$ . Since $f_2$ is a simple reducible element of $\theta$ , it follows from Corollary 1 that $N(f_1,f_3)\leq \deg f_2\leq \deg f_1$ and by Lemma 2 .iii) the pair $f_3,f_1$ is 2-reduced. Without loss of generality, we can assume $g_2$ in (Equation 13 ) to be unreducible in $\phi =(f_1,g_2,f_3)$ ; then $\phi$ satisfies the conditions of one of Propositions 4 , 5 . The rest of the proof can now be fulfilled similarly to that of Lemma 15 .
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Lemma 17.
If $\bar{f_1}\in \langle \bar{f_3}\rangle$ and $\deg f_1>\deg f_3$ , then $\tau$ is a simple automorphism.
Proof.
Put $n=\deg f_3$ , $\bar{f_1}=\alpha \bar{f_3}^k$ , $k\geq 2$ . Then $\deg f_1=nk$ . Consider the simple automorphism $\phi =(f_1,g_2,f_3)$ defined by (Equation 13 ). Since $\deg \phi <\deg \theta$ , Lemma 9 implies that $\psi =(f_1-\alpha f_3^k,g_2,f_3)$ is also a simple automorphism. Therefore, by Lemma 9 again, the sequence
$$\begin{eqnarray*} \psi \rightarrow (f_1-\alpha f_3^k,f_2,f_3)\rightarrow \theta \end{eqnarray*}$$
proves that $f_1$ is a simple reducible element of $\theta$ . Interchanging the elements $f_1,f_2$ , by Lemmas 14 , 15 , 16 , we can restrict ourselves to the case when $\bar{f_2}\in \langle \bar{f_3}\rangle$ , $\deg f_2>\deg f_3$ . Thus, we can take $\deg f_2=nr$ , $r\geq 2$ . By Lemma 14 , we can assume that $f_1,f_2$ is a $\ast$ -reduced pair, with $\deg f_1=mp,\ \deg f_2=ms,\ (p,s)=1,\ m\geq n$ . Then $N(f_1,f_2)>m(ps-p-s)>n\geq \deg a$ . Hence by Corollary 1 , $\bar{a}\in \langle \bar{f}_1,\bar{f}_2\rangle$ , which evidently implies that $a\in F$ .
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This finishes the proof of Theorem 1 .