On solvable elements in the Weyl algebra

TL;DR

This paper explores the properties of solvable elements in the Weyl algebra and relates their classification to the open Dixmier problem, providing conditions for unsolvability and analyzing their characteristics.

Contribution

It introduces new sufficient conditions for elements to be unsolvable in the Weyl algebra and studies properties of solvable elements, advancing understanding of the Dixmier problem.

Findings

Sufficient conditions for unsolvability are established.

Properties of solvable elements are characterized.

The work relates solvability to the Dixmier open question.

Abstract

We show that to determine all solvable elements in the Weyl algebra is closely related to the Dixmier's open question. Sufficient conditions for an elements being unsolvable are given, and properties of solvable elements are obtained.

Equations

$$[p,q]=pq-qp=1.$$

$$[p_i,p_j]=[q_i,q_j]=0,[p_i,q_j]={\delta }_{ij},$$

$$(\mathrm{ad}x)y=:[x,y]=1,$$

$$A_s=⟨p^i{q}^j|j-i=s⟩.$$

$$[h,p]=-p,[h,q]=q.$$

$$f(h)p^n=p^{n}f(h-n)\text{and}f(h)q^n=q^{n}f(h+n)$$

$$x=f_{-s}(h)p^s+\cdots +f_0(h)+\cdots +f_t(h)q^t,f_i(X)\in \mathbb{F}[X],s,t\in \mathbb{N}.$$

$$\omega (p)=-q\text{and}\omega (q)=p.$$

$$V(x)$$

$$N(x)$$

$$C(x)$$

$$D(x)$$

$$\mathbb{F}[x]$$

$$A-\mathbb{F}={\cup }_{i=1}^5{\Delta }_i,$$

$${\Delta }_1$$

$${\Delta }_2$$

$${\Delta }_3$$

$${\Delta }_4$$

$${\Delta }_5$$

$$v_{\rho ,\sigma }(f)={\text{sup}}_{(i,j)\in E(f)}(i\rho +j\sigma ),$$

$$\sum _{(i,j)\in E_{\rho ,\sigma }(x)}{\alpha }_{ij}{X}^i{Y}^j$$

$$\sum _{(i,j)\in E_{\rho ,\sigma }(x)}{\alpha }_{ij}{p}^i{q}^j$$

$$E_{k\rho ,k\sigma }(x)=E_{\rho ,\sigma }(x),v_{k\rho ,k\sigma }(x)=v_{\rho ,\sigma }(x).$$

$$E_{1,1}(x)=\{(4,0),(3,1),(2,2)\},E_{1,2}(x)=\{(2,2),(0,3)\},$$

$$V(x)\ne C(x)$$

$$v_{1,1}(x)=2k+i,v_{1,1}(y)=2l+j,$$

$$(c{X}^k{Y}^{k+i}{)}^{v_{1,1}(y)}=(c^{\prime }{X}^{l+j}{Y}^l{)}^{v_{1,1}(x)},$$

$$v_{\rho ,\sigma }([x,y])$$

$$=v_{\rho ,\sigma }(x)+v_{\rho ,\sigma }(y)-(\rho +\sigma )$$

$$\ge v_{\rho ,\sigma }(y)$$

$$>0,$$

$$g^{v_{\rho ,\sigma }(x)}=f^{v_{\rho ,\sigma }(y)},$$

$$x=c_0+c_d{q}^d+c_{d+1}{q}^{d+1}+\cdots +c_n{q}^n,c_i\in \mathbb{F},c_d\ne 0$$

$$y=y_{-s}+\cdots +y_0+\cdots +y_t,y_i\in A_i,s,t\in \mathbb{N}.$$

$$\begin{array}{rl}[x,y]& =[c_d{q}^d+\cdots +c_n{q}^n,y_{-s}+\cdots +y_0+\cdots +y_t]\\ & =[c_d{q}^d,y_{-s}]+u,\end{array}$$

$$y=b_{-1}p+f_0(h)+\cdots +f_t(h)q^t.$$

Full text

On solvable elements in the Weyl algebra

Chaowen Zhang

Department of Mathematics,

China University of Mining and Technology,

Xuzhou, 221116, Jiang Su, P. R. China

Mathematics Subject Classification (2010): 16S32; 16W50.

1 Introduction

Let $\mathbb{F}$ be a field of characteristic zero. The Weyl algebra $A$ over $\mathbb{F}$ is an associative algebra with generators $p,q$ satisfying the relation ([5, 11])

[TABLE]

This algebra was first introduced by H. Weyl. The $n$-th Weyl algebra $A_{n}$ over $\mathbb{F}$ is defined by $2n$ generators $p_{1},q_{1},\cdots,p_{n},q_{n},$ subject to the relations

[TABLE]

where $\delta_{ij}$ is the Kronecker symbol (see [4, 6, 9, 11]).

In [5], Dixmier proposed the following question: Is an algebra endomorphism of $A$ necessarily an automorphism? This question is generalized to all the Weyl algebras, known as the Dixmier Conjecture. It was shown that the Dixmier conjecture is stably equivalent to the Jacobian Conjecture (See [1, 2, 4, 9, 11]). But currently, the Dixmier Conjecture remains open even for the case $n=1$.

For convenience, we say that an element $x\in A$ is solvable if there is an element $y\in A$ such that

[TABLE]

and $x$ is unsolvable if it is not solvable. Since  $[y,-x]=[x,y]$  for $x,y\in A$, $y$ is also solvable if $x$ is so. Let $\text{Aut}(A)$ denote the group of all the algebra automorphisms of $A$. Then it is clear that if $x$ is solvable, so is $\Phi(x)$ for any $\Phi\in\text{Aut}(A)$. Properties of solvable elements are studies in [6, 7] and especially, a necessary condition for an element being solvable is given in [7].

In this paper we show that to determine all solvable elements in $A$ is closely related to the Dixmier’s above question. We give some sufficient conditions for an element in $A$ being unsolvable, and we also study the properties of solvable elements.

The paper is arranged as follows. In Section 2 we give preliminaries. In Section 3 we study the properties of solvable elements. We show that the Dixmier’s open question is equivalent to the statement that each element in $\Delta_{2}$ is unsolvable. In Section 4 we prove two reduction theorems which give sufficient conditions for an element being unsolvable. Finally, in Section 5 we give a geometric description of solvable elements.

2 Preliminaries

According to [5], $A$ has a basis $p^{i}q^{j},\ i,j\geq 0.$ For each $s\in\mathbb{Z}$, define a subspace $A_{s}$ of $A$ by

[TABLE]

It is easy to check that $A_{i}A_{j}\subseteq A_{i+j}$ for all $i,j$. Therefore $A$ becomes a $\mathbb{Z}$-graded algebra $A=\oplus_{i\in\mathbb{Z}}A_{i}$.

Denote the element $pq\in A$ by $h$. Then we have

[TABLE]

A short computation shows that (See [7, Lemma 2.1] or [8, Lemma 5.1])

[TABLE]

for $f(X)\in\mathbb{F}[X]$ and $n\in\mathbb{N}$. Then each element of $A_{i}$ for $i>0$ (resp. $i<0$; $i=0$) can be uniquely expressed in the form $f(h)q^{i}$ (resp. $f(h)p^{i}$; $f(h)$), and hence each element in $A$ can be written uniquely as

[TABLE]

Define an automorphism $\omega$ of $A$ by letting

[TABLE]

Then we have  $\omega(h)=-qp=1-h$, and hence  $\omega(A_{i})=A_{-i}$  for all $i$.

In the following we draw the standard terminologies and notation from [5].

For $x\in A$, let

[TABLE]

where $\bar{\mathbb{F}}$ is an algebraic closure of $\mathbb{F}$. Then by [5, Corollary 6.7], the set $A-\mathbb{F}$ has the following partition:

[TABLE]

where

[TABLE]

It is clear that each $\Delta_{i}$ is invariant under every $\Phi\in\text{Aut}(A)$. An element $x$ is called nilpotent (resp. strictly nilpotent) if $x\in\Delta_{1}\cup\Delta_{2}$ (resp. $\Delta_{1}$). By definition it is easy to see that $x$ is nilpotent if and only if $N(x)\neq C(x)$.

Let $f=\sum\alpha_{ij}X^{i}Y^{j}\in\mathbb{F}[X,Y]$, and let $E(f)$ denote the set of pairs $(i,j)$ such that $\alpha_{ij}\neq 0$. For two real numbers $\rho,\sigma$ we denote

[TABLE]

where we assign $-\infty$ to be $v_{\rho,\sigma}(0)$. We denote by $E_{\rho,\sigma}(f)$ the set of pairs $(i,j)$ in $E(f)$ such that $i\rho+j\sigma=v_{\rho,\sigma}(f)$. Then $E_{\rho,\sigma}(f)$ is nonempty if $f\neq 0$. We say that $f$ is $(\rho,\sigma)$-homogeneous of $(\rho,\sigma)$-degree $v_{\rho,\sigma}(f)$ if $E(f)=E_{\rho,\sigma}(f)$.

Let $x=\sum\alpha_{ij}p^{i}q^{j}\in A$. Then we define $E(x),\ v_{\rho,\sigma}(x),\ E_{\rho,\sigma}(x)$ to be those for the polynomial $f=\sum\alpha_{ij}X^{i}Y^{j}$. In particular, the polynomial

[TABLE]

is called the $(\rho,\sigma)$-polynomial of $x$, and the element

[TABLE]

is called the $(\rho,\sigma)$-term of $x$.

Lemma 2.1**.**

([5, Lemma 2.4]) Let $x,y\in A$, and let $\rho,\sigma$ be real numbers such that $\rho+\sigma>0$. Then

(1) the $(\rho,\sigma)$-polynomial of $xy$ is the product of $(\rho,\sigma)$-polynomials of $x$ and $y$.

(2) $v_{\rho,\sigma}(xy)=v_{\rho,\sigma}(x)+v_{\rho,\sigma}(y).$

Let $x\in A$, and let $\rho,\sigma,k$ be three positive numbers. Then it is clear that

[TABLE]

It follows that $\rho_{1}/\sigma_{1}\neq\rho_{2}/\sigma_{2}$, if $E_{\rho_{1},\sigma_{1}}(x)\neq E_{\rho_{2},\sigma_{2}}(x)$, where $\rho_{1},\sigma_{1},\rho_{2},\sigma_{2}$ are all positive numbers.

We assume in the sequel that $(\rho,\sigma)\in\mathcal{P}$, where $\mathcal{P}$ is the set of all pairs of relatively prime positive integers. We call $E_{\rho,\sigma}(x)$ an edge of $x$ if it contains more than one point, and a vertex of $x$ if it is a singleton. Thus, $E_{\rho,\sigma}(x)$ is an edge if and only if the $(\rho,\sigma)$-polynomial of $x$ is not a monomial. If the intersection of two edges of $x$ is a vertex, we say that the vertex joins these two edges, and the two edges are adjacent. These terminologies can be geometrically described (see [6, Introduction]).

Example: Let $x=p^{4}+p^{3}q+p^{2}q^{2}+q^{3}+q$. Then $x$ has two edges

[TABLE]

and they are joined by a vertex $E_{2,3}(x)=\{(2,2)\}$.

Since $E(x)$ is finite set, $x$ can only have finitely many edges and vertices. It is also possible that $x$ has no edges. For example: $x=p^{2}q^{2}+pq+1$. Then $E_{\rho,\sigma}(x)=\{(2,2)\}$ for all $(\rho,\sigma)\in\mathcal{P}$.

3 Properties of solvable elements

Assume that $x\in A$ is solvable, and let $y\in A$ be an element such that $[x,y]=1$. Then we have $(\mathrm{ad}\ x)^{2}y=0$ and hence $y\in V(x)$. But $y\notin C(x)$. So we have

[TABLE]

and hence $x\notin\Delta_{5}$. Since $y\in N(x)$, so that $N(x)\neq C(x)$, we have $x\notin\Delta_{3}\cup\Delta_{4}$. Thus, we obtain the following corollary.

Corollary 3.1**.**

If $x\in A$ is solvable, then $x$ is nilpotent.

Lemma 3.2**.**

[5, Lemma 2.7]** Assume that $\rho$ and $\sigma$ are integers such that $\rho+\sigma>0$. Let $x$ and $y$ be elements in $A$, and let $f$ and $g$ be respectively their $(\rho,\sigma)$-polynomials. Then there exist elements $\mathfrak{t},u\in A$ such that

(a) $[x,y]=\mathfrak{t}+u$;

(b) $E(\mathfrak{t})=E(\mathfrak{t},\rho,\sigma)$ and $v_{\rho,\sigma}(\mathfrak{t})=v_{\rho,\sigma}(x)+v_{\rho,\sigma}(y)-(\rho+\sigma)$;

(c) $v_{\rho,\sigma}(u)<v_{\rho,\sigma}(x)+v_{\rho,\sigma}(y)-(\rho+\sigma)$.

In addition, the following conditions are equivalent:

(1) $\mathfrak{t}=0$;

(2) $g^{v_{\rho,\sigma}(x)}=cf^{v_{\rho,\sigma}(y)}$ for some nonzero scalar $c$.

Lemma 3.3**.**

Let $x\in A_{i}$ and $y\in A_{j}$ with $i>0$ and $j<0$. If  $[x,y]=0$, then either  $x=0$ or $y=0$.

Proof.

Suppose on the contrary that $x=f_{i}(h)q^{i}\neq 0$ and $y=f_{-j}(h)p^{j}\neq 0$, where $f_{i}(X)$ and $g_{-j}(X)$ are polynomials of degree respectively $k$ and $l$. Choose $\rho=\sigma=1$. Then we have

[TABLE]

and the $(\rho,\sigma)$-polynomials of $x$ and $y$ are respectively $cX^{k}Y^{k+i}$ and $c^{\prime}X^{l+j}Y^{l}$, where $c,c^{\prime}\in\mathbb{C}\setminus 0$.

Since $[x,y]=0$, we have by Lemma 3.2 that

[TABLE]

implying that $k/(k+i)=(l+j)/l$, a contradiction. Thus, we must have $x=0$ or $y=0$. ∎

Lemma 3.4**.**

Let $x\in A$ be solvable, let $y\in A$ be an element such that $[x,y]=1$, and let $(\rho,\sigma)\in\mathcal{P}$ be a pair such that $v_{\rho,\sigma}(x)\geq\rho+\sigma$. Write $[x,y]=\mathfrak{t}+u$ as in Lemma 3.2. Then $\mathfrak{t}=0$.

Proof.

By assumption we have $y\in A-\mathbb{F}$, which gives $v_{\rho,\sigma}(y)>0$.

Suppose on the contrary that $\mathfrak{t}\neq 0$. By Lemma 3.2 we have

[TABLE]

contrary to the fact that $v_{\rho,\sigma}(1)=0$. So we must have $\mathfrak{t}=0$. ∎

Lemma 3.5**.**

Let $x\in A_{i}$ with $i>0$. If $v_{\rho,\sigma}(x)\geq\rho+\sigma$ for some $(\rho,\sigma)\in\mathcal{P}$, then $x$ is unsolvable.

Proof.

Suppose on the contrary that $x$ is solvable. Let $y\in A$ be an element such that $[x,y]=1$. Since $1\in A_{0}$, we must have $y\in A_{-i}$.

Let $f$ and $g$ be respectively the $(\rho,\sigma)$-polynomials of $x$ and $y$. Write $[x,y]=\mathfrak{t}+u$ as in Lemma 3.2. Then we have by Lemma 3.4 that $\mathfrak{t}=0$. Using Lemma 3.2 once again we get

[TABLE]

which leads to a contradiction by a similar argument as that used in the proof of Lemma 3.3. Thus, $x$ is unsolvable. ∎

Lemma 3.6**.**

Let $x\in\Delta_{1}$ be a solvable element, and let $y\in A$ be an element such that $[x,y]=1$. Then $y\in\Delta_{1}$.

Proof.

Since $x\in\Delta_{1}$, by [5, Theorem 9.1] there exists $\Phi\in\text{Aut}(A)$ such that $\Phi(x)\in\mathbb{F}[q]$. Since $\Phi$ stabilizes $\Delta_{1}$, we may assume that $x\in\mathbb{F}[q]$.

Let

[TABLE]

and

[TABLE]

Since $[c_{i}q^{i},y_{j}]\in A_{i+j}$ for all $i,j$, we have

[TABLE]

where $u\in\sum_{i>d-s}A_{i}$.

If $s\neq d$, since $1\in A_{0}$, we have $[c_{d}q^{d},y_{-s}]=0$ and hence $y_{-s}=0$ by Lemma 3.3. It follows that $y=y_{-d}+\cdots+y_{0}+\cdots+y_{t}$.

If $d>1$, then since $v_{1,1}(x)\geq d\geq 2$, we have by Lemma 3.5 that $x$ is unsolvable, contrary to the assumption. Therefore, we must have $d=1$, which implies that $[x_{1},y_{-1}]=1$, so that $y_{-1}$ is solvable. Now Lemma 3.5 implies that $v_{1,1}(y_{-1})=1$. Thus, we have $y_{-1}=b_{-1}p$ for some $b_{-1}\in\mathbb{F}\setminus 0$ and hence

[TABLE]

If $t>1$, then $[x,y]=1$ implies that $[q^{n},f_{t}(h)q^{t}]=0$, which gives

[TABLE]

Since $A$ has no zero divisors, we have $f_{t}(h-n)-f_{t}(h)=0$, implying that $f_{t}(X)$ is a scalar (denoted by $b_{t}$). It follows that

[TABLE]

and hence $[x,y-y_{t}]=1$.

Apply the above discussion to $y-y_{t}$ in place of $y$. Continue this process. Then we obtain

[TABLE]

and hence $y=b_{-1}p+b_{0}+\cdots+b_{t}q^{t}$.

By [5, Theorem 8.10], the group $\text{Aut}(A)$ is generated by elements

[TABLE]

It follows that $\text{exp}(\text{ad}\ f(q))\in\text{Aut}(A)$ for any polynomial $f(X)\in\mathbb{F}[X]$ with $\text{deg}f(X)>0$.

Since $p\in\Delta_{1}$, we have

[TABLE]

This completes the proof. ∎

Lemma 3.7**.**

Let $x\in A-\mathbb{F}$, and let $u=c_{0}+c_{1}x+\cdots+c_{n}x^{n}\in\mathbb{F}[x]$ with $c_{n}\neq 0$. If $u$ is solvable, then $n=1$.

Proof.

Since $u$ is solvable, we have $C(u)=\mathbb{F}[u]$ by [7, Theorem 2.11]. Therefore $x$ is a polynomial of $u$, since $x\in C(u)$. It follows that if $n>1$, there exists a nonzero polynomial $f(X)\in\mathbb{F}[X]$ such that $f(x)=0$, contrary to [5, Proposition 2.5]. So we must have $n=1$.

∎

Theorem 3.8**.**

The following statements are equivalent:

(1) Each endomorphism of $A$ is an automorphism.

(2) Each solvable element in $A$ is strictly nilpotent.

(3) Each element in $\Delta_{2}$ is unsolvable.

Proof.

The equivalence of (2) and (3) is immediate from Corollary 3.1.

$(1)\Rightarrow(2)$: Let $x\in A$ be solvable, and let $y\in A$ be an element such that $[x,y]=1$. Then there exists a unique endomorphism $\varphi$ of $A$ such that $\varphi(p)=x$ and $\varphi(q)=y$. Hence $\varphi\in\text{Aut}(A)$ by (1), which shows that $x$ is strictly nilpotent since $p$ is so.

$(2)\Rightarrow(1)$: Let $\varphi$ be an endomorphism of $A$. Then $\varphi(q)$ is solvable and hence $\varphi(q)\in\Delta_{1}$ by (2). By [5, Corollary 6.7 (1)], we have $N(\varphi(q))=A$. Then [5, Lemma 8.9 (1)] says that there is $\Phi\in\text{Aut}(A)$ such that $\Phi\varphi(q)\in\mathbb{F}[q]$. Hence we have by Lemma 3.7 that

[TABLE]

Since  $[\Phi\varphi(p),\Phi\varphi(q)]=[p,q]=1$, the proof of Lemma 3.6 shows that

[TABLE]

for some $t\in\mathbb{N}$. Clearly we have $b_{-1}=c_{1}^{-1}$.

Let

[TABLE]

Then we have $\Phi\varphi(p)=\Phi^{\prime}(b_{-1}p)$ and hence $\Phi^{\prime-1}\Phi\varphi(p)=b_{-1}p$. Since

[TABLE]

we have

[TABLE]

It is easily seen that the endomorphism $\Phi^{\prime-1}\Phi\varphi$ (of $A$) is indeed an automorphism. Therefore, $\varphi\in\text{Aut}(A)$. ∎

Let us see what Lemma 3.7 implies. Recall the partition: $A-\mathbb{F}=\cup^{5}_{i=1}\Delta_{i}$. Given a polynomial $f(X)\in\mathbb{F}[X]$ with $\text{deg}f(X)>1$, set

[TABLE]

By Dixmier’s problem 6 (proved in [8, Theorem 1.4] and [3, p.4]), we have $f(\Delta_{5})\subseteq\Delta_{5}$. Therefore, $f(x)$ is unsolvable for $x\in\Delta_{5}$.

Let $x\in A-\mathbb{F}$. Then we have $C(x)=C(f(x))$ by [5, Corollary 4.5], whereas [5, Proposition 10.3] says that $N(x)=N(f(x))$. Therefore $x$ is nilpotent (resp. strictly nilpotent) if and only if $f(x)$ is nilpotent (resp. strictly nilpotent); that is (see [3, p.3]),

[TABLE]

Thus, Lemma 3.7 tells us that $f(\Delta_{2})$ is a subset of $\Delta_{2}$ consisting of unsolvable elements. In view of Theorem 3.8, this reduces the solution to the Dixmier’s open question to determining the solvability of elements in the set

[TABLE]

4 Reduction theorems

In this section we establish two reduction theorems which give sufficient conditions for an element in $A$ being unsolvable.

First, we determine the solvability of  $x\in A$  if $v_{\rho,\sigma}(x)<\rho+\sigma$ for some $(\rho,\sigma)\in\mathcal{P}$. In view of $\omega\in\text{Aut}(A)$, it suffices to assume that $\rho\geq\sigma$.

By the Division Algorithm, there exists unique integers $l$ and $r$ such that $\rho=l\sigma+r$, and $0\leq r<\sigma$. Then $x$ is of the form

[TABLE]

If $c_{-1}\neq 0$, then we have

[TABLE]

and hence $x$ is solvable since $c_{-1}p$ is so; if $c_{-1}=0$, so that $x$ is a polynomial of $q$, then the solvability of $x$ is given by Lemma 3.7.

In the sequel we assume that $v_{\rho,\sigma}(x)\geq\rho+\sigma$ for all $(\rho,\sigma)\in\mathcal{P}$.

4.1 The first reduction theorem

We give a sufficient condition for an element in $A$ being unsolvable in this subsection. The first lemma says that the majority of non-negatively graded elements are unsolvable.

Lemma 4.1**.**

Let $x=x_{k}+x_{k+1}+\cdots+x_{n}\in A,\ x_{i}\in A_{i}$. If $k>1$, then $x$ is unsolvable.

Proof.

Suppose on the contrary that $x$ is solvable. Let

[TABLE]

be an element such that $[x,y]=1$.

Choose $\rho=\sigma=1$. Recall from Section 2 that each element in $A_{i}$ is of the form $f_{i}(h)q^{i}$ if $i\geq 0$ or $f_{i}(h)p^{i}$ if $i<0$, so that its $(\rho,\sigma)$-degree is at least $|i|$ by Lemma 2.1.

It is no loss to assume that $x_{k}\neq 0$. First, we claim that $y_{-s}=0$ if $s>0$.

Suppose that $y_{-s}\neq 0$ and $s>0$. Write

[TABLE]

If $s\neq k$, then the fact that $[x_{k},y_{-s}]\in A_{k-s}$ and $1\in A_{0}$ yields $[x_{k},y_{-s}]=0$, contrary to Lemma 3.3. If $s=k$, then since $[x_{k},y_{-k}]\in A_{0}$ and $y^{\prime}\in\sum_{i>0}A_{i}$, we have $[x_{k},y_{-k}]=1$, so that $x_{k}$ is solvable. This contradicts Lemma 3.5 since

[TABLE]

Then the claim holds. It follows that

[TABLE]

implying that $[x,y]\in\sum_{i>0}A_{i}$, contrary to the fact that $1\in A_{0}$.

Therefore, $x$ is unsolvable. ∎

By applying the automorphism $\omega$ of $A$, we obtain that an element of the form

[TABLE]

is also unsolvable.

Define the height of an element $x=\sum x_{i},\ x_{i}\in A_{i}$ by  $\text{ht}(x)=\text{max}\{i|x_{i}\neq 0\}$.

Theorem 4.2**.**

Let $x=x_{s}+x_{s+1}+\cdots+x_{t}\in A$, $x_{i}\in A_{i}$, $s,t\in\mathbb{Z}$. If $x_{t}\neq 0$, $t>1$, and $C(x_{t})=\mathbb{F}[x_{t}]$ or, if $x_{s}\neq 0$, $s<-1$, and $C(x_{-s})=\mathbb{F}[x_{-s}]$, then $x$ is unsolvable.

Proof.

In view of the automorphism $\omega$ of $A$, it suffices to prove the theorem under the assumptions $x_{t}\neq 0$, $t>1$, and $C(x_{t})=\mathbb{F}[x_{t}]$.

Suppose on the contrary that $x$ is solvable. Then there is

[TABLE]

such that $[x,y]=1$. Now that $y$ is solvable, we have $\text{ht}(y)=t^{\prime}\geq-1$ by the conclusion following Lemma 4.1.

Write

[TABLE]

Since $t+t^{\prime}\geq 1$ and $1\in A_{0}$, we obtain $[x_{t},y_{t^{\prime}}]=0$ and hence $y_{t^{\prime}}\in C(x_{t})=\mathbb{F}[x_{t}]$. Let $f(X)\in\mathbb{F}[X]$ be the polynomial such that $y_{t^{\prime}}=f(x_{t})$. Since $x_{t}^{k}\in A_{tk}$ for all $k\in\mathbb{N}$, we have

[TABLE]

and hence $y_{t^{\prime}}=cx_{t}^{n}$, implying that $\text{ht}(y)=t^{\prime}=nt\geq 0$.

If $t^{\prime}=0$, so that $n=0$ since $t>1$, then $y_{0}=c\in\mathbb{F}$. It follows that

[TABLE]

a contradiction.

If $t^{\prime}>0$, since

[TABLE]

we have $\text{ht}(y-f(x))<t^{\prime}=\text{ht}(y)$. Since $[x,y-f(x)]=1$, the above discussion applies to $y-f(x)$ in place of $y$ as long as $\text{ht}(y-f(x))>0$. Continue this process. Then we will obtain an element $y^{\prime}$ such that $[x,y^{\prime}]=1$ and $\text{ht}(y^{\prime})\leq 0$, a contradiction.

Thus, $x$ must be unsolvable. ∎

4.2 The second reduction theorem

In this subsection we establish the second sufficient condition for $x$ being unsolvable.

Lemma 4.3**.**

([5, Proposition 7.4]) Let $\rho,\sigma$ be positive integers such that $\rho\nmid\sigma$ and $\sigma\nmid\rho$. Let $x\in A$ with $v_{\rho,\sigma}(x)>\rho+\sigma$. If the $(\rho,\sigma)$-polynomial of $x$ is not a monomial, then $x\in\Delta_{5}$.

By the lemma, if an element $x\in A$ is solvable, so that $x\notin\Delta_{5}$, and if $x$ has an edge $E_{\rho,\sigma}(x)$ with $(\rho,\sigma)\in\mathcal{P}$, then the assumption $v_{\rho,\sigma}(x)>\rho+\sigma$ yields either $\rho|\sigma$ or $\sigma|\rho$; that is, either $\rho=1$ or $\sigma=1$.

In view of the automorphism $\omega$ of $A$, we assume that $\rho=n\geq 1$ and $\sigma=1$ in the remainder of this subsection.

Lemma 4.4**.**

Let $f,g\in\mathbb{F}[X,Y]$ be two $(\rho,\sigma)$-homogeneous polynomials with $(\rho,\sigma)$-degrees respectively $w$ and $v$. Assume that $f$ is not equal to $f_{1}^{m}$ ($m>1$) for any polynomial $f_{1}\in\mathbb{F}[X,Y]$. If $f^{v}=g^{w}$, then $w|v$ and hence $g=f^{v/w}$.

Proof.

Since $\rho=n$ and $\sigma=1$, the $(\rho,\sigma)$-degree of a $(\rho,\sigma)$-homogeneous polynomial is divisible by $n$. Assume that $w=w_{1}n$ and $v=v_{1}n$ for some integers $v_{1}$ and $w_{1}$. Then we get

[TABLE]

with $s_{1},\dots,s_{k}$, $r_{1},\dots r_{l}\in\mathbb{N}\setminus 0$ and $c,c^{\prime}\in\mathbb{F}\setminus 0$, where $\lambda_{1},\dots,\lambda_{k}$ (resp. $\mu_{1},\dots,\mu_{l}$) are distinct numbers in $\bar{\mathbb{F}}$.

Since $f^{v}=g^{w}$, we have $k=l$ and, by an rearrangement of indices, $\lambda_{i}=\mu_{i}$ for all $i$. It follows that $s_{i}v=r_{i}w$ for $1\leq i\leq k$.

Since $f$ is not equal to $f_{1}^{m}$ ($m>1$) for any polynomial $f_{1}$, we get $(s_{1},\dots,s_{k})=1$. Then since $w$ divides $s_{i}v$ for all $i$,  $w$ divides

[TABLE]

implying that $g=f^{v/w}$, as desired. ∎

Theorem 4.5**.**

Let $x\in A$ be an element such that $v_{\rho,\sigma}(x)\geq\rho+\sigma=n+1$, and let $f$ be its $(\rho,\sigma)$-polynomial. If $f$ is not equal to $f_{1}^{m}$ ($m>1$) for any $f_{1}\in\mathbb{F}[X,Y]$, then $x$ is unsolvable.

Proof.

Suppose on the contrary that $x$ is solvable. Let $y\in A$ be an element such that $[x,y]=1$, and let $g$ be its $(\rho,\sigma)$-polynomial.

Write $[x,y]=\mathfrak{t}+u$ as in Lemma 3.2. Then we have $\mathfrak{t}=0$ by Lemma 3.4. Hence, Lemma 3.2 says that

[TABLE]

so that $g=c_{l}f^{l}$ ($c_{l}\neq 0$) by Lemma 4.4, where $l=v_{\rho,\sigma}(y)/v_{\rho,\sigma}(x)\in\mathbb{N}\setminus 0$.

Note that

[TABLE]

But $v_{\rho,\sigma}(y-c_{l}x^{l})<v_{\rho,\sigma}(y)$, since $v_{\rho,\sigma}(y)-v_{\rho,\sigma}(c_{l}x^{l})=g-c_{l}f^{l}=0$.

Apply the above discussion to $y-c_{l}x^{l}$ in place of $y$, if $y-c_{l}x^{l}$ is not a scalar. Continue the process. Since $\rho>0$ and $\sigma>0$, so that the $(\rho,\sigma)$-degree of every element in $A-\mathbb{F}$ is positive, ultimately we will obtain a polynomial $h(X)\in\mathbb{F}[X]$ such that $y-h(x)$ is a scalar. It follows that $[x,y-h(x)]=0$, contrary to the fact that $[x,y]=[x,y-h(x)]$.

Therefore, $x$ must be unsolvable. ∎

5 A geometric characterization of solvable elements

In this section we give a necessary condition for an element $x\in A$ being solvable and having more than one edges.

Lemma 5.1**.**

Let  $i,j,i_{0},j_{0},a,b$ be real numbers such that $a<b$. If  $ia+j\leq i_{0}a+j_{0}$  and  $ib+j<i_{0}b+j$, then  $iX+j<i_{0}X+j_{0}$ for all  $X\in(a,b)$.

Proof.

Define a function

[TABLE]

Then the graph of $Y=f(X)$ is a line, on which the segment with $X$-coordinates between $a$ and $b$ lies below the $X$-axis, since $f(a)\leq 0$ and $f(b)<0$ by assumption. Thus, we have $f(X)<0$ and hence $iX+j<i_{0}X+j_{0}$  for all $X\in(a,b)$. . ∎

Let $x=\sum\alpha_{ij}p^{i}q^{j}\in A$, and let $E_{\rho_{1},\sigma_{1}}(x)$ and $E_{\rho_{2},\sigma_{2}}(x)$, $(\rho_{1},\sigma_{1}),(\rho_{2},\sigma_{2})\in\mathcal{P}$,  be two distinct edges of $x$. Then we have the following lemma.

Lemma 5.2**.**

Then intersection $E_{\rho_{1},\sigma_{1}}(x)\cap E_{\rho_{2},\sigma_{2}}(x)$ contains at most one pair of integers.

Proof.

Recall from Section 2 that $\rho_{1}/\sigma_{1}\neq\rho_{2}/\sigma_{2}$. If there are two pair of integers $(i_{0},j_{0}),(i,j)$ contained in $E_{\rho_{1},\sigma_{1}}(x)\cap E_{\rho_{2},\sigma_{2}}(x)$, then we have

[TABLE]

so that

[TABLE]

The assumption $\rho/\sigma_{1}\neq\rho_{2}/\sigma_{2}$ implies that $i-i_{0}=j-j_{0}=0$, and hence $(i,j)=(i_{0},j_{0})$. This completes the proof. ∎

Let $x=\sum\alpha_{ij}p^{i}q^{j}\in A$, and assume that

[TABLE]

are two adjacent edges of $x$ joined by a vertex $(i_{0},j_{0})$. By definition, the $(\rho_{1},\sigma_{1})$-term of $x$ is of the form

[TABLE]

and the $(\rho_{2},\sigma_{2})$-term of $x$ is of the form

[TABLE]

Lemma 5.3**.**

(see [6, Proposition 3.7(3)]) There exists $(\rho,\sigma)\in\mathcal{P}$ such that the $(\rho,\sigma)$-term of $x$ is $\alpha_{i_{0},j_{0}}p^{i_{0}}q^{j_{0}}.$

Proof.

Without loss of generality we assume that $\rho_{1}/\sigma_{1}<\rho_{2}/\sigma_{2}$. Let $(\rho,\sigma)\in\mathcal{P}$ be pair such that $\rho/\sigma\in(\rho_{1}/\sigma_{1},\rho_{2}/\sigma_{2})$. We show that $(\rho,\sigma)$ is a desired pair.

By definition we need to show that

[TABLE]

Let $(i,j)\in E(x)\setminus(i_{0},j_{0})$. Then since $(i_{0},j_{0})$ joins $E_{\rho_{1},\sigma_{1}}(x)$ and $E_{\rho_{2},\sigma_{2}}(x)$, meaning that $i_{0}\rho_{l}+j_{0}\sigma_{l}=v_{\rho_{l},\sigma_{l}}(x)$ for $l=1,2$, we have by Lemma 5.2

[TABLE]

Without loss of generality we assume that

[TABLE]

or, equivalently,

[TABLE]

Then Lemma 5.1 says that $i\rho/\sigma+j<i_{0}\rho/\sigma+j_{0}$ and hence $i\rho+j\sigma<i_{0}\rho+j_{0}\sigma$. Therefore, the $(\rho,\sigma)$-term of $x$ is $\alpha_{ij}p^{i_{0}}q^{j_{0}}$.

∎

Note: The above proof shows that for any pair $(\rho,\sigma)\in\mathcal{P}$ such that

[TABLE]

the $(\rho,\sigma)$-term of $x$ is the same.

Lemma 5.4**.**

Keep the assumptions before Lemma 5.3 on $x$. In addition, assume that $v_{\rho,\sigma}(x)\geq\rho+\sigma$ for all $(\rho,\sigma)\in\mathcal{P}$. If $x$ is solvable with $y\in A$ such that $[x,y]=1$, then $y$ also has adjacent edges $E_{\rho_{1},\sigma_{1}}(y)$, $E_{\rho_{2},\sigma_{2}}(y)$.

Proof.

By our assumptions, the $(\rho_{1},\sigma_{1})$-polynomial and the $(\rho_{2},\sigma_{2})$-polynomial of $x$ are respectively

[TABLE]

and

[TABLE]

Since $x$ is solvable, by Theorem 4.5 we may write these two polynomials respectively as $f_{1}^{r_{1}}$ ($r_{1}>1$) and $f_{2}^{r_{2}}$ ($r_{2}>1$) for some $f_{1},f_{2}\in\mathbb{F}[X,Y]$. We may further assume that, for $l=1,2$, $f_{l}$ is no longer equal to $\kappa_{l}^{k}$ ($k>1$) for any $\kappa_{l}\in\mathbb{F}[X,Y]$. Since $E_{\rho_{1},\sigma_{1}}(x)$ and $E_{\rho_{2},\sigma_{2}}(x)$ are edges of $x$, neither $f_{1}$ nor $f_{2}$ is a monomial.

For each pair $(\rho_{l},\sigma_{l})$, $l=1,2$, write $[x,y]=\mathfrak{t}+u$ as in Lemma 3.2. Then we have $\mathfrak{t}=0$ by Lemma 3.4. Hence, by Lemma 3.2 and Lemma 4.4 the $(\rho_{l},\sigma_{l})$-polynomial of $y$ is of the form

[TABLE]

Thus, $y$ has the edges $E_{\rho_{1},\sigma_{1}}(y)$ and $E_{\rho_{2},\sigma_{2}}(y)$, which we are to show are adjacent.

Let $(\rho,\sigma)\in\mathcal{P}$ be a pair of integers such that $\rho/\sigma\in(\rho_{1}/\sigma_{1},\rho_{2}/\sigma_{2})$. By the proof of Lemma 5.3, the $(\rho,\sigma)$-polynomial of $x$ is $\alpha_{i_{0},j_{0}}X^{i_{0}}Y^{j_{0}}$. Let $y=\sum\beta_{ij}p^{i}q^{j}$ and let $\tau$ be its $(\rho,\sigma)$-polynomial. Write $[x,y]=\mathfrak{t}+u$ as in Lemma 3.2. Then we have $\mathfrak{t}=0$ by Lemma 3.4. Hence Lemma 3.2 says that

[TABLE]

Therefore $\tau$ is a monomial, which we denote by $\beta_{st}X^{s}Y^{t}$. Then

[TABLE]

implying that $s/t=i_{0}/j_{0}$, so that $(s,t)=(Ki_{0},Kj_{0})$ for some $K>0$. Since $\tau$ is the $(\rho,\sigma)$-polynomial of $y$, we have

[TABLE]

which implies that

[TABLE]

so that $(s,t)$ is independent of the choice of $(\rho,\sigma)$ as above. In other words, the element $\beta_{st}p^{s}q^{t}$ is the $(\rho,\sigma)$-term of $y$ for any $(\rho,\sigma)\in\mathcal{P}$ such that $\rho/\sigma\in(\rho_{1}/\sigma_{1},\rho_{2}/\sigma_{2})$.

We now show that the vertex $(s,t)$ joins $E_{\rho_{1},\sigma_{1}}(y)$ and $E_{\rho_{2},\sigma_{2}}(y)$.

Choose a sequence of pairs $(\rho_{(n)},\sigma_{(n)})\in\mathcal{P},\ n=1,2,\dots$, such that

[TABLE]

For each $n$, since $\beta_{st}p^{s}q^{t}$ is the $(\rho_{(n)},\sigma_{(n)})$-term of $y$, we have

[TABLE]

and hence

[TABLE]

Taking the limit as $n\rightarrow\infty$, we get

[TABLE]

so that

[TABLE]

implying that $(s,t)\in E_{\rho_{1},\sigma_{1}}(y)$.

Similarly we obtain $(s,t)\in E_{\rho_{2},\sigma_{2}}(y)$. Then Lemma 5.2 says that

[TABLE]

so that the edges $E_{\rho_{1},\sigma_{1}}(y)$ and $E_{\rho_{2},\sigma_{2}}(y)$ are adjacent. ∎

Let $x\in A$ be an element having edges

[TABLE]

Then $\rho_{1}/\sigma_{1},\dots,\rho_{k}/\sigma_{k}$ are $k$ distinct rational numbers. It’s no loss of generality to assume that

[TABLE]

For each $l$, $1\leq l<k$, take $(\rho,\sigma)\in\mathcal{P}$ such that $\rho/\sigma\in(\rho_{l}/\sigma_{l},\ \rho_{l+1}/\sigma_{l+1})$.  Then $E_{\rho,\sigma}(x)$ is no longer an edge, hence it must be a vertex which, by a similar argument as in the proof of Lemma 5.4, joins $E_{\rho_{l},\sigma_{l}}(x)$ and $E_{\rho_{l+1},\sigma_{l+1}}(x)$. Therefore, the edges $E_{\rho_{l},\sigma_{l}}(x)$ and $E_{\rho_{l+1},\sigma_{l+1}}(x)$ are adjacent.

We claim that $E_{\rho_{l},\sigma_{l}}(x)$ and $E_{\rho_{l+j},\sigma_{l+j}}(x)$ are not adjacent if $j>1$. Suppose on the contrary that they are adjacent. Then by the proof of Lemma 5.3, $E_{\rho,\sigma}(x)$ is a vertex for any $(\rho,\sigma)\in\mathcal{P}$ such that $\rho/\sigma\in(\rho_{l}/\sigma_{l},\rho_{l+j}/\sigma_{l+j})$, contrary to the fact that $E_{\rho_{l+1},\sigma_{l+1}}(x)$ is an edge. Therefore, two distinct edges

[TABLE]

are adjacent if and only if $i=j-1$.

We are now ready to study solvable elements in $A$ using these conclusions.

In the following, assume that $x\in A$ is a solvable element having edges

[TABLE]

where $\rho_{1}/\sigma_{1}<\cdots<\rho_{k}/\sigma_{k}$. We also assume that

[TABLE]

Then by Theorem 4.5, for each $l$, $1\leq l\leq k$, the $(\rho_{l},\sigma_{l})$-polynomial of $x$ is of the form $f_{l}^{r_{l}}$ ($r_{l}>1$) for some $f_{l}\in\mathbb{F}[X,Y]$. We assume that each $f_{l}$ is no longer equal to $\kappa^{n}$ ($n>1$) for any polynomial $\kappa\in\mathbb{F}[X,Y]$.

Lemma 5.5**.**

With these assumptions on $x$, we have

[TABLE]

Proof.

It suffices to show that $(r_{1},r_{2})>1$. The other inequalities can be proved similarly.

We assume that $(r_{1},r_{2})=1$ and derive a contradiction.

Let $x=\sum\alpha_{ij}p^{i}q^{j}$, and let $(i_{0},j_{0})$ be the vertex of $x$ joining $E_{\rho_{1},\sigma_{1}}(x)$ and $E_{\rho_{2},\sigma_{2}}(x)$. Then we have

[TABLE]

and

[TABLE]

Let $y\in A$ be an element such that $[x,y]=1$. By Lemma 5.4, $y$ also has adjacent edges $E_{\rho_{l},\sigma_{l}}(y)$, $l=1,2$. In particular, the proof of Lemma 5.4 shows that, for $l=1,2$, the $(\rho_{l},\sigma_{l})$-polynomial of $y$ is also of the form $c_{l}f_{l}^{R_{l}}$ for some $R_{l}\in\mathbb{N}\setminus 0$ and $c_{l}\in\mathbb{F}\setminus 0$.

Let $(s,t)$ be the vertex joining $E_{\rho_{1},\sigma_{1}}(y)$ and $E_{\rho_{2},\sigma_{2}}(y)$. Then we have

[TABLE]

and

[TABLE]

From the proof of Lemma 5.4 we also obtain $i_{0}/j_{0}=s/t$.

For brevity, set

[TABLE]

Since both $x$ and $y$ are solvable, so that $x,y\in A-\mathbb{F}$,   $v_{1}$ and $v_{2}$ are both nonzero.

From above we have

[TABLE]

implying that

[TABLE]

and hence

[TABLE]

Write the rational number $\theta$ as $m/n$ such that $(m,n)\in\mathcal{P}$. Then we get

[TABLE]

and hence  $n|r_{1},\ n|r_{2}$,  since $(n,m)=1$. It follows that $n=1$, which yields

[TABLE]

Set $y^{\prime}=y-c_{1}x^{m}$. Then since $c_{1}f_{1}^{R_{1}}-c_{1}(f_{1}^{r_{1}})^{m}=0$, the $(\rho_{1},\sigma_{1})$-degree of the $(\rho_{1},\sigma_{1})$-polynomial of $y^{\prime}$ is strictly less than that of $y$; that is, $v_{\rho_{1},\sigma_{1}}(y^{\prime})<v_{\rho_{1},\sigma_{1}}(y)$. Since $[x,y^{\prime}]=1$, the above discussion applies to $y^{\prime}$ in place of $y$ unless $y^{\prime}$ is a scalar. Continue this process. Then we will obtain that $y$ is a polynomial of $x$, so that $[x,y]=0$, a contradiction.

Therefore, we must have $(r_{1},r_{2})>1$, as desired. ∎

Theorem 5.6**.**

Keep the assumptions on $x$ before Lemma 5.5. Then

[TABLE]

Proof.

The case $k=2$ is immediate from Lemma 5.5. Thus we assume that $k>2$.

We assume that $(r_{1},r_{2},\dots,r_{k})=1$ and derive a contradiction.

Let $y\in A$ be an element such that $[x,y]=1$. Then Lemma 5.4 says that $y$ also has $k$ edges

[TABLE]

such that, for each $l$, $1\leq l<k$, the edges $E_{\rho_{l},\sigma_{l}}(y)$ and $E_{\rho_{l+1},\sigma_{l+1}}(y)$ are joined by a vertex which we denote now by $(s_{l},t_{l})$. In addition, from the proof of Lemma 5.5 we have that, for $1\leq l\leq k$, the $(\rho_{l},\sigma_{l})$-polynomial of $y$ equals  $c_{l}f_{l}^{R_{l}}$ for some $R_{l}\in\mathbb{N}\setminus 0$ and $c_{l}\in\mathbb{F}\setminus 0$.

Set $v_{l}=:v_{\rho_{l},\sigma_{l}}(f_{l})$ for $1\leq l\leq k$. In view of the proof of Lemma 5.5, we have $v_{l}\neq 0$ for all $l$.

Since for all $l$ with $1\leq l<k$, $(i_{l},j_{l})$ joins the edges $E_{\rho_{l},\sigma_{l}}(x)$, $E_{\rho_{l+1},\sigma_{l+1}}(x)$, and $(s_{l},t_{l})$ joins the edges $E_{\rho_{l},\sigma_{l}}(y)$, $E_{\rho_{l+1},\sigma_{l+1}}(y)$, we have

[TABLE]

and

[TABLE]

From the proof of Lemma 5.5 we have  $s_{l}/i_{l}=t_{l}/j_{l}$  for $1\leq l<k$. Then by the identities $i_{1}\rho_{2}+j_{1}\sigma_{2}=i_{2}\rho_{2}+j_{2}\sigma_{2}$ obtained from (1) and $s_{1}\rho_{2}+t_{1}\sigma_{2}=s_{2}\rho_{2}+t_{2}\sigma_{2}$ obtained from (2), we have

[TABLE]

Inductively we obtain

[TABLE]

Denote this rational number by $m/n$ such that $(m,n)\in\mathcal{P}$. By comparing the identities (1) with the identities (2) we obtain

[TABLE]

Then the assumption $(r_{1},\dots,r_{k})=1$ leads to a contradiction by a similar argument as that used in the proof of Lemma 5.5. Therefore, we must have $(r_{1},\dots,r_{k})>1,$ as desired. ∎