The Waring problem for upper triangular matrix algebras

Abstract

Our goal of the paper is to investigate the Waring problem for upper triangular matrix algebras, which gives a complete solution of a conjecture proposed by Panja and Prasad in 2023.

1 Introduction

The classical Waring problem proposed by Edward Waring in 1770 asserted that for every positive integer k there exists a positive integer $g(k)$ such that every positive integer can be expressed as a sum of $g(k)$ kth powers of nonnegative integers. In 1909, David Hilbert solved the problem. Various extensions and variations of this problem have been studied by different groups of mathematicians (see [2–4, 9, 11, 13, 14, 16, 17, 18]).

In 2009, Shalev [18] proved that given a word $w\neq 1$ , every element in any finite non-abelian simple group G of sufficiently high order can be written as the product of three elements from $w(G)$ , the image of the word map induced by w. In 2011, Larsen, Shalev, and Tiep [14] proved that, under the same assumptions, every element in G is the product of two elements from $w(G)$ , which gave a definitive solution of the Waring problem for finite simple groups.

Let $n\geq 2$ be an integer. Let K be a field, and let $K\langle X\rangle $ be the free associative algebra over K, freely generated by the countable set $X=\{x_1,x_2,\ldots \}$ of noncommutative variables. We refer to the elements of $K\langle X\rangle $ as polynomials.

Let $p(x_1,\ldots ,x_m)\in K\langle X\rangle $ . Let $\mathcal {A}$ be an algebra over K. The set

$$\begin{align*}p(\mathcal{A})=\{p(a_1,\ldots,a_m)~|~a_1,\ldots,a_m\in \mathcal{A}\} \end{align*}$$

is called the image of p (on $\mathcal {A}$ ).

In 2020, Brešar [2] initiated the study of various Waring’s problems for matrix algebras. He proved that if $\mathcal {A}=M_n(K)$ , where $n\geq 2$ and K is an algebraically closed field with characteristic $0$ , and f is a noncommutative polynomial which is neither an identity nor a central polynomial of $\mathcal {A}$ , then every trace zero matrix in $\mathcal {A}$ is a sum of four matrices from $f(\mathcal {A})-f(\mathcal {A})$ [2, Corollary 3.19]. In 2023, Brešar and Šemrl [3] proved that any traceless matrix can be written as sum of two matrices from $f(M_n(\mathcal {C}))-f(M_n(\mathcal {C}))$ , where $\mathcal {C}$ is the complex field and f is neither an identity nor a central polynomial for $M_n(\mathcal {C})$ . Recently, they [4] have proved that if $\alpha _1, \alpha _2,\alpha _3\in \mathcal {C}\setminus \{0\}$ and $\alpha _1+\alpha _2+\alpha _3=0$ , then any traceless matrix over $\mathcal {C}$ can be written as $\alpha _1A_1+\alpha _2A_2+\alpha _3A_3$ , where $A_i\in f(M_n(\mathcal {C}))$ .

By $T_n(K),$ we denote the set of all $n\times n$ upper triangular matrices over K. By $T_n(K)^{(0)}$ , we denote the set of all $n\times n$ strictly upper triangular matrices over K. More generally, if $t\geq 0$ , the set of all upper triangular matrices whose entries $(i,j)$ are zero, for $j- i\leq t$ , will be denoted by $T_n(K)^{(t)}$ . It is easy to check that $J^t=T_n(K)^{(t-1)}$ , where $t\geq 1$ and J is the Jacobson radical of $T_n(K)$ (see [1, Example 5.58]).

Let $p(x_1,\ldots ,x_m)$ be a noncommutative polynomial with zero constant term over K. We define its order as the least positive integer r such that $p(T_r(K))=\{0\}$ but $p(T_{r+1}(K))\neq \{0\}$ . Note that $T_1(K)=K$ . We say that p has order $0$ if $p(K)\neq \{0\}$ . We denote the order of p by ord $(p)$ . For a detailed introduction of the order of polynomials, we refer the reader to the book [7, Chapter 5].

In 2023, Panja and Prasad [16] discussed the image of polynomials with zero constant term and Waring-type problems on upper triangular matrix algebras over an algebraically closed field, which generalized two results in [6, 19]. More precisely, they obtained the following main result.

Theorem 1.1 [16, Theorem 5.18]

Let $n\geq 2$ and $m\geq 1$ be integers. Let $p(x_1,\ldots ,x_m)$ be a polynomial with zero constant term in noncommutative variables over an algebraically closed field K. Set $r=$ ord $(p)$ . Then one of the following statements holds.

1. (i) Suppose that $r=0$ . We have that $p(T_n(K))$ is a dense subset of $T_n(K)$ (with respect to the Zariski topology).

2. (ii) Suppose that $r=1$ . We have that $p(T_n(K))=T_n(K)^{(0)}$ .

3. (iii) Suppose that $1<r<n-1$ . We have that $p(T_n(K))\subseteq T_n(K)^{(r-1)}$ , and equality might not hold in general. Furthermore, for every n and $r,$ there exists d such that each element of $T_n(K)^{(r-1)}$ can be written as a sum of d many elements from $p(T_n(K))$ .

4. (iv) Suppose that $r=n-1$ . We have that $p(T_n(K))=T_{n}(K)^{(n-2)}$ .

5. (v) Suppose that $r\geq n$ . We have that $p(T_n(K))=\{0\}$ .

They proposed the following conjecture.

Conjecture 1.1 [16, Conjecture]

Let $p(x_1,\ldots ,x_m)$ be a polynomial with zero constant term in noncommutative variables over an algebraically closed field K. Suppose ord ${(p)=r}$ , where $1<r<n-1$ . Then $p(T_n(K))+p(T_n(K))=T_n(K)^{(r-1)}$ .

We note that if p is a multilinear polynomial and K is an infinite field, then $p(T_n(K))=T_n(K)^{(r-1)}$ (see [8, 10, 15]).

In the present paper, we shall prove the following main result of the paper, which gives a complete solution of Conjecture 1.1.

Theorem 1.2 Let $n\geq 2$ and $m\geq 1$ be integers. Let $p(x_1,\ldots ,x_m)$ be a polynomial with zero constant term in noncommutative variables over an infinite field K. Suppose ord $(p)=r$ , where $1<r<n-1$ . We have that $p(T_n(K))+p(T_n(K))=T_n(K)^{(r-1)}$ . Furthermore, if $r=n-2$ , we have that $p(T_n(K))=T_n(K)^{(n-3)}$ .

We organize the paper as follows: In Section 2, we shall give some preliminaries. We shall modify some results in [5, 8, 12], which will be used in the proof of Theorem 1.2. In Section 3, we shall give the proof of Theorem 1.2 by using some new arguments (for example, compatible variables in polynomials and recursive polynomials).

2 Preliminaries

Let $\mathcal {N}$ be the set of all positive integers. Let $m\in \mathcal {N}$ . Let K be a field. Set $K^*=K\setminus \{0\}$ . For any $k\in \mathcal {N}$ , we set

$$\begin{align*}T_m^k=\left\{(i_1,\ldots,i_k)\in \mathcal{N}^k~|~1\leq i_1,\ldots,i_k\leq m\right\}. \end{align*}$$

Let $p(x_1,\ldots ,x_m)$ be a polynomial with zero constant term in noncommutative variables over K. We can write

(1) $$ \begin{align} p(x_1,\ldots,x_m)=\sum\limits_{k=1}^d\left(\sum\limits_{(i_1,i_2,\ldots,i_k)\in T_m^k}\lambda_{i_1i_2\dots i_k}x_{i_1}x_{i_2}\dots x_{i_k}\right), \end{align} $$

where $\lambda _{i_1i_2\dots i_k}\in K$ and d is the degree of p.

We begin with the following result, which is slightly different from [5, Lemma 3.2]. We give its proof for completeness.

Lemma 2.1 For any $u_i=(a_{jk}^{(i)})\in T_n(K)$ , $i=1,\ldots ,m$ , we set

$$\begin{align*}\bar{a}_{jj}=(a_{jj}^{(1)},\ldots,a_{jj}^{(m)}), \end{align*}$$

where $j=1,\ldots ,n$ . We have that

(2) $$ \begin{align} p(u_1,\ldots,u_m)=\left( \begin{array}{cccc} p(\bar{a}_{11}) & p_{12} & \ldots & p_{1n}\\ 0 & p(\bar{a}_{22}) & \ldots & p_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \ldots & p(\bar{a}_{nn}) \end{array} \right), \end{align} $$

where

$$\begin{align*}p_{st}=\sum\limits_{k=1}^{t-s}\left(\sum\limits_{\substack{s=j_1<j_2<\cdots <j_{k+1}=t\\ (i_1,\ldots,i_k) \in T_m^k}} p_{i_1\dots i_k}(\bar{a}_{j_1j_1},\ldots,\bar{a}_{j_{k+1}j_{k+1}})a_{j_1j_2}^{(i_1)}\dots a_{j_{k}j_{k+1}}^{(i_k)}\right) \end{align*}$$

for all $1\leq s<t\leq n$ , where $p_{i_1,\ldots ,i_k}(z_1,\ldots ,z_{m(k+1)})$ , $1\leq i_1,i_2,\ldots ,i_k\leq m$ , $k=1,\ldots , n-1$ , is a polynomial in commutative variables over K.

Proof Let $u_i=(a_{jk}^{(i)})\in T_n(K)$ , where $i=1,\ldots ,m$ . For any $1\leq i_1,\ldots ,i_k\leq m$ , we easily check that

$$\begin{align*}u_{i_1}\dots u_{i_k}=\left( \begin{array}{cccc} m_{11} & m_{12} & \ldots & m_{1n}\\ 0 & m_{22} & \ldots & m_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \ldots & m_{nn} \end{array} \right), \end{align*}$$

where

$$\begin{align*}m_{st}=\sum\limits_{s=j_1\leq j_2\leq \cdots \leq j_{k+1}=t}a_{j_1j_2}^{(i_1)}\dots a_{j_{k}j_{k+1}}^{(i_k)} \end{align*}$$

for all $1\leq s\leq t\leq n$ . It follows from (1) that

$$ \begin{align*} \begin{aligned} p(u_1,\ldots,u_m)&=\sum\limits_{k=1}^d\left(\sum\limits_{(i_1,\ldots,i_k)\in T_m^k}\lambda_{i_1\dots i_k}u_{i_1}\dots u_{i_k}\right)\\ &=\sum\limits_{k=1}^d\left(\sum\limits_{(i_1,\ldots,i_k)\in T_m^k}\lambda_{i_1\dots i_k}\left( \begin{array}{cccc} m_{11} & m_{12} & \ldots & m_{1n}\\ 0 & m_{22} & \ldots & m_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \ldots & m_{nn} \end{array} \right)\right)\\ &=\left( \begin{array}{cccc} p_{11} & p_{12} & \ldots & p_{1n}\\ 0 & p_{22} & \ldots & p_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \ldots & p_{nn} \end{array} \right), \end{aligned} \end{align*} $$

where

$$ \begin{align*} \begin{aligned} p_{st}&=\sum\limits_{k=1}^d\left(\sum\limits_{(i_1,\ldots,i_k)\in T_m^k}\lambda_{i_1\dots i_k}m_{st}\right)\\ &=\sum\limits_{k=1}^d\left(\sum\limits_{(i_1,\ldots,i_k)\in T_m^k}\lambda_{i_1\dots i_k}\left( \sum\limits_{s=j_1\leq j_2\leq \cdots \leq j_{k+1}=t}a_{j_1j_2}^{(i_1)}\dots a_{j_{k}j_{k+1}}^{(i_k)}\right)\right)\\ &=\sum\limits_{k=1}^{d}\left(\sum\limits_{\substack{s=j_1\leq j_2\leq \cdots \leq j_{k+1}=t\\ (i_1,\ldots,i_k)\in T_m^k}}\lambda_{i_1i_2\dots i_k}a_{j_1j_2}^{(i_1)}\dots a_{j_{k}j_{k+1}}^{(i_k)}\right), \end{aligned} \end{align*} $$

where $1\leq s\leq t\leq n$ . In particular,

$$ \begin{align*} \begin{aligned} p_{ss}&=\sum\limits_{k=1}^{d}\left(\sum\limits_{(i_1,\ldots,i_k)\in T_m^k}\lambda_{i_1i_2\dots i_k}a_{ss}^{(i_1)}\dots a_{ss}^{(i_k)}\right)\\ &=p(\bar{a}_{ss}) \end{aligned} \end{align*} $$

for all $s=1,\ldots ,n$ , and

$$ \begin{align*} \begin{aligned} p_{st}&=\sum\limits_{k=1}^{d}\left(\sum\limits_{\substack{s=j_1\leq j_2\leq \cdots \leq j_{k+1}=t\\ (i_1,\ldots,i_k)\in T_m^k}}\lambda_{i_1i_2\dots i_k}a_{j_1j_2}^{(i_1)}\dots a_{j_{k}j_{k+1}}^{(i_k)}\right)\\ &=\sum\limits_{k=1}^{t-s}\left(\sum\limits_{\substack{s=j_1<j_2<\dots <j_{k+1}=t\\ (i_1,\ldots,i_k)\in T_m^k}} p_{i_1i_2\dots i_k}(\bar{a}_{j_1j_1},\ldots,\bar{a}_{j_{k+1}j_{k+1}})a_{j_1j_2}^{(i_1)}\dots a_{j_{k}j_{k+1}}^{(i_k)}\right) \end{aligned} \end{align*} $$

for all $1\leq s<t\leq n$ , where $p_{i_1,\ldots ,i_k}(z_1,\ldots ,z_{m(k+1)})$ is a polynomial in commutative variables over K. This proves the result.

The following result will be used in the proof of our main result.

Lemma 2.2 Let $m\geq 1$ be an integer. Let $p(x_1,\ldots ,x_m)$ be a polynomial with zero constant term in noncommutative variables over K. Let $p_{i_1,\ldots ,i_k}(z_1,\ldots ,z_{m(k+1)})$ be a polynomial in commutative variables over K in (2), where $1\leq i_1,\ldots ,i_k\leq m$ , $1\leq k\leq n-1$ . Suppose that ord $(p)=r$ , $1<r<n-1$ . We have that:

1. (i) $p(K)=\{0\}$ .

2. (ii) $p_{i_1,\ldots ,i_k}(K)=\{0\}$ for all $1\leq i_1,\ldots ,i_k\leq m$ , where $k=1,\ldots ,r-1$ .

3. (iii) $p_{i_1^{\prime },\ldots ,i_r^{\prime }}(K)\neq \{0\}$ for some $1\leq i_1^{\prime },\ldots ,i_r^{\prime }\leq m$ .

Proof The statement (i) is clear. We now claim that the statement (ii) holds true. Suppose on the contrary that

$$\begin{align*}p_{i_1^{\prime}\dots i_s^{\prime}}(K)\neq \{0\} \end{align*}$$

for some $1\leq i_1^{\prime },\ldots ,i_s^{\prime }\leq m$ , where $1\leq s\leq r-1$ . Then there exist $\bar {b}_{j}\in K^m$ , where ${j=1,\ldots ,s+1}$ such that

$$\begin{align*}p_{i_1^{\prime}\dots i_s^{\prime}}(\bar{b}_{1},\ldots,\bar{b}_{s+1})\neq 0. \end{align*}$$

We take $u_i=(a_{jk}^{(i)})\in T_{s+1}(K)$ , $i=1,\ldots ,m$ , where

$$\begin{align*}\left\{ \begin{aligned} \bar{a}_{jj}&=\bar{b}_{j},\quad j=1,\ldots,s+1,\\ a_{k,k+1}^{(i_k^{\prime})}&=1,\quad k=1,\ldots,s,\\ a_{jk}^{(i)}&=0,\quad\mbox{otherwise}. \end{aligned} \right. \end{align*}$$

It follows from (2) that

$$\begin{align*}p_{1,s+1}=p_{i_1^{\prime}\dots i_s^{\prime}}(\bar{b}_{1},\ldots,\bar{b}_{s+1})\neq 0. \end{align*}$$

This implies that $p(T_{s+1}(K))\neq \{0\}$ , a contradiction. This proves the statement (ii).

We finally claim that the statement (iii) holds true. Note that $p(T_{1+r}(K))\neq \{0\}$ . Thus, we have that there exist $u_{i}=(a_{jk}^{(i)})\in T_{1+r}(K)$ , $i=1,\ldots ,m$ , such that

$$\begin{align*}p(u_1,\ldots,u_m)=(p_{st})\neq 0. \end{align*}$$

In view of the statement (ii), we get that

$$\begin{align*}p_{1,r+1}=\sum\limits_{\substack{1=j_1<j_2< \cdots <j_{r+1}=r+1\\ (i_1,\ldots,i_r)\in T_m^r}}p_{i_1i_2\dots i_r}(\bar{a}_{j_1j_1},\ldots,\bar{a}_{j_{r+1}j_{r+1}})a_{j_1j_2}^{(i_1)}\dots a_{j_{r}j_{r+1}}^{(i_r)}\neq 0. \end{align*}$$

This implies that $p_{i_1^{\prime },\ldots ,i_r^{\prime }}(K)\neq \{0\}$ for some $1\leq i_1^{\prime },\ldots ,i_r^{\prime }\leq m$ . This proves the statement (iii). The proof of the result is complete.

The following well-known result will be used in the proof of the rest results.

Lemma 2.3 [12, Theorem 2.19]

Let K be an infinite field. Let $f(x_1,\ldots ,x_m)$ be a nonzero polynomial in commutative variables over K. Then there exist $a_1,\ldots ,a_m\in K$ such that $f(a_1,\ldots ,a_m)\neq 0$ .

Lemma 2.4 Let $n,s$ be integers with $1\leq s\leq n$ . Let $p(x_1,\ldots ,x_s)$ be a nonzero polynomial in commutative variables over an infinite field K. We have that there exist $a_1,\ldots ,a_n\in K$ such that

$$ \begin{align*} p(a_{i_1},\ldots,a_{i_s})\neq 0 \end{align*} $$

for all $1\leq i_1<\cdots <i_s\leq n$ .

Proof We set

$$\begin{align*}f(x_1,\ldots,x_n)=\prod_{1\leq i_1<\cdots<i_s\leq n}p(x_{i_1},\ldots,x_{i_s}). \end{align*}$$

It is clear that $f\neq 0$ . In view of Lemma 2.3, we have that there exist $a_1,\ldots ,a_n\in K$ such that

$$\begin{align*}f(a_1,\ldots,a_n)\neq 0. \end{align*}$$

This implies that

$$\begin{align*}p(a_{i_1},\ldots,a_{i_s})\neq 0 \end{align*}$$

for all $1\leq i_1<\cdots <i_s\leq n$ . This proves the result.

The following technical result is a generalized form of [8, Lemma 2.11], which discusses compatible variables in polynomials.

Lemma 2.5 Let $t\geq 1$ . Let $U_i=\{i_1,\ldots ,i_s\}\subseteq \mathcal {N}$ , $i=1,\ldots ,t$ . Let $p_i(x_{i_1},\ldots ,x_{i_s})$ be a nonzero polynomial in commutative variables over an infinite field K, where $i=1,\ldots ,t$ . Then there exist $a_{k}\in K$ with $k\in \bigcup _{i=1}^tU_i$ such that

$$\begin{align*}p_i(a_{i_1},\ldots,a_{i_s})\neq 0 \end{align*}$$

for all $i=1,\ldots ,t$ .

Proof Without loss of generality, we assume that

$$\begin{align*}\{1,2,\ldots,n\}=\bigcup_{i=1}^tU_i. \end{align*}$$

We set

$$\begin{align*}f(x_1,\ldots,x_n)=\prod_{i=1}^{t}p_i(x_{i_1},\ldots,x_{i_s}). \end{align*}$$

It is clear that $f\neq 0$ . In view of Lemma 2.3, we have that there exist $a_1,\ldots ,a_n\in K$ such that

$$\begin{align*}f(a_{1},\ldots,a_n)\neq 0. \end{align*}$$

This implies that

$$\begin{align*}p_i(a_{i_1},\ldots,a_{i_s})\neq 0 \end{align*}$$

for all $i=1,\ldots ,t$ . This proves the result.

The following technical result will be used in the proof of the main result of the paper.

Lemma 2.6 Let $s\geq 1$ and $t\geq 2$ be integers. Let K be an infinite field. Let $a_{ij}\in K$ , where $1\leq i\leq t$ , $1\leq j\leq s$ with $a_{11}\in K^*$ and $b\in K^*$ . For any $2\leq i\leq t$ , there exists a nonzero element in $\{a_{i1},\ldots ,a_{is}\}$ . Then there exist $c_i\in K$ , $i=1,\ldots ,s$ , such that

$$\begin{align*}\left\{ \begin{aligned} a_{11}c_1+\cdots +a_{1s}c_s&=b;\\ a_{i1}c_1+\cdots +a_{is}c_s&\neq 0 \end{aligned} \right. \end{align*}$$

for all $i=2,\ldots ,t$ .

Proof Suppose first that $s=1$ . Note that $a_{i1}\in K^*$ , $i=1,\ldots ,t$ . Take $c_1=a_{11}^{-1}b$ . It is clear

$$\begin{align*}\left\{ \begin{aligned} a_{11}c_1&=b;\\ a_{i1}c_1&\neq 0 \end{aligned} \right. \end{align*}$$

for all $2\leq i\leq t$ . Suppose next that $s\geq 2$ . Suppose first that $a_{i1}\neq 0$ for all $i=2,\ldots ,t$ . We define the following polynomials:

$$\begin{align*}\left\{ \begin{aligned} f_1(x_2,\ldots,x_s)&=b-a_{12}x_2-\cdots -a_{1s}x_s;\\ f_i(x_2,\ldots,x_s)&=a_{i1}a_{11}^{-1}b+(a_{i2}-a_{i1}a_{11}^{-1}a_{12})x_2+\cdots +(a_{is}-a_{i1}a_{11}^{-1}a_{1s})x_s \end{aligned} \right. \end{align*}$$

for all $2\leq i\leq t$ . Since $b,a_{i1}\in K^*$ , $i=1,\ldots ,t$ , we note that $f_i\neq 0$ for all $i=1,\ldots ,t$ . In view of Lemma 2.5, we get that there exist $c_2,\ldots ,c_s\in K$ such that

$$\begin{align*}f_i(c_2,\ldots,c_s)\neq 0 \end{align*}$$

for all $i=1,\ldots ,t$ . This implies that

(3) $$ \begin{align} \left\{ \begin{aligned} b-a_{12}c_2-\cdots -a_{1s}c_s&\neq 0;\\ a_{i1}a_{11}^{-1}b+(a_{i2}-a_{i1}a_{11}^{-1}a_{12})c_2+\cdots +(a_{is}-a_{i1}a_{11}^{-1}a_{1s})c_s&\neq 0 \end{aligned} \right. \end{align} $$

for all $2\leq i\leq t$ . We set

$$\begin{align*}c_1=a_{11}^{-1}(b-a_{12}c_2-\cdots -a_{1s}c_s). \end{align*}$$

It follows from (3) that

$$\begin{align*}\left\{ \begin{aligned} a_{11}c_1+\cdots +a_{1s}c_s&=b;\\ a_{i1}c_1+\cdots +a_{is}c_s&\neq 0 \end{aligned} \right. \end{align*}$$

for all $2\leq i\leq t$ , as desired.

Suppose next that $a_{i1}=0$ , $i=2,\ldots ,t$ . Note that $a_{il(i)}\neq 0$ , for some $2\leq l(i)\leq s$ for all $i=2,\ldots ,t$ . We define the following polynomials:

$$\begin{align*}\left\{ \begin{aligned} f_1(x_2,\ldots,x_s)&=a_{12}x_2+\cdots +a_{1s}x_s-b;\\ f_i(x_2,\ldots,x_s)&=a_{i2}x_2+\cdots +a_{is}x_s \end{aligned} \right. \end{align*}$$

for all $2\leq i\leq t$ . Note that $f_i\neq 0$ for all $i=1,\ldots ,t$ . In view of Lemma 2.5, we get that there exist $c_i\in K$ , $i=2,\ldots ,s$ , such that

$$\begin{align*}f_i(c_2,\ldots,c_s)\neq 0 \end{align*}$$

for all $i=1,\ldots ,t$ . That is

$$\begin{align*}\left\{ \begin{aligned} &a_{12}c_2+\cdots +a_{1s}c_s-b\neq 0;\\ &a_{i2}c_2+\cdots +a_{is}c_s\neq 0 \end{aligned} \right. \end{align*}$$

for all $2\leq i\leq t$ . Since $a_{11}\neq 0$ we get that there exists $c_{1}\in K$ such that

$$\begin{align*}a_{11}c_1=b-a_{12}c_2-\cdots -a_{1s}c_s. \end{align*}$$

This implies that

$$\begin{align*}\left\{ \begin{aligned} a_{11}c_1+&a_{12}c_2+\cdots +a_{1s}c_s=b;\\ &a_{i2}c_2+\cdots +a_{is}c_s\neq 0 \end{aligned} \right. \end{align*}$$

for all $2\leq i\leq t$ , as desired.

We finally assume that there exist $a_{i1}\neq 0$ and $a_{j1}=0$ for some $i,j\in \{2,\ldots ,t\}$ . Without loss of generality, we assume that $a_{i1}\neq 0$ for all $i=2,\ldots ,t_1$ and $a_{i1}=0$ for all $i=t_1+1,\ldots ,t$ . We define the following polynomials:

$$\begin{align*}\left\{ \begin{aligned} f_1(x_2,\ldots,x_s)&=b-a_{12}x_2-\cdots -a_{1s}x_s;\\ f_{i}(x_2,\ldots,x_s)&=a_{i1}a_{11}^{-1}b+(a_{i2}-a_{i1}a_{11}^{-1}a_{12})x_2+\cdots +(a_{is}-a_{i1}a_{11}^{-1}a_{1s})x_s;\\ f_j(x_2,\ldots,x_s)&=a_{j2}x_2+\cdots +a_{js}x_s \end{aligned} \right. \end{align*}$$

for all $2\leq i\leq t_1$ and $t_1+1\leq j\leq t$ . Note that $b,a_{i1}\in K^*$ , $i=1,\ldots ,t_1$ , $a_{jl(j)}\neq 0$ where $2\leq l(j)\leq s$ for all $j=t_1+1,\ldots t$ . It is clear that $f_i\neq 0$ for all $i=1,\ldots ,t$ . In view of Lemma 2.5, we get that there exist $c_i\in K$ , $i=2,\ldots ,s$ , such that

$$\begin{align*}f_i(c_2,\ldots,c_s)\neq 0, \end{align*}$$

where $i=1,\ldots ,t$ . This implies that

(4) $$ \begin{align} \left\{ \begin{aligned} b-a_{12}c_2-\cdots -a_{1s}c_s&\neq 0;\\ a_{i1}a_{11}^{-1}b+(a_{i2}-a_{i1}a_{11}^{-1}a_{12})c_2+\cdots +(a_{is}-a_{i1}a_{11}^{-1}a_{1s})c_s&\neq 0;\\ a_{j2}c_2+\cdots +a_{js}c_s&\neq 0 \end{aligned} \right. \end{align} $$

for all $2\leq i\leq t_1$ and $t_1+1\leq j\leq t$ . We set

$$\begin{align*}c_1=a_{11}^{-1}(b-a_{12}c_2-\cdots -a_{1s}c_s). \end{align*}$$

It follows from (4) that

$$\begin{align*}\left\{ \begin{aligned} a_{11}c_1+\cdots +a_{1s}c_s&=b;\\ a_{i1}c_1+\cdots +a_{is}c_s&\neq 0;\\ a_{j1}c_2+\cdots +a_{js}c_s&\neq 0 \end{aligned} \right. \end{align*}$$

for all $2\leq i\leq t_1$ and $t_1+1\leq j\leq t$ , as desired. The proof of the result is now complete.

3 The proof of Theorem 1.2

Let $n\geq 2$ and $m\geq 1$ be integers. Let $p(x_1,\ldots ,x_m)$ be a polynomial with zero constant term in noncommutative variables over an infinite field K. Suppose that $1<r<n-1$ , where $r=ord(p)$ .

Take any $u_i=(a_{jk}^{(i)})\in T_n(K)$ , $i=1,\ldots ,m$ . In view of both Lemma 2.1 and Lemma 2.2, we have that

(5) $$ \begin{align} p(u_1,\ldots,u_m)=(p_{s,r+s+t}), \end{align} $$

where

$$\begin{align*}p_{s,r+s+t}=\sum\limits_{k=r}^{r+t}\left(\sum\limits_{\substack{s=j_1<\cdots <j_{k+1}=r+s+t\\ (i_1,\ldots,i_k)\in T_m^k}} p_{i_1\dots i_k}(\bar{a}_{j_1j_1},\ldots,\bar{a}_{j_{k+1}j_{k+1}})a_{j_1j_2}^{(i_1)}\dots a_{j_{k}j_{k+1}}^{(i_k)}\right) \end{align*}$$

for all $1\leq s<r+s+t\leq n$ and

$$\begin{align*}p_{i_1^{\prime}\dots i_r^{\prime}}(K)\neq \{0\} \end{align*}$$

for some $1\leq i_1^{\prime },\ldots ,i_r^{\prime }\leq m$ . It follows from Lemma 2.4 that there exist $\bar {c}_1,\ldots ,\bar {c}_n\in K^m$ such that

(6) $$ \begin{align} p_{i_1^{\prime}\dots i_r^{\prime}}(\bar{c}_{j_1},\ldots,\bar{c}_{j_{r+1}})\neq 0 \end{align} $$

for all $1\leq j_1<\dots <j_{r+1}\leq n$ . We set

$$\begin{align*}\left\{ \begin{aligned} \bar{a}_{jj}&=\bar{c}_j,\quad j=1,\ldots,n;\\ a_{i,i+1}^{(k)}&=a_{i,i+1}^{(k)},\quad i=1,\ldots,r-1 \text{ and } k=1,\ldots,m;\\ a_{r+s-1,r+s+t}^{(i_k^{\prime})}&=x_{r+s-1,r+s+t}^{(i_k^{\prime})},\quad 1\leq s<r+s+t\leq n, k=1,\ldots,r;\\ a_{ij}^{(k)}&=0,\quad\mbox{otherwise}. \end{aligned} \right. \end{align*}$$

For any $1\leq s<r+s+t\leq n$ , we set

$$\begin{align*}U_{s,r+s+t}=\left\{(r+u-1,r+u+w,i_k^{\prime})~|~x_{r+u-1,r+u+w}^{(i_k^{\prime})}\quad\mbox{in } p_{s,r+s+t}\right\} \end{align*}$$

and

$$\begin{align*}\overline{U}_{s,r+s+t}=\left\{(r+u-1,r+u,i_k^{\prime})~|~(r+u-1,r+u,i_k^{\prime})\in U_{s,r+s+t}\right\}. \end{align*}$$

We define an order on the set

$$\begin{align*}\{(s,r+s+t)~|~1\leq s<r+s+t\leq n\} \end{align*}$$

as follows:

1. (i) $(s,r+s+t)<(s_1,r+s_1+t_1)$ if $t<t_1$ ;

2. (ii) $(s,r+s+t)<(s_1,r+s_1+t_1)$ if $t=t_1$ and $s<s_1$ .

That is,

(7) $$ \begin{align} (1,r+1)<\cdots <(n-r,n)<(1,r+2)<\cdots <(n-r-1,n)<\cdots <(1,n). \end{align} $$

For any $1\leq s<r+s+t\leq n$ , we set

$$\begin{align*}W_{s,r+s+t}=\bigcup_{(1,r+1)\leq (i,r+i+j)\leq (s,r+s+t)}U_{i,r+i+j}, \end{align*}$$

and

$$\begin{align*}\overline{W}_{s,r+s+t}=\bigcup_{(1,r+1)\leq (i,r+i+j)\leq (s,r+s+t)}\overline{U}_{i,r+i+j}. \end{align*}$$

We begin with the following lemmas, which will be used in the proof of our main result.

Lemma 3.1 Let $1\leq s<r+s\leq n$ . Suppose that $(s,r+s)\neq (1,r+1)$ . We claim that

(8) $$ \begin{align} \overline{W}_{s,r+s}\setminus\left\{(r+s-1,r+s,i_k^{\prime})~|~1\leq k\leq r\right\}=\overline{W}_{s-1,r+s-1}. \end{align} $$

Proof We first claim that

$$\begin{align*}\overline{W}_{s,r+s}\setminus\left\{(r+s-1,r+s,i_k^{\prime})~|~1\leq k\leq r\right\}\subseteq \overline{W}_{s-1,r+s-1}. \end{align*}$$

Take any $(r+i-1,r+i,i_k^{\prime })\in \overline {W}_{s,r+s}\setminus \left \{(r+s-1,r+s,i_k^{\prime })~|~1\leq k\leq r\right \}$ . We have that

$$\begin{align*}(r+i-1,r+i,i_k^{\prime})\in \overline{U}_{s_2,r+s_2} \end{align*}$$

for some $(1,r+1)\leq (s_2,r+s_2)\leq (s,r+s)$ . This implies that

$$\begin{align*}r+i\leq r+s_2\leq r+s. \end{align*}$$

We get that $i\leq s$ . Suppose that $i=s$ . It follows that

$$\begin{align*}(r+i-1,r+i,i_k^{\prime})\in\left\{(r+s-1,r+s,i_k^{\prime})~|~1\leq k\leq r\right\}, \end{align*}$$

a contradiction. Hence $i\leq s-1$ . It is clear that

$$\begin{align*}(r+i-1,r+i,i_k^{\prime})\in \overline{U}_{i,r+i}, \end{align*}$$

where $(1,r+1)\leq (i,r+i)\leq (s-1,r+s-1)$ . It follows that

$$\begin{align*}(r+i-1,r+i,i_k^{\prime})\in \overline{W}_{s-1,r+s-1}. \end{align*}$$

We obtain that

$$\begin{align*}\overline{W}_{s,r+s}\setminus\left\{(r+s-1,r+s,i_k^{\prime})~|~1\leq k\leq r\right\}\subseteq \overline{W}_{s-1,r+s-1}, \end{align*}$$

as desired. We next claim that

$$\begin{align*}\overline{W}_{s-1,r+s-1}\subseteq\overline{W}_{s,r+s}\setminus\left\{(r+s-1,r+s,i_k^{\prime})~|~1\leq k\leq r\right\}. \end{align*}$$

If $(r+s-1,r+s,i_k^{\prime })\in \overline {W}_{s-1,r+s-1}$ for $1\leq k\leq r$ , we have that

$$\begin{align*}r+s\leq r+s-1, \end{align*}$$

a contradiction. Hence

$$\begin{align*}\left\{(r+s-1,r+s,i_k^{\prime})~|~1\leq k\leq r\right\}\bigcap \overline{W}_{s-1,r+s-1}=\emptyset. \end{align*}$$

Since $\overline {W}_{s-1,r+s-1}\subseteq \overline {W}_{s,r+s}$ we get that

$$\begin{align*}\overline{W}_{s-1,r+s-1}\subseteq\overline{W}_{s,r+s}\setminus\left\{(r+s-1,r+s,i_k^{\prime})~|~1\leq k\leq r\right\}, \end{align*}$$

as desired. We obtain that

$$\begin{align*}\overline{W}_{s-1,r+s-1}=\overline{W}_{s,r+s}\setminus\left\{(r+s-1,r+s,i_k^{\prime})~|~1\leq k\leq r\right\}. \end{align*}$$

This proves the result.

Lemma 3.2 Let $1\leq s<r+s+t\leq n$ . Suppose that $t>0$ . We claim that

$$\begin{align*}\overline{W}_{s_1,r+s_1+t_1}=\overline{W}_{s,r+s+t}, \end{align*}$$

where

$$\begin{align*}(s_1,r+s_1+t_1)=\max\{(i,r+i+j)~|~(1,r+1)\leq (i,r+i+j)<(s,r+s+t)\}. \end{align*}$$

Proof We first claim that

$$\begin{align*}\overline{W}_{s,r+s+t}=\overline{W}_{n-r,n}. \end{align*}$$

Since $t>0$ , we note that

$$\begin{align*}(s,r+s+t)>(n-r,n). \end{align*}$$

This implies that $\overline {W}_{s,r+s+t}\supseteq \overline {W}_{n-r,n}$ . Take any $(r+u-1,r+u,i_k^{\prime })\in \overline {W}_{s,r+s+t}$ . It is clear that

$$\begin{align*}(r+u-1,r+u,i_k^{\prime})\in \overline{U}_{u,r+u}\subseteq\overline{W}_{n-r,n}. \end{align*}$$

This implies that $\overline {W}_{s,r+s+t}\subseteq \overline {W}_{n-r,n}$ . Hence, $\overline {W}_{s,r+s+t}=\overline {W}_{n-r,n}$ as desired.

Since $(n-r,n)<(s,r+s+t)$ we get that

$$\begin{align*}(n-r,n)\leq (s_1,r+s_1+t_1)<(s,r+s+t). \end{align*}$$

This implies that

$$\begin{align*}\overline{W}_{n-r,n}\subseteq\overline{W}_{s_1,r+s_1+t_1}\subseteq \overline{W}_{s,r+s+t}. \end{align*}$$

Since $\overline {W}_{s,r+s+t}=\overline {W}_{n-r,n}$ we obtain that $\overline {W}_{s_1,r+s_1+t_1}=\overline {W}_{s,r+s+t}$ . This proves the result.

The following technical result will be used in the proof of the next result.

Lemma 3.3 Let $1\leq s<r+s+t\leq n$ . If $(r+i-1,r+i+j,i_k^{\prime })\in U_{s,r+s+t}$ , we have that $j\leq t$ .

Proof Suppose that $(r+i-1,r+i+j,i_k^{\prime })\in U_{s,r+s+t}$ . That is, $x_{r+i-1,r+i+j}^{(i_k^{\prime })}$ appears in $p_{s,r+s+t}$ . In view of (5), we note that every monomial in $p_{s,r+s+t}$ is made up of at least r elements multiplied together. This implies that

$$\begin{align*}((r+s+t)-s)-((r+i+j)-(r+i-1))\geq r-1. \end{align*}$$

We obtain that $j\leq t$ . This proves the result.

Lemma 3.4 Let $1\leq s<r+s+t\leq n$ and $t>0$ . We claim that

$$\begin{align*}W_{s_1,r+s_1+t_1}=W_{s,r+s+t}\setminus\{(r+s-1,r+s+t,i_k^{\prime})~|~1\leq k\leq r\}, \end{align*}$$

where

$$\begin{align*}(s_1,r+s_1+t_1)=\max\{(i,r+i+j)~|~(1,r+1)\leq (i,r+i+j)<(s,r+s+t)\}. \end{align*}$$

Proof We first claim that

$$\begin{align*}W_{s_1,r+s_1+t_1}\subseteq W_{s,r+s+t}\setminus\{(r+s-1,r+s+t,i_k^{\prime})~|~1\leq k\leq r\}. \end{align*}$$

If $(r+s-1,r+s+t,i_k^{\prime })\in W_{s_1,r+s_1+t_1}$ for some $1\leq k\leq r$ , we get that

(9) $$ \begin{align} (r+s-1,r+s+t,i_k^{\prime})\in U_{s_2,r+s_2+t_2} \end{align} $$

for some $(1,r+1)\leq (s_2,r+s_2+t_2)\leq (s_1,r+s_1+t_1)$ . It is clear that

$$\begin{align*}t_2\leq t_1\leq t. \end{align*}$$

In view of Lemma 3.3, we get that $t\leq t_2$ . It follows that

$$\begin{align*}t_1=t_2=t. \end{align*}$$

Since $(s_1,r+s_1+t_1)<(s,r+s+t)$ we get that $s_1<s$ . Since $(s_2,r+s_2+t_2)\leq (s_1,r+s_1+t_1)$ we get that $s_2\leq s_1$ . Thus, we obtain that $s_2<s$ . It follows from (9) that

$$\begin{align*}r+s+t\leq r+s_2+t_2. \end{align*}$$

This implies that $s\leq s_2$ , a contradiction. Hence, we have that

$$\begin{align*}(r+s-1,r+s+t,i_k^{\prime})\not\in W_{s_1,r+s_1+t_1} \end{align*}$$

for all $1\leq k\leq r$ . It is clear that $W_{s_1,r+s_1+t_1}\subseteq W_{s,r+s+t}$ . We obtain that

$$\begin{align*}W_{s_1,r+s_1+t_1}\subseteq W_{s,r+s+t}\setminus\{(r+s-1,r+s+t,i_k^{\prime})~|~1\leq k\leq r\}, \end{align*}$$

as desired. We next claim that

$$\begin{align*}W_{s,r+s+t}\setminus\{(r+s-1,r+s+t,i_k^{\prime})~|~1\leq k\leq r\}\subseteq W_{s_1,r+s_1+t_1}. \end{align*}$$

For any $(r+i-1,r+i+j,i_k^{\prime })\in W_{s,r+s+t}\setminus \{(r+s-1,r+s+t,i_k^{\prime })~|~1\leq k\leq r\}$ , we have

$$\begin{align*}(r+i-1,r+i+j,i_k^{\prime})\in U_{s_2,r+s_2+t_2} \end{align*}$$

for some $(1,r+1)\leq (s_2,r+s_2+t_2)\leq (s,r+s+t)$ . This implies that $t_2\leq t$ . In view of Lemma 3.3, we note that $j\leq t_2$ . We have that $j\leq t$ . It is clear that

$$\begin{align*}(r+i-1,r+i+j,i_k^{\prime})\in U_{i,r+i+j}, \end{align*}$$

where $(1,r+1)\leq (i,r+i+j)\leq (s,r+s+t)$ . Note that

$$\begin{align*}(r+i-1,r+i+j,i_k^{\prime})\not\in\{(r+s-1,r+s+t,i_k^{\prime})~|~1\leq k\leq r\}. \end{align*}$$

We get that

$$\begin{align*}(i,r+i+j)\neq (s,r+s+t). \end{align*}$$

This implies that

$$\begin{align*}(1,r+1)\leq (i,r+i+j)\leq (s_1,r+s_1+t_1)\leq (s,r+s+t). \end{align*}$$

It follows that $U_{i,r+i+j}\subseteq W_{s_1,r+s_1+t_1}$ . We have that

$$\begin{align*}(r+i-1,r+i+j,i_k^{\prime})\in W_{s_1,r+s_1+t_1}. \end{align*}$$

We obtain that

$$\begin{align*}W_{s,r+s+t}\setminus\{(r+s-1,r+s+t,i_k^{\prime})~|~1\leq k\leq r\}\subseteq W_{s_1,r+s_1+t_1}, \end{align*}$$

as desired. Thus, we obtain that

$$\begin{align*}W_{s_1,r+s_1+t_1}=W_{s,r+s+t}\setminus\{(r+s-1,r+s+t,i_k^{\prime})~|~1\leq k\leq r\}. \end{align*}$$

This proves the result.

We set

$$\begin{align*}\hat{c}_{s,t}=(\bar{c}_s,\bar{c}_{s+1},\ldots,\bar{c}_{r+s-1},\bar{c}_{r+s+t}). \end{align*}$$

It follows from (6) that

(10) $$ \begin{align} p_{i_1^{\prime}\dots i_r^{\prime}}(\hat{c}_{s,t})\neq 0. \end{align} $$

For any $1\leq s<r+s\leq n$ and $s\leq r-1$ , we set

$$\begin{align*}f_{s,r}=\sum\limits_{(i_1,\ldots,i_{r-s})\in T_m^{r-s}}p_{i_1\dots i_{r-s}i_{r-s+1}^{\prime}\dots i_r^{\prime}}(\hat{c}_{s,t})a_{s,s+1}^{(i_1)}\dots a_{r-1,r}^{(i_{r-s})}. \end{align*}$$

We set

$$\begin{align*}V_{s,r}=\{(i,i+1,k)~|~i=s,\ldots,r-1,\quad k=1,\ldots,m\}, \end{align*}$$

where $1\leq s<r+s\leq n$ and $s\leq r-1$ . It is clear that $f_{s,r}$ is a polynomial on commutative variables indexed by elements from $V_{s,r}$ .

For any $1\leq s<r+s\leq n$ and $s\geq r$ , we set

$$\begin{align*}f_{s,r}=p_{i_1^{\prime}\dots i_{r}^{\prime}}(\hat{c}_{s,t}). \end{align*}$$

We claim that $f_{s,r}(K)\neq \{0\}$ for all $1\leq s<r+s\leq n$ . In view of (10), it suffices to prove that $f_{s,r}(K)\neq 0$ , where $1\leq s<r+s\leq n$ and $s\leq r-1$ .

We take $a_{i,i+1}^{(k)}\in K$ , $(i,i+1,k)\in V_{s,r}$ such that

$$\begin{align*}\left\{ \begin{aligned} a_{s+i,s+i+1}^{(i_{i+1}^{\prime})}&=1,\quad i=0,\ldots,r-s-1;\\ a_{i,i+1}^{(k)}&=0,\quad\mbox{otherwise}. \end{aligned} \right. \end{align*}$$

It follows from (10) that

$$\begin{align*}f_{s,r}(a_{i,i+1}^{(k)})=p_{i_1^{\prime}\dots i_r^{\prime}}(\hat{c}_{s,t})\neq 0, \end{align*}$$

as desired. In view of Lemma 2.5, we get that there exist $a_{i,i+1}^{(k)}\in K$ , $(i,i+1,k)\in \bigcup _{s=1}^{\min\{n-r,r-1\}}V_{s,r}$ such that

$$\begin{align*}f_{s,r}(a_{i,i+1}^{(k)})\neq 0 \end{align*}$$

for all $1\leq s<r+s\leq n$ and $s\leq r-1$ .

For any $2\leq s\leq r+s\leq n$ , we define

(11) $$ \begin{align} f_{s,r+s-i}=\sum\limits_{(i_1,\ldots,i_{r-i})\in T_m^{r-i}} p_{i_1\dots i_{r-i}i_{r-i+1}^{\prime}\dots i_r^{\prime}}(\hat{c}_{s,t})a_{s,s+1}^{(i_1)}\dots a_{r+s-i-1,r+s-i}^{(i_{r-i})} \end{align} $$

for all $1\leq i\leq \min\{s-1,r-1\}$ . It is clear that $f_{s,r+s-i}$ is a polynomial over K on commutative variables indexed by elements from $\overline {W}_{s-i,r+s-i}$ , where $1\leq i\leq \min\{s-1, r-1\}$ .

The following result implies that $f_{s,r+s-i}$ , where $1\leq i\leq \min\{s-1,r-1\}$ , is a recursive polynomial.

Lemma 3.5 For any $2\leq s<r+s\leq n$ , we claim that

$$\begin{align*}f_{s,r+s-i}=f_{s,r+s-i-1}x_{r+s-i-1,r+s-i}^{(i_{r-i}^{\prime})}+\sum\limits_{\substack{1\leq k\leq r\\i_k^{\prime}\neq i_{r-i}^{\prime}}}\alpha_{s,r+s-i-1,k}x_{r+s-i-1,r+s-i}^{(i_{k}^{\prime})} \end{align*}$$

for all $1\leq i\leq \min\{s-1,r-1\}$ , where both $f_{s,r+s-i-1}$ and $\alpha _{s,r+s-i-1,k}$ are polynomials over K on commutative variables indexed by elements from $\overline {W}_{s-i-1,r+s-i-1}$ .

Proof We get from (11) that

(12) $$ \begin{align} \begin{aligned} f&_{s,r+s-i}=\left(\sum\limits_{(i_1,\ldots,i_{r-i-1})\in T_m^{r-i-1}} p_{i_1\dots i_{r-i-1}i_{r-i}^{\prime}\dots i_r^{\prime}}(\hat{c}_{s,t})a_{s,s+1}^{(i_1)}\dots a_{r+s-i-2,r+s-i-1}^{(i_{r-i-1})}\right)x_{r+s-i-1,r+s-i}^{(i_{r-i}^{\prime})}\\ &\ \ \ +\sum\limits_{\substack{1\leq k\leq r\\i_k^{\prime}\neq i_{r-i}^{\prime}}}\left(\sum\limits_{(i_1,\ldots,i_{r-i-1})\in T_m^{r-i-1}} p_{i_1\dots i_{r-i-1}i_k^{\prime}i_{r-i+1}^{\prime}\dots i_r^{\prime}}(\hat{c}_{s,t})a_{s,s+1}^{(i_1)}\dots a_{r+s-i-2,r+s-i-1}^{(i_{r-i-1})}\right)x_{r+s-i-1,r+s-i}^{(i_{k}^{\prime})} \end{aligned} \end{align} $$

for all $1\leq i\leq \min\{s-1,r-1\}$ . It follows from (11) that

$$\begin{align*}f_{s,r+s-i-1}=\sum\limits_{(i_1,\ldots,i_{r-i-1})\in T_m^{r-i-1}} p_{i_1\dots i_{r-i-1}i_{r-i}^{\prime}\dots i_r^{\prime}}(\hat{c}_{s,t})a_{s,s+1}^{(i_1)}\dots a_{r+s-i-2,r+s-i-1}^{(i_{r-i-1})}. \end{align*}$$

We set

$$\begin{align*}\alpha_{s,r+s-i-1,k}=\sum\limits_{(i_1,\ldots,i_{r-i-1})\in T_m^{r-i-1}} p_{i_1\dots i_{r-i-1}i_k^{\prime}i_{r-i+1}^{\prime}\dots i_r^{\prime}}(\hat{c}_{s,t})a_{s,s+1}^{(i_1)}\dots a_{r+s-i-2,r+s-i-1}^{(i_{r-i-1})} \end{align*}$$

for all $1\leq i\leq \min\{s-1,r-1\}$ and $k=1,\ldots ,r$ . It follows from both (11) and (12) that

$$\begin{align*}f_{s,r+s-i}=f_{s,r+s-i-1}x_{r+s-i-1,r+s-i}^{(i_{r-i}^{\prime})}+\sum\limits_{\substack{1\leq k\leq r\\i_k^{\prime}\neq i_{r-i}^{\prime}}}\alpha_{s,r+s-i-1,k}x_{r+s-i-1,r+s-i}^{(i_{k}^{\prime})} \end{align*}$$

for all $1\leq i\leq \min\{s-1,r-1\}$ . It is clear that both $f_{s,r+s-i-1}$ and $\alpha _{s,r+s-i-1,k}$ are polynomials over K on commutative variables indexed by elements from

$$\begin{align*}\overline{W}_{s-i,r+s-i}\setminus\{(r+s-i-1,r+s-i,i_k^{\prime})~|~k=1,\ldots r\}. \end{align*}$$

In view of Lemma 3.1, we note that

$$\begin{align*}\overline{W}_{s-i-1,r+s-i-1}=\overline{W}_{s-i,r+s-i}\setminus\{(r+s-i-1,r+s-i,i_k^{\prime})~|~k=1,\ldots r\}. \end{align*}$$

We have that both $f_{s,r+s-i-1}$ and $\alpha _{s,r+s-i-1,k}$ are polynomials over K on commutative variables indexed by elements from $\overline {W}_{s-i-1,r+s-i-1}$ . This proves the result.

Lemma 3.6 For any $1\leq s<r+s\leq n$ , we have that

$$\begin{align*}p_{s,r+s+t}=f_{s,r+s-1}x_{r+s-1,r+s+t}^{(i_r^{\prime})}+\sum\limits_{\substack{1\leq k\leq r\\i_k^{\prime}\neq i_r^{\prime}}}\beta_{s,r+s-1,k}x_{r+s-1,r+s+t}^{(i_k^{\prime})}+\beta_{s,r+s+t}, \end{align*}$$

where $f_{1,r}\in K^*$ , $\beta _{1,r,k}\in K$ , $k=1,\ldots ,r$ with $i_k^{\prime }\neq i_r^{\prime }$ , $f_{s,r+s-1},\beta _{s,r+s-1,k}$ , $s\geq 2$ , $1\leq k\leq r$ with $i_k^{\prime }\neq i_r^{\prime }$ are polynomials on some commutative variables in $\overline {W}_{s_1,r+s_1+t_1}$ and $\beta _{s,r+s+t}$ , where $t>0$ , is a polynomial over K in some commutative variables in $W_{s_1,r+s_1+t_1}$ , where

$$\begin{align*}(s_1,r+s_1+t_1)=\max\{(i,r+i+j)~|~(1,r+1)\leq (i,r+i+j)<(s,r+s+t)\}. \end{align*}$$

Moreover, $\beta _{s,r+s}=0$ .

Proof It follows from (5) that

(13) $$ \begin{align} \begin{aligned} p&_{s,r+s+t}=\left(\sum\limits_{(i_1,\ldots,i_{r-1})\in T_m^{r-1}}p_{i_1\dots i_{r-1}i_r^{\prime}}(\hat{c}_{s,t})a_{s,s+1}^{(i_1)}\dots a_{r+s-2,r+s-1}^{(i_{r-1})}\right)x_{r+s-1,r+s+t}^{(i_r^{\prime})}\\ &+\sum\limits_{\substack{1\leq k\leq r\\i_k^{\prime}\neq i_r^{\prime}}}\left(\sum\limits_{(i_1,\ldots,i_{r-1})\in T_m^{r-1}}p_{i_1\dots i_{r-1}i_k^{\prime}}(\hat{c}_{s,t})a_{s,s+1}^{(i_1)}\dots a_{r+s-2,r+s-1}^{(i_{r-1})}\right)x_{r+s-1,r+s+t}^{(i_k^{\prime})}\\ &+\sum\limits_{k=r}^{r+t}\left(\sum\limits_{\substack{s=j_1<\cdots <j_{k+1}=r+s+t\\ (j_k,j_{k+1})\neq (r+s-1,r+s+t)\\ (i_1,\ldots,i_k)\in T_m^k}} p_{i_1\dots i_k}(\bar{c}_{j_1},\ldots,\bar{c}_{j_{k+1}})a_{j_1j_2}^{(i_1)}\cdots a_{j_{k}j_{k+1}}^{(i_k)}\right). \end{aligned} \end{align} $$

It follows from (11) that

$$\begin{align*}f_{s,r+s-1}=\sum\limits_{(i_1,\ldots,i_{r-1})\in T_m^{r-1}}p_{i_1\dots i_{r-1}i_r^{\prime}}(\hat{c}_{s,t})a_{s,s+1}^{(i_1)}\dots a_{r+s-2,r+s-1}^{(i_{r-1})}. \end{align*}$$

We set

$$\begin{align*}\beta_{s,r+s-1,k}=\sum\limits_{(i_1,\ldots,i_{r-1})\in T_m^{r-1}}p_{i_1\dots i_{r-1}i_k^{\prime}}(\hat{c}_{s,t})a_{s,s+1}^{(i_1)}\dots a_{r+s-2,r+s-1}^{(i_{r-1})} \end{align*}$$

for $k=1,\ldots ,r$ with $i_k^{\prime }\neq i_r^{\prime }$ , and

$$\begin{align*}\beta_{s,r+s+t}=\sum\limits_{k=r}^{r+t}\left(\sum\limits_{\substack{s=j_1<\cdots <j_{k+1}=r+s+t\\ (j_k,j_{k+1})\neq (r+s-1,r+s+t)\\ (i_1,\ldots,i_k)\in T_m^k}} p_{i_1\dots i_k}(\bar{c}_{j_1},\ldots,\bar{c}_{j_{k+1}})a_{j_1j_2}^{(i_1)}\dots a_{j_{k}j_{k+1}}^{(i_k)}\right). \end{align*}$$

It follows from (13) that

(14) $$ \begin{align} p_{s,r+s+t}=f_{s,r+s-1}x_{r+s-1,r+s+t}^{(i_r^{\prime})}+\sum\limits_{\substack{1\leq k\leq r\\i_k^{\prime}\neq i_r^{\prime}}}\beta_{s,r+s-1,k}x_{r+s-1,r+s+t}^{(i_k^{\prime})}+\beta_{s,r+s+t}, \end{align} $$

where $f_{1,r}\in K^*, \beta _{1,r,k}\in K$ , $k=1,\ldots ,r$ with $i_k^{\prime }\neq i_r^{\prime }$ , $f_{s,r+s-1},\beta _{s,r+s+t,k}$ , where $s\geq 2$ , $1\leq k\leq r$ with $i_k^{\prime }\neq i_r^{\prime }$ , are polynomials on some commutative variables indexed by elements from

(15) $$ \begin{align} \overline{W}_{s,r+s+t}\setminus\left\{(r+s-1,r+s+t,i_k^{\prime}),\quad k=1,\ldots,r\right\} \end{align} $$

and $\beta _{s,r+s+t}$ , where $t>0$ , is a polynomial over K in some commutative variables indexed by elements from

(16) $$ \begin{align} W_{s,r+s+t}\setminus\left\{(r+s-1,r+s+t, i_k^{\prime}),\quad k=1,\ldots,r\right\}. \end{align} $$

Suppose first that $t=0$ . In view of Lemma 3.1, we note that

$$\begin{align*}\overline{W}_{s-1,r+s-1}=\overline{W}_{s,r+s+t}\setminus\left\{(r+s-1,r+s, i_k^{\prime}),\quad k=1,\ldots,r\right\}. \end{align*}$$

We get from (15) that $f_{s,r+s-1},\beta _{s,r+s+t,k}$ , where $s\geq 2$ , $1\leq k\leq r$ with $i_k^{\prime }\neq i_r^{\prime }$ , are polynomials on some commutative variables indexed by elements from $\overline {W}_{s-1,r+s-1}$ . It is clear that $\beta _{s,r+s}=0$ . Suppose next that $t>0$ . In view of Lemma 3.2, we note that

$$\begin{align*}\overline{W}_{s_1,r+s_1+t_1}=\overline{W}_{s,r+s+t}. \end{align*}$$

We get from (15) that $f_{s,r+s-1},\beta _{s,r+s+t,k}$ , where $s\geq 2$ , $1\leq k\leq r$ with $i_k^{\prime }\neq i_r^{\prime }$ , are polynomials on some commutative variables indexed by elements from $\overline {W}_{s_1,r+s_1+t_1}$ . In view of Lemma 3.4, we note that

$$\begin{align*}W_{s_1,r+s_1+t_1}=W_{s,r+s+t}\setminus\left\{(r+s-1,r+s+t, i_k^{\prime}),\quad k=1,\ldots,r\right\}. \end{align*}$$

We get from (16) that $\beta _{s,r+s+t}$ is a polynomial over K in some commutative variables indexed by elements from $W_{s_1,r+s_1+t_1}$ . This proves the result.

The following result is crucial for the proof of the main result.

Lemma 3.7 Let $p(x_1,\ldots ,x_m)$ be a polynomial with zero constant term in noncommutative variables over an infinite field K. Suppose ord $(p)=r$ , where $1<r<n-1$ . For any $A'=(a_{s,r+s+t}^{\prime })\in T_n(K)^{(r-1)}$ , where $a_{s,r+s}^{\prime }\neq 0$ for all $1\leq s<r+s+t\leq n$ , we have that $A'\in p(T_n(K))$ .

Proof Take any $A'=(a_{s,r+s+t}^{\prime })\in T_n(K)^{(r-1)}$ , where $a_{s,r+s}^{\prime }\neq 0$ for all ${1\leq s<r+s\leq n}$ . For any $1\leq s<r+s+t\leq n$ , we claim that there exist $c_{r+u-1,r+u+w}^{(i_k^{\prime })}\in K$ with

$$\begin{align*}(r+u-1,r+u+w,k)\in W_{s,r+s+t} \end{align*}$$

such that

$$\begin{align*}p_{i,r+i+j}(c_{r+u-1,r+u+w}^{(i_k^{\prime})})=a_{i,r+i+j} \end{align*}$$

for all $(1,r+1)\leq (i,r+i+j)\leq (s,r+s+t)$ and

$$\begin{align*}f_{s',r+s'-v}(c_{r+u-1,r+u}^{(i_k^{\prime})})\neq 0 \end{align*}$$

for all $f_{s',r+s'-v}$ on commutative variables in $\overline {W}_{s,r+s+t}$ , where $s'\geq 2$ and $1\leq v\leq \min\{s'-1,r-1\}$ .

We prove the claim by induction on $(s,r+s+t)$ . Suppose first that $(s,r+s+t)=(1,r+1)$ . Note that

$$\begin{align*}W_{1,r+1}=\overline{W}_{1,r+1}=\{(r,r+1,i_k^{\prime})~|~k=1,\ldots,r\}. \end{align*}$$

In view of Lemma 3.6, we get that

(17) $$ \begin{align} p_{1,r+1}=f_{1,r}x_{r,r+1}^{(i_r^{\prime})}+\sum\limits_{\substack{1\leq k\leq r\\i_k\neq i_r^{\prime}}}\beta_{1,r,k}x_{r,r+1}^{(i_k^{\prime})}, \end{align} $$

where $f_{1,r}\in K^*$ , $\beta _{1,r,k}\in K$ , $k=1,\ldots ,r$ with $i_k^{\prime }\neq i_r^{\prime }$ .

Take any $f_{s',r+s'-v}$ on $x_{r,r+1}^{(i_k^{\prime })}$ , where $k=1,\ldots ,r$ , $s'\geq 2$ , and $1\leq v\leq \min\{s'-1,r-1\}$ , we get from Lemma 3.5 that

$$\begin{align*}r+s'-v-1=r \end{align*}$$

and so $v=s'-1$ . It follows that

(18) $$ \begin{align} f_{s',r+s'-v}=f_{s',r}x_{r,r+1}^{(i_{r-v}^{\prime})}+\sum\limits_{\substack{1\leq k\leq r\\i_k^{\prime}\neq i_{r-v}^{\prime}}}\alpha_{s',r,k}x_{r,r+1}^{(i_{k}^{\prime})}. \end{align} $$

Note that $f_{s',r}\in K^*$ and $\alpha _{s',r,k}\in K$ , $k=1,\ldots ,r$ with $i_k^{\prime }\neq i_{r-v}$ . Note that $a_{1,r+1}^{\prime }\in K^*$ . In view of Lemma 2.6, we get from both (17) and (18) that there exist $c_{r,r+1}^{(i_k^{\prime })}\in K$ , $k=1,\ldots ,r$ , such that

$$\begin{align*}\left\{ \begin{aligned} p_{1,r+1}(c_{r,r+1}^{(i_k^{\prime})})&=a_{1,r+1}^{\prime},\\ f_{s',r+s'-v}(c_{r,r+1}^{(i_k^{\prime})})&\neq 0, \end{aligned} \right. \end{align*}$$

where $2\leq s'\leq r$ and $v=s'-1$ , as desired.

Suppose next that $(s,r+s+t)\neq (1,r+1)$ . We rewrite (7) as follows:

$$\begin{align*}(1,r+1)<\cdots<(s_1,r+s_1+t_1)<(s,r+s+t)<\cdots<(1,n), \end{align*}$$

where

$$\begin{align*}(s_1,r+s_1+t_1)=\max\{(i,r+i+j)~|~(1,r+1)\leq (i,r+i+j)<(s,r+s+t)\}. \end{align*}$$

By induction on $(s_1,r+s_1+t_1),$ we have that there exist $c_{r+u-1,r+u+w}^{(i_k^{\prime })}\in K$ with

$$\begin{align*}(r+u-1,r+u+w,k)\in W_{s_1,r+s_1+t_1} \end{align*}$$

such that

$$\begin{align*}p_{i,r+i+j}(c_{r+u-1,r+u+w}^{(i_k^{\prime})})=a_{i,r+i+j}^{\prime} \end{align*}$$

for all $(1,r+1)\leq (i,r+i+j)\leq (s_1,r+s_1+t_1)$ and

$$\begin{align*}f_{s',r+s'-v}(c_{r+u-1,r+u}^{(i_k^{\prime})})\neq 0 \end{align*}$$

for any $f_{s',r+s'-v}$ with commutative variables in $\overline {W}_{s_1,r+s_1+t_1}$ , where $s'\geq 2$ , and $1\leq v\leq \min\{s'-1,r-1\}$ . We now divide the proof into the following two cases.

Suppose first that $t=0$ . Note that

$$\begin{align*}(s_1,r+s_1+t_1)=(s-1,r+s-1). \end{align*}$$

That is, $s_1=s-1$ and $t_1=0$ . In view of Lemma 3.6, we get that

(19) $$ \begin{align} p_{s,r+s}=f_{s,r+s-1}x_{r+s-1,r+s}^{(i_r^{\prime})}+\sum\limits_{\substack{1\leq k\leq r\\i_k^{\prime}\neq i_r^{\prime}}}\beta_{s,r+s-1,k}x_{r+s-1,r+s}^{(i_k^{\prime})}, \end{align} $$

where $f_{s,r+s-1}, \beta _{s,r+s-1,k}$ , where $k=1,\ldots ,r$ with $i_k^{\prime }\neq i_r^{\prime }$ , are polynomials in commutative variables in $\overline {W}_{s_1,r+s_1}$ . By induction hypothesis, we get that $f_{s,r+s-1}\in K^*$ and $\beta _{s,r+s-1,k}\in K$ .

Take any $f_{s',r+s'-v}$ on commutative variables indexed by elements from $\overline {W}_{s,r+s}$ , where $s'\geq 2$ and $1\leq v\leq \min\{s'-1,r-1\}$ . Suppose first that $f_{s',r+s'-v}$ is a polynomial on commutative variables indexed by elements from $\overline {W}_{s_1,r+s_1}$ . By induction hypothesis we have that $f_{s',r+s'-v}\in K^*$ . Suppose next that $f_{s',r+s'-v}$ is not a polynomial on commutative variables indexed by elements from $\overline {W}_{s_1,r+s_1}$ . In view of Lemma 3.1, we note that

$$\begin{align*}\overline{W}_{s,r+s}\setminus \overline{W}_{s-1,r+s-1}=\left\{(r+s-1,r+s,i_k^{\prime})~|~k=1,\ldots,r\right\}. \end{align*}$$

This implies that $x_{r+s-1,r+s}^{(i_k^{\prime })}$ appears in $f_{s',r+s'-v}$ for $k=1,\ldots , r$ . In view of Lemma 3.5 we get that

$$\begin{align*}(r+s'-v-1,r+s'-v)=(r+s-1,r+s) \end{align*}$$

and so $v=s'-s$ . We get that

(20) $$ \begin{align} f_{s',r+s'-v}=f_{s',r+s'-v-1}x_{r+s-1,r+s}^{(i_{r-v}^{\prime})}+\sum\limits_{\substack{1\leq k\leq r\\i_k^{\prime}\neq i_{r-v}^{\prime}}}\alpha_{s',r+s'-v-1,k}x_{r+s-1,r+s}^{(i_{k}^{\prime})}, \end{align} $$

where $f_{s',r+s'-v-1}$ and $\alpha _{s',r+s'-v-1,k}$ , $k=1,\ldots ,r$ with $i_k^{\prime }\neq i_{r-v}^{\prime }$ , are polynomials over K on commutative variables indexed by elements from $\overline {W}_{s_1,r+s_1}$ . By induction hypothesis, we have that $f_{s',r+s'-v-1}\in K^*$ and $\alpha _{s',r+s'-v-1,k}\in K$ , where $k=1,\ldots ,r$ with $i_k^{\prime }\neq i_{r-v}^{\prime }$ .

Note that $a_{s,r+s}^{\prime }\in K^*$ . In view of Lemma 2.6, we get from both (19) and (20) that there exist $c_{r+s-1,r+s}^{(i_k^{\prime })}\in K$ , $k=1,\ldots ,r$ , such that

$$\begin{align*}\left\{ \begin{aligned} p_{s,r+s}(c_{r+s-1,r+s}^{(i_k^{\prime})})&=a_{s,r+s}^{\prime};\\ f_{s',r+s'-v}(c_{r+s-1,r+s}^{(i_k^{\prime})})&\neq 0, \end{aligned} \right. \end{align*}$$

as desired.

Suppose next that $t>0$ . It follows from Lemma 3.6 that

(21) $$ \begin{align} p_{s,r+s+t}=f_{s,r+s-1}x_{r+s-1,r+s+t}^{(i_r^{\prime})}+\sum\limits_{\substack{1\leq k\leq r\\i_k^{\prime}\neq i_r^{\prime}}}\beta_{s,r+s-1,k}x_{r+s-1,r+s+t}^{(i_k^{\prime})}+\beta_{s,r+s+t}, \end{align} $$

where $f_{s,r+s-1},\beta _{s,r+s-1,k}$ , where $k=1,\ldots ,r$ with $i_k^{\prime }\neq i_r^{\prime }$ , are polynomials over K in commutative variables indexed by elements from $\overline {W}_{r+s_1+t_1}$ , and $\beta _{s,r+s+t}$ is a polynomial over K in commutative variables indexed by elements from $W_{s_1,r+s_1+t_1}$ . By induction hypothesis, we have that $f_{s,r+s-1}\in K^*$ , $\beta _{s,r+s-1,k}\in K$ for all $k=1,\ldots ,r$ with $i_k^{\prime }\neq i_r^{\prime }$ , and $\beta _{s,r+s+t}\in K$ .

Take $c_{r+s-1,r+s+t}^{(i_k^{\prime })}\in K$ , where $k=1,\ldots ,r$ in (21) such that

$$\begin{align*}\left\{ \begin{aligned} c_{r+s-1,r+s+t}^{(i_r^{\prime})}&=f_{s,r+s-1}^{-1}(a_{s,r+s+t}^{\prime}-\beta_{s,r+s+t});\\ c_{r+s-1,r+s+t}^{(i_k^{\prime})}&=0,\quad\mbox{for all } 1\leq k\leq r \text{ with } i_k^{\prime}\neq i_r^{\prime}. \end{aligned} \right. \end{align*}$$

We get that

$$\begin{align*}p_{s,r+s+t}(c_{r+s-1,r+s+t}^{(i_k^{\prime})})=a_{s,r+s+t}^{\prime}. \end{align*}$$

Take any $f_{s',r+s'-v}$ on commutative variables indexed by elements from $\overline {W}_{s,r+s+t}$ , where $s'\geq 2$ and $1\leq v\leq \min\{s'-1,r-1\}$ . In view of Lemma 3.2, we note that

$$\begin{align*}\overline{W}_{s,r+s+t}=\overline{W}_{s_1,r+s_1+t_1}. \end{align*}$$

This implies that $f_{s',r+s'-v}$ is a commutative polynomial over K on some commutative variables indexed by elements from $\overline {W}_{s_1,r+s_1+t_1}$ . By induction hypothesis, we get that

$$\begin{align*}f_{s',r+s'-v}\in K^*, \end{align*}$$

where $s'\geq 2$ and $1\leq v\leq \min\{s'-1,r-1\}$ , as desired. This proves the claim.

Let $(s,r+s+t)=(1,n)$ . We have that there exist $c_{r+u-1,r+u+w}^{(i_k^{\prime })}\in K$ , $k=1,\ldots ,r$ , with

$$\begin{align*}(r+u-1,r+u+w,k)\in W_{1,n}, \end{align*}$$

such that

(22) $$ \begin{align} p_{i,r+i+j}(c_{r+u-1,r+u+w}^{(i_k^{\prime})})=a_{i,r+i+j}^{\prime} \end{align} $$

for all $(1,r+1)\leq (i,r+i+j)\leq (1,n)$ and

$$\begin{align*}f_{s',r+s'-v}(c_{r+u-1,r+u}^{(i_k^{\prime})})\neq 0 \end{align*}$$

for all $f_{s',r+s'-v}$ on commutative variables indexed by elements from $\overline {W}_{1,n}$ , where $s'\geq 2$ and $1\leq v\leq \min\{s'-1,r-1\}$ . It follows from both (5) and (22) that

$$\begin{align*}p(u_1,\ldots,u_m)=(p_{s,r+s+t})=(a_{s,r+s+t}^{\prime})=A'. \end{align*}$$

This implies that $A'\in p(T_n(K))$ . The proof of the result is complete.

Lemma 3.8 Let $n\geq 4$ and $m\geq 1$ be integers. Let $p(x_1,\ldots ,x_m)$ be a polynomial with zero constant term in noncommutative variables over an infinite field K. Suppose that ord $(p)=n-2$ . We have that $p(T_n(K))=T_n(K)^{(n-3)}$ .

Proof In view of Lemma 2.2(ii), we note that $p(T_n(K))\subseteq T_n(K)^{(n-3)}$ . It suffices to prove that $T_n(K)^{(n-3)}\subseteq p(T_n(K))$ .

For any $u_i=(a_{jk}^{(i)})\in T_n(K)$ , $i=1,\ldots ,m$ , in view of Lemma 2.2(ii), we get from (2) that

(23) $$ \begin{align} p(u_1,\ldots,u_m)=\left( \begin{array}{ccccc} 0 & 0 & \ldots & p_{1,n-1} & p_{1n}\\ 0 & 0 & \ldots & 0 & p_{2n}\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & \ldots & 0 & 0 \end{array} \right), \end{align} $$

where

$$\begin{align*}\left\{ \begin{aligned} p_{1,n-1}&=\sum\limits_{(i_1,\ldots,i_{n-2})\in T_m^{n-2}}p_{i_1\dots i_{n-2}}(\bar{a}_{11},\ldots,\bar{a}_{n-1,n-1})a_{12}^{(i_1)}\dots a_{n-2,n-1}^{(i_{n-2})};\\ p_{2n}&=\sum\limits_{(i_1,\ldots,i_{n-2})\in T_m^{n-2}} p_{i_1\dots i_{n-2}}(\bar{a}_{22},\ldots,\bar{a}_{n,n})a_{23}^{(i_1)}\dots a_{n-1,n}^{(i_{n-2})};\\ p_{1n}&=\sum\limits_{(i_1,\ldots,i_{n-1})\in T_m^{n-1}}p_{i_1\dots i_{n-1}}(\bar{a}_{11},\ldots,\bar{a}_{nn})a_{12}^{(i_1)}\dots a_{n-1,n}^{(i_{n-1})}\\ &\ \ \ +\sum\limits_{\substack{1=j_1<\cdots <j_{n-1}=n\\ (i_1,\ldots,i_{n-2})\in T_m^{n-2}}} p_{i_1\dots i_{n-2}}(\bar{a}_{j_1j_1},\ldots,\bar{a}_{j_{n-1}j_{n-1}})a_{j_1j_2}^{(i_1)}\dots a_{j_{n-2}j_{n-1}}^{(i_{n-2})}. \end{aligned} \right. \end{align*}$$

In view of Lemma 2.2(iii), we have that

$$\begin{align*}p_{i_1^{\prime},\ldots,i_{n-2}^{\prime}}(K)\neq\{0\}, \end{align*}$$

for some $i_1^{\prime },\ldots ,i_{n-2}^{\prime }\in \{1,\ldots ,m\}$ . It follows from Lemma 2.4 that there exist $\bar {b}_1,\ldots ,\bar {b}_n\in K^m$ such that

$$\begin{align*}p_{i_1^{\prime},\ldots,i_{n-2}^{\prime}}(\bar{b}_{j_1},\ldots,\bar{b}_{j_{n-1}})\neq 0 \end{align*}$$

for all $1\leq j_1<\cdots <j_{n-1}\leq n$ .

For any $A'=(a_{s,n-2+s+t}^{\prime })\in T_n(K)^{(n-3)}$ , where $1\leq s<n-2+s+t\leq n$ , we claim that there exist $u_i=(a_{jk}^{(i)})\in T_n(K)$ , $i=1,\ldots ,m$ , such that

$$\begin{align*}p(u_1,\ldots,u_m)=(p_{s,n-2+s+t})=A'. \end{align*}$$

That is,

$$\begin{align*}\left\{ \begin{aligned} p_{1,n-1}&=a_{1,n-1}^{\prime};\\ p_{2n}&=a_{2n}^{\prime};\\ p_{1n}&=a_{1n}^{\prime}. \end{aligned} \right. \end{align*}$$

We prove the claim by the following two cases:

Case 1. Suppose that $a_{1,n-1}^{\prime }\neq 0$ . We take

$$\begin{align*}\left\{ \begin{aligned} \bar{a}_{jj}&=\bar{b}_j, \quad\mbox{for all } j=1,\ldots,n;\\ a_{12}^{(i_1^{\prime})}&=x_{12}^{(i_1^{\prime})};\\ a_{12}^{(k)}&=0\quad\mbox{for all } k=1,\ldots,m \text{ with } k\neq i_1^{\prime};\\ a_{n-1,n}^{(i_{n-2}^{\prime})}&=x_{n-1,n}^{(i_{n-2}^{\prime})};\\ a_{n-1,n}^{(k)}&=0\quad\mbox{for all } k=1,\ldots,m \text{ with } k\neq i_{n-2}^{\prime};\\ a_{n-2,n}^{(i_{n-2}^{\prime})}&=x_{n-2,n}^{(i_{n-2}^{\prime})};\\ a_{j,j+2}^{(i)}&=0\quad\mbox{for all } 1\leq i\leq m, 3\leq j+2\leq n \text{ with } (j,j+2,i)\neq (n-2,n,i_{n-2}^{\prime}). \end{aligned} \right. \end{align*}$$

It follows from (23) that

(24) $$ \begin{align} \left\{\! \begin{aligned} p_{1,n-1}&=\left(\sum\limits_{\substack{(i_2,\ldots,i_{n-2})\in T_m^{n-3}}}p_{i_1^{\prime}i_2\dots i_{n-2}}(\bar{b}_{1},\ldots,\bar{b}_{n-1})a_{23}^{(i_2)}\dots a_{n-2,n-1}^{(i_{n-2})}\right)x_{12}^{(i_1^{\prime})};\\ p_{2n}&=\left(\sum\limits_{\substack{(i_1,\ldots,i_{n-3})\in T_m^{n-3}}}p_{i_1\dots i_{n-3}i_{n-2}^{\prime}}(\bar{b}_{2},\ldots,\bar{b}_{n})a_{23}^{(i_1)}\dots a_{n-2,n-1}^{(i_{n-3})}\right)x_{n-1,n}^{(i_{n-2}^{\prime})};\\ p_{1n}&=\left(\sum\limits_{(i_2,\ldots,i_{n-2})\in T_m^{n-3}}p_{i_1^{\prime}i_2\dots i_{n-2}i_{n-2}^{\prime}}(\bar{b}_{1},\ldots,\bar{b}_{n})a_{23}^{(i_2)}\dots a_{n-2,n-1}^{(i_{n-2})}\right)x_{12}^{(i_1^{\prime})}x_{n-1,n}^{(i_{n-2}^{\prime})}\\ &\ \ \ \left(\sum\limits_{\substack{(i_2,\ldots,i_{n-3})\in T_m^{n-4}}}p_{i_1^{\prime}i_2\dots i_{n-3}i_{n-2}^{\prime}}(\bar{b}_{1},\ldots,\bar{b}_{n-2},\bar{b}_{n})a_{23}^{(i_2)}\dots a_{n-3,n-2}^{(i_{n-3})}\!\right)\!x_{12}^{(i_1^{\prime})}x_{n-2,n}^{(i_{n-2}^{\prime})}\kern-1pt. \end{aligned} \right. \end{align} $$

We set

(25) $$ \begin{align} \left\{ \begin{aligned} f_{1,n-1}&=\sum\limits_{\substack{(i_2,\ldots,i_{n-2})\in T_m^{n-3}}}p_{i_1^{\prime}i_2\dots i_{n-2}}(\bar{b}_{1},\ldots,\bar{b}_{n-1})a_{23}^{(i_2)}\dots a_{n-2,n-1}^{(i_{n-2})},\\ f_{2n}&=\sum\limits_{\substack{(i_1,\ldots,i_{n-3})\in T_m^{n-3}}}p_{i_1\dots i_{n-3}i_{n-2}^{\prime}}(\bar{b}_{2},\ldots,\bar{b}_{n})a_{23}^{(i_1)}\dots a_{n-2,n-1}^{(i_{n-3})},\\ f_{1n}&=\sum\limits_{\substack{(i_2,\ldots,i_{n-3})\in T_m^{n-4}}}p_{i_1^{\prime}i_2\dots i_{n-3}i_{n-2}^{\prime}}(\bar{b}_{1},\ldots,\bar{b}_{n-2},\bar{b}_{n})a_{23}^{(i_2)}\dots a_{n-3,n-2}^{(i_{n-3})}, \end{aligned} \right. \end{align} $$

and

$$ \begin{align*} \begin{aligned} V_{1,n-1}&=\{(i,i+1,k)~|~i=2,\ldots,n-2,k=1,\ldots,m\};\\ V_{2n}&=V_{1,n-1};\\ V_{1n}&=\{(i,i+1,k)~|~i=2,\ldots,n-3,k=1,\ldots,m\}. \end{aligned} \end{align*} $$

Note that $f_{1,n-1},f_{2n},f_{1n}$ are polynomials over K on commutative variables indexed by elements from $V_{1,n-1},V_{2n}, V_{1n}$ , respectively.

We claim that $f_{1,n-1},f_{2n},f_{1n}\neq 0$ . Indeed, we take $a_{jk}^{(i)}\in K$ , $(j,k,i)\in V_{1,n-1}$ such that

$$\begin{align*}\left\{ \begin{aligned} a_{s,s+1}^{(i_s^{\prime})}&=1,\quad\mbox{for all } s=2,\ldots,n-2;\\ a_{jk}^{(i)}&=0,\quad\mbox{otherwise}. \end{aligned} \right. \end{align*}$$

It follows from (25) that

$$\begin{align*}f_{1,n-1}(a_{jk}^{(i)})=p_{i_1^{\prime}\dots i_{n-2}^{\prime}}(\bar{b}_{1},\ldots,\bar{b}_{n-1})\neq 0, \end{align*}$$

as desired. Next, we take $a_{jk}^{(i)}\in K$ , $(j,k,i)\in V_{2n}$ such that

$$\begin{align*}\left\{ \begin{aligned} a_{s,s+1}^{(i_{s-1}^{\prime})}&=1,\quad\mbox{for all } s=2,\ldots,n-2;\\ a_{jk}^{(i)}&=0,\quad\mbox{otherwise}. \end{aligned} \right. \end{align*}$$

It follows from (25) that

$$\begin{align*}f_{2n}(a_{jk}^{(i)})=p_{i_1^{\prime}\dots i_{n-2}^{\prime}}(\bar{b}_{2},\ldots,\bar{b}_{n})\neq 0, \end{align*}$$

as desired. Finally, we take $a_{jk}^{(i)}\in K$ , $(j,k,i)\in V_{1n}$ such that

$$\begin{align*}\left\{ \begin{aligned} a_{s,s+1}^{(i_s^{\prime})}&=1,\quad\mbox{for all } s=2,\ldots,n-3;\\ a_{jk}^{(i)}&=0,\quad\mbox{otherwise}. \end{aligned} \right. \end{align*}$$

It follows from (25) that

$$\begin{align*}f_{1n}(a_{jk}^{(i)})=p_{i_1^{\prime}\dots i_{n-2}^{\prime}}(\bar{b}_{1},\ldots,\bar{b}_{n-2},\bar{b}_n)\neq 0, \end{align*}$$

as desired. In view of Lemma 2.5, we get that there exist $a_{jk}^{(i)}\in K$ , where $(j,k,i)\in V_{1,n-1}\cup V_{2n}\cup V_{1n}$ such that

$$\begin{align*}\left\{ \begin{aligned} f_{1,n-1}(a_{jk}^{(i)})&\neq 0;\\ f_{2n}(a_{jk}^{(i)})&\neq 0;\\ f_{1n}(a_{jk}^{(i)})&\neq 0. \end{aligned} \right. \end{align*}$$

We set

$$\begin{align*}\alpha=\sum\limits_{(i_2,\ldots,i_{n-2})\in T_{n-3}}p_{i_1^{\prime}i_2\dots i_{n-2}i_{n-2}^{\prime}}(\bar{b}_{1},\ldots,\bar{b}_{n})a_{23}^{(i_2)}\dots a_{n-2,n-1}^{(i_{n-2})}. \end{align*}$$

It follows from (24) that

(26) $$ \begin{align} \left\{ \begin{aligned} p_{1,n-1}&=f_{1,n-1}x_{12}^{(i_1^{\prime})};\\ p_{2n}&=f_{2n}x_{n-1,n}^{(i_{n-2}^{\prime})};\\ p_{1n}&=f_{1n}x_{12}^{(i_1^{\prime})}x_{n-2,n}^{(i_{n-2}^{\prime})}+\alpha x_{12}^{(i_1^{\prime})}x_{n-1,n}^{(i_{n-2}^{\prime})}. \end{aligned} \right. \end{align} $$

We take

$$\begin{align*}\left\{ \begin{aligned} x_{12}^{(i_1^{\prime})}&=f_{1,n-1}^{-1}a_{1,n-1}^{\prime};\\ x_{n-1,n}^{(i_{n-2}^{\prime})}&=f_{2n}^{-1}a_{2n}^{\prime};\\ x_{n-2,n}^{(i_{n-2}^{\prime})}&=f_{1n}^{-1}f_{1,n-1}(a_{1,n-1}^{\prime})^{-1}\left(a_{1n}^{\prime}-\alpha f_{1,n-1}^{-1}a_{1,n-1}^{\prime}f_{2n}^{-1}a_{2n}^{\prime}\right). \end{aligned} \right. \end{align*}$$

It follows from (26) that

$$\begin{align*}\left\{ \begin{aligned} p_{1,n-1}&=a_{1,n-1}^{\prime};\\ p_{2n}&=a_{2n}^{\prime};\\ p_{1n}&=a_{1n}^{\prime}, \end{aligned} \right. \end{align*}$$

as desired.

Case 2. Suppose that $a_{1,n-1}^{\prime }=0$ . We take

$$\begin{align*}\left\{ \begin{aligned} \bar{a}_{jj}&=\bar{b}_j, \quad\mbox{for all } j=1,\ldots,n;\\ a_{12}^{(k)}&=0,\quad\mbox{for all } k=1,\ldots,m;\\ a_{23}^{(i_{1}^{\prime})}&=x_{23}^{(i_{1}^{\prime})};\\ a_{23}^{(k)}&=0,\quad\mbox{for all } k=1,\ldots,m \text{ with } k\neq i_{1}^{\prime};\\ a_{13}^{(i_{1}^{\prime})}&=x_{13}^{(i_{1}^{\prime})};\\ a_{j,j+2}^{(k)}&=0,\quad\mbox{for all } 1\leq j<j+2\leq n \text{ with } (j,j+2,k)\neq (1,3,i_1^{\prime}). \end{aligned} \right. \end{align*}$$

It follows from (23) that

(27) $$ \begin{align} \left\{ \begin{aligned} p_{1,n-1}&=0;\\ p_{2n}&=\left(\sum\limits_{\substack{(i_2,\ldots,i_{n-2})\in T_m^{n-3}}}p_{i_1^{\prime}i_2\dots i_{n-2}}(\bar{b}_{2},\ldots,\bar{b}_{n})a_{34}^{(i_2)}\dots a_{n-1,n}^{(i_{n-2})}\right)x_{23}^{(i_{1}^{\prime})};\\ p_{1n}&=\left(\sum\limits_{\substack{(i_2,\ldots,i_{n-2})\in T_m^{n-3}}}p_{i_1^{\prime}i_2\dots i_{n-2}}(\bar{b}_{1},\bar{b}_3,\ldots,\bar{b}_{n})a_{34}^{(i_2)}\dots a_{n-1,n}^{(i_{n-2})}\right)x_{13}^{(i_1^{\prime})}. \end{aligned} \right. \end{align} $$

We set

(28) $$ \begin{align} \left\{ \begin{aligned} g_{2n}&=\sum\limits_{\substack{(i_2,\ldots,i_{n-2})\in T_m^{n-3}}}p_{i_1^{\prime}i_2\dots i_{n-2}}(\bar{b}_{2},\ldots,\bar{b}_{n})a_{34}^{(i_2)}\dots a_{n-1,n}^{(i_{n-2})};\\ g_{1n}&=\sum\limits_{\substack{(i_2,\ldots,i_{n-2})\in T_m^{n-3}}}p_{i_1^{\prime}i_2\dots i_{n-2}}(\bar{b}_{1},\bar{b}_3,\ldots,\bar{b}_{n})a_{34}^{(i_2)}\dots a_{n-1,n}^{(i_{n-2})}; \end{aligned} \right. \end{align} $$

and

$$\begin{align*}V=\{(i,i+1,k)~|~i=3,\ldots,n-1,k=1,\ldots,m\}. \end{align*}$$

Note that both $g_{2n}$ and $g_{1n}$ are polynomials over K on some commutative variables indexed by elements from V. We claim that $g_{2n},g_{1n}\neq 0$ . Indeed, we take $a_{jk}^{(i)}\in K$ , $(j,k,i)\in V$ such that

$$\begin{align*}\left\{ \begin{aligned} a_{s,s+1}^{(i_{s-1}^{\prime})}&=1,\quad\mbox{for all } s=3,\ldots,n-1;\\ a_{jk}^{(i)}&=0,\quad\mbox{otherwise}. \end{aligned} \right. \end{align*}$$

It follows from (28) that

$$ \begin{align*} \begin{aligned} g_{2n}&=p_{i_1^{\prime}\dots i_{n-2}^{\prime}}(\bar{b}_{2},\ldots,\bar{b}_{n})\neq 0;\\ g_{1n}&=p_{i_1^{\prime}\dots i_{n-2}^{\prime}}(\bar{b}_{1},\bar{b}_3,\ldots,\bar{b}_{n})\neq 0, \end{aligned} \end{align*} $$

as desired. It follows from (27) that

(29) $$ \begin{align} \left\{ \begin{aligned} p_{1,n-1}&=0;\\ p_{2n}&=g_{2n}x_{23}^{(i_{1}^{\prime})};\\ p_{1n}&=g_{1n}x_{13}^{(i_1^{\prime})}. \end{aligned} \right. \end{align} $$

We take

$$\begin{align*}\left\{ \begin{aligned} x_{23}^{(i_1^{\prime})}&=g_{2n}^{-1}a_{2n}^{\prime};\\ x_{13}^{(i_1^{\prime})}&=g_{1n}^{-1}a_{1n}^{\prime}. \end{aligned} \right. \end{align*}$$

It follows from (29) that

$$\begin{align*}\left\{ \begin{aligned} p_{1,n-1}&=0;\\ p_{2n}&=a_{2,n}^{\prime};\\ p_{1n}&=a_{1n}^{\prime}, \end{aligned} \right. \end{align*}$$

as desired. We obtain that

$$\begin{align*}p(u_1,\ldots,u_m)=(p_{s,n-2+s+t})=(a_{s,n-2+s+t}^{\prime})=A'. \end{align*}$$

This implies that $T_n(K)^{(n-3)}\subseteq p(T_n(K))$ . Hence $p(T_n(K))=T_n(K)^{(n-3)}$ .

We are ready to give the proof of the main result of the paper.

Proof The proof of Theorem 1.2

For any $A=(a_{s,r+s+t})\in T_n(K)^{(r-1)}$ , we set

$$\begin{align*}\left\{ \begin{aligned} f_{s,r+s}(x_{s,r+s})&=a_{s,r+s}-x_{s,r+s};\\ g_{s,r+s}(x_{s,r+s})&=x_{s,r+s} \end{aligned} \right. \end{align*}$$

for all $1\leq s<r+s\leq n$ . It is clear that both $f_{s,r+s}$ and $g_{s,r+s}$ are nonzero polynomials in commutative variables over K, where $1\leq s<r+s\leq n$ . It follows from Lemma 2.5 that there exist $b_{s,r+s}\in K$ , $1\leq s<r+s\leq n$ , such that

$$\begin{align*}\left\{ \begin{aligned} f_{s,r+s}(b_{s,r+s})&\neq 0;\\ g_{s,r+s}(b_{s,r+s})&\neq 0 \end{aligned} \right. \end{align*}$$

for all $1\leq s<r+s\leq n$ . That is,

$$\begin{align*}\left\{ \begin{aligned} a_{s,r+s}-b_{s,r+s}&\neq 0;\\ b_{s,r+s}&\neq 0 \end{aligned} \right. \end{align*}$$

for all $1\leq s<r+s\leq n$ . We set

$$\begin{align*}b_{s,r+s+t}=a_{s,r+s+t} \end{align*}$$

for all $1\leq s<r+s+t\leq n$ and $t>0$ and

$$\begin{align*}\left\{ \begin{aligned} c_{s,r+s}&=a_{s,r+s}-b_{s,r+s},\quad\mbox{for all } 1\leq s<r+s\leq n;\\ c_{s,r+s+t}&=0,\quad\mbox{for all } 1\leq s<r+s+t\leq n \text{ and } t>0. \end{aligned} \right. \end{align*}$$

We set

$$\begin{align*}B=(b_{s,r+s+t})\quad\mbox{and}\quad C=(c_{s,r+s+t}). \end{align*}$$

It is clear that

$$\begin{align*}A=B+C, \end{align*}$$

where $B,C\in T_n(K)^{(r-1)}$ with $b_{s,r+s},c_{s,r+s}\in K^*$ for all $1\leq s<r+s\leq n$ . In view of Lemma 3.7, we get that there exist $u_i,v_i\in T_n(K)$ , $i=1,\ldots ,m$ , such that

$$\begin{align*}p(u_1,\ldots,u_m)=B\quad\mbox{and}\quad p(v_1,\ldots,v_m)=C. \end{align*}$$

It follows that

$$\begin{align*}p(u_1,\ldots,u_m)+p(v_1,\ldots,v_m)=A. \end{align*}$$

This implies that

$$\begin{align*}T_n(K)^{(r-1)}\subseteq p(T_n(K))+p(T_n(K)). \end{align*}$$

In view of Lemma 2.2(ii), we note that $p(T_n(K))\subseteq T_n(K)^{(r-1)}$ . Since $T_n(K)^{(r-1)}$ is a subspace of $T_n(K),$ we get that

$$\begin{align*}p(T_n(K))+p(T_n(K))\subseteq T_n(K)^{(r-1)}. \end{align*}$$

We obtain that

$$\begin{align*}p(T_n(K))+p(T_n(K))=T_n(K)^{(r-1)}. \end{align*}$$

In particular, if $r=n-2,$ we get from Lemma 3.8 that

$$\begin{align*}p(T_n(K))=T_n(K)^{(n-3)}. \end{align*}$$

The proof of the result is complete.

We conclude the paper with following example.

Example 3.1 Let $n\geq 5$ and $1<r<n-2$ be integers. Let K be an infinite field. Let

$$\begin{align*}p(x,y)=[x,y]^r. \end{align*}$$

We have that ord $(p)=r$ and $p(T_n(K))\neq T_n(K)^{(r-1)}$ .

Proof It is easy to check that $p(T_r(K))=\{0\}$ . Set

$$\begin{align*}f(x,y)=[x,y]. \end{align*}$$

Note that f is a multilinear polynomial over K. It is clear that ord $(f)=1$ . In view of [10, Theorem 4.3] or [15, Theorem 1.1], we have that

$$\begin{align*}f(T_{r+1}(K))=T_{r+1}(K)^{(0)}. \end{align*}$$

It implies that there exist $A,B\in T_{r+1}(K)$ such that

$$\begin{align*}[A,B]=e_{12}+e_{23}+\cdots +e_{r,r+1}. \end{align*}$$

We get that

$$\begin{align*}p(A,B)=[A,B]^r=e_{1,r+1}\neq 0. \end{align*}$$

This implies that $p(T_{r+1}(K))\neq \{0\}$ . We obtain that ord $(p)=r$ .

Suppose on contrary that $p(T_n(K))=T_n(K)^{(r-1)}$ for some $n\geq 5$ and $1<r<n-2$ . For $e_{1,r+1}+e_{3,r+3}\in T_n(K)^{(r-1)}$ , we get that there exists $B,C\in T_n(K)$ such that

$$\begin{align*}p(B,C)=[B,C]^r=e_{1,r+1}+e_{3,r+3}. \end{align*}$$

It is clear that $[B,C]\in T_n(K)^{(0)}$ . We set

$$\begin{align*}[B,C]=(a_{s,1+s+t}). \end{align*}$$

It follows that

$$\begin{align*}[B,C]^r=e_{1,r+1}+e_{3,r+3}. \end{align*}$$

We get from the last relation that

$$\begin{align*}\left\{ \begin{aligned} (a_{12}a_{23}\dots a_{r,r+1})e_{1,r+1}&=e_{1,r+1};\\ (a_{23}a_{34}\dots a_{r+1,r+2})e_{2,r+2}&=0;\\ (a_{34}a_{45}\dots a_{r+2,r+3})e_{3,r+3}&=e_{3,r+3}. \end{aligned} \right. \end{align*}$$

This is a contradiction. We obtain that $p(T_n(K))\neq T_n(K)^{(r-1)}$ for all $n\geq 5$ and ${1<r<n-2}$ . This proves the result.

We remark that [16, Example 5.7] is a special case of Example 3.1 ( $r=2$ and $n=5$ ).