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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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 $p(T_n(K))=T_n(K)^{(n-3)}$ .

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.

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 [Reference Gargate and de Mello10, Theorem 4.3] or [Reference Luo and Wang15, 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.