Images of Generalized Multilinear Polynomials on Upper Triangular Matrix Algebras

Abstract

A polynomial is called a generalized multilinear polynomial if it is a sum of some multilinear polynomials over a field. The goal of this paper is to give a description of the images of generalized multilinear polynomials on upper triangular matrix algebras, generalizing all results of the Fagundes–Mello conjecture proposed by Fagundes and Mello in 2019.

1. Introduction

Let F be a field. Let $p(x_1,\dots ,x_m)$ be a polynomial in noncommutative variables over F. Let $\mathcal{R}$ be a noncommutative algebra over F. The set

$$p\left(\mathcal{R}\right)=\left\{p(b_1,\dots ,b_m)\right|b_1,\dots ,b_m\in \mathcal{R}\}$$

is said to be the image of p on $\mathcal{R}$.

In recent years, the images of polynomials on noncommutative algebras have been a relatively active research direction in the theory of noncommutative algebras (see [1] for details).

By $U_n\left(F\right)$ we denote the set of all $n\times n$ upper triangular matrices over F, where $n\ge 2$. For any $s\ge 0$, by $U_n{\left(F\right)}^{\left(s\right)}$ we denote the set of all $n\times n$ matrices over F whose entries $(j,k)$ are zero, for $k-j\le s$. We set $U_n{\left(F\right)}^{(-1)}=U_n\left(F\right)$.

In 2019, Fagundes [2] proved that the image of every multilinear polynomial on $U_n{\left(F\right)}^{\left(0\right)}$ over any field F is a vector space. In 2019, Fagundes and Mello [3] investigated the images of multilinear polynomials of lower degrees on $U_n\left(F\right)$. They presented the following:

Conjecture 1

([3]). The image of every multilinear polynomial on $U_n\left(F\right)$ is always a vector space.

Since Conjecture 1 was proposed, several results on Conjecture 1 have been obtained (see [4,5,6,7,8] for details). In 2025, under the weakest hypothesis $\left|F\right|\ge n$, Chen, Luo, and Wang [9] gave a positive solution for Conjecture 1.

It should be remarked that Conjecture 1 remains open for $\left|F\right|<n$.

We now give a generalization of the well-known definition of multilinear polynomials as follows:

Definition 1.

A polynomial over F is called a generalized multilinear polynomial over F if it is a sum of some multilinear polynomials over F.

It is clear that a multilinear polynomial is a type of generalized multilinear polynomial. A generalized multilinear polynomial is not a multilinear polynomial in general. For example, $p(x_1,x_2,x_3)=x_1{x}_2{x}_3+x_1{x}_2+x_2{x}_1+x_2$ is a generalized multilinear polynomial, but it is not a multilinear polynomial.

In the present paper, we shall give a description of the images of generalized multilinear polynomials on $U_n\left(F\right)$.

Theorem 1.

Let F be a field with $\left|F\right|\ge n$. Let $p(x_1,\dots ,x_m)$ be a generalized multilinear polynomial over F. Then $p\left(U_n\left(F\right)\right)=U_n{\left(F\right)}^{(r-1)}$, where r is the order of p.

We remark that Theorem 1.3 [9] is a consequence of Theorem 1.

In Section 2, we give the definition of the order of polynomials. In Section 3, we give some results on polynomials in commutative variables over finite fields, which will be used in the proof of Theorem 1. In Section 4, we present a proof of Theorem 1.

2. The Order of Polynomials

By $\mathcal{I}$ we denote the set of all positive integers. By $F^{\ast }$ we denote the set of all nonzero elements in F. We write

$$U_n\left(F\right)=\left(\begin{array}{cc}F& F^{n-1}\\ & U_{n-1}\left(F\right)\end{array}\right).$$

Let $\mathcal{R}$ be an algebra over F. Denote by $\mathcal{T}\left(\mathcal{R}\right)$ the set of all polynomial identities of $\mathcal{R}$. It is easy to check that

$$\mathcal{T}\left(F\right)\supset \mathcal{T}\left(U_2\left(F\right)\right)\supset \mathcal{T}\left(U_3\left(F\right)\right)\supset \cdots .$$

A polynomial p in noncommutative variables over F has order 0 if $p\notin \mathcal{T}\left(F\right)$. If $p\in \mathcal{T}\left(F\right)$, we define its order as the lowest integer r such that $p\in \mathcal{T}\left(U_r\left(F\right)\right)$ but $p\notin \mathcal{T}\left(U_{r+1}\left(F\right)\right)$, where $U_1\left(F\right)=F$.

Let $m,k\ge 1$ be integer. Set

$$V_m^k=\left\{(j_1,\dots ,j_k)\in {\mathcal{I}}^k|1\le j_1,\dots ,j_k\le m,j_u\ne j_v \mathrm{for} \mathrm{all} u\ne v\right\}.$$

Let $p(x_1,\dots ,x_m)$ be a generalized multilinear polynomial over F. By d we denote the degree of p. We can write

$$p(x_1,\dots ,x_m)=\sum _{k=1}^d\left(\sum _{(j_1,\dots ,j_k)\in V_m^k}{\lambda }_{j_1\cdots j_k}{x}_{j_1}\cdots x_{j_k}\right),$$

(1)

where ${\lambda }_{i_1\cdots i_k}\in F$.

For any ${\left(b_{jk}^{\left(i\right)}\right)}_{n\times n}\in U_n\left(F\right)$, where $1\le i\le m$, we set

$${\overline{b}}_{ii}=(b_{ii}^{\left(1\right)},\dots ,b_{ii}^{\left(m\right)})$$

for $i=1,\dots ,n$.

3. Some Results on Polynomials in Commutative Variables over Finite Fields

Let $F[y_1,\dots ,y_m]$ be the polynomial ring in m-commutative variables over F. An element $f(y_1,\dots ,y_m)\in F[y_1,\dots ,y_m]$ is said to be a polynomial in commutative variables over F. By $\partial \left(f\right)$ we denote the degree of f. By $\mathrm{L}(F^m,F)$ we denote the set of all functions from $F^m$ to F. Note that $\mathrm{L}(F^m,F)$ is a commutative ring in which operations are defined by

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill (f+g)(a_1,\dots ,a_m)& =f(a_1,\dots ,a_m)+g(a_1,\dots ,a_m)\hfill \\ \hfill \left(fg\right)(a_1,\dots ,a_m)& =f(a_1,\dots ,a_m)g(a_1,\dots ,a_m)\hfill \end{array}\end{array}$$

Define

$$\Psi :F[y_1,\dots ,y_m]\to \mathrm{L}(F^m,F)$$

by

$$\Psi \left(f\right)(a_1,\dots ,a_m)=f(a_1,\dots ,a_m)$$

for all $a_1,\dots ,a_m\in F$. It is straightforward to see that $\Psi$ is a homomorphism of rings. Note that $\Psi$ is injective if F is an infinite field (see [10] Theorem 2.19).

Although the following result must be well-known, we cannot find the source of the result. Hence, we give its proof for completeness.

Theorem 2.

Let F be a finite field with $\left|F\right|=q$. Then

$$Ker(\Psi )=⟨y_1^q-y_1,\dots ,y_m^q-y_m⟩,$$

the ideal of $F[y_1,\dots ,y_m]$ generalized by $\{y_i^q-y_i|i=1,\dots ,m\}$.

Proof. 

We set $K=Ker(\Psi )$ and

$$J=⟨y_1^q-y_1,\dots ,y_m^q-y_m⟩.$$

Since $a^q-a=0$ for all $a\in F$, we find that $J\subseteq K$. We first claim that every $f\in \mathrm{L}(F^m,F)$ is given by a reduced polynomial. Indeed, we define

$$P_f(y_1,\dots ,y_m)=\sum _{(a_1,\dots ,a_m)\in F^m}f(a_1,\dots ,a_m)\prod _{i=1}^m\left(1-{(y_i-a_i)}^{q-1}\right).$$

It is clear that $P_f(y_1,\dots ,y_m)$ is a reduced polynomial. Moreover, $P_f(a_1,\dots ,a_m)=f(a_1,\dots ,a_m)$ for all $(a_1,\dots ,a_m)\in F^m$. That is, f is given by $P_f$, as desired. This implies that $\Phi$ is surjective.

For any $r>q$, we note that

$$a^r=a^{q+(r-q)}=a^q{a}^{r-q}=a{a}^{r-q}=a^{r-q+1}.$$

Thus, from the last operation, for any polynomial f we find that there exists a reduced polynomial $\tilde{f}$ such that $f-\tilde{f}\in J$.

It is clear that the total number of all reduced monomials in $F[y_1,\dots ,y_m]$ is $q^m$. Since the set of all reduced polynomials forms an F-subspace of $F[y_1,\dots ,y_m]$ with the basis given by the reduced monomials, we obtain that the total number of all reduced polynomials is $q^{q^m}$. It is clear that $|\mathrm{L}(F^m,F)|=q^{q^m}$. From the surjectivity of $\Psi$, we know that

$$|F[y_1,\dots ,y_m]/K|=|\mathrm{L}(F^m,F)|=q^{q^m}.$$

It follows that the quotient map

$$F[y_1,\dots ,y_m]/J\to F[y_1,\dots ,y_m]/K$$

is a bijection. Since $J\subseteq K$, we finally obtain that $J=K$. □

Using Theorem 2 we give a generalization of Proposition 3.2 [9].

Theorem 3.

Let F be a finite field with $\left|F\right|=q$. Let $f(y_1,\dots ,y_m)$ be a nonzero polynomial in commutative variables over F. Suppose that $f\left(F\right)=\left\{0\right\}$. Then every monomial with the largest degree in f must contain a variable whose exponent is not less than q.

Proof. 

In view of Theorem 2, we have

$$f(y_1,\dots ,y_m)=f_1(y_1,\dots ,y_m)(y_1^q-y_1)+\cdots +f_s(y_1,\dots ,y_m)(y_s^q-y_s),$$

where every $f_i(y_i^q-y_i)\ne 0$. We can write

$$f_i(y_1,\dots ,y_m)=\sum _{j=1}^{s_i}{m}_{ij}(y_i^q-y_i),$$

where $m_{i1},\dots ,m_{i{s}_i}$ are F-independent monomials in $f_i$, $i=1,\dots ,s$ and $j=1,\dots ,s_i$. It follows that

$$f(y_1,\dots ,y_m)=\sum _{i=1}^s\sum _{j=1}^{s_i}{m}_{ij}(y_i^q-y_i).$$

(2)

Let M be a monomial with the largest degree in f. We obtain from (2) that either $M=\lambda m_{ij}{y}_i^q$ or $M=\lambda m_{ij}{y}_i$ for some $\lambda \in F^{\ast }$, $i\in \{1,\dots ,s\}$. If $M=\lambda m_{ij}{y}_i^q$, there is nothing to prove. We now assume that $M=\lambda m_{ij}{y}_i$.

Since M is a monomial with the largest degree in f and $\partial \left(m_{ij}{y}_i^q\right)>deg\left(M\right)$, we find that $m_{ij}{y}_i^q$ does not appear in f. This implies that $m_{ij}{y}_i^q$ is eliminated by a sum of monomials in $f_k$, where $k=1,\dots ,s$. It follows that

$$m_{ij}{y}_i^q+\lambda m_{ku}{y}_k^q=0or{m}_{ij}{y}_i^q+\lambda m_{ku}{y}_k=0,$$

where $\lambda \in F^{\ast }$ and $k\in \{1,\dots ,s\}$.

We now assume that $m_{ij}{y}_i^q+\lambda m_{ku}{y}_k^q=0$. If $i=k$, we obtain that $m_{ij}+\lambda m_{iu}=0$. This implies that $m_{ij}$ and $m_{iu}$ are F-dependent, which is a contradiction. Hence $i\ne k$. It follows that $m_{ij}=m^{\prime }{y}_k^q$, where $m^{\prime }$ is a monomial. Hence, $M=m_{ij}{y}_i=m^{\prime }{x}_k^q{y}_i$, as desired.

We next assume that $m_{ij}{y}_i^q+\lambda m_{ku}{y}_k=0$. We have that $m_{ku}{y}_k\ne 0$ in $f_k$. Hence, $m_{ku}{y}_k^q=m_{ku}(y_k^q-y_k)$ appears in $f_k$. Note that

$$\partial \left(m_{ku}{y}_k^q\right)>\partial \left(m_{ku}{y}_k\right)=\partial \left(m_{ij}{y}_i^q\right)>\partial \left(m_{ij}{y}_i\right)=\partial \left(M\right),$$

which is a contradiction. This proves the result. □

Lemma 1.

Suppose that $1\le r\le n-1$. Let F be a field. Let $f({\overline{y}}_1,\dots ,{\overline{y}}_{r+1})$ be a nonzero polynomial in commutative variables over F such that the degree of every variable in all monomials of f is 1. If $\left|F\right|>n-r$, then there exist ${\overline{a}}_1,\dots ,{\overline{a}}_n\in F^m$ such that

$$f({\overline{a}}_u,\dots ,{\overline{a}}_{r+u-1},{\overline{a}}_{r+u+v})\ne 0$$

for all $1\le u<r+u+v\le n$.

Proof. 

Set

$$h_{u,r+u+v}=f({\overline{y}}_u,\dots ,{\overline{y}}_{r+u-1},{\overline{y}}_{r+u+v}),$$

where $1\le u<r+u+v\le n$. Note that the degree of every variable in all nonzero monomials of $h_{u,r+u+v}$ is 1. We set

$$h({\overline{y}}_1,\dots ,{\overline{y}}_n)=\prod _{1\le u<r+u+v\le n}{h}_{u,r+u+v}.$$

We prove that $h\left(F\right)\ne \left\{0\right\}$.

If F is infinite, we obtain from [10] Theorem 2.19 that $h\left(F\right)\ne \left\{0\right\}$, as desired. We now assume that F is finite.

Suppose that $h\left(F\right)=\left\{0\right\}$. Let $P_{1,r+1}$ be a monomial with the largest degree in $h_{1,r+1}$. We may assume that

$$P_{1,r+1}=\mu y_{j_1}^{\left(i_1\right)}\cdots y_{j_d}^{\left(i_d\right)}$$

for some $\mu \in F^{\ast }$, $1\le j_1\le \cdots \le j_d\le r+1$ and $(i_1,\dots ,i_d)\in V_m^d$. We first discuss the case of $1\le j_1\le \cdots \le j_d<r+1$. We set

$$P_{u,r+u+v}=\mu y_{j_1+u-1}^{\left(i_1\right)}\cdots y_{j_d+u-1}^{\left(i_d\right)},$$

where $1\le s<r+u+v\le n$. We set

$$P=\prod _{1\le u<r+u+v\le n}{P}_{u,r+u+v}.$$

Note that P is a nonzero monomial with the largest degree in h. Take $y_{j_w+u-1}^{\left(i_w\right)}\in P$. Note that $y_{j_w+u-1}^{\left(i_w\right)}\notin P_{u^{\prime },r+u^{\prime }+v^{\prime }}$ for $1\le u^{\prime }<u^{\prime }+r+v^{\prime }\le n$ and $u^{\prime }\ne u$. It follows that the exponent of $y_{j_w+u-1}^{\left(i_w\right)}\in P$ is less than or equal to $n-r$. We next discuss the case of $j_1=\cdots =j_d=r+1$. Set

$$P_{s,r+u+v}=\mu y_{r+u+v}^{\left(i_1\right)}\cdots y_{r+u+v}^{\left(i_d\right)},$$

where $1\le u<r+u+v\le n$. Set $P={\prod }_{1\le u<r+u+v\le n}{P}_{u,r+u+v}$. It is easy to check that P is a monomial with the largest degree in h and the exponent of each variable in P is less than or equal to $n-r$.

Finally, we discuss the case of $1\le t\le d$ such that $j_t<r+1$ and $j_{t+1}=\cdots =j_d=r+1$. Set

$$P_{u,r+u+v}=\mu y_{j_1+u-1}^{\left(i_1\right)}\cdots y_{j_t+u-1}^{\left(i_t\right)}{y}_{r+u+v}^{\left(i_{t+1}\right)}\cdots y_{r+u+v}^{\left(i_d\right)}$$

for all $1\le u<r+u+v\le n$ and

$$P=\prod _{1\le u<r+u+v\le n}{P}_{u,r+u+v}.$$

It is easy to check that P is a monomial with the largest degree in h and the exponent of each variable in P is less than or equal to $n-r$.

From Theorem 3 we find that P contains a variable whose exponent is not less than $\left|F\right|$, which is a contradiction. Therefore, $h\left(F\right)\ne \left\{0\right\}$. It follows that there exist ${\overline{a}}_1,\dots ,{\overline{a}}_n\in F^m$ such that

$$h({\overline{a}}_1,\dots ,{\overline{a}}_n)=\prod _{1\le u<r+u+v\le n}{h}_{u,r+u+v}({\overline{a}}_u,\dots ,{\overline{a}}_{r+u-1},{\overline{a}}_{r+u+v})\ne 0.$$

This proves the result. □

4. The Proof of the Main Result

We will give the proof of our main result using the following two lemmas.

Lemma 2.

Suppose that $p\left(F\right)\ne \left\{0\right\}$. We have that $p\left(U_n\left(F\right)\right)=U_n\left(F\right)$.

Proof. 

It is clear that $p\left(U_n\left(F\right)\right)\subseteq U_n\left(F\right)$. We claim that $U_n\left(F\right)\subseteq p\left(U_n\left(F\right)\right)$. For any $A_1,\dots ,A_m\in U_n\left(F\right)$ with $A_i{A}_j=A_j{A}_i$ for all $i,j=1,\dots ,m$, we obtain from (1) that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill p(A_1,\dots ,A_m)& =\sum _{k=1}^d\left(\sum _{(i_1,\dots ,i_k)\in V_m^k}{\lambda }_{i_1\cdots i_k}{A}_{i_1}\cdots A_{i_k}\right)\hfill \\ & =\sum _{k=1}^d\left(\sum _{1\le i_1<\cdots <i_k\le m}{\beta }_{i_1\cdots i_k}{A}_{i_1}\cdots A_{i_k}\right),\hfill \end{array}\end{array}$$

(3)

where ${\beta }_{i_1\cdots i_k}\in F$. Since $p\left(F\right)\ne \left\{0\right\}$, we have that there exists a minimum integer $r\ge 1$ such that

$${\beta }_{i_1^{\prime }\cdots i_r^{\prime }}\ne 0.$$

We rewrite (3) as follows:

$$p(A_1,\dots ,A_m)=\sum _{k=r}^d\left(\sum _{1\le i_1<\cdots <i_k\le m}{\beta }_{i_1\cdots i_k}{A}_{i_1}\cdots A_{i_k}\right).$$

(4)

For any $A^{\prime }\in U_n\left(F\right)$, we take $A_i\in U_n\left(F\right)$, $i=1,\dots ,m$, where

$$\left\{\begin{array}{cc}\hfill A_{i_1^{\prime }}& ={\beta }_{i_1^{\prime }\cdots i_r^{\prime }}^{-1}{A}^{\prime };\hfill \\ \hfill A_{i_k^{\prime }}& =1_{U_n\left(F\right)},k=2,\dots ,r;\hfill \\ \hfill A_i& =0,otherwise.\hfill \end{array}\right.$$

Note that $A_i{A}_j=A_j{A}_i$ for all $i,j$. It follows from (4) that

$$p(A_1,\dots ,A_m)=A^{\prime }.$$

Hence $U_n\left(F\right)\subseteq p\left(U_n\left(F\right)\right)$, as desired. □

By using the same arguments as Lemma 4.2 [9], we give the following result.

Lemma 3.

Let $1\le r\le n-1$, where r is the order of p. Then $p\left(U_n\left(F\right)\right)=U_n{\left(F\right)}^{(r-1)}$.

Proof. 

In view of [11] Lemma 3.2, we have $p\left(U_n\left(F\right)\right)\subseteq U_n{\left(F\right)}^{(r-1)}$. We claim that $U_n{\left(F\right)}^{(r-1)}\subseteq p\left(U_n\left(F\right)\right)$. For any $B_i=\left(b_{jk}^{\left(i\right)}\right)\in U_n\left(F\right)$, $i=1,\dots ,m$, we obtain from [11] Lemma 3.2 that

$$p(B_1,\dots ,B_m)={\left(p_{u,r+u+v}\right)}_{n\times n}\in U_n{\left(F\right)}^{(r-1)},$$

where

$$p_{u,r+u+v}=\sum _{k=r}^{r+v}\left(\sum _{\begin{array}{c}u=j_1<\cdots <j_{k+1}=r+u+v\\ (i_1,\dots ,i_k)\in V_m^k\end{array}}{f}_{i_1\cdots i_k}({\overline{b}}_{j_1{j}_1},\dots ,{\overline{b}}_{j_{k+1}{j}_{k+1}})b_{j_1{j}_2}^{\left(i_1\right)}\cdots b_{j_k{j}_{k+1}}^{\left(i_k\right)}\right)$$

(5)

for all $1\le u<r+u+v\le n$. It is easy to check that the degree of every variable in all monomials of $f_{i_1,\dots ,i_k}$ is 1. In particular, $f_{i_1^{\prime }\cdots i_r^{\prime }}\ne 0$ for some $(i_1^{\prime },\dots ,i_r^{\prime })\in V_m^r$.

It follows from Lemma 1 that there exist ${\overline{a}}_1,\dots ,{\overline{a}}_n\in F^m$ such that

$$f_{i_1^{\prime }\cdots i_r^{\prime }}({\overline{a}}_u,\dots ,{\overline{a}}_{r+u-1},{\overline{a}}_{r+u+v})\ne 0,$$

(6)

where $1\le u<r+u+v\le n$. By taking ${\overline{b}}_{jj}={\overline{a}}_j$ in (5) for every j, we obtain

$$p(B_1,\dots ,B_m)={\left(p_{u,r+u+v}\right)}_{n\times n}$$

where

$$\begin{array}{c}\hfill \begin{array}{c}{p}_{u,r+u+v}=\sum _{k=r}^{r+v}\left(\sum _{\begin{array}{c}u=j_1<\cdots <j_{k+1}=r+u+v\\ (i_1,\dots ,i_k)\in V_m^k\end{array}}{f}_{i_1\cdots i_k}({\overline{a}}_{j_1},\dots ,{\overline{a}}_{j_{k+1}})b_{j_1{j}_2}^{\left(i_1\right)}\cdots b_{j_k{j}_{k+1}}^{\left(i_k\right)}\right)\hfill \\ =\sum _{\begin{array}{c}u=j_1<\cdots <j_{r+1}=r+u+v\\ (i_1,\dots ,i_r)\in V_m^r\end{array}}{f}_{i_1\cdots i_r}({\overline{a}}_{j_1},\dots ,{\overline{a}}_{j_{r+1}})b_{j_1{j}_2}^{\left(i_1\right)}\cdots b_{j_r{j}_{r+1}}^{\left(i_r\right)}\hfill \\ +\sum _{k=r+1}^{r+v}\left(\sum _{\begin{array}{c}u=j_1<\cdots <j_{k+1}=r+u+v\\ (i_1,\dots ,i_k)\in V_m^k\end{array}}{f}_{i_1\cdots i_k}({\overline{a}}_{j_1},\dots ,{\overline{a}}_{j_{k+1}})b_{j_1{j}_2}^{\left(i_1\right)}\cdots b_{j_k{j}_{k+1}}^{\left(i_k\right)}\right)\hfill \end{array}\end{array}$$

(7)

for all $1\le u\le r+u+v\le n$. Set

$$S=\left\{y_{r+u-1,r+u+v}^{\left(i_r^{\prime }\right)}|1\le u\le r+u+v\le n\right\}.$$

We define an order on S. If $v<v^{\prime }$, we set

$$y_{r+u-1,r+u+v}^{\left(i_r^{\prime }\right)}<y_{r+u^{\prime }-1,r+u^{\prime }+v^{\prime }}^{\left(i_r^{\prime }\right)}.$$

If $v=v^{\prime }$ and $u<u^{\prime }$, we set

$$y_{r+u-1,r+u+v}^{\left(i_r^{\prime }\right)}<y_{r+u^{\prime }-1,r+u^{\prime }+v^{\prime }}^{\left(i_r^{\prime }\right)}.$$

We first consider the case of $r=1$. Set

$${\overline{g}}_{u,1+u+v}=f_{i_1^{\prime }}\left({\widehat{a}}_{u,v}\right),$$

where

$${\widehat{a}}_{u,v}=({\overline{a}}_u,\dots ,{\overline{a}}_{r+u-1},{\overline{a}}_{r+u+v})$$

for all $1\le 1+u+v\le n$. We next consider the case of $r>1$. Set

$$g_{u,r+u+v}=\sum _{(i_1,\dots ,i_{r-1},i_r^{\prime })\in V_m^r}{f}_{i_1\cdots i_{r-1}{i}_r^{\prime }}\left({\overline{a}}_{u,v}\right)y_{u,u+1}^{\left(i_1\right)}\cdots y_{r+u-2,r+u-1}^{\left(i_{r-1}\right)}$$

for all $1\le u<r+u+v\le n$. In view of Proposition 3.4 [9], we find that there exist ${\overline{a}}_{12},\dots ,{\overline{a}}_{n-2,n-1}\in F^m$ such that

$$g_{u,r+u+v}({\overline{a}}_{u,u+1},\dots ,{\overline{a}}_{r+u-2.r+u-1})\ne 0$$

for all $1\le u<r+u+v\le n$. Set

$${\overline{g}}_{u,r+u+v}=g_{u,r+u+v}({\overline{a}}_{s,s+1},\dots ,{\overline{a}}_{r+s-2.r+s-1})$$

for all $1\le u<r+u+v\le n$. It follows from (7) that

$$\begin{array}{c}\hfill \begin{array}{c}{p}_{u,r+u+v}=\left(\sum _{\begin{array}{c}(i_1,\dots ,i_r)\in V_m^r\\ i_r=i_r^{\prime }\end{array}}{f}_{i_1\cdots i_r}\left({\widehat{a}}_{u,v}\right)b_{s,s+1}^{\left(i_1\right)}\cdots b_{r+u-2,r+u-1}^{\left(i_{r-1}\right)}\right)b_{r+u-1,r+u+v}^{\left(i_r^{\prime }\right)}\hfill \\ +\left(\sum _{\begin{array}{c}(i_1,\dots ,i_r)\in V_m^r\\ i_r\ne i_r^{\prime }\end{array}}{f}_{i_1\cdots i_r}\left({\widehat{a}}_{u,v}\right)b_{s,s+1}^{\left(i_1\right)}\cdots b_{r+u-2,r+u-1}^{\left(i_{r-1}\right)}\right)b_{r+u-1,r+u+v}^{\left(i_r\right)}\hfill \\ +\sum _{\begin{array}{c}u\le j_1<\cdots j_{r+1}\le r+u+v\\ (j_r,j_{r+1})\ne (r+u-1,r+u+v)\\ (i_1,\dots ,i_r)\in V_m^r\end{array}}{f}_{i_1\cdots i_r}({\overline{a}}_{j_1},\dots ,\overline{{a|}_{j_{r+1}}})b_{j_1{j}_2}^{\left(i_1\right)}\cdots b_{j_r{j}_{r+1}}^{\left(i_r\right)}\hfill \\ +\sum _{k=r+1}^{r+v}\left(\sum _{\begin{array}{c}u=j_1<j_2<\cdots <j_{k+1}=r+u+v\\ (i_1,\dots ,i_k)\in V_m^k\end{array}}{f}_{i_1\cdots i_k}({\overline{a}}_{j_1},\dots ,{\overline{a}}_{j_{k+1}})b_{j_1{j}_2}^{\left(i_1\right)}\cdots b_{j_k{j}_{k+1}}^{\left(i_k\right)}\right)\hfill \end{array}\end{array}$$

(8)

for all $1\le u<r+u+v\le n$. By taking

$$\left\{\begin{array}{cc}\hfill b_{j,j+1}^{\left(i\right)}& =a_{j,j+1}^{\left(i\right)}\mathrm{where} 1\le j\le n-2,1\le i\le m,\mathrm{and} i\ne i_r^{\prime };\hfill \\ \hfill b_{r+u-1,r+u+v}^{\left(i_r^{\prime }\right)}& =y_{r+u-1,r+u+v}^{\left(i_r^{\prime }\right)}\mathrm{where} 1\le u<r+u+v\le n;\hfill \\ \hfill b_{jk}^{\left(i\right)}& =0,\mathrm{otherwise}\hfill \end{array}\right.$$

in (8), we find that

$$p_{u,r+u+v}={\overline{g}}_{u,r+u+v}{y}_{r+u-1,r+u+v}^{\left(i_r^{\prime }\right)}+{\beta }_{u,r+u+v}$$

(9)

for all $1\le u\le r+u+v\le n$, where ${\overline{g}}_{u,r+u+v}\in F^{\ast }$ and ${\beta }_{u,r+u+v}$ is a polynomial in some variables that rank ahead of $y_{r+u-1,r+u+v}^{\left(i_r^{\prime }\right)}$. Note that ${\beta }_{1,r+1}\in K$.

For any $B^{\prime }={\left(b_{u,r+u+v}^{\prime }\right)}_{n\times n}\in U_n{\left(F\right)}^{(r-1)}$, we have the following equations:

$${\overline{g}}_{u,r+u+v}{y}_{r+u-1,r+u+v}^{\left(i_r^{\prime }\right)}+{\beta }_{u,r+u+v}=b_{u,r+u+v}^{\prime }$$

(10)

for all $1\le u<r+u+v\le n$. Note that ${\overline{g}}_{u,r+u+v}\in F^{\ast }$ for all $1\le u<r+u+v\le n$ and ${\beta }_{1,r+1}\in F$. It is easy to check that (10) has the following solution:

$$\left\{a_{r+u-1,r+u+v}^{\left(i_r^{\prime }\right)}\in F|1\le u<r+u+v\le n\right\}.$$

Set

$$y_{r+u-1,r+u+v}^{\left(i_r^{\prime }\right)}=a_{r+u-1,r+u+v}^{\left(i_r^{\prime }\right)}$$

for all $1\le u<r+u+v\le n$ in (10). We obtain from both (9) and (10) that

$$p_{u,r+u+v}\left(a_{r+u-1,r+u+v}^{\left(i_r^{\prime }\right)}\right)=b_{u,r+u+v}^{\prime }$$

for all $1\le u<r+u+v\le n$. We obtain that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill p(B_1,\dots ,B_m)& ={\left(p_{u,r+u+v}\left(a_{r+u-1,r+u+v}^{\left(i_r^{\prime }\right)}\right)\right)}_{n\times n}\hfill \\ & ={\left(b_{u,r+u+v}^{\prime }\right)}_{n\times n}\hfill \\ & =B^{\prime }.\hfill \end{array}\end{array}$$

It follows that $U_n{\left(F\right)}^{(r-1)}\subseteq p\left(U_n\left(F\right)\right)$. This proves the result. □

We are ready to give the proof of Theorem 1.

Proof of Theorem 1.

If $r=0$, from Lemma 2 we find that $p\left(U_n\left(F\right)\right)=U_n{\left(F\right)}^{(-1)}$. If $1\le r\le n-1$, from Lemma 3 we find that $p\left(U_n\left(F\right)\right)=U_n{\left(F\right)}^{(r-1)}$. If $r\ge n$, we find that $p\left(U_n\left(F\right)\right)=\left\{0\right\}$ from the definition of the order of polynomials. □

5. Conclusions

In the present paper, we provide a new concept of generalized multilinear polynomials over a field, which is a generalization of the well known concept of multilinear polynomials over a field. Under the same mild hypothesis as our previous result, we give a description of the images of generalized multilinear polynomials on upper triangular matrix algebras, which extends all results from the Fagundes–Mello conjecture proposed by Fagundes and Mello in 2019. It should be remarked that the hypothesis in our main result (Theorem 1) is currently the weakest hypothesis that ensures the Fagundes–Mello conjecture holds. By using some methods from reference [9], we complete the proof of our main result (Theorem 1.2). Further research work could be carried out to discuss the images of generalized multilinear polynomials on upper triangular matrix algebras over any finite fields.

It should be noted that studying the images of generalized multilinear polynomials on other algebras is also an interesting topic.