The image of polynomials in one variable on $2\times 2$ upper triangular matrix algebras

1.   Introduction

Let $K$ be a field. By $K\langle x_1, \ldots, x_n\rangle$ we denote the free $K$-algebra generated by non-commuting indeterminate $x_1, \ldots, x_n$ and refer to the elements of $K\langle x_1, \ldots, x_n\rangle$ as polynomials. Without special explanation we always assume that every polynomial over $K$ is a polynomial with zero constant term.

Images of polynomials evaluated on algebras play an important role in non-commutative algebras. The old and famous Lvov-Kaplansky conjecture asserts: The image of a multilinear polynomial in non-commutative variables over a field $K$ on the matrix algebra $M_n(K)$ is a vector space (see ^[6] for details). The parallel topic in group theory (the images of words in groups) has been studied extensively (see ^[2,17]).

In 2012, Kanel-Belov, Malev and Rowen ^[6] made a major breakthrough and solved the Lvov-Kaplansky conjecture for $n = 2$. In 2016, Kanel-Belov, Malev and Rowen ^[7] solved the Lvov-Kaplansky conjecture for $n = 3$. We remark that in ^[6,7] the authors considered multilinear elements only. Some results on the Lvov-Kaplansky conjecture have been obtained in ^[8,9,12].

We remark that the images of multilinear polynomials of small degree on Lie algebras ^[1,13] and Jordan algebras ^[11] have been discussed.

In 2019, Fagundes ^[3] gave a complete description of the image of multilinear polynomials on strictly upper triangular matrix algebras. In 2019, Fagundes and Mello ^[4] discussed the image of multilinear polynomials of degree up to four on upper triangular matrix algebras. They proposed the following variation of the Lvov-Kaplansky conjecture:

Conjecture 1.1. The image of a multilinear polynomial over a field $K$ on the uppertrinagular matrix algebra $T_n(K)$ is always a vector space.

In 2019, Wang ^[14] gave a positive answer of Conjecture 1 for $n = 2$. We remark that there exists a gap in the proof of [14,Theorem 1]. In 2019, Wang, Liu and Bai ^[15] gave a correct proof of [14,Theorem 1]. It should be mentioned that Conjecture 1 has been answered (see ^[5,10]).

In 2021, Zhou and Wang ^[18] gave a complete description of the image of completely homogeneous polynomials on $2\times 2$ upper triangular matrix algebras over an algebraically closed field. In the same year, Wang, Zhou and Luo ^[16] gave the Zariski topology structure of the image of polynomials on $2\times 2$ upper triangular matrix algebras over an algebraically closed field.

In the present paper, we give a description of the image of polynomials in one variable on $2\times 2$ upper triangular matrix algebras over an algebraically closed field. As consequences, we give concrete descriptions of the images of polynomials of degrees up to $4$ in one variable on $2\times 2$ upper triangular matrix algebras over an algebraically closed field.

2.   Single elements of polynomials in one variable

Let $K$ be a field. Let $p(x)$ be a polynomial in one variable over $K$. We now give the following definition, which is crucial for the proof of our main result.

Definition 2.1. Let $K$ be a field. Let $p(x)$ be a polynomial in one variable over $K$. An element $c\in K$ is said to be a single element of $p$ if $p(x)-c$ has a simple root in $K$.

By $S(p)$ we denote the set of all simple elements of $p$.

The following examples give complete descriptions of the set of all simple elements of polynomials of degree up to $4$ in one variable. We omit the proofs of both Examples 2.1 and 2.2.

Example 2.1. Let $K$ be an algebraically closed field. Let

where $\beta\in K$. Then one of the following statements holds:

(i) Suppose char$(K)\neq 2$. Then $S(p) = K\setminus\{-\frac{1}{4}\beta^2\}$;

(ii) Suppose char$(K) = 2$ and $\beta = 0$. Then $S(p) = \emptyset$;

(iii) Suppose char$(K) = 2$ and $\beta\neq 0$. Then $S(p) = K$.

Example 2.2. Let $K$ be an algebraically closed field. Let

where $\beta_1, \beta_2\in K$. Then one of the following statements holds:

(i) Suppose char$(K)\neq 3$ and $\beta_1^2 = 3\beta_2$. Then $S(p) = K\setminus\{-\frac{1}{27}\beta_1^3\}$;

(ii) Suppose char$(K)\neq 3$ and $\beta_1^2\neq 3\beta_2$. Then $S(p) = K$;

(iii) Suppose char$(K) = 3$ and $\beta_1 = \beta_2 = 0$. Then $S(p) = \emptyset$;

(iv) Suppose char$(K) = 3$ and either $\beta_1\neq 0$ or $\beta_2\neq 0$. Then $S(p) = K$.

Example 2.3. Let $K$ be an algebraically closed field. Let

where $\beta_1, \beta_2, \beta_3\in K$. Then one of the following statements holds:

(i) Suppose char$(K) = 2$ and $\beta_1 = \beta_3 = 0$. Then $S(p) = \emptyset$;

(ii) Suppose char$(K) = 2$ and either $\beta_1\neq 0$ or $\beta_3\neq 0$. Then $S(p) = K$;

(iii) Suppose char$(K)\neq 2$ and $\beta_1 = \beta_3 = 0$. Then $S(p) = \emptyset$;

(iv) Suppose char$(K)\neq 2$, $\beta_1 = 0$ and $\beta_3\neq 0$. Then $S(p) = K$;

(v) Suppose char$(K)\neq 2$, $\beta_1\neq 0$, $\beta_2 = \frac{1}{4}\beta_1^2+2\beta_3\beta_1^{-1}$. Then $S(p) = K\setminus\{-(\beta_1^{-1}\beta_3)^2\}$;

(vi) Suppose char$(K)\neq 2$, $\beta_1\neq 0$, $\beta_2\neq\frac{1}{4}\beta_1^2+2\beta_3\beta_1^{-1}$. Then $S(p) = K$.

Proof. We just give the proof of (i). The other statements can be proved analogously. For $c\in K$, we set

It is easy to check that $f(x)$ has no simple roots in $K$ if and only if

where $\alpha, \beta\in K$. Expanding (2.1) and comparing the coefficients of (2.1) we obtain

Suppose first that char$(K) = 2$ and $\beta_1 = \beta_3 = 0$. Let $\omega_1, \omega_2\in K$ be a solution of the following equation:

It follows that

Let $\gamma_1\in K$ be a solution of $x^2 = \omega_1$. Let $\gamma_2\in K$ be a solution of $x^2 = \omega_2$. We have

It follows that

In view of Definition 1, we get that $c\not\in S(p)$. Hence $S(p) = \emptyset$. This proves (i).

3.   The main result

Set $K^* = K\setminus\{0\}$. Let $T_2(K)$ be the set of all $2\times 2$ upper triangular matrices over $K$.

We give a description of the image of polynomials in one variable on $2\times 2$ upper triangular matrix algebras over an algebraically closed field.

Theorem 3.1. Let $d\geq 1$ be integer. Let $K$ be an algebraically closed field.Let

where $\beta_i\in K$ for $i = 1, \ldots, d$ with $\beta_d\neq 0$. We have

In particular, $p(T_2(K))$ is not a vector space if $S(p)\neq K$.

Proof. For $\left(\begin{array}{cc} a & m\\ & a \end{array} \right)\in T_2(K)$, we claim that

where $p'(x)$ is the derivation of $p(x)$. Indeed, we have

For $\left(\begin{array}{cc} a & m\\ & b \end{array} \right)\in T_2(K)$, where $a\neq b$, we claim that

Indeed, we have

For any $\left(\begin{array}{cc} a' & m'\\ & b' \end{array} \right)\in T_2(K)$, where $a'\neq b'$, we have that there exist $a, b\in K$ such that

Note that $a\neq b$. Set

Take $u = \left(\begin{array}{cc} a & \lambda m'\\ & b \end{array} \right)$. It follows from (3.2) that

This implies that

For any $\left(\begin{array}{cc} a' & 0\\ & a' \end{array} \right)\in T_2(K)$, we have that there exists $a\in K$ such that $p(a) = a'$. Take $u = \left(\begin{array}{cc} a & 0\\ & a \end{array} \right)\in T_2(K)$. It follows from (3.1) that

This implies that

where $I_2$ is the identity matrix of $T_2(K)$. For $c\in S(p)$, we set

In view of Definition 2.1, we have that $f(x)$ has a simple root $\omega\in K$. We have

and

where $f'$ and $p'$ are the derivations of $f$ and $p$, respectively. Set

for $m\in K$. It follows from (3.1) that

This implies that

We get from (3.4)–(3.6) that

For any $u = \left(\begin{array}{cc} a & m\\ & a \end{array} \right)\in T_2(K)$, we get from (3.1) that

Suppose first $p'(a) = 0$. We have

Suppose next $p'(a)\neq 0$. Set

It is clear that

This implies that $a\in K$ is a simple root of $f(x)$.

In view of Definition 2.1, we get $p(a)\in S(p)$. We have

For any $u = \left(\begin{array}{cc} a & m\\ & b \end{array} \right)\in T_2(K)$, where $a\neq b$, we get from (3.2) that

Suppose first $p(a)\neq p(b)$. We have

Suppose next $p(a) = p(b)$. We have that

We get from (3.3)–(3.10) that

We obtain

Suppose $S(p)\neq K$. We claim that $p(T_2(K))$ is not a vector space.

Take $a\in K\setminus S(p)$. Suppose first $0\in S(p)$. Take

We have

This implies that $p(T_2(K))$ is not a vector space. Suppose next $0\not\in S(p)$. Take $\left(\begin{array}{cc} 1 & 1\\ & 0 \end{array}\right), \left(\begin{array}{cc} -1 & 0\\ & 0 \end{array}\right)\in p(T_2(K))$. We have

This implies that $p(T_2(K))$ is not a vector space. The proof of the result is complete.

As a consequence, we give a concrete description of the image of polynomials of degree up to $4$ in one variable on $2\times 2$ upper triangular matrix algebras over an algebraically closed field.

Corollary 3.1. Let $K$ be an algebraically closed field. We have

(1) Let $p(x) = x^2+\beta x$, where $\beta\in K$. Then one of the following statementsholds:

(i) Suppose char$(K)\neq 2$. Then

is not a vector space;

(ii) Suppose char$(K) = 2$ and $\beta = 0$. Then

is not a vector space;

(iii) Suppose char$(K) = 2$ and $\beta\neq 0$. Then $p(T_2(K)) = T_2(K)$.

(2) Let $p(x) = x^3+\beta_1x^2+\beta_2x$, where $\beta_1, \beta_2\in K$. Then one of the following statementsholds:

(i) Suppose char$(K)\neq 3$ and $\beta_1^2 = 3\beta_2$. Then

is not a vector space;

(ii) Suppose char$(K)\neq 3$ and $\beta_1^2\neq 3\beta_2$. Then $p(T_2(K)) = T_2(K)$;

(iii) Suppose char$(K) = 3$ and $\beta_1 = \beta_2 = 0$. Then

is not a vector space;

(iv) Suppose char$(K) = 3$ and either $\beta_1\neq 0$ or $\beta_2\neq 0$.Then $p(T_2(K)) = T_2(K)$.

(3) Let $p(x) = x^4+\beta_1x^3+\beta_2x^2+\beta_3x$, where $\beta_1, \beta_2, \beta_3\in K$. Then one of the following statements holds:

(i) Suppose char$(K) = 2$ and $\beta_1 = \beta_3 = 0$. Then

is not a vector space

(ii) Suppose char$(K) = 2$ and either $\beta_1\neq 0$ or $\beta_3\neq 0$. Then $p(T_2(K)) = T_2(K)$;

(iii) Suppose char$(K)\neq 2$ and $\beta_1 = \beta_3 = 0$. Then

is not a vector space;

(iv) Suppose char$(K)\neq 2$, $\beta_1 = 0$ and $\beta_3\neq 0$. Then $p(T_2(K)) = T_2(K)$;

(v) Suppose char$(K)\neq 2$, $\beta_1\neq 0$, $\beta_2 = \frac{1}{4}\beta_1^2+2\beta_3\beta_1^{-1}$. Then

is not a vector space;

(vi) Suppose char$(K)\neq 2$, $\beta_1\neq 0$, $\beta_2\neq\frac{1}{4}\beta_1^2+2\beta_3\beta_1^{-1}$. Then $p(T_2(K)) = T_2(K)$.

Proof. The statement (1) follows from both Example 2.1 and Theorem 3.1. The statement (2) follows from both Example 2.2 and Theorem 3.1. The statement (3) follows from both Example 2.3 and Theorem 3.1.

4.   Conclusions

In this paper, we first defined the set of all single elements of polynomials in one variable. We next gave a description of the image of polynomials in one variable on $2\times 2$ upper triangular matrix algebras over an algebraically closed field. As an application of our main result, we gave concrete descriptions of the images of polynomials of degrees up to $4$ in one variable on $2\times 2$ upper triangular matrix algebras over an algebraically closed field.

Conflict of interest

The authors declare no conflicts of interest in this paper.