Central polynomials with involution for the algebra of $2\times 2$ upper triangular matrices

TL;DR

This paper characterizes all *-central polynomials for the algebra of 2x2 upper triangular matrices over a field with involution of the first kind, expanding understanding of polynomial identities in this algebra.

Contribution

It provides a complete description of *-central polynomials for UT_2(F) with involution of the first kind, a new result in the theory of polynomial identities.

Findings

Describes all *-central polynomials for UT_2(F) with involution of the first kind.

Extends the understanding of polynomial identities in upper triangular matrix algebras.

Contributes to the classification of polynomial identities with involution.

Abstract

Let F be a field of characteristic different from $2$, and let $UT_2(F)$ be the algebra of $2\times 2$ upper triangular matrices over $F$. For every involution of the first kind on $UT_2(F)$, we describe the set of all $*$-central polynomials for this algebra.

Equations

$$f(a_1,\dots ,a_n)=0$$

$$f(a_{1},\ldots,a_{n})\in Z(A)\ \mbox{(center of $A$)}$$

$$C(A)\supseteq Id(A)+F.$$

$$[x_1,x_2{]}^2$$

$$C(U{T}_n(F))=Id(U{T}_n(F))+F$$

$$f(a_1,a_1^{⊛},\dots ,a_n,a_n^{⊛})=0$$

$$f(a_1,a_1^{⊛},\dots ,a_n,a_n^{⊛})\in Z(A)$$

$$C(U{T}_2(F),⊛)\ne Id(U{T}_2(F),⊛)+F.$$

$$a=(1/2)(a+a^{\ast })+(1/2)(a-a^{\ast }).$$

$$\left(\begin{array}[]{cc}a&c\\ 0&b\end{array}\right)^{\star}=\left(\begin{array}[]{cc}b&c\\ 0&a\end{array}\right)\ \ \ \mbox{and}\ \ \ \left(\begin{array}[]{cc}a&c\\ 0&b\end{array}\right)^{s}=\left(\begin{array}[]{cc}b&-c\\ 0&a\end{array}\right)$$

$$(U{T}_2(F),\ast )\simeq (U{T}_2(F),\star )or(U{T}_2(F),\ast )\simeq (U{T}_2(F),s).$$

$$(x_1{x}_2{x}_3^{\ast }+2x_1^{\ast }{x}_4{)}^{\ast }=x_3{x}_2^{\ast }{x}_1^{\ast }+2x_4^{\ast }{x}_1.$$

$$Y=\{y_1,y_2,\dots \},Z=\{z_1,z_2,\dots \},y_i=x_i+x_i^{\ast },z_i=x_i-x_i^{\ast }$$

$$f(a_1,\dots ,a_n,b_1,\dots ,b_m)=0$$

$$\phi (f^{\ast })=(\phi (f){)}^{\ast }$$

$$f(g_1,\dots ,g_n,h_1,\dots ,h_m)\in Id(A,⊛)$$

$$[a_1,a_2]=a_1{a}_2-a_2{a}_1\text{mbox}and[a_1,\dots ,a_{n-1},a_n]=[[a_1,\dots ,a_{n-1}],a_n]$$

$$z_1^{r_1}\cdots z_m^{r_m}{f}_1\cdots f_t$$

$$y_1^{s_1}\cdots y_n^{s_n}{g}_{(s_1,\dots ,s_n)}$$

$$\{w+(B\cap I):w\in W\}$$

$$y_1^{s_1}\cdots y_n^{s_n}w+I,$$

$$g(a_1,a_1^{⊛},\dots ,a_n,a_n^{⊛})\in Z(A)$$

$$f(a_1,\dots ,a_n,b_1,\dots ,b_m)\in Z(A)$$

$$f(g_1,\dots ,g_n,h_1,\dots ,h_m)$$

$$f=\sum _{i=0}^{d_w}{f}^{(i)}$$

$$⟨f{⟩}^{TS(\ast )}=⟨f^{(0)},f^{(1)},\dots ,f^{(d_w)}{⟩}^{TS(\ast )}.$$

$$f=f^{+}+f^{-}$$

$$f(Y_1,\dots ,Y_n,Z_1,\dots ,Z_m)\in A^{-}$$

$$f(Y_1,\dots ,Y_n,Z_1,\dots ,Z_m)-f^{-}(Y_1,\dots ,Y_n,Z_1,\dots ,Z_m)=f^{+}(Y_1,\dots ,Y_n,Z_1,\dots ,Z_m)$$

$$f^{+}(Y_1,\dots ,Y_n,Z_1,\dots ,Z_m)=0$$

Full text

Keywords: Involution, Upper triangular matrices, Identities with involution, Central polynomials with involution, PI-algebra.

2010 AMS MSC Classification: 16R10, 16R50, 16W10.

1 Introduction

Let $F$ be a field. In this paper, every algebra is unitary associative over $F$. Let $F\langle X\rangle$ be the free unitary associative algebra, freely generated over $F$ by the infinite set $X=\{x_{1},x_{2},\ldots\}$.

A polynomial $f(x_{1},\ldots,x_{n})\in F\langle X\rangle$ is a polynomial identity for an algebra $A$ if

[TABLE]

for all $a_{1},\ldots,a_{n}\in A$. Denote by $Id(A)$ the set of all polynomial identities for $A$. It is known that $Id(A)$ is a T-ideal, that is, an ideal closed under all endomorphisms of $F\langle X\rangle$.

If $S\subseteq F\langle X\rangle$, we denote by $\langle S\rangle^{T}$ the T-ideal generated by $S$, that is, the intersection of all T-ideals containing $S$. Given a T-ideal $I$, if there exists a finite set $S$ such that $I=\langle S\rangle^{T}$, we say that $I$ is finitely generated as a T-ideal.

In 1950, Specht [25] posed the following problem:

Problem 1.1** (Specht’s problem).**

Is $Id(A)$ finitely generated, as a T-ideal, for every algebra $A$ ?

The answers to this question are:

a) Yes, if char$(F)=0$. Kemer [16].

b) No, if char$(F)\neq 0$. Belov ([4]), Grishin ([14]) and Shchigolev ([21]).

In general, the description of $Id(A)$ is a hard problem. The algebra $UT_{n}(F)$ of $n\times n$ upper triangular matrices plays an important role in the theory of PI-algebras. Maltsev [17] described $Id(UT_{n}(F))$ when char$(F)=0$, and Siderov [23] when $F$ is any field. In particular, they proved that $Id(UT_{n}(F))$ is finitely generated, as a T-ideal.

A T-space is a vector subspace of $F\langle X\rangle$ closed under all endomorphisms of $F\langle X\rangle$. Every T-ideal is a T-space. Another important T-space is the set of all central polynomials for an algebra $A$, denoted by $C(A)$. A polynomial $f(x_{1},\ldots,x_{n})\in F\langle X\rangle$ is a central polynomial for an algebra $A$ if

[TABLE]

for all $a_{1},\ldots,a_{n}\in A$. Note that

[TABLE]

Thus, some authors don’t include $Id(A)+F$ in the definition of $C(A)$. In this paper, if $f\in Id(A)+F$, we say that $f$ is a trivial central polynomial for $A$.

Let $M_{n}(F)$ be the $n\times n$ matrix algebra. It is known that

[TABLE]

is a non-trivial central polynomial for $M_{2}(F)$. Here, $[x_{1},x_{2}]=x_{1}x_{2}-x_{2}x_{1}$ is the commutator of $x_{1}$ and $x_{2}$. In 1956, Kaplansky [15] posed the following problem:

Problem 1.2**.**

Does there exist a non-trivial central polynomial for $M_{n}(F)$ for all $n\geq 3$ ?

Formanek ([13]) and Razmyslov ([19]) answer “yes” to the question, and this was very important for ring theory. Let $\tau(n)$ be the minimal degree of the non-trivial central polynomial for $M_{n}(F)$ when char$(F)=0$. We known that $\tau(1)=1$ and $\tau(2)=4$. Drensky and Kasparian [11, 12] proved that $\tau(3)=8$. We don’t known $\tau(n)$ when $n\geq 4$. It is an open problem.

If $S\subseteq F\langle X\rangle$, we denote by $\langle S\rangle^{TS}$ the T-space generated by $S$, that is, the intersection of all T-spaces containing $S$. Given a T-space $I$, if there exists a finite set $S$ such that $I=\langle S\rangle^{TS}$, we say that $I$ is finitely generated as a T-space. Shchigolev [22] proved the following theorem:

Theorem 1.3**.**

If $char(F)=0$ then every T-space is finitely generated.

If $F$ is an infinite field of char$(F)>2$, we have an important example of non-finitely generated T-space: it is $C(G)$, where $G$ is the infinite dimensional Grassmann algebra. See [3, 6].

It is well known that

[TABLE]

for all $n\geq 2$. See [20, Exercise 1.4.2] and [10, Example 3.2]. For the algebra $M_{2}(F)$, the T-space $C(M_{2}(F))$ was described when $F$ is an infinite field of char$(F)\neq 2$. See [7, 18].

From now on, $F$ will be a field of char$(F)\neq 2$. Furthermore, we will consider algebras with involution of the first kind only.

Let $X=\{x_{1},x_{2},\ldots\}$, $X^{*}=\{x_{1}^{*},x_{2}^{*},\ldots\}$ be two disjoint infinite sets. Denote by $F\langle X\cup X^{*}\rangle$ the free unitary associative algebra, freely generated by $X\cup X^{*}$. Let $A$ be an algebra with involution $\circledast$. A polynomial $f(x_{1},x_{1}^{*},\ldots,x_{n},x_{n}^{*})\in F\langle X\cup X^{*}\rangle$ is a polynomial identity with involution (or $*$-polynomial identity) for $(A,\circledast)$ if

[TABLE]

for all $a_{1},\ldots,a_{n}\in A$. Denote by $Id(A,\circledast)$ the set of all $*$-polynomial identities for $(A,\circledast)$. It is known that $Id(A,\circledast)$ is a $T(*)$-ideal, that is, an $*$-ideal of $F\langle X\cup X^{*}\rangle$ closed under all $*$-endomorphisms of $F\langle X\cup X^{*}\rangle$.

If $S\subseteq F\langle X\cup X^{*}\rangle$, we denote by $\langle S\rangle^{T(*)}$ the $T(*)$-ideal generated by $S$, that is, the intersection of all $T(*)$-ideals containing $S$. Given a $T(*)$-ideal $I$, if there exists a finite set $S$ such that $I=\langle S\rangle^{T(*)}$, we say that $I$ is finitely generated as a $T(*)$-ideal.

Recently, Aljadeff, Giambruno, Karasik ([1]) and Sviridova ([26]) proved the following:

Theorem 1.4**.**

Let $F$ be a field of $char(F)=0$. If $A$ is an algebra with involution $\circledast$, then $Id(A,\circledast)$ is finitely generated as a $T(*)$-ideal

Di Vincenzo, Koshlukov, La Scala [28] described the involutions of the first kind on $UT_{n}(F)$. They proved that there exist two classes of inequivalent involutions when $n$ is even and a single class otherwise. They also described:

a) $Id(UT_{2}(F),\circledast)$ when $F$ is infinite,

b) $Id(UT_{3}(F),\circledast)$ when char$(F)=0$,

for all involutions of the first kind on $UT_{2}(F)$ and $UT_{3}(F)$ respectively. Urure and Gonçalves [27] described $Id(UT_{2}(F),\circledast)$ when $F$ is finite. In particular, $Id(UT_{2}(F),\circledast)$ is finitely generated as a $T(*)$-ideal (see [27, 28]). It is an open problem to describe $Id(UT_{n}(F),\circledast)$ in other cases.

Now, a $T(*)$-space is a vector subspace of $F\langle X\cup X^{*}\rangle$ closed under all $*$-endomorphisms of $F\langle X\cup X^{*}\rangle$. Every $T(*)$-ideal is a $T(*)$-space. Another important $T(*)$-space is the set of all $*$-central polynomials for an algebra with involution $(A,\circledast)$, denoted by $C(A,\circledast)$. A polynomial $f(x_{1},x_{1}^{*},\ldots,x_{n},x_{n}^{*})\in F\langle X\cup X^{*}\rangle$ is a central polynomial with involution (or $*$-central polynomial) for $(A,\circledast)$ if

[TABLE]

for all $a_{1},\ldots,a_{n}\in A$.

If $W\subseteq F\langle X\cup X^{*}\rangle$, we denote by $\langle W\rangle^{TS(*)}$ the $T(*)$-space generated by $W$, that is, the intersection of all $T(*)$-spaces containing $W$. Given a $T(*)$-space $I$, if there exists a finite set $W$ such that $I=\langle W\rangle^{TS(*)}$, we say that $I$ is finitely generated as a $T(*)$-space.

If $F$ is an infinite field, Brandão and Koshlukov [5] decribed $C(M_{2}(F),\circledast)$ for every involution $\circledast$ on $M_{2}(F)$. Silva [24] studied $C(M_{1,1}(G),\circledast)$.

In this paper, we describe $C(UT_{2}(F),\circledast)$ for every involution of the first kind $\circledast$ and every field $F$ (finite or infinite) with char$(F)\neq 2$. In particular, we prove that

[TABLE]

Compare this information with (1). Moreover, we prove the following theorem:

Theorem 1.5**.**

Let $F$ be a field of $char(F)\neq 2$. If $\circledast$ is an involution of the first kind on $UT_{2}(F)$, then $C(UT_{2}(F),\circledast)$ is finitely generated as a $T(*)$-space.

2 Involution

From now on $F$ will be a field of char$(F)\neq 2$. Let $A$ be an unitary associative algebra over $F$. A map $*:A\rightarrow A$ is an involution on $A$ if

a) $(a+b)^{*}=a^{*}+b^{*}$ for all $a,b\in A$,

b) $(ab)^{*}=b^{*}a^{*}$ for all $a,b\in A$,

c) $(a^{*})^{*}=a$ for all $a\in A$.

Let $Z(A)$ be the center of $A$. If $a^{*}=a$ for all $a\in Z(A)$, then $*$ is called an involution of the first kind on $A$. Otherwise $*$ is called an involution of the second kind on $A$. From now on we consider involutions of the first kind only. In this case $(\lambda a)^{*}=\lambda(a^{*})$ for all $\lambda\in F$, $a\in A$. An element $a\in A$ is said to be symmetric if $a^{*}=a$. It’s skew-symmetric if $a^{*}=-a$. Denote by $A^{+}$ and $A^{-}$ the following vector spaces: $A^{+}=\{a\in A:\ a^{*}=a\}$ and $A^{-}=\{a\in A:\ a^{*}=-a\}$. If $a\in A$ then

[TABLE]

Therefore $A=A^{+}\oplus A^{-}$ as a vector space.

Let $(A,*)$ and $(B,\circ)$ be algebras with involutions $*$ and $\circ$ respectively. We say that they are isomorphic as algebras with involution if there exists an algebra isomorphism $\varphi:A\rightarrow B$ such that $\varphi(a^{*})=(\varphi(a))^{\circ}$ for all $a\in A$. In this case we denote $(A,*)\simeq(B,\circ)$.

Denote by $\star$ and $s$ the following involutions on $UT_{2}(F)$:

[TABLE]

for all $a,b,c\in F$. By [28, Propositions 2.5 and 2.6] we have the next corollary:

Corollary 2.1**.**

If $*$ is an involution of the first kind on $UT_{2}(F)$, then

[TABLE]

Moreover, $(UT_{2}(F),\star)$ and $(UT_{2}(F),s)$ are not isomorphic as algebras with involution.

3 $*$-polynomial identities and $*$-central polynomials

Let $X=\{x_{1},x_{2},\ldots\}$ and $X^{*}=\{x_{1}^{*},x_{2}^{*},\ldots\}$ be two disjoint infinite sets. Denote by $F\langle X\cup X^{*}\rangle$ the free unitary associative algebra freely generated by $X\cup X^{*}$ over $F$. This algebra has an involution $*:F\langle X\cup X^{*}\rangle\rightarrow F\langle X\cup X^{*}\rangle$ induced by the map $X\cup X^{*}\rightarrow X\cup X^{*}$ defined by $x_{i}\rightarrow x_{i}^{*}$ and $x_{i}^{*}\rightarrow x_{i}$. For example

[TABLE]

Let $F\langle Y\cup Z\rangle$ be the free unitary associative algebra freely generated by $Y\cup Z$ over $F$, where

[TABLE]

for all $i\geq 1$. Note that $F\langle X\cup X^{*}\rangle=F\langle Y\cup Z\rangle$, $y_{i}$ is symmetric and $z_{i}$ is skew-symmetric. Thus $f(y_{1},\ldots,y_{n},z_{1},\ldots,z_{m})$ is a $*$-polynomial identity for an algebra with involution $(A,\circledast)$ if

[TABLE]

for all $a_{1},\ldots,a_{n}\in A^{+}$ and $b_{1},\ldots,b_{m}\in A^{-}$. Denote by $Id(A,\circledast)$ the set of all $*$-polynomial identities for $(A,\circledast)$. This set is a $T(*)$-ideal that is an ideal invariant under all $*$-endomorphisms of $F\langle X\cup X^{*}\rangle$. Here $*$-endomorphism means an endomorphism $\varphi$ of the algebra $F\langle X\cup X^{*}\rangle$ such that

[TABLE]

for all $f\in F\langle X\cup X^{*}\rangle$. In particular, if $f(y_{1},\ldots,y_{n},z_{1},\ldots,z_{m})\in Id(A,\circledast)$ then

[TABLE]

for all $g_{1},\ldots,g_{n}\in F\langle Y\cup Z\rangle^{+}$ and $h_{1},\ldots,h_{m}\in F\langle Y\cup Z\rangle^{-}$.

We denote by $\langle W\rangle^{T(*)}$ the $T(*)$-ideal of $F\langle Y\cup Z\rangle$ generated by $W$ that is the smallest $T(*)$-ideal of $F\langle Y\cup Z\rangle$ containing $W$.

If $A$ is an algebra we denote the commutators as follows:

[TABLE]

for all $a_{1},\ldots,a_{n}\in A$, $n\geq 2$. A polynomial $f(y_{1},\ldots,y_{n},z_{1},\ldots,z_{m})\in F\langle Y\cup Z\rangle$ is called $Y$-proper if $f$ is a linear combination of polynomials

[TABLE]

where $t\geq 0$, $r_{1},\ldots,r_{m}\geq 0$ and $f_{i}\in F\langle Y\cup Z\rangle$ is a commutator of lenght $\geq 2$ for all $i=1,\ldots,t$. Denote by $B$ the vector space of all $Y$-proper polynomials. By the Poincaré-Birkhoff-Witt theorem every element $g(y_{1},\ldots,y_{n},z_{1},\ldots,z_{m})\in F\langle Y\cup Z\rangle$ is a linear combination of polynomials

[TABLE]

where $s_{1},\ldots,s_{n}\geq 0$ and $g_{(s_{1},\ldots,s_{n})}\in B$.

Using [9, Lemma 2.1] and similar arguments as in [8, Proposition 4.3.11] we state the following:

Proposition 3.1**.**

Let $F$ be an infinite field of char$(F)\neq 2$. Let $I$ be a $T(*)$-ideal of $F\langle Y\cup Z\rangle$. Consider $W\subset B$ such that

[TABLE]

is a basis for the quotient vector space $B/(B\cap I)$. Then the set of all polynomials

[TABLE]

where $s_{1},\ldots,s_{n}\geq 0,$  $n\geq 1$ and $w\in W$, is a basis for the quotient vector space $F\langle Y\cup Z\rangle/I$.

Recall that a polynomial $g(x_{1},x_{1}^{*},\ldots,x_{n},x_{n}^{*})\in F\langle X\cup X^{*}\rangle$ is a $*$-central polynomial for an algebra with involution $(A,\circledast)$ if

[TABLE]

for all $a_{1},\ldots,a_{n}\in A$. Note that $f(y_{1},\ldots,y_{n},z_{1},\ldots,z_{m})\in F\langle Y\cup Z\rangle$ is a $*$-central polynomial for $(A,\circledast)$ if

[TABLE]

for all $a_{1},\ldots,a_{n}\in A^{+}$ and $b_{1},\ldots,b_{m}\in A^{-}$. Denote by $C(A,\circledast)$ the set of all $*$-central polynomials for $(A,\circledast)$. This set is a $T(*)$-space that is a vector space invariant under all $*$-endomorphisms of $F\langle X\cup X^{*}\rangle$.

If $W\subseteq F\langle Y\cup Z\rangle$ then we denote by $\langle W\rangle^{TS(*)}$ the $T(*)$-space generated by $W$ that is the smallest $T(*)$-space of $F\langle Y\cup Z\rangle$ containing $W$. It’s the vector space generated by the polynomials

[TABLE]

where $f(y_{1},\ldots,y_{n},z_{1},\ldots,z_{m})\in W$ ,  $g_{1},\ldots,g_{n}\in F\langle Y\cup Z\rangle^{+}$ and $h_{1},\ldots,h_{m}\in F\langle Y\cup Z\rangle^{-}$.

Using similar arguments as in [8, Proposition 4.2.3] we state the following:

Proposition 3.2**.**

Let $F$ be a field (finite or infinite) with $|F|\geq q$. Let $f\in F\langle Y\cup Z\rangle$ and $w\in Y\cup Z$. Write

[TABLE]

where $f^{(i)}$ is the homogeneous component of $f$ with $\deg_{w}f^{(i)}=i$. If $d_{w}<q$ then

[TABLE]

Using Proposition 3.2 and similar arguments as in [8, Proposition 4.2.3] we state the following:

Proposition 3.3**.**

Let $I$ be a $T(*)$-space of $F\langle Y\cup Z\rangle$.

• a)

If $F$ is an infinite field then $I$ is generated, as a $T(*)$-space, by its multihomogeneous elements.

• b)

If $F$ is a field of char$(F)=0$* then $I$ is generated, as a $T(*)$-space, by its multilinear elements.*

Lemma 3.4**.**

Let $f=f(y_{1},\ldots,y_{n},z_{1},\ldots,z_{m})\in F\langle Y\cup Z\rangle$ and write

[TABLE]

where $f^{+}\in F\langle Y\cup Z\rangle^{+}$ and $f^{-}\in F\langle Y\cup Z\rangle^{-}$. Consider an algebra with involution $(A,\circledast)$. If

[TABLE]

for all $Y_{1},\ldots,Y_{n}\in A^{+}$ and $Z_{1},\ldots,Z_{m}\in A^{-}$ then $f^{+}\in Id(A,\circledast)$.

Proof.

Let $Y_{1},\ldots,Y_{n}\in A^{+}$ and $Z_{1},\ldots,Z_{m}\in A^{-}$. Since $f^{+}\in F\langle Y\cup Z\rangle^{+}$ we have $f^{+}(Y_{1},\ldots,Y_{n},Z_{1},\ldots,Z_{m})\in A^{+}$. Since

[TABLE]

we have $f^{+}(Y_{1},\ldots,Y_{n},Z_{1},\ldots,Z_{m})\in A^{-}$ too. Therefore

[TABLE]

as desired. ∎

4 $*$-central polynomials for $(UT_{2}(F),\star)$

In this section, we describe the $*$-central polynomials for $(UT_{2}(F),\star)$ where

[TABLE]

for all $a,b,c\in F$. Note that $\{e_{11}+e_{22},\ e_{12}\}$ and $\{e_{11}-e_{22}\}$ form a basis for the vector spaces $(UT_{2}(F))^{+}$ and $(UT_{2}(F))^{-}$ respectively.

The next lemma is proved in [28, Theorem 3.1]. See [27, Lemma 5.2] too.

Lemma 4.1**.**

Let $F$ be a field of $char(F)\neq 2$. If $I=Id(UT_{2}(F),\star)$ then

[TABLE]

for all $\sigma\in S_{m+1}$.

Lemma 4.2**.**

Let $F$ be a field of $char(F)\neq 2$. Let

[TABLE]

where $m\geq 0$. If $m$ is even then $f\in C(UT_{2}(F),\star)$. If $m$ is odd then $f\notin C(UT_{2}(F),\star)$.

Proof.

Suppose $m=2n$ for some $n\geq 0$. If $Z_{i}=\lambda_{i}(e_{11}-e_{22})$, where $\lambda_{i}\in F$, then

[TABLE]

Thus $f\in C(UT_{2}(F),\star)$.

Suppose $m=2n+1$ for some $n\geq 0$. Then

[TABLE]

Therefore $f\notin C(UT_{2}(F),\star)$. ∎

Proposition 4.3**.**

Let $F$ be a field of $char(F)\neq 2$. Let

[TABLE]

be a polynomial where $f$ is a homogeneous polynomial in the variable $z_{m}$. If $g\in C(UT_{2}(F),\star)$ then

[TABLE]

Proof.

Let $Y_{1},\ldots,Y_{n}\in UT_{2}(F)^{+}$ and $Z_{1},\ldots,Z_{m}\in UT_{2}(F)^{-}$. Then

[TABLE]

for some $a,b,c,d\in F$. Since $f(y_{1},\ldots,y_{n},z_{1},\ldots,z_{m})z_{m}\in C(UT_{2}(F),\star)$, we obtain

[TABLE]

Thus $cd=0$ and $ad=-bd$.

We have two cases:

Case 1. $\deg_{z_{m}}f=0$.

In this case, $f(Y_{1},\ldots,Y_{n},Z_{1},\ldots,Z_{m})=f(Y_{1},\ldots,Y_{n},Z_{1},\ldots,Z_{m-1})$. If $d=1$ then $a=-b$ and $c=0$. Thus $f(Y_{1},\ldots,Y_{n},Z_{1},\ldots,Z_{m})\in UT_{2}(F)^{-}$.

Case 2. $\deg_{z_{m}}f\geq 1$.

In this case, if $d=0$ then $f(Y_{1},\ldots,Y_{n},Z_{1},\ldots,Z_{m})=0\in UT_{2}(F)^{-}$. If $d\neq 0$ then $a=-b$ and $c=0$. Thus $f(Y_{1},\ldots,Y_{n},Z_{1},\ldots,Z_{m})\in UT_{2}(F)^{-}$.

By the two cases we have $f(Y_{1},\ldots,Y_{n},Z_{1},\ldots,Z_{m})\in UT_{2}(F)^{-}$ for all $Y_{1},\ldots,Y_{n}\in UT_{2}(F)^{+}$ and $Z_{1},\ldots,Z_{m}\in UT_{2}(F)^{-}$. By Lemma 3.4 we can write $f=f^{+}+f^{-}$ where $f^{+}\in Id(UT_{2}(F),\star)$ and $f^{-}\in F\langle Y\cup Z\rangle^{-}$. Thus

[TABLE]

The proof is complete. ∎

Lemma 4.4**.**

Let $F$ be a field of $char(F)\neq 2$. If $n\geq 1$ then

[TABLE]

Proof.

This is a direct consequence of Lemma 4.2 and Lemma 4.3. ∎

4.1 $C(UT_{2}(F),\star)$ when char$(F)=0$

The next theorem was proved in [28]. See Proposition 3.1 and [28, Theorem 3.1] for details.

Theorem 4.5**.**

Let $F$ be an infinite field of char$(F)\neq 2$. Consider the involution $\star$ defined in (4). Denote by $I$ the $T(*)$-ideal generated by the polynomials

[TABLE]

Then $Id(UT_{2}(F),\star)=I$. Moreover, the quotient vector space $F\langle Y\cup Z\rangle/I$ has a basis consisting of all polynomials of the form

[TABLE]

where $n\geq 1$,   $m\geq 1$,   $s_{1},\ldots,s_{n},r_{1},\ldots,r_{m}\geq 0$,   $k\geq 1$.

Now we will prove the first main theorem of this paper.

Theorem 4.6**.**

Let $F$ be a field of char$(F)=0$. Consider the involution $\star$ defined in (4). The set of all $*$-central polynomials of $(UT_{2}(F),\star)$ is

[TABLE]

Proof.

Denote $I=Id(UT_{2}(F),\star)$ and $C=C(UT_{2}(F),\star)$. By Lemma 4.2, we have

[TABLE]

Let $f(y_{1},\ldots,y_{n},z_{1},\ldots,z_{m})\in C$ be a multilinear polynomial. We shall prove that $f\in(I+\langle z_{1}z_{2}\rangle^{TS(*)}+F)$. By Theorem 4.5, we have $f+I=\overline{f}+I$ where

[TABLE]

for some $\alpha,\alpha_{k}\in F$. Thus, there exists $g\in I$ such that $f=\overline{f}+g$. In particular, $\overline{f}=(f-g)\in C$.

Case 1. $n=0$ and $m=0$.

In this case, $\overline{f}=\alpha$ and so $f\in(F+I)\subset(I+\langle z_{1}z_{2}\rangle^{TS(*)}+F)$.

Case 2. $n=0$ and $m>0$.

In this case,

[TABLE]

By Lemma 4.2, we have that $\alpha=0$ or $m$ is even. By Lemma 4.4, we obtain $f\in(I+\langle z_{1}z_{2}\rangle^{TS(*)}+F)$.

Case 3. $n>0$ and $m=0$.

In this case,

[TABLE]

If $\alpha\neq 0$ then

[TABLE]

Thus $\overline{f}\notin C$, which is a contradiction. Therefore $\alpha=0$ and $f\in I\subset(I+\langle z_{1}z_{2}\rangle^{TS(*)}+F)$.

Case 4. $n>0$ and $m=1$.

In this case,

[TABLE]

Since $\overline{f}\in C$ we have $\overline{f}(1,\ldots,1,e_{11}-e_{22})=\alpha(e_{11}-e_{22})\in Z(UT_{2}(F))$. Thus $\alpha=0$ and

[TABLE]

Since $\overline{f}\in C$ we obtain $\overline{f}(1,\ldots,1,e_{12},1,\ldots,1,e_{11}-e_{22})=2\alpha_{k}e_{12}\in Z(UT_{2}(F))$. Thus $\alpha_{k}=0$ for all $k=1,\ldots,n$. We prove that $f\in I\subset(I+\langle z_{1}z_{2}\rangle^{TS(*)}+F)$.

Case 5. $n>0$ and $m\geq 2$.

By Lemma 4.1, we have $f+I=\overline{f}+I=\overline{\overline{f}}+I$ where

[TABLE]

Since $I\subset C$ we have $\overline{\overline{f}}\in C$. By Proposition 4.3, we obtain $\overline{\overline{f}}\in(I+\langle z_{1}z_{2}\rangle^{TS(*)})$. Thus $f\in(I+\langle z_{1}z_{2}\rangle^{TS(*)})\subset(I+\langle z_{1}z_{2}\rangle^{TS(*)}+F)$.

By the five cases and by Proposition 3.3 we have $C=I+\langle z_{1}z_{2}\rangle^{TS(*)}+F$ as desired. ∎

4.2 $C(UT_{2}(F),\star)$ when $F$ is an infinite field of char$(F)>2$

We start this section with the next proposition. Similar result is obtained in [2, Theorem 6 in Chapter 4] when we consider $T$-ideals of the free Lie algebra. Moreover, similar result is obtained when we consider $T$-spaces of the free associative algebra.

Proposition 4.7**.**

Let $F$ be an infinite field of $char(F)=p>2$. If $H$ is a $T(*)$-space then $H$ is generated, as a $T(*)$-space, by its multihomogeneous elements $f(y_{1},\ldots,y_{n},z_{1},\ldots,z_{m})\in H$ with multidegree $(p^{a_{1}},\ldots,p^{a_{n}},p^{b_{1}},\ldots,p^{b_{m}})$ where $a_{1},\ldots,a_{n},b_{1},\ldots,b_{m}\geq 0$.

Proof.

Denote by $H_{M}$ the set of all multihomogeneous elements of $H$, and by $H_{PM}$ the set of all multihomogeneous elements $f(y_{1},\ldots,y_{n},z_{1},\ldots,z_{m})\in H$ with multidegree $(p^{a_{1}},\ldots,p^{a_{n}},p^{b_{1}},\ldots,p^{b_{m}})$ where $a_{1},\ldots,a_{n},b_{1},\ldots,b_{m}\geq 0$, $n\geq 0$ and $m\geq 0$.

By Proposition 3.3 it follows that

[TABLE]

We have to prove $\langle H_{M}\rangle^{TS(*)}=\langle H_{PM}\rangle^{TS(*)}$. It’s clear that $\langle H_{M}\rangle^{TS(*)}\supseteq\langle H_{PM}\rangle^{TS(*)}$. Note that

[TABLE]

Let $g(y_{1},\ldots,y_{n},z_{1},\ldots,z_{m})\in H_{M}$.

a) If $g\in H_{PM}$ then $g\in\langle H_{PM}\rangle^{TS(*)}$.

b) Suppose $g\notin H_{PM}$. Denote by

[TABLE]

the multidegree of $g$. Without loss of generality, we may assume that $\deg_{y_{1}}g=d_{y_{1}}$ is not a power of $p$. Let $\deg_{y_{1}}g=p^{k}q$ where $(p,q)=1$. Denote by $\overline{g}(y_{1},y_{2},\ldots,y_{n},y_{n+1},z_{1},\ldots,z_{m})$ the multihomogeneous component of

[TABLE]

with multidegree

[TABLE]

where

[TABLE]

Since $F$ is an infinite field, we have $\overline{g}\in\langle g\rangle^{TS(*)}$. It is known that

[TABLE]

Thus, since

[TABLE]

we have $g\in\langle\overline{g}\rangle^{TS(*)}$. We prove that $\langle g\rangle^{TS(*)}=\langle\overline{g}\rangle^{TS(*)}$. Now we can use the same arguments in $\overline{g}$. After a few steps, we will obtain $\langle g\rangle^{TS(*)}=\langle h\rangle^{TS(*)}$ for some $h\in H_{PM}$. Thus $g\in\langle H_{PM}\rangle^{TS(*)}$. We prove that $\langle H_{M}\rangle^{TS(*)}\subseteq\langle H_{PM}\rangle^{TS(*)}$ as desired. ∎

Lemma 4.8**.**

Let $F$ be an infinite field of char$(F)=p>2$. Let $L$ be the $T(*)$-space

[TABLE]

a) Then $L\subseteq C(UT_{2}(F),\star)$.

b) Let $a_{1},\ldots,a_{n}\geq 1$. If $f(y_{1},\ldots,y_{n})$ is a multihomogeneous polynomial with multidegree $(p^{a_{1}},\ldots,p^{a_{n}})$ then $f\in L$.

Proof.

a) Let $Y\in UT_{2}(F)^{+}$. Thus $Y=\begin{pmatrix}a&b\\ 0&a\end{pmatrix}$ for some $a,b\in F$. If $i\geq 1$ then

[TABLE]

Therefore $Y^{p}\in Z(UT_{2}(F))$ and $y_{1}^{p}\in C(UT_{2}(F),\star)$. By Lemma 4.2, we have $z_{1}z_{2}\in C(UT_{2}(F),\star)$. Therefore $L\subseteq C(UT_{2}(F),\star)$.

b) Denote $I=Id(UT_{2}(F),\star)$. By Theorem 4.5,

[TABLE]

for some $\alpha\in F$. Since $[y_{i},y_{j}]\in I$ (see Theorem 4.5), we obtain

[TABLE]

where $g=1/2(y_{1}^{p^{a_{1}-1}}\cdots y_{n}^{p^{a_{n}-1}}+y_{n}^{p^{a_{n}-1}}\cdots y_{1}^{p^{a_{1}-1}})$. Since $g$ is a symmetric polynomial, we have $\alpha g^{p}\in\langle y_{1}^{p}\rangle^{TS(*)}$ and therefore $f\in(I+\langle y_{1}^{p}\rangle^{TS(*)})\subseteq L$. The proof is complete. ∎

Theorem 4.9**.**

Let $F$ be an infinite field of char$(F)=p>2$. Consider the involution $\star$ defined in (4). The set of all $*$-central polynomials of $(UT_{2}(F),\star)$ is

[TABLE]

Proof.

Denote $I=Id(UT_{2}(F),\star)$ and $C=C(UT_{2}(F),\star)$. By Lemma 4.8, we have

[TABLE]

Let $f(y_{1},\ldots,y_{n},z_{1},\ldots,z_{m})\in C$ be a multihomogeneous polynomial with multidegree $(p^{a_{1}},\ldots,p^{a_{n}},p^{b_{1}},\ldots,p^{b_{m}})$ where $a_{1},\ldots,a_{n},b_{1},\ldots,b_{m}\geq 0$. We shall prove that $f\in(I+\langle z_{1}z_{2}\rangle^{TS(*)}+\langle y_{1}^{p}\rangle^{TS(*)})$. By Theorem 4.5 we obtain $f+I=\overline{f}+I$ where

[TABLE]

for some $\alpha_{1},\ldots,\alpha_{n},\alpha\in F$.

Case 1. $n=0$ and $m=0$.

In this case, $f=\alpha$ and so

[TABLE]

Case 2. $n\geq 1$ and $m=0$.

In this case,

[TABLE]

If $a_{i}=0$, for some $i$, then

[TABLE]

and $\overline{f}(1,\ldots,1,y_{i},1,\ldots,1)=\alpha y_{i}\in C$. Thus $\alpha=0$, $\overline{f}=0$ and $f\in I\subset(I+\langle z_{1}z_{2}\rangle^{TS(*)}+\langle y_{1}^{p}\rangle^{TS(*)})$.

Suppose $a_{1},\ldots,a_{n}\geq 1$. By Lemma 4.8,

[TABLE]

and therefore $f\in(I+\langle z_{1}z_{2}\rangle^{TS(*)}+\langle y_{1}^{p}\rangle^{TS(*)})$.

Case 3. $m=1$ and $b_{m}=0$.

In this case,

[TABLE]

Since $\overline{f}\in C$ we have $\overline{f}(1,\ldots,1,z_{1})=\alpha z_{1}\in C$. Thus $\alpha=0$ and

[TABLE]

If $Y_{1}=\ldots=Y_{k-1}=Y_{k+1}=\ldots=Y_{n}=e_{11}+e_{22}$, $Y_{k}=e_{11}+e_{22}+e_{12}$ and $Z_{1}=e_{11}-e_{22}$, then

[TABLE]

Thus $\alpha_{k}=0$ for all $k=1,\ldots,n$ and $\overline{f}=0$. Therefore

[TABLE]

Case 4. $m\geq 2$ and $b_{m}=0$.

In this case,

[TABLE]

By Lemma 4.1 we have

[TABLE]

Thus $\overline{f}+I=\widetilde{f}z_{m}+I$ where

[TABLE]

Since $\widetilde{f}z_{m}\in C$, by Proposition 4.3 we have that $\widetilde{f}z_{m}\in(I+\langle z_{1}z_{2}\rangle^{TS(*)}+\langle y_{1}^{p}\rangle^{TS(*)})$. Therefore

[TABLE]

Case 5. $m\geq 1$ and $b_{m}\geq 1$.

By Lemma 4.1,

[TABLE]

Thus $\overline{f}+I=\widetilde{f}z_{m}+I$ where

[TABLE]

Since $\widetilde{f}z_{m}\in C$, by Proposition 4.3 we have

[TABLE]

Therefore $f\in(I+\langle z_{1}z_{2}\rangle^{TS(*)}+\langle y_{1}^{p}\rangle^{TS(*)}).$

By the five cases it follows that

[TABLE]

as desired. The proof is complete. ∎

4.3 $C(UT_{2}(F),\star)$ when $F$ is a finite field

Let $F$ be a finite field with $|F|=q$ elements and char$(F)=p\neq 2$. Since $(\ F-\{0\}\ ,\ \cdot\ )$ is a group, we have $a^{q-1}=1$ for all $a\in F-\{0\}$. In particular,

[TABLE]

for all $a\in F$. Hence, if

[TABLE]

then

[TABLE]

for all $a,b\in F$ and $i\geq 1$.

The next lemma is direct consequence of [8, Proposition 4.2.3]. See [27, Lemma 2.1] too.

Lemma 4.10**.**

Let $F$ be a finite field with $|F|=q$. Let $f\in F\langle X\rangle$ be a polynomial given by

[TABLE]

where $\alpha_{(d_{1},\ldots,d_{n})}\in F$. If $f$ is a polynomial identity for $F$ then $\alpha_{(d_{1},\ldots,d_{n})}=0$ for all $(d_{1},\ldots,d_{n})$.

Let $\Lambda_{n}$ be the set of all elements $(s_{1},\ldots,s_{n})\in\mathbb{Z}^{n}$ such that:

• a)

$0\leq s_{1},\ldots,s_{n}<2q$,

• b)

If $s_{i}\geq q$ for some $i$, then $s_{j}<q$ for all $j\neq i$.

The next theorem was proved in [27, Theorem 5.9].

Theorem 4.11**.**

Let $F$ be a finite field with $|F|=q$ elements and char$(F)\neq 2$. Consider the involution $\star$ defined in (4). Denote by $J$ the $T(*)$-ideal generated by the polynomials

[TABLE]

Then $Id(UT_{2}(F),\star)=J$. Moreover, the quotient vector space $F\langle Y\cup Z\rangle/J$ has a basis consisting of all polynomials of the form

[TABLE]

Proposition 4.12**.**

Let $F$ be a finite field with $|F|=q$ and char$(F)\neq 2$. Then

[TABLE]

is a $*$-central polynomial for $(UT_{2}(F),\star)$ for all $l\geq 0$.

Proof.

Denote $f(y_{1},y_{2})=ly_{1}(y_{2}^{q+l-1}-y_{2}^{l})+y_{1}^{q}y_{2}^{l}$ and consider

[TABLE]

where $a_{i},b_{i}\in F$, $i=1,2$. By (7), we have

[TABLE]

Thus

[TABLE]

By (7), we have

[TABLE]

Therefore

[TABLE]

as desired. The proof is complete. ∎

Denote by $V$ the following $T(*)$-space:

[TABLE]

By Proposition 4.12, we obtain

[TABLE]

Since char$(F)=p$, if $l=kp$ and $y_{1}=1$ then $ly_{1}(y_{2}^{q+l-1}-y_{2}^{l})+y_{1}^{q}y_{2}^{l}=y_{2}^{kp}.$ Hence

[TABLE]

for all $k\geq 0$.

From now on we write

[TABLE]

Lemma 4.13**.**

Let $F$ be a finite field with $|F|=q$ and char$(F)\neq 2$. If $l,n\geq 0$ then

[TABLE]

Proof.

The case $n=0$ is consequence of Proposition 4.12. In fact, substituting in

[TABLE]

the variable $y_{1}$ by $1$, we will have $l(y_{2}^{q+l-1}-y_{2}^{l})+y_{2}^{l}\equiv 0$.

Suppose $n\geq 1$. Denote $u=(1/2)(y_{1}y_{2}\cdots y_{n}+y_{n}\cdots y_{2}y_{1})$ and $J=Id(UT_{2}(F),\star)$. Since $u$ is a symmetric polynomial it follows that

[TABLE]

Since $[y_{i},y_{j}]\in J$ (see Theorem 4.11), we have

[TABLE]

Thus $u+J=y_{1}\cdots y_{n}+J$ and $u^{q}+J=y_{1}^{q}\cdots y_{n}^{q}+J.$ Hence

[TABLE]

Now we use (10) to finish the proof. ∎

Corollary 4.14**.**

Let $F$ be a finite field with $|F|=q$ and char$(F)=p\neq 2$. If $f(y_{1},\ldots,y_{n})\in F\left\langle Y\cup Z\right\rangle$ and $p\nmid l$ then there exists $g(y_{1},\ldots,y_{n})\in F\left\langle Y\cup Z\right\rangle$ such that

[TABLE]

Proof.

By Lemma 4.13, we have

[TABLE]

Since $fy_{n+1}^{q+l-1}$ is a linear combination of polynomials

[TABLE]

we finish the proof. ∎

Proposition 4.15**.**

Let $F$ be a finite field with $|F|=q$ and char$(F)=p\neq 2$. Consider

[TABLE]

where $\alpha_{i}\in F$ for all $i$. If $f\in C(UT_{2}(F),\star)$ then $f\equiv 0$.

Proof.

Let $g$ be given by

[TABLE]

By Corollary 4.14, we have

[TABLE]

for some $\beta_{i}\in F$. Thus, there exist $\gamma_{i}\in F$ such that

[TABLE]

By (9), we obtain

[TABLE]

It follows from Proposition 4.12 that

[TABLE]

Since $f\in C(UT_{2}(F),\star)$, by (11) and (12) we have that $h\in C(UT_{2}(F),\star)$ where

[TABLE]

If $a\in F$ and

[TABLE]

then by (7),

[TABLE]

Since $h(Y)\in Z(UT_{2}(F))$ it follows that

[TABLE]

for all $a\in F$. By Lemma 4.10, we have

[TABLE]

Thus $h=0$ and $f\equiv 0$ as desired. ∎

Lemma 4.16**.**

Let $F$ be a finite field with $|F|=q$ and char$(F)=p\neq 2$. Then $y_{1}^{pq}-y_{1}^{p}\in Id(UT_{2}(F),\star)$.

Proof.

If $a,b\in F$ and $Y=\begin{pmatrix}a&b\\ 0&a\end{pmatrix},$ then by (7),

[TABLE]

Thus $Y^{pq}-Y^{p}=0$ as desired. ∎

Lemma 4.17**.**

Let $F$ be a finite field with $|F|=q$ and char$(F)=p\neq 2$. If $i\geq 0$ then

[TABLE]

Proof.

Denote $u=(1/2)(y_{1}y_{3}^{p}+y_{3}^{p}y_{1})$ and $J=Id(UT_{2}(F),\star)$. Since $u$ is a symmetric polynomial, we have

[TABLE]

Since $y_{i}y_{j}+J=y_{j}y_{i}+J$ it follows that

[TABLE]

Thus, by Lemma 4.16,

[TABLE]

Now we use (14) and (15) to obtain

[TABLE]

By (13) and (16) it follows that $\left(iy_{1}(y_{2}^{q+i-1}-y_{2}^{i})+y_{1}^{q}y_{2}^{i}\right)y_{3}^{p}\equiv 0$ as desired. ∎

Corollary 4.18**.**

Let $F$ be a finite field with $|F|=q$ and char$(F)=p\neq 2$. If $f(y_{1},\ldots,y_{n})\equiv 0$ then $f(y_{1},\ldots,y_{n})y_{n+1}^{lp}\equiv 0$ for all $l\geq 0$.

Proof.

Write $f=f_{V}+f_{I}$, where $f_{V}\in V$ and $f_{I}\in Id(UT_{2}(F),\star)$. Since

[TABLE]

we can suppose $f(y_{1},\ldots,y_{n})\in V$. In this case, $f$ is a linear combination of polynomials

[TABLE]

where $g_{1},g_{2}\in F\left\langle Y\cup Z\right\rangle^{+}$ and $i\geq 0$. By Lemma 4.17 we have

[TABLE]

Thus $f(y_{1},\ldots,y_{n})y_{n+1}^{p}\equiv 0$. Since $V+Id(UT_{2}(F),\star)$ is a $T(*)$-space we have $f(y_{1},\ldots,y_{n})y_{n+1}^{lp}=f(y_{1},\ldots,y_{n})(y_{n+1}^{l})^{p}\equiv 0$ for all $l\geq 0$. ∎

Proposition 4.19**.**

Let $F$ be a finite field with $|F|=q$ and char$(F)=p\neq 2$. If $f(y_{1},\ldots,y_{n})\in C(UT_{2}(F),\star)$ then $f(y_{1},\ldots,y_{n})\equiv 0$.

Proof.

By Theorem 4.11 we can suppose

[TABLE]

where $s=(s_{1},\ldots,s_{n})$, $\alpha_{s}\in F$.

We will prove the proposition using induction on $n$.

The case $n=1$ is consequence of Proposition 4.15.

Suppose $n\geq 2$. Write

[TABLE]

Note that:

a) If $0\leq i\leq q-1$ then

[TABLE]

b) If $q\leq i\leq 2q-1$ then

[TABLE]

Write

[TABLE]

By Corollary 4.14, there exist polynomials $g_{i}(y_{1},\ldots,y_{n-1})$ such that

[TABLE]

Thus, there exist polynomials $h_{i}(y_{1},\ldots,y_{n-1})$ such that

[TABLE]

Denote

[TABLE]

Since $f\in C(UT_{2}(F,\star))$ and $V+Id(UT_{2}(F,\star))\subseteq C(UT_{2}(F,\star))$, we have by (18) that $h\in C(UT_{2}(F,\star))$.

Consider the following matrices

[TABLE]

where $a_{j}\in F$ for all $j$. By (7) we have

[TABLE]

In particular, if $p|i$ then

[TABLE]

Hence

[TABLE]

where

[TABLE]

Since $h(Y_{1},\ldots,Y_{n})\in Z(UT_{2}(F))$ we obtain $\overline{h}(a_{1},\ldots,a_{n})=0$ for all $a_{1},\ldots,a_{n}\in F$. By Lemma 4.10 it follows that

[TABLE]

for all $i=1,\ldots,q$ where $p|i$ ; and

[TABLE]

for all $i=1,\ldots,q-1$ where $p\nmid i$.

By (17), (19) and Lemma 4.10, we have $f_{q+i-1}(y_{1},\ldots,y_{n-1})=0$ for all $i=1,\ldots,q$ where $p|i$. Thus

[TABLE]

Since $h\in C(UT_{2}(F),\star)$, by Proposition 3.2 it follows that $h_{i}y_{n}^{i}\in C(UT_{2}(F),\star)$ for all $i=0,\ldots,q-1$. Substituting in $h_{i}y_{n}^{i}$ the variable $y_{n}$ by $1$, it follows that

[TABLE]

for all $i=0,\ldots,q-1$. We have two cases:

a) Case $p\nmid i$.

Consider

[TABLE]

where $a_{k},b_{k}\in F$ and $k=1,\ldots,n-1$. We have

[TABLE]

for some $\beta\in F$. Since $h_{i}\in C(UT_{2}(F,\star))$ it follows that $\beta=0$. By (20), we have $h_{i}(a_{1},\ldots,a_{n-1})=0$ too. Thus $h_{i}(Y_{1},\ldots,Y_{n-1})=0$ and $h_{i}(y_{1},\ldots,y_{n-1})\in Id(UT_{2}(F),\star)$ for all $i=0,\ldots,q-1$ where $p\nmid i$.

b) Case $p|i$.

Since $h_{i}(y_{1},\ldots,y_{n-1})\in C(UT_{2}(F),\star)$, we obtain, by induction, that $h_{i}\equiv 0$. Thus, by Corollary 4.18, it follows that $h_{i}y_{n}^{i}\equiv 0$ for all $i=0,\ldots,q-1$ where $p|i$.

By (18), (21) and two cases above, we have

[TABLE]

as desired. ∎

Theorem 4.20**.**

Let $F$ be a finite field with $|F|=q$ and char$(F)=p\neq 2$. Then

[TABLE]

Proof.

Firstly, we will prove the following claim :

Claim 1. The set $C(UT_{2}(F),\star)$ equals

[TABLE]

Proof of Claim 1. Denote $J=Id(UT_{2}(F),\star)$ and

[TABLE]

Since $z_{1}z_{2}\in C(UT_{2}(F),\star)$, we have $C(UT_{2}(F),\star)\supseteq\left\langle z_{1}z_{2}\right\rangle^{TS(*)}$. Therefore, by Proposition 4.12, we obtain

[TABLE]

Consider $f\in C(UT_{2}(F),\star)$. We will prove that $f\in\left(V+\left\langle z_{1}z_{2}\right\rangle^{TS(*)}+J\right)$. By Theorem 4.11, $f=f_{J}+f_{\Upsilon}$ where $f_{J}\in J$ and $f_{\Upsilon}$ is a linear combination of polynomials

[TABLE]

Since $f_{J}\in J$, we have

[TABLE]

Thus we can suppose $f=f_{\Upsilon}$. Since $\deg_{z_{i}}f<q$, we can suppose $f$ a homogeneous polynomial in the variable $z_{i}$ for all $i=1,\ldots,m$ (see Proposition 3.2). Denote $\deg_{z_{i}}f=r_{i}$.

If $r_{1}=\ldots=r_{m}=0$ then by Proposition 4.19

[TABLE]

Suppose $r_{i}\neq 0$ for some $i$. Renumbering the indices if necessary, we may assume that $r_{i}\geq 1$ for all $i=1,\ldots,m$. Since $f$ is a linear combination of polynomials in $\Upsilon_{1}$ and $\Upsilon_{2}$, we have $f=f_{1}+f_{2}$ where

[TABLE]

with $0\leq s_{1},\ldots,s_{n}<q$,   $1\leq r_{1},\ldots,r_{m}<q$, $\ \ n\geq 1,\ \ m\geq 1,\ \ k\geq 1$,   $s=(s_{1},\ldots,s_{n})$,   $\alpha_{(n,k,s)}\in F$; and

[TABLE]

with $0\leq s_{1},\ldots,s_{n}<q$,    $1\leq r_{1},\ldots,r_{m}<q$, $\ \ n\geq 1,\ \ m\geq 1$,   $s=(s_{1},\ldots,s_{n}),$     $\beta_{(n,s)}\in F$.

We have three cases:

Case 1. $m=1$ and $r_{m}=1$.

In this case, $f=f_{1}+f_{2}$ where

[TABLE]

and

[TABLE]

Let

[TABLE]

where $a_{i}\in F$. We have

[TABLE]

where $\theta\in F$. Since $f(Y_{1},Y_{2},\ldots,Z_{1})\in Z(UT_{2}(F))$, it follows that $\theta=0$ and

[TABLE]

for all $a_{1},\ldots,a_{n}\in F$. Since $0\leq s_{1},\ldots,s_{n}<q$ we have, by Lemma 4.10, that $\beta_{(n,s)}=0$ for all $n,s$. Thus $f=f_{1}$. If $\overline{Y}_{1},\overline{Y}_{2},\ldots\in UT_{2}(F)^{+}$ and $\overline{Z}_{1}\in UT_{2}(F)^{-}$ then

[TABLE]

for some $\alpha\in F$. Since $f(\overline{Y}_{1},\overline{Y}_{2},\ldots,\overline{Z}_{1})\in Z(UT_{2}(F))$, we obtain $\alpha=0$, that is $f\in J$. Therefore $f\in\left(V+\left\langle z_{1}z_{2}\right\rangle^{TS(*)}+J\right)$.

Case 2. $m\geq 2$ and $r_{m}=1$.

In this case, $f=f_{1}+f_{2}$ where

[TABLE]

and

[TABLE]

By Lemma 4.1, we have

[TABLE]

Thus $f+J=\widetilde{f}z_{m}+J$ where

[TABLE]

Since $\widetilde{f}z_{m}\in C(UT_{2}(F),\star)$ we have, by Proposition 4.3, that $\widetilde{f}z_{m}\in(\langle z_{1}z_{2}\rangle^{TS(*)}+J)$. Therefore

[TABLE]

Case 3. $m\geq 1$ and $r_{m}\geq 2$.

By Lemma 4.1, we have

[TABLE]

Thus $f+J=\widetilde{f}z_{m}+J$ where

[TABLE]

Since $\widetilde{f}z_{m}\in C(UT_{2}(F),\star)$ we have, by Proposition 4.3, that $\widetilde{f}z_{m}\in(\langle z_{1}z_{2}\rangle^{TS(*)}+J)$. Therefore

[TABLE]

We prove that

[TABLE]

By (22) we finish the proof of Claim 1.

Claim 2. If $l\geq 0$ then

[TABLE]

Proof of Claim 2. If $l\geq 0$, let $k,r$ be the integers such that $l=kp+r$ and $0\leq r<p$. We have

[TABLE]

Case 1. $r=0$.

In this case, by (23), it follows that

[TABLE]

Note that

[TABLE]

This case is done.

Case 2. $1\leq r<p$.

Denote $u=(1/2)(y_{1}y_{2}^{kp}+y_{2}^{kp}y_{1})$. Since

[TABLE]

we have $u+J=y_{1}y_{2}^{kp}+J$. Moreover, by Lemma 4.16,

[TABLE]

Thus, by (23), we obtain

[TABLE]

Therefore

[TABLE]

as desired.

By the Claims 1 and 2 we complete the proof of theorem. ∎

5 $*$-central polynomials for $(UT_{2}(F),s)$

In this section, we describe the $*$-central polynomials for $(UT_{2}(F),s)$ where

[TABLE]

for all $a,b,c\in F$. Note that $\{e_{11}+e_{22}\}$ and $\{e_{11}-e_{22},e_{12}\}$ form a basis for the vector spaces $(UT_{2}(F))^{+}$ and $(UT_{2}(F))^{-}$ respectively.

The next theorem was proved in [28, Theorem 3.2]:

Theorem 5.1**.**

Let $F$ be an infinite field with $char(F)\neq 2$. Then $Id(UT_{2}(F),s)$ is the $T(*)$-ideal generated by the polynomials

[TABLE]

The next theorem was proved in [27, Theorem 6.15]:

Theorem 5.2**.**

Let $F$ be a finite field with $|F|=q$ elements and char$(F)\neq 2$. Then $Id(UT_{2}(F),s)$ is the $T(*)$-ideal generated by the polynomials

[TABLE]

Theorem 5.3**.**

Let $F$ be a field of char$(F)\neq 2$. The set of all $*$-central polynomials of $(UT_{2}(F),s)$ is

[TABLE]

Proof.

Since $(UT_{2}(F))^{+}=Z(UT_{2}(F))$ we have

[TABLE]

Let $f=f(y_{1},\ldots,y_{n},z_{1},\ldots,z_{m})\in C(UT_{2}(F),s)$ and write

[TABLE]

where $f^{+}\in F\langle Y\cup Z\rangle^{+}$ and $f^{-}\in F\langle Y\cup Z\rangle^{-}$. If $Y_{1},\ldots,Y_{n}\in UT_{2}(F)^{+}$ and $Z_{1},\ldots,Z_{m}\in UT_{2}(F)^{-}$ then $f^{-}(Y_{1},\ldots,Y_{n},Z_{1},\ldots,Z_{m})\in UT_{2}(F)^{-}$ and

[TABLE]

Thus $f^{-}(Y_{1},\ldots,Y_{n},Z_{1},\ldots,Z_{m})=0$ and $f^{-}\in Id(UT_{2}(F),s)$. Since $f^{+}\in\langle y_{1}\rangle^{TS(*)}$ it follows that $f\in Id(UT_{2}(F),s)+\langle y_{1}\rangle^{TS(*)}$. We prove that

[TABLE]

as desired. ∎

6 Conclusion

Let $F$ be a field of char$(F)\neq 2$, and let $*$ be an involution of the first kind on $UT_{2}(F)$. By Corollary 2.1, we have

[TABLE]

Thus, by Theorem 4.6, Theorem 4.9, Theorem 4.20 and Theorem 5.3 we have described $C(UT_{2}(F),*)$.

Since

[TABLE]

it follows that

[TABLE]

Moreover, by Theorem 4.5, Theorem 4.11, Theorem 5.1, Theorem 5.2 it follows that there exists a finite set $S$ such that

[TABLE]

Since

[TABLE]

we prove the Theorem 1.5.

Acknowledgments

The first author was supported by Ph.D. grant from CAPES. The second author was partially supported by FAPESP grant No. 2014/09310-5, and by CNPq grant No. 406401/2016-0.