An identity on generalized derivations involving multilinear polynomials in prime rings

Abstract

Let R be a prime ring of characteristic different from 2 with its Utumi ring of quotients U, extended centroid C, $f(x_1,\ldots ,x_n)$ a multilinear polynomial over C, which is not central-valued on R and d a nonzero derivation of R. By f(R), we mean the set of all evaluations of the polynomial $f(x_1,\ldots ,x_n)$ in R. In the present paper, we study $b[d(u),u]+p[d(u),u]q+[d(u),u]c=0$ for all $u\in f(R)$, which includes left-sided, right-sided as well as two-sided annihilating conditions of the set $\{[d(u),u] : u\in f(R)\}$. We also examine some consequences of this result related to generalized derivations and we prove that if F is a generalized derivation of R and d is a nonzero derivation of R such that

$$\begin{aligned} F^2([d(u), u])=0 \end{aligned}$$

for all $u\in f(R)$, then there exists $a\in U$ with $a^2=0$ such that $F(x)=xa$ for all $x\in R$ or $F(x)=ax$ for all $x\in R$.

1 Introduction

A ring R is said to be prime if for any $a,b\in R$, $aRb= \{0\}$ implies either $a =0$ or $b = 0$ and is said to be semiprime if for any $a\in R$, $aRa=\{0\}$ implies $a=0.$ Let Z(R) denote the center of R and U be the Utumi ring of quotients of R and $C=Z(U)$. The symbols [x, y] denote the Lie commutator $xy-yx$ for any $x, y \in R$. By a derivation, we mean an additive mapping $d: R\rightarrow R$ such that $d(xy)=d(x)y+xd(y)$ for all x, $y \in R$.

Several authors found a number of results investigating the relationship between the behaviour of additive mappings defined on a prime (or semiprime) ring R and the structure of R. Posner [17] proved that if R is a prime ring and d a nonzero derivation on R such that $[d(r),r]\in Z(R)$, then R is commutative. Several authors have generalized the Posner’s result.

Lee and Lee in [13] proved that if $[d(f(x_1, \ldots , x_n)), f(x_1, \ldots , x_n)]_k =0$ for all $x_1, \ldots , x_n$ in some nonzero ideal of R, then $f(x_1, \ldots , x_n)$ is central-valued on R, except when char($R)=2$ and R satisfies $s_4(x_1,x_2,x_3,x_4)$, the standard identity in four variables. Later on, De Filippis and Di Vincenzo [5] considered the situation $\delta ([d(f(x_1,\ldots ,x_n)),f(x_1,\ldots ,x_n)])=0$ for all $x_1,\ldots ,x_n\in R$, where d and $\delta$ are two derivations of R. The statement of De Filippis and Di Vincenzo’s theorem is the following:

Let K be a noncommutative ring with unity, R a prime K-algebra of characteristic different from 2, d and $\delta$ nonzero derivations of R,  and $f(x_1,\ldots ,x_n)$ a multilinear polynomial over K. If $\delta ([d(f(x_1,\ldots ,x_n)),f(x_1,\ldots ,x_n)])=0$ for all $x_1,\ldots ,x_n\in R$, then $f(x_1,\ldots ,x_n)$ is central-valued on R.

It is natural to consider the situation when derivation $\delta$ is replaced by $\delta ^2$, that is, $\delta ^2([d(f(x_1,\ldots ,x_n)),f(x_1,\ldots ,x_n)])=0$ for all $x_1,\ldots ,x_n\in R$. In the present paper, we investigate a more general case replacing $\delta ^2$ with $F^2$, where F is a generalized derivation of R.

On the other hand, Dhara [7] studied $[d^2(f(x_1,\ldots ,x_n)),f(x_1,\ldots ,x_n)]=0$ for all $x_1,\ldots ,x_n\in \rho$ in prime ring R, where d is a derivation of R and $\rho$ is a nonzero right ideal of R.

We will continue the study of analogue problems involving generalized derivations on the appropriate subsets of prime rings. An additive mapping $F: R\rightarrow R$ associated with a derivation d on R such that $F(xy)=F(x)y+xd(y)$ for all $x, y \in R$, is said to be generalized derivation. For some fixed $a, b \in U,$ an additive mapping $F:R \rightarrow R$ defined as $F(x)=ax+xb$ for all $x \in R$ is an example of generalized derivation. In [2], the following result was obtained:

Let R be a prime ring of characteristic different from 2 with extended centroid C, $f(x_1,\ldots ,x_n)$ be a multilinear polynomial over C, which is not central valued on R. If d is a derivation of R,  and F is a generalized derivation of R such that $F([d(f(x_1,\ldots ,x_n)), f(x_1,\ldots ,x_n)])=0$ for all $x_1,\ldots ,x_n\in R$, then either $F=0$ or $d=0$.

In this line of investigation, in [4], De Filippis and Di Vincenzo proved the following:

Let R be a prime algebra over a commutative ring K with unity, and $f(x_1,\ldots , x_n)$ be a multilinear polynomial over K, not central valued on R. Suppose that d is a nonzero derivation of R,  and F is a nonzero generalized derivation of R such that $d([F(f(r_1,\ldots ,r_n)), f(r_1,\ldots ,r_n )])=0$ for all $r_1,\ldots , r_n\in R$. If the characteristic of R is different from 2, then one of the following holds:

• (1) there exists $\lambda \in C$, the extended centroid of R, such that $F(x) = \lambda x$, for all $x\in R$;

• (2) there exists $a\in U$, the Utumi quotient ring of R, and $\lambda \in C = Z(U)$ such that $F(x) = ax + xa + \lambda x$ for all $x\in R$, and $f(x_1,\ldots , x_n)^2$ is central-valued on R.

Furthermore, Tiwari et al. [18] investigated $d([F^2(f(r_1,\ldots ,r_n)), f(r_1,\ldots ,r_n )])=0$ for all $r_1,\ldots , r_n\in R$, where d is a nonzero derivation of R, and F is a generalized derivation of R.

In the present paper, we prove the following:

Main Theorem

Let R be a prime ring of characteristic different from 2 with Utumi quotient ring U and $f(x_1,\ldots ,x_n)$ a non-central multilinear polynomial over the extended centroid C. If d is a nonzero derivation of R and F is a generalized derivation of R such that

$$\begin{aligned} F^2([d(f(x_1, \ldots , x_n)), f(x_1, \ldots , x_n)])=0 \end{aligned}$$

for all $x_1, \ldots , x_n \in R$, then there exists $a\in U$ with $a^2=0$ such that $F(x)=xa$ for all $x\in R$ or $F(x)=ax$ for all $x\in R$.

Here we give an example which shows that in our result, the primeness of the ring is essential.

Example

Define $R=\Bigg \{\left( \begin{array}{c@{\quad }c} x &{} y \\ 0 &{} 0 \\ \end{array} \right) : x,y\in \mathbb {Z} \Bigg \}$ and a multilinear polynomial $f(r,s)=rs$. We see that R is a ring under usual operations and f(r, s) is not central valued on R. Also, note that R is not a prime ring. Now we define maps $d,F,g: R \rightarrow R$ such that $d\left( \begin{array}{c@{\quad }c} x &{} y \\ 0 &{} 0 \\ \end{array}\right) =\left( \begin{array}{c@{\quad }c} 0 &{} y \\ 0 &{} 0 \\ \end{array}\right)$, $F\left( \begin{array}{c@{\quad }c} x &{} y \\ 0 &{} 0 \\ \end{array}\right) =\left( \begin{array}{c@{\quad }c} x &{} 0 \\ 0 &{} 0 \\ \end{array}\right)$ and $g\left( \begin{array}{c@{\quad }c} x &{} y \\ 0 &{} 0 \\ \end{array}\right) =\left( \begin{array}{c@{\quad }c} 0 &{} -y \\ 0 &{} 0 \\ \end{array} \right)$. Notice that d is a nonzero derivation on R and F is a generalized derivation associated to the derivation g on R. It can easily be seen that $F^2([d(f(r,s)), f(r,s)])=0$ for all $r,s\in R$. Thus R satisfies the hypothesis of the main theorem. However, the conclusion of the main theorem does not hold as g is a nonzero derivation of R.

2 Preliminaries

In what follows, R always denotes a prime ring and U denotes the Utumi ring of quotients of R. $f(x_1,\ldots ,x_n)$ denotes the multilinear polynomial over C which is in the form

$$\begin{aligned} f(x_1,\ldots ,x_n)=x_1x_2\cdots x_n+\sum \limits _{\sigma \in S_n, \sigma \ne id} \alpha _\sigma x_{\sigma (1)}x_{\sigma (2)} \ldots x_{\sigma (n)}, \end{aligned}$$

for some $\alpha _\sigma \in C$ and $S_n$ the symmetric group of degree n.

The definition and axiomatic formulation of Utumi quotient ring U can be found in [1] and [3].

We have the following properties which we need:

• (1) $R\subseteq U$;

• (2) U is a prime ring with identity;

• (3) The center of U is denoted by C and is called the extended centroid of R. C is a field.

Moreover, we recall some known facts.

Fact 1. Let $\mathcal {K}$ be an algebra over a field $\mathbb {E}.$ A generalized polynomial identity (GPI) of $\mathcal {K}$ is a polynomial expression g in non commutative indeterminates and fixed coefficients from $\mathcal {K}$ between the indeterminates such that g vanishes on all replacements by elements of $\mathcal {K}$. The generalized polynomial in the context of Utumi quotient ring U is defined as follows:

Suppose that V is a set of C-independent vectors of U and $Y=\{y_1,y_2,y_3, \ldots \}$ is a countable set, where $y_i$ are non commuting indeterminates. Let $C\langle Y \rangle$ be the free algebra over C in the set Y. Consider $\mathcal {W}= U_{^{*}C} C\langle Y \rangle ,$ the free product of U and $C\langle Y \rangle$ over C. The elements of $\mathcal {W}$ are called generalized polynomials. An element $h \in \mathcal {W}$ of the form $h = s_0x_1s_1x_2s_2 \ldots x_ns_n,$ where $\{s_0, \ldots ,s_n\} \subseteq U$ and $\{x_1, \ldots ,x_n\} \subseteq Y$ is said to be a monomial. Therefore, each $g\in \mathcal {W}$ can be represented as a finite sum of monomials. A V-monomial is of the form $e = v_0x_1v_1x_2v_2 \ldots x_nv_n,$ where $\{v_0, \ldots ,v_n\} \subseteq V$ and $\{x_1, \ldots ,x_n\} \subseteq Y.$ Thus an element $g \in \mathcal {W}$ can be written as $g = \sum _{i} \beta _{i}e_{i},$ where $\beta _{i} \in C$ and $e_{i}$ are V-monomials. An element $g \in \mathcal {W}$ is trivial if and only if $\beta _{i}=0$ for each i. For more details, we refer the reader to [1, 3].

Fact 2. If I is a two-sided ideal of R, then R, I and U satisfy the same generalized polynomial identities (GPIs) with coefficients in U (see [3]).

Fact 3. Every derivation d of R can be uniquely extended to a derivation of U (see Proposition 2.5.1 in [1]).

Fact 4. If I is a two-sided ideal of R, then R, I and U satisfy the same differential identities (see [14]).

Fact 5. Let d be a derivation on R. By $f^d(x_1,\ldots ,x_n)$, $f^{d^2}(x_1,\ldots ,x_n)$ and $f^{d^3}(x_1,\ldots ,x_n)$, we denote the polynomials obtained from $f(x_1,\ldots ,x_n)$ by replacing each coefficient $\alpha _\sigma$ with $d(\alpha _\sigma )$, $d^2(\alpha _\sigma )$ and $d^3(\alpha _\sigma )$, respectively. Then we have

$$\begin{aligned} d(f(x_1,\ldots ,x_n))=&f^d(x_1,\ldots ,x_n)+\sum \limits _i f(x_1,\ldots ,d(x_i),\ldots ,x_n), \\ d^2(f(x_1,\ldots ,x_n))=&f^{d^2}(x_1,\ldots ,x_n)+2\sum \limits _i f^d(x_1,\ldots ,d(x_i),\ldots ,x_n)\\&+\sum \limits _i f(x_1,\ldots ,d^2(x_i),\ldots ,x_n) \\&+\sum \limits _{i\ne j} f(x_1, \ldots , d(x_i), \ldots , d(x_j), \ldots , x_n) \end{aligned}$$

and

$$\begin{aligned} d^3(f(x_1,\ldots ,x_n))&=f^{d^3}(x_1,\ldots ,x_n) + 3\sum \limits _i f^{d^2}(x_1,\ldots ,d(x_i),\ldots ,x_n)\\&\quad +\,3\sum \limits _i f^{d}(x_1,\ldots ,d(x_i),\ldots ,d(x_j),\ldots ,x_n)\\&\quad +\, 3\sum \limits _i f^{d}(x_1,\ldots ,d^2(x_i),\ldots ,x_n)\\&\quad +\,\sum \limits _{i\ne j\ne k} f(x_1,\ldots ,d(x_i),\ldots ,d(x_j),\ldots ,d(x_k),\ldots ,x_n)\\&\quad +\,2 \sum \limits _{i\ne j} f(x_1,\ldots ,d^2(x_i),\ldots ,d(x_j),\ldots ,x_n)\\&\quad +\, 2\sum \limits _{i\ne j} f(x_1,\ldots ,d(x_i),\ldots ,d^2(x_j),\ldots ,x_n)\\&\quad +\,\sum \limits _i f(x_1,\ldots ,d^3(x_i),\ldots ,x_n). \end{aligned}$$

3 The case when F is inner

In this section, we study all the possible situation of annihilating condition of the set $\{ [d(x),x] | x\in f(R)\}$, where d is a derivation of R. For any subset S of R, denote by $r_R (S)$ the right annihilator of S in R, that is, $r_R (S) = \{x \in R |Sx = 0\}$ and $l_R (S)$ the left annihilator of S in R,  that is, $l_R (S) = \{x \in R | x S = 0\}$. If $r_R (S) = l_R (S)$, then $r_R (S)$ is called an annihilator ideal of R and is written as $\mathrm{ann}_R (S)$.

In [6], De Filippis and Di Vincenzo studied the left annihilating condition of the set $\{ [d(x),x] | x\in f(R)\}$. More precisely, they proved that if R is a prime ring of $\mathrm{char}(R)\ne 2$ and d is a nonzero derivation of R satisfying $a[d(x),x]=0$ for all $x\in f(R)$, then $a=0$.

Now we will study a more general situation, involving left sided, right sided as well as two-sided annihilating conditions. More specifically, we study the situation $b[d(x),x]+p[d(x),x]q+[d(x),x]c=0$ for all $x\in f(R)$, where $b,c,p,q\in R$.

First we consider that d is an inner derivation of R, that is, $d(x)=[a,x]$ for all $x\in R$. Then

$$\begin{aligned} b[d(f(r)),f(r)]+p[d(f(r)),f(r)]q+[d(f(r)),f(r)]c=0 \end{aligned}$$

gives

$$\begin{aligned}&b\big (af(r)^2 - 2f(r)af(r) + f(r)^2a) + p(af(r)^2 \\&\quad -\,2f(r)af(r)+f(r)^2a\big )q\\&\quad +\,\big (af(r)^2 - 2f(r)af(r) + f(r)^2a\big )c=0, \end{aligned}$$

that is,

$$\begin{aligned}&baf(r)^2 - 2bf(r)af(r) + bf(r)^2a + paf(r)^2q \\&\quad -\,2pf(r)af(r)q+pf(r)^2aq \\&\quad +\,af(r)^2c-2f(r)af(r)c+f(r)^2ac=0 \end{aligned}$$

for any $r=(r_1,\ldots ,r_n) \in R^n$. We rewrite it as

$$\begin{aligned}&a_1f(r)^2 - 2a_2f(r)a_3f(r) + a_2f(r)^2a_3 + a_4f(r)^2a_5 \\&\quad -\, 2a_6f(r)a_3f(r)a_5 + a_6f(r)^2a_7\\&\quad +\, a_3f(r)^2a_8 - 2f(r)a_3f(r)a_8 + f(r)^2a_9=0 \end{aligned}$$

for any $r=(r_1,\ldots ,r_n) \in R^n$, where $a_1=ba, a_2=b, a_3=a, a_4=pa, a_5=q, a_6=p, a_7=aq, a_8=c, a_9=ac$. Now we study this situation in a matrix ring.

We need the following:

Lemma 3.1

[4, Lemma 1]. Let F be an infinite field and $k \ge 2$. If $A_1, \ldots , A_n$ are not scalar matrices in $M_k(F)$ then there exists some invertible matrix $P \in M_k(F)$ such that any matrices $PA_1P^{-1}, \ldots , PA_nP^{-1}$ have all non-zero entries.

PROPOSITION 3.2

Let $R=M_k(F)$ be the ring of all $k \times k$ matrices over the infinite field F, $f(x_1, \ldots , x_n)$ a non-central multilinear polynomial over F and $a_1, a_2, \ldots , a_9 \in R$. If

$$\begin{aligned}&a_1f(r)^2 - 2a_2f(r)a_3f(r) + a_2f(r)^2a_3 + a_4f(r)^2a_5 \\&\quad -\, 2a_6f(r)a_3f(r)a_5 + a_6f(r)^2a_7\\&\quad +\, a_3f(r)^2a_8 - 2f(r)a_3f(r)a_8 + f(r)^2a_9=0 \end{aligned}$$

for all $r=(r_1, \ldots , r_n)\in R^n$, then either $a_3$ or $a_5$ or $a_6$ is central.

Proof

By the hypothesis, we have

$$\begin{aligned} \begin{aligned}&a_1f(r_1, \ldots , r_n)^2-2a_2f(r_1, \ldots , r_n)a_3f(r_1, \ldots , r_n)+a_2f(r_1, \ldots , r_n)^2a_3\\&\quad +\,a_4f(r_1, \ldots , r_n)^2a_5-2a_6f(r_1, \ldots , r_n)a_3f(r_1, \ldots , r_n)a_5 \\&\quad +\,a_6f(r_1, \ldots , r_n)^2a_7\\&\quad +\,a_3f(r_1, \ldots , r_n)^2a_8-2f(r_1, \ldots , r_n)a_3f(r_1, \ldots , r_n)a_8 \\&\quad +\,f(r_1, \ldots , r_n)^2a_9=0. \end{aligned} \end{aligned}$$

Suppose that $a_3\notin Z(R)$, $a_5\notin Z(R)$ and $a_6\notin Z(R)$. Then we shall prove that this case leads to a contradiction.

Since $a_3\notin Z(R)$, $a_5\notin Z(R)$ and $a_6\notin Z(R)$, by Lemma 3.1, there exists a F-automorphism $\phi$ of $M_k(F)$ such that $\phi (a_3)$, $\phi (a_5)$ and $\phi (a_6)$ have all nonzero entries. Clearly, R satisfies the GPI,

$$\begin{aligned} \begin{aligned}&\phi (a_1)f(r_1, \ldots , r_n)^2-2\phi (a_2)f(r_1, \ldots , r_n)\phi (a_3)f(r_1, \ldots , r_n)\\&\quad +\,\phi (a_2)f(r_1, \ldots , r_n)^2\phi (a_3)+\phi (a_4)f(r_1, \ldots , r_n)^2\phi (a_5)\\&\quad -\,2\phi (a_6)f(r_1, \ldots , r_n)\phi (a_3)f(r_1, \ldots , r_n)\phi (a_5) \\&\quad +\,\phi (a_6)f(r_1, \ldots , r_n)^2\phi (a_7)\\&\quad +\,\phi (a_3)f(r_1, \ldots , r_n)^2\phi (a_8)-2f(r_1, \ldots , r_n)\phi (a_3)f(r_1, \ldots , r_n)\phi (a_8)\\&\quad +\,f(r_1, \ldots , r_n)^2\phi (a_9)=0. \end{aligned} \end{aligned}$$

(1)

As usual, by $e_{ij}$, $1\le i,j\le k$, we denote the matrix unit whose (i, j)-entry is equal to 1 and all its other entries are equal to 0. Since $f(x_1, \ldots , x_n)$ is non-central, by [13] (see also [15]), there exist $s_1, \ldots , s_n \in M_k(F)$ and $\beta \in F {\setminus } \{0\}$ satisfying $f(s_1, \ldots , s_n)=\beta e_{\mathrm{st}}$ with $s \ne t$. Moreover, since the set $\{f(y_1, \ldots , y_n) : y_1, \ldots , y_n \in M_k(F)\}$ is invariant under the action of all F-automorphisms of $M_k(F)$, for any $i \ne j$, there exists $u_1, \ldots , u_n \in M_k(F)$ such that $f(u_1, \ldots , u_n)=e_{ij}$. Hence by (1) we have

$$\begin{aligned} \begin{aligned}&\phi (a_1)e_{ij}^2-2\phi (a_2)e_{ij}\phi (a_3)e_{ij}+\phi (a_2)e_{ij}^2\phi (a_3) +\phi (a_4)e_{ij}^2\phi (a_5)\\&\quad -\,2\phi (a_6)e_{ij}\phi (a_3)e_{ij}\phi (a_5)+\phi (a_6)e_{ij}^2\phi (a_7)\\&\quad +\,\phi (a_3)e_{ij}^2\phi (a_8)-2e_{ij}\phi (a_3)e_{ij}\phi (a_8)+e_{ij}^2\phi (a_9)=0. \end{aligned} \end{aligned}$$

Multiplying left side and right side by $e_{ij}$, we obtain $2e_{ij}\phi (a_6)e_{ij}\phi (a_3)e_{ij}\phi (a_5)e_{ij}=0$. Since char$(R)\ne 2$, we have $\phi (a_6)_{ji}\phi (a_3)_{ji}\phi (a_5)_{ji}=0$. This is a contradiction as $\phi (a_3)$, $\phi (a_5)$ and $\phi (a_6)$ have all nonzero entries. Thus we conclude that either $a_3$ or $a_5$ or $a_6$ is central. $\square$

PROPOSITION 3.3

Let $R=M_k(F)$ be the ring of all matrices over the field F with $\mathrm{char} (R)\ne 2$, $f(x_1, \ldots , x_n)$ a non-central multilinear polynomial over F and $a_1, a_2, \ldots , a_9 \in R$. If

$$\begin{aligned}&a_1f(r)^2-2a_2f(r)a_3f(r)+a_2f(r)^2a_3+a_4f(r)^2a_5 \\&\quad -2a_6f(r)a_3f(r)a_5+a_6f(r)^2a_7\\&\quad +\,a_3f(r)^2a_8-2f(r)a_3f(r)a_8+f(r)^2a_9=0 \end{aligned}$$

for all $r=(r_1, \ldots , r_n) \in R^n$, then either $a_3$ or $a_5$ or $a_6$ is central.

Proof

If F is an infinite field, then by Proposition 3.2, we get the desired result. Next, we assume that F is finite.

Let E be an infinite field extension of the field F. Suppose that $\bar{R} = M_{k}(E)\,\cong R\otimes _{F} E$. Note that the multilinear polynomial $f(r_{1}, \ldots , r_{n})$ is central-valued on R if and only if it is central-valued on $\bar{R}$. R satisfies the GPI,

$$\begin{aligned} \Psi (r_1,\ldots ,r_n)&= a_1f(r_1, \ldots , r_n)^2-2a_2f(r_1, \ldots , r_n)a_3f(r_1, \ldots , r_n)\\&\quad +\,a_2f(r_1, \ldots , r_n)^2a_3+a_4f(r_1, \ldots , r_n)^2a_5\\&\quad -\,2a_6f(r_1, \ldots , r_n)a_3f(r_1, \ldots , r_n)a_5+a_6f(r_1, \ldots , r_n)^2a_7\\&\quad +\,a_3f(r_1, \ldots , r_n)^2a_8-2f(r_1, \ldots , r_n)a_3f(r_1, \ldots , r_n)a_8\\&\quad +\,f(r_1, \ldots , r_n)^2a_9=0 \end{aligned}$$

which is multi-homogeneous of multi-degree $(2,\ldots , 2)$ in the indeterminates $r_{1},\ldots , r_{n}$. Thus the complete linearization of $\Psi (r_{1},\ldots , r_{n})$ is a multilinear generalized polynomial $\Phi (r_{1},\ldots , r_{n},r_{1},\ldots , r_{n})$ in 2n indeterminates. Clearly, $\Phi (r_{1},\ldots , r_{n},r_{1},\ldots , r_{n}) = 2^{n}\Psi (r_{1},\ldots ,r_{n})$.

Note that the multilinear polynomial $\Phi (r_{1},\ldots , r_{n},r_{1},\ldots , r_{n})$ is a generalized polynomial identity for both R and $\bar{R}$. Since char($F) \ne 2$, we obtain $\Psi (r_{1},\ldots , r_{n})=0$ for all $r_{1},\ldots , r_{n} \in \bar{R}$. Hence by Proposition 3.2, the proof of proposition follows. $\square$

Lemma 3.4

Let R be a prime ring of characteristic different from 2 with extended centroid C and $f(x_1, \ldots , x_n)$ a non-central multilinear polynomial over C. Suppose that for some $a_1, a_2, \ldots , a_9 \in R$,

$$\begin{aligned}&a_1f(r)^2 - 2a_2f(r)a_3f(r) + a_2f(r)^2a_3 + a_4f(r)^2a_5 \\&\quad -\, 2a_6f(r)a_3f(r)a_5 + a_6f(r)^2a_7\\&\quad +\,a_3f(r)^2a_8 - 2f(r)a_3f(r)a_8 + f(r)^2a_9=0 \end{aligned}$$

for all $r=(r_1, \ldots , r_n)\in R^n$, then either $a_3$ or $a_5$ or $a_6$ is central.

Proof

Since R satisfies the generalized polynomial identity (GPI),

$$\begin{aligned} \begin{aligned} g(x_1,\ldots ,x_n)&= a_1f(x_1,\ldots ,x_n)^2-2a_2f(x_1,\ldots ,x_n)a_3f(x_1,\ldots ,x_n)\\&+\,a_2f(x_1,\ldots ,x_n)^2a_3+a_4f(x_1,\ldots ,x_n)^2a_5\\&-\,2a_6f(x_1,\ldots ,x_n)a_3f(x_1,\ldots ,x_n)a_5+a_6f(x_1,\ldots ,x_n)^2a_7\\&+\,a_3f(x_1,\ldots ,x_n)^2a_8-2f(x_1,\ldots ,x_n)a_3f(x_1,\ldots ,x_n)a_8\\&+\,vf(x_1,\ldots ,x_n)^2a_9=0 \end{aligned} \end{aligned}$$

(2)

for all $x_1, \ldots , x_n \in R$. Assume that $a_3\notin C$, $a_5 \notin C$ and $a_6\notin C$. By Fact 2, R and U satisfy the same GPI, U satisfies $g(x_1,\ldots ,x_n)=0.$ Suppose that $g(x_1,\ldots ,x_n)$ is a trivial GPI for U. Let $\mathcal {W}=U*_CC\{x_1,x_2,\ldots ,x_n\}$, the free product of U and $C\{x_1,\ldots ,x_n\}$, the free C-algebra in noncommuting indeterminates $x_1,x_2,\ldots , x_n$. So $g(x_1,\ldots ,x_n)$ is a zero element in $\mathcal {W}=U*_CC\{x_1,\ldots ,x_n\}$. In equation (2), the term $-2a_6f(x_1,\ldots ,x_n)a_3f(x_1,\ldots ,x_n)a_5$ appears nontrivially, implying that

$$\begin{aligned} -2a_6f(x_1,\ldots ,x_n)a_3f(x_1,\ldots ,x_n)a_5=0\in \mathcal {W}. \end{aligned}$$

This implies that either $a_3$ or $a_5$ or $a_6$ is central.

Now assume that $g(x_1,\ldots ,x_n)$ is a non-trivial GPI for U. In case C is infinite, we have $g(r_{1},\ldots , r_{n})= 0$ for all $r_{1},\ldots , r_{n} \in U\otimes _{C} \bar{C}$, where $\bar{C}$ is the algebraic closure of C. Moreover, both U and $U\otimes _{C} \bar{C}$ are prime and centrally closed algebras [8]. Hence, we substitute U or $U\otimes _{C} \bar{C}$ in place of R according to C finite or infinite respectively. Without loss of generality, we may suppose that $C = Z(R)$ and R is a centrally closed C-algebra. Using Martindale’s theorem [16], R is then a primitive ring having nonzero Socle $\mathrm{soc}(R)$ with C as the associated division ring. Hence by Jacobson’s theorem [10, p. 75], R is isomorphic to a dense ring of linear transformations of some vector space V over C.

First, suppose that V is finite dimensional over C, that is, $\mathrm{dim}_CV = k$. By density of R, we have $R\cong M_{k}(C)$. Since $f(r_{1},\ldots , r_{n})$ is not central-valued on R, R must be noncommutative and so $k \ge 2$. In this case, by Proposition 3.3, we get that either $a_3$ or $a_5$ or $a_6$ is in C, a contradiction.

If V is infinite dimensional over C, then for any $e^2=e\in \mathrm{soc}(R)$, we have $eRe\cong M_t(C)$ with $t\,{=}\,$dim$_CVe$. Since $a_3$, $a_5$ and $a_6$ are not in C, there exist $h_1,h_2,h_3\in \mathrm{soc}(R)$ such that $[a_3,h_1]\ne 0$, $[a_5,h_2]\ne 0$ and $[a_6,h_3]\ne 0$. By Litoff’s theorem [9], there exists idempotent $e\in \mathrm{soc}(R)$ such that $a_3h_1, h_1a_3, a_5h_2, h_2a_5, a_6h_3, h_3a_6, h_1, h_2,$

$h_3\in eRe$. Since R satisfies GPI, it follows that

$$\begin{aligned}&e\{a_1f(ex_1e,\ldots ,ex_ne)^2-2a_2f(ex_1e,\ldots ,ex_ne)a_3f(ex_1e,\ldots ,ex_ne) \\&+\,a_2f(ex_1e,\ldots ,ex_ne)^2a_3+a_4f(ex_1e,\ldots ,ex_ne)^2a_5 \\&-\,2a_6f(ex_1e,\ldots ,ex_ne)a_3f(ex_1e,\ldots ,ex_ne)a_5+a_6f(ex_1e,\ldots ,ex_ne)^2a_7 \\&+va_3f(ex_1e,\ldots ,ex_ne)^2a_8-2f(ex_1e,\ldots ,ex_ne)a_3f(ex_1e,\ldots ,ex_ne)a_8 \\&+\,f(ex_1e,\ldots ,ex_ne)^2a_9\}e=0, \end{aligned}$$

where the subring eRe satisfies

$$\begin{aligned}&ea_1ef(x_1,\ldots ,x_n)^2-2ea_2ef(x_1,\ldots ,x_n)ea_3ef(x_1,\ldots ,x_n) \\&+\,ea_2ef(x_1,\ldots ,x_n)^2ea_3e+ea_4ef(x_1,\ldots ,x_n)^2ea_5e \\&-\,2ea_6ef(x_1,\ldots ,x_n)ea_3ef(x_1,\ldots ,x_n)ea_5e+ea_6ef(x_1,\ldots ,x_n)^2ea_7e \\&+\,ea_3ef(x_1,\ldots ,x_n)^2ea_8e-2f(x_1,\ldots ,x_n)ea_3ef(x_1,\ldots ,x_n)ea_8e \\&+\,f(x_1,\ldots ,x_n)^2ea_9e=0. \end{aligned}$$

Then by the above finite dimensional case, either $ea_3e$ or $ea_5e$ or $ea_6e$ is the central element of eRe. This leads to a contradiction, since $a_3h_1=(ea_3e)h_1=h_1ea_3e=h_1a_3$, $a_5h_2=(ea_5e)h_2=h_2(ea_5e)=h_2a_5$ and $a_6h_3=(ea_6e)h_3=h_3(ea_6e)=h_3a_6$.

Hence we have proved that either $a_3$ or $a_5$ or $a_6$ is in C. $\square$

Theorem 3.5

Let R be a prime ring of characteristic different from 2, $f(x_1,\ldots ,x_n)$ a non-central multilinear polynomial over C and d a nonzero derivation of R. Suppose that for some $b,c,p,q\in R$, $b[d(u),u]+p[d(u),u]q+[d(u),u]c=0$ for all $u\in f(R)$. Then one of the following holds:

• (1) b, p, $pq+c\in C$ and $b+pq+c=0$;

• (2) $b+pq$, q, $c\in C$ and $b+pq+c=0$.

Proof

Let d be an inner derivation of R, that is, $d(x)=[a,x]$ for all $x\in R$. By hypothesis, R satisfies

$$\begin{aligned} b[[a,f(r)],f(r)]+p[[a,f(r)],f(r)]q+[[a,f(r)],f(r)]c=0, \end{aligned}$$

(3)

that is,

$$\begin{aligned}&baf(r)^2-2bf(r)af(r)+bf(r)^2a+paf(r)^2q \\&\quad -\,2pf(r)af(r)q+pf(r)^2aq\\&\quad +\,af(r)^2c-2f(r)af(r)c+f(r)^2ac=0 \end{aligned}$$

for all $r=(r_1,\ldots ,r_n) \in R^n$. Since d is nonzero, $a\notin C$. In this case, by Lemma 3.4, we have either $p\in C$ or $q\in C$.

Case i. Let $p\in C$. Then by hypothesis, R satisfies

$$\begin{aligned} b[[a,f(r)],f(r)]+[[a,f(r)],f(r)](pq+c)=0. \end{aligned}$$

By Lemma 3.3 in [2], $b, pq+c$ and $(b+pq+c)a$ are in C. Since $a\notin C$, we conclude that $b+pq+c=0$. This is our conclusion (1).

Case ii. Let $q\in C$. By hypothesis, R satisfies

$$\begin{aligned} (b+pq)[[a,f(r)],f(r)]+[[a,f(r)],f(r)]c=0. \end{aligned}$$

By Lemma 3.3 in [2], $b+pq, c$ and $(b+pq+c)a$ are in C. Since $a\notin C$, we conclude that $b+pq+c=0$. This is our conclusion (2).

Next, suppose that d is an outer derivation of R. By using Fact 5 and Kharchenko’s theorem [11], we can replace $d(x_{i})$ with $y_{i}$ and then R satisfies

$$\begin{aligned}&b[f^d(x_1,\ldots ,x_n)+\sum \limits _i f(x_1,\ldots ,y_i,\ldots ,x_n), f(x_1,\ldots ,x_n)]\\&\qquad +\,p[f^d(x_1,\ldots ,x_n)+\sum \limits _i f(x_1,\ldots ,y_i,\ldots ,x_n), f(x_1,\ldots ,x_n)]q\\&\qquad +\,[f^d(x_1,\ldots ,x_n)+\sum \limits _i f(x_1,\ldots ,y_i,\ldots ,x_n), f(x_1,\ldots ,x_n)]c=0. \end{aligned}$$

In particular, R satisfies blended component

$$\begin{aligned} \begin{aligned}&b[\underset{i}{\Sigma } f(x_1,\ldots ,y_i,\ldots , x_n), f(x_1,\ldots ,x_n)] \\&\quad +p[\underset{i}{\Sigma } f(x_1,\ldots ,y_i,\ldots , x_n), f(x_1,\ldots ,x_n)]q\\&\quad +[\underset{i}{\Sigma } f(x_1,\ldots ,y_i,\ldots , x_n), f(x_1,\ldots ,x_n)]c=0. \end{aligned} \end{aligned}$$

(4)

Since R is noncommutative, we choose $a'\in R$ such that $a'\notin C$. Replacing $[a', x_i]$ in place of $y_i$ in equation (4), we get

$$\begin{aligned} b[[a',f(r)],f(r)]+p[[a',f(r)],f(r)]q+[[a',f(r)],f(r)]c=0 \end{aligned}$$

for all $r=(r_1,\ldots ,r_n) \in R^n$, which is the same as equation (3). Then by the same argument as above, we have our conclusions. $\square$

In particular, for right-sided annihilator condition, we have the following.

COROLLARY 3.6

Let R be a prime ring of characteristic different from 2, $f(x_1,\ldots ,x_n)$ a non-central multilinear polynomial over C and d a nonzero derivation of R. Suppose that for some $a\in R$, $[d(u),u]a=0$ for all $u\in f(R)$. Then $a=0$.

In particular, for two-sided annihilator condition, we have the following.

COROLLARY 3.7

Let R be a prime ring of characteristic different from 2, $f(x_1,\ldots ,x_n)$ a non-central multilinear polynomial over C and d a nonzero derivation of R. Suppose that for some $a,b\in R$, $a[d(u),u]b=0$ for all $u\in f(R)$. Then either $a=0$ or $b=0$.

Putting $p=0$ and $q=0$ in Theorem 3.5, we have the inner part of Theorem 5.3 of [2]. More precisely, we obtain the following.

COROLLARY 3.8

Let R be a prime ring of characteristic different from 2, $f(x_1,\ldots ,x_n)$ a non-central multilinear polynomial over C and d a nonzero derivation of R. Suppose that for some $b,c\in R$, $b[d(u),u]+[d(u),u]c=0$ for all $u\in f(R)$. Then $b=-c\in C$.

Replacing b by $s^{2}$, c by $t^{2}$, $p=2s$ and $q=t$ in Theorem 3.5, we obtain the following.

COROLLARY 3.9

Let R be a prime ring of characteristic different from 2 and $f(x_1,\ldots ,x_n)$ a non-central multilinear polynomial over C. If d is a nonzero derivation of R, and F is an inner generalized derivation of R such that

$$\begin{aligned} F^2([d(f(x_1,\ldots ,x_n)), f(x_1,\ldots ,x_n)])=0 \end{aligned}$$

for all $x_1,\ldots ,x_n\in R$, then there exists $a\in U$ such that $F(x)=xa$ for all $x\in R$ or $F(x)=ax$ for all $x\in R$, with $a^2=0$.

In the next section, we will extend Corollary 3.9 to the arbitrary generalized derivation. Now we are ready to prove the main theorem.

4 The proof of the main theorem

Lee [12] proved that every generalized derivation can be uniquely extended to a generalized derivation of U, and thus all generalized derivations of R will be implicitly assumed to be defined on the whole U. In particular, Lee proved that every generalized derivation g on a dense right ideal of R can be uniquely extended to U and has the form $g(x) = ax + d(x)$ for some $a\in U$ and a derivation d of R.

Theorem 4.1

Suppose that R is a prime ring of characteristic different from 2 and $f(x_1,\ldots ,x_n)$ is a non-central multilinear polynomial over C. If d is a nonzero derivation of R, and F is a generalized derivation of R such that

$$\begin{aligned} F^2([d(f(x_1,\ldots ,x_n)), f(x_1,\ldots ,x_n)])=0 \end{aligned}$$

for all $x_1,\ldots ,x_n\in R$, then there exists $a\in U$ such that $F(x)=xa$ for all $x\in R$ or $F(x)=ax$ for all $x\in R$, with $a^2=0$.

Proof

In light of [12, Theorem 3], we may assume that there exist $b \in U$ and derivation $\delta$ of U such that $F(x) = bx + \delta (x)$ and so, $F^2(x)=b^2x+2b\delta (x)+\delta (b)x+\delta ^2(x)$. Since R and U satisfy the same generalized polynomial identities (see Fact 2) as well as the same differential identities (see Fact 4), without loss of generality, we have

$$\begin{aligned} F^2[d(f(r_1,\ldots ,r_n)), f(r_1,\ldots ,r_n)]=0 \end{aligned}$$

for all $r_1,\ldots ,r_n \in U$. If F is an inner generalized derivation of R, then assume that $F(x)=bx+xc$ for all $x\in R$, with some $b,c\in U$. In this case, by the hypothesis

$$\begin{aligned}b^2[d(r), r]+2b[d(r), r]c+[d(r), r]c^2=0\end{aligned}$$

for all $r\in f(R)$. Then by Theorem 3.5, one of the following holds:

• (i) $b^2,b, 2bc+c^2\in C$ and $b^2+2bc+c^2=0$, that is $(b+c)^2=0$. In this case, $F(x)=x(b+c)$ for all $x\in R$ with $(b+c)^2=0$.

• (ii) $b^2+2bc, c, c^2\in C$ and $b^2+2bc+c^2=0$, that is, $(b+c)^2=0$. In this case, $F(x)=(b+c)x$ for all $x\in R$ with $(b+c)^2=0$.

Now, we assume that F is outer. By the hypothesis, U satisfies

$$\begin{aligned} b^2[d(r),r]+2b\delta ([d(r),r])+\delta (b)[d(r),r]+\delta ^2([d(r),r])=0 \end{aligned}$$

(5)

for all $r\in f(R)$.

Case I. Let d and $\delta$ be C-dependent modulo inner derivations of U, that is, $\alpha d+\beta \delta =ad_q$, where $\alpha ,\beta \in C$, $q\in U$ and $ad_q(x)=[q,x]$ for all $x\in U$. If $\alpha =0$, then $\delta$ must be inner and so F is inner, a contradiction. Hence $\alpha \ne 0$, and hence $d=\lambda \delta +ad_p$, where $\lambda =-\alpha ^{-1}\beta$ and $p=\alpha ^{-1}q$.

Then by the hypothesis, it follows that

$$\begin{aligned} \begin{aligned}&b^2[\lambda \delta (r)+[p,r], r]+2b\delta ([\lambda \delta (r)+[p,r],r]) \\&\quad + \delta (b)[\lambda \delta (r) + [p,r],r] \\&\quad +\delta ^2([\lambda \delta (r)+[p,r],r])=0 \end{aligned} \end{aligned}$$

(6)

for all $r\in f(R)$.

Using Fact 5, substitute the values of $\delta (f(r_1,\ldots ,r_n))$, $\delta ^2(f(r_1,\ldots ,r_n))$ and $\delta ^3(f(r_1,\ldots ,r_n))$ in equation (6). Then by Kharchenko’s theorem [11], we can replace $\delta (r_i)$ with $y_i$, $\delta ^2(r_i)$ with $w_i$ and $\delta ^3(r_i)$ with $z_i$ in equation (6) and then U satisfies the blended component

$$\begin{aligned}{}[\lambda \underset{i}{\Sigma } f(r_1,\ldots ,z_i,\ldots ,r_n), f(r_1,\ldots ,r_n)]=0. \end{aligned}$$

We choose $q\in U$ such that $q\notin C$ and replace $z_i$ by $[q,r_i]$. Then U satisfies

$$\begin{aligned}{}[\lambda q, f(r_1,\ldots ,r_n)]_2=0. \end{aligned}$$

By [13, Theorem], $\lambda q\in C$. Since $q\notin C$, $\lambda =0$. Hence by equation (6),

$$\begin{aligned} b^2[[p,r], r]+2b\delta ([[p,r],r])+\delta (b)[[p,r],r]+\delta ^2([[p,r],r])=0 \end{aligned}$$

(7)

for all $r\in f(R)$.

Putting the values of $\delta (f(r_1,\ldots ,r_n))$ and $\delta ^2(f(r_1,\ldots ,r_n))$ in equation (7), then again by Kharchenko’s theorem [11], we can replace $\delta (r_i)$ with $y_i$ and $\delta ^2(r_i)$ with $w_i$ in (7), and then U satisfies the blended component

$$\begin{aligned}&{[[}p, \underset{i}{\Sigma } f(r_1,\ldots ,w_i,\ldots ,r_n)], f(r_1,\ldots ,r_n)]\\&+[[p, f(r_1,\ldots ,r_n)], \underset{i}{\Sigma } f(r_1,\ldots ,w_i,\ldots ,r_n)]=0. \end{aligned}$$

By taking $w_1=r_1$ and $w_2=\cdots =w_n=0$, U satisfies

$$\begin{aligned} 2[[p, f(r_1,\ldots ,r_n)], f(r_1,\ldots ,r_n)]=0. \end{aligned}$$

Since char($R)\ne 2$, by [13, Theorem] $p\in C$. This gives that $d=0$, a contradiction.

Case II. Let d and $\delta$ be C-independent modulo inner derivations of U. Then by applying Fact 5 and Kharchenko’s theorem [11] to equation (5), we can replace $d(r_i)$ with $y_i$, $\delta (r_i)$ with $z_i$, $\delta d(r_i)$ with $s_i$, $\delta ^2(r_i)$ with $t_i$ and $\delta ^2 d(r_i)$ with $u_i$. Then U satisfies the blended component

$$\begin{aligned}{}[\underset{i}{\Sigma } f(r_1,\ldots ,u_i,\ldots ,r_n), f(r_1,\ldots ,r_n)]=0. \end{aligned}$$

In particular, replacing $u_i$ with $[q, r_i]$ for some $q\notin C$, U satisfies

$$\begin{aligned}{}[q, f(r_1,\ldots ,r_n)]_2=0. \end{aligned}$$

Again by [13, Theorem], $q\in C$, a contradiction. $\square$

COROLLARY 4.2

Let R be a prime ring of characteristic different from 2 with extended centroid C and $f(x_1,\ldots ,x_n)$ a multilinear polynomial over C. If d and $\delta$ are two nonzero derivations of R such that

$$\begin{aligned} \delta ^2([d(f(x_1,\ldots ,x_n)), f(x_1,\ldots ,x_n)])=0\end{aligned}$$

for all $x_1,\ldots ,x_n\in R$, then $f(x_1,\ldots ,x_n)$ is central-valued on R.