Commutativity of rings with constraints involving a subset

Moharram A. Khan

Czechoslovak Mathematical Journal (2003)

  • Volume: 53, Issue: 3, page 545-559
  • ISSN: 0011-4642

Abstract

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Suppose that R is an associative ring with identity 1 , J ( R ) the Jacobson radical of R , and N ( R ) the set of nilpotent elements of R . Let m 1 be a fixed positive integer and R an m -torsion-free ring with identity 1 . The main result of the present paper asserts that R is commutative if R satisfies both the conditions (i) [ x m , y m ] = 0 for all x , y R J ( R ) and (ii) [ ( x y ) m + y m x m , x ] = 0 = [ ( y x ) m + x m y m , x ] , for all x , y R J ( R ) . This result is also valid if (i) and (ii) are replaced by (i) ' [ x m , y m ] = 0 for all x , y R N ( R ) and (ii) ' [ ( x y ) m + y m x m , x ] = 0 = [ ( y x ) m + x m y m , x ] for all x , y R N ( R ) . Other similar commutativity theorems are also discussed.

How to cite

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Khan, Moharram A.. "Commutativity of rings with constraints involving a subset." Czechoslovak Mathematical Journal 53.3 (2003): 545-559. <http://eudml.org/doc/30798>.

@article{Khan2003,
abstract = {Suppose that $R$ is an associative ring with identity $1$, $J(R)$ the Jacobson radical of $R$, and $N(R)$ the set of nilpotent elements of $R$. Let $m \ge 1$ be a fixed positive integer and $R$ an $m$-torsion-free ring with identity $1$. The main result of the present paper asserts that $R$ is commutative if $R$ satisfies both the conditions (i) $[x^m,y^m] = 0$ for all $x,y \in R \setminus J(R)$ and (ii) $[(xy)^m + y^mx^m, x] = 0 = [(yx)^m + x^my^m, x]$, for all $x,y \in R \setminus J(R)$. This result is also valid if (i) and (ii) are replaced by (i)$^\{\prime \}$$[x^m,y^m] = 0$ for all $x,y \in R \setminus N(R)$ and (ii)$^\{\prime \}$$[(xy)^m + y^m x^m, x] = 0 = [(yx)^m + x^m y^m, x]$ for all $x,y \in R\backslash N(R) $. Other similar commutativity theorems are also discussed.},
author = {Khan, Moharram A.},
journal = {Czechoslovak Mathematical Journal},
keywords = {commutativity theorems; Jacobson radicals; nilpotent elements; periodic rings; torsion-free rings; commutativity theorems; Jacobson radical; nilpotent elements; periodic rings; torsion-free rings; polynomial constraints; commutator constraints},
language = {eng},
number = {3},
pages = {545-559},
publisher = {Institute of Mathematics, Academy of Sciences of the Czech Republic},
title = {Commutativity of rings with constraints involving a subset},
url = {http://eudml.org/doc/30798},
volume = {53},
year = {2003},
}

TY - JOUR
AU - Khan, Moharram A.
TI - Commutativity of rings with constraints involving a subset
JO - Czechoslovak Mathematical Journal
PY - 2003
PB - Institute of Mathematics, Academy of Sciences of the Czech Republic
VL - 53
IS - 3
SP - 545
EP - 559
AB - Suppose that $R$ is an associative ring with identity $1$, $J(R)$ the Jacobson radical of $R$, and $N(R)$ the set of nilpotent elements of $R$. Let $m \ge 1$ be a fixed positive integer and $R$ an $m$-torsion-free ring with identity $1$. The main result of the present paper asserts that $R$ is commutative if $R$ satisfies both the conditions (i) $[x^m,y^m] = 0$ for all $x,y \in R \setminus J(R)$ and (ii) $[(xy)^m + y^mx^m, x] = 0 = [(yx)^m + x^my^m, x]$, for all $x,y \in R \setminus J(R)$. This result is also valid if (i) and (ii) are replaced by (i)$^{\prime }$$[x^m,y^m] = 0$ for all $x,y \in R \setminus N(R)$ and (ii)$^{\prime }$$[(xy)^m + y^m x^m, x] = 0 = [(yx)^m + x^m y^m, x]$ for all $x,y \in R\backslash N(R) $. Other similar commutativity theorems are also discussed.
LA - eng
KW - commutativity theorems; Jacobson radicals; nilpotent elements; periodic rings; torsion-free rings; commutativity theorems; Jacobson radical; nilpotent elements; periodic rings; torsion-free rings; polynomial constraints; commutator constraints
UR - http://eudml.org/doc/30798
ER -

References

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