On group graded rings satisfying polynomial identities

Abstract

A number of classical theorems of ring theory deal with nilness and nilpotency of the Jacobson radical of various ring constructions (see [10], [18] This paper is devoted to the radicals of group graded rings, which have been actively investigated by many authors (see [10], [14]). Let G be a group. An associative ring R = 0 ^g is said to be G-graded {stronglyFirst, we consider algebras over a field of characteristic zero. In this case our result will be also of interest in connection with the well-known problem of finding necessary and sufficient conditions for the Jacobson radical to be homogeneous. An ideal / of R = 0 R g is said to be homogeneous if / = 0 IC\R g . This problem has not been solved Pi-algebra over a field of characteristic zero. If the Jacobson radical J(R e ) is nil, then J(R) is a homogeneous nil ideal of R.The following corollary to the main theorem is worth mentioning.COROLLARY 2. Let G be a group with identity e, and let R = 0 R g be a strongly g &G G-graded Pi-algebra over a field of characteristic zero. IfJ(R e ) is nilpotent, then J(R) is nilpotent.It is impossible to replace strongly graded algebras by ordinary group graded algebras in Corollary 2. Indeed, let A be the free commutative algebra with free generators a u a 2 , Denote by / the ideal of A generated by a u a\, a\, Then All is positively graded, and so All = 0 A z where Z is the infinite cyclic group. Although A o = 0, it is zeZ clear that J(A/I) = A/1 is not nilpotent.One cannot omit the restriction on the characteristic of the field neither in Theorem Glasgow Math.

On graded nil good rings

“…According to [7, Proposition 2.5], J(R e ) is nil. Now, by [16,Theorem 3] we have that that J(R) is nil since R is by assumption P I. On the other hand, R is by assumption Jacobson radical ring.…”

“…Now, according to [5,Theorem 4.4] we have that J g (R) = J(R). On the other hand, [16,Theorem 3] implies that J(R) is nil, and hence J g (R) is graded-nil. Let H be the homogeneous part of R/J g (R).…”

“…We know by Amitsur-Levitski theorem (see [1]) that M 2 (R) is P I-ring. Now according to [16,Theorem 3] Proof. Using mathematical induction on n, we will prove that M n (R)(σ) is graded fine where R = R/J g (R).…”

On graded nil clean rings

“…Therefore, our assumption yields that J(R e ) is nil and that R e /J(R e ) is nil clean. R is by assumption PI, and hence, Theorem 3 in [17] implies that J(R) is nil. Since G is finite, we have that J(R e ) = J g (R) ∩ R e , according to Corollary 4.2 in [4].…”

“…Therefore, our assumption yields that J(R e ) is nil. Now, Theorem 3 in [17] tells us that J(R) is nil since R is by assumption PI. However, R is by assumption Jacobson radical ring.…”