Symmetric Elements Under Oriented Involutions in Group Rings

has characterized those rings with involution J, with 1, and such that every symmetric element has an inverse under the additional hypothesis that S~~, the subring generated by ,S (the set of symmetric elements), is A, the ring itself.In this paper we raise the question as to when S~ = A under a weaker hypothesis and prove the following theorem.(A henceforth is such that 2A =A and 2x=0 implies x=0 for any xEA.)Theorem 1.Let A be a ring with 1 and suppose that S is a simple Jordan ring under the Jordan multiplication, s o t = st+ts for all s and t in S. Then either S~ = A or SEZ, the center of A, or K, the set o/ skew elements, is an ideal o/ A with K2EZ, K3 = (8), and S is an associative ring under Jordan multiplication.We note that Osborn proves that under his hypothesis, either A is a division ring; the direct sum of two division rings which are antiisomorphic, and / interchanges the summands; or 51 is a field under Jordan multiplication, K is an ideal of A where K2 = (d).A corollary to our result is Corollary 1.Let A be as in Theorem 1. Suppose /urther that A contains no nilpotent ideals.Then either A is simple or A is a direct sum o/ two simple rings which are anti-isomorphic and such that J interchanges the direct summands.

Let R be a 2-torsion free commutative ring with involution, and δ a nonzero derivation of R. Let S be the set of symmetric elements in R, and let K be the set of anti-symmetric elements in R. In this article, we investigate the semiprimeness of the Lie rings Sδ when δ is symmetric and Kδ when δ is anti-symmetric.

Ordered ∗ -Rings

Herstein's Lie theory revisited

Two questions on semi-clean group rings

For a Noetherian ‐algebra , there is a canonical inclusion , and each element in the image satisfies a certain compatibility condition. We call compatible if the image coincides with the set of all compatible elements. For example, for a Dynkin quiver and a commutative Noetherian ring , the path algebra is compatible. In this paper, we prove that is compatible when is an extended Dynkin quiver and is either a Dedekind domain or a Noetherian semilocal normal ring of dimension 2.

A commutative ring R is called J-rad clean in case, for any r ∈ R, there is an idempotent e ∈ R such that r−e ∈ U(R) and re ∈ J(R), where U(R) and J(R) denote the set of units and the Jacobson radical of R, respectively. Also, a ring R is called semiregular if R/J(R) is regular in the sense of von Neumann and idempotents lift modulo J(R). We demonstrate that these two concepts are, actually, equivalent and we portray a portion of the properties of this class of rings. In particular, as a direct application, we prove that the commutative group ring RG is J-rad clean if, and only if, R is a commutative J-rad clean ring and G is a torsion abelian group, provided that J(R) is nil.

Let $KG$ be the group ring of a group $G$ over a commutative ring $K$ with unity. The rings $KG$ are described for which $xx^\sigma =x^\sigma x$ for all $x=\sum _{g\in G}\alpha _gg\in KG$, where $x\mapsto x^\sigma =~\sum _{g\in G}\alpha _gf(g)\sigma (g)$ is an involution of $KG$; here $f: G\to U(K)$ is a homomorphism and $\sigma$ is an antiautomorphism of order two of $G$.

Let B be any nite commutative ring with identity. In this case, B(X; 1 p kZ0) B(X; 1 p 2Z0) B(X; 1 p Z0), where p is a prime number and k 1, is the descending chain of commutative semigroup rings. All these semigroup rings are containing the polynomial ring B(X;Z0). In this paper, we introduced a construction technique of cyclic codes through the semi- group ring B(X; 1 p kZ0) instead of a polynomial ring.

Lie properties of symmetric elements in group rings

The algebraic hypercenter of an algebra over a commutative ring

Let S be a cancellative monoid with quotient group of torsion-free rank a.We show that the monoid ring R[S] is a Hilbert ring if and only if the polynomial ring R[{ X, },s/] is a Hilbert ring, where |/| = a.Assume that R is a commutative unitary ring and G is an abelian group.The first research problem listed in [K, Chapter 7] is that of determining equivalent conditions in order that the group ring R[G] should be a Hilbert ring.Matsuda has considered this question in [Ml, 6].His results show that for G finitely generated, R[G] and R are simultaneously Hilbert rings; if G is not finitely generated, then R[G] a Hilbert ring implies that R is a Hilbert ring, but the converse fails.In [M2, 5] Matsuda considers briefly the corresponding Hilbert-ring-characterization problem for a monoid ring R[S], where S is torsion-free and cancellative.Let a be the torsion-free rank of G and let {Xj)lsl be a set of indeterminates over R of cardinality a.In Corollary 1 we show that R[G] and P[{^,},e/] are simultaneously Hilbert rings.While the problem of determining equivalent conditions for P[{Ar,}/e/] to be a Hilbert ring has not been completely resolved, it has been worked on extensively [Kr, Go, L, Gl, H], and a significant body of positive results exists concerning this problem.The proof of Theorem 1 suggests the following conjecture: if S is a cancellative commutative monoid with quotient group G, then R[S] is a Hilbert ring if and only if R[G] is a Hilbert ring; this result is established in Theorem 2.All monoids considered are assumed to be commutative, and rings are assumed to be commutative and unitary.The statement of Theorem 1 uses the following terminology.For a cardinal number a, an extension ring T of R is said to be a-generated over R if T is generated over R by a set of cardinality at most a.Theorem 1. Assume that X = {XA is a set of indeterminates of cardinality a over the ring R. Denote by Za the direct sum of a copies of the additive group Z of integers.The following conditions are equivalent.(1) R[X] is not a Hilbert ring.(2) There exists a prime ideal P of R such that R/P admits an a-generated extension ring that is a G-domain, but not afield.(3) The group ring R[Za] is not a Hilbert ring._

Let G be an abelian group, R a commutative ring of prime characteristic p with identity and R t G a commutative twisted group ring of G over R. Suppose p is a fixed prime, G p and S(R t G) are the p-components of G and of the unit group U(R t G) of R t G, respectively. Let R* be the multiplicative group of R and let f α(S) be the α-th Ulm-Kaplansky invariant of S(R t G) where α is any ordinal. In the paper the invariants f n (S), n ∈ ℕ ∪ {0}, are calculated, provided G p = 1. Further, a commutative ring R with identity of prime characteristic p is said to be multiplicatively p-perfect if (R*)p = R*. For these rings the invariants f α(S) are calculated for any ordinal α and a description, up to an isomorphism, of the maximal divisible subgroup of S(R t G) is given.

Let G be an abelian group and let R be a commutative ring with identity. Denote by R t G a commutative twisted group algebra (a commutative twisted group ring) of G over R, by ℬ(R) and ℬ(R t G) the nil radicals of R and R t G, respectively, by G p the p-component of G and by G 0 the torsion subgroup of G. We prove that: i. If R is a ring of prime characteristic p, the multiplicative group R* of R is p-divisible and ℬ(R) = 0, then there exists a twisted group algebra R t 1 (G/G p ) such that R t G/ℬ(R t G) ≅ R t 1 (G/G p ) as R-algebras; ii. If R is a ring of prime characterisitic p and R* is p-divisible, then ℬ(R t G) = 0 if and only if ℬ(R) = 0 and G p = 1; and iii. If B(R) = 0, the orders of the elements of G 0 are not zero divisors in R, H is any group and the commutative twisted group algebra R t G is isomorphic as R-algebra to some twisted group algebra R t 1 H, then R t G 0 ≅ R t 1 H 0 as R-algebras.

In this paper some conditions for a skew group ring or a crossed product to have finite weak global dimension are given.Using these results we obtain some necessary conditions and some sufficient conditions for a skew group ring or a crossed product to be a Dubrovin valuation ring.If R*G is a skew group ring, where the coefficient ring R is a commutative ring and G is a finite group, then we prove that the conditions we obtained become necessary and sufficient conditions.In particular, if R is a commutative valuation ring, then R*G is a Dubrovin valuation ring if and only if G T=<1>,where G T is the inertial group of R.