5 research outputs found
Cofactor matrices
AbstractThe function which maps a square matrix A to its cofactor matrix cof(A) is examined. A characterization is given for the image of the function. Its injective properties on the general linear group of nonsingular matrices are also addressed
Elementary sequences
Explicit finite algebraic formulas are given for some well-known sequences, including complements of polynomial sequences
On the mapping xy→(xy)n in an associative ring
We consider the following condition (*) on an associative ring
R:(*). There exists a function f from R into R such that f is a group homomorphism of (R,+), f is
injective on R2, and f(xy)=(xy)n(x,y) for some
positive integer n(x,y)>1. Commutativity and structure are
established for Artinian rings R satisfying (*), and a
counterexample is given for non-Artinian rings. The results
generalize commutativity theorems found elsewhere. The case n(x,y)=2 is examined in detail
© Hindawi Publishing Corp. ON THE MAPPING xy → (xy) n IN AN ASSOCIATIVE RING
We consider the following condition (*) on an associative ring R: (*). There exists a function f from R into R such that f is a group homomorphism of (R,+), f is injective on R 2,and f(xy) = (xy) n(x,y) for some positive integer n(x,y)> 1. Commutativity and structure are established for Artinian rings R satisfying (*), and a counterexample is given for non-Artinian rings. The results generalize commutativity theorems found elsewhere. The case n(x,y) = 2 is examined in detail. 2000 Mathematics Subject Classification: 16D70, 16P20. Let R be an associative ring, not necessarily with unity, and let R + denote the additive group of R. In[3], it was shown that R is commutative if it satisfies the following condition. (I) For each x and y in R, there exists n = n(x,y)> 1 such that (xy) n = xy. We generalize this result by considering the condition below. (II) There exists a function f from R into R such that f is a group homomorphism of R +, f is injective on R 2,andf(xy) = (xy) n(x,y) for some positive integer n = n(x,y)> 1 depending on x and y