tailieunhanh - Extensions and s-comparability of exchange rings

Let S be a ring extension of R. In this note, for any positive integer s we study s-comparability related to ring extensions. We show that if S is an excellent extension of R, R and S are exchange rings, and R has the n-unperforation property. | Turk J Math 36 (2012) , 544 – 549. ¨ ITAK ˙ c TUB doi: Extensions and s-comparability of exchange rings Chaoling Huang Abstract Let S be a ring extension of R . In this note, for any positive integer s we study s -comparability related to ring extensions. We show that if S is an excellent extension of R , R and S are exchange rings, and R has the n -unperforation property. R satisfies s -comparability if and we only if so does S , and we prove that for a 2-sided ideal J of S , and an exchange subring R of the exchange ring S , which contains J as a direct summand, then R satisfies s -comparability if and only if so does R/J . Key Words: Exchange rings, excellent extensions, s -comparability 1. Introduction We say that S is a ring extension of R if there is a (unital) ring homomorphism f : R → S . Let S be a ring and let R be a subring of S (with the same 1). S is called a finite normalizing extension of R if there exist elements a1 , . . . , an ∈ S such that a1 = 1, S = Ra1 + · · · + Ran , ai R = Rai for all i = 1, . . . , n. Finite normalizing extensions have been studied in many papers such as [4, 8, 9, 10, 13]. S is called a free normalizing extension of R if a1 = 1, S = Ra1 + · · · + Ran is finite normalizing extension and S is free with basis {a1 , . . . , an } as both a right R -module and a left R -module. S is said to be an excellent extension of R in case S is a free normalizing extension of R and S is right R -projective (that is, if MS is a right S -module and NS is a submodule of MS , then NR | MR implies NS | MS , where N |M means N is a direct summand of M ). For any right R -module M , Crawley and J´ onsson defined M to have the exchange property if for every right R -module A and any 2 decompositions of A, A = M ⊕ N = ⊕i∈I Ai where M ∼ = M , there are submodules A i ⊆ Ai such that A = M ⊕ (⊕i∈I A i ). It follows from the modular law that A i must be a direct summand of Ai for all i. Warfield [12] called a ring R an .