Question

6. Let R be a ring, and let 11 and 12 be ideals of R. We define the product of 11 and 12 to be 1112 = {TER:r => aibi, with k > 1, Q1, ..., ak € 11, b1,..., bk € 12 In other words, an element of the product 1.12 is a finite sum of products a;bi, where a, comes from I and bi comes from 12. (a) Prove that 11 12 is an ideal of R, contained in the intersection 11 12. (b) Given any positive integers a and b, show that (aZ)(bZ) = abz.

Answer 1

aj let - cibi e I, Iz and a aj bj e Zitz Then, a; e I, and bi & Iz w i = 1,2, ...,K and by aj f I, and bj f I V j = 1,2,..., I as ai e I, and aj & I, and Also, bigla and bj f Iz Now, let VER and È aibi & I, Iz r Ěaibi - Ž crai) bi since I, is ideal, so, rod, and rER so rai E I, Also, bie Iz =) 8 crai) bi E I, Iz rrow , b) vet a, b e z . To show, caz) (bZ) = (abZ) Let ak, e caz) and bake (62) > Then, cak (6K2) e caz) (b ) = cab) CK K2) E let abk E aba where kez Then, cak) (by G abk = (ak) (b.1) E (az) (bz) o C KE 2 and irr test cak) car and lake let laki) (bK2) Elaz) (b ZC2), where, K, , K 26 Z = (ab) CK, K₂) E car ch Z) Then o) (as (K, K2) c alab ) C. KIK2 E 2) = a (a) (bZ) = Cab 2) Yow,