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Sample text

Then xm • .!!!!. , xm is a unity modulo pmZ(qi 1 li E 1), and so pmZ(qi 1 liE I) is a modular ideal. Let I~ 0 be a proper ideal in mZ(qi 1 1i E I)~ Q, and let mr be the smallest positive integer in I. Clearly I= rmZ(qi 1 1i E 1), and (rm, qi) = 1 for all i E I. Since I. is a proper ideal in mZ(qi 1 1i E 1), r ~ 1. Let p be a prime such that plr. Then p ¢ {qi liE I}, and I= pmZ(qi 1 li E I). Hence every maximal ideal M in mZ(qi 1 1i E I) is of the form M= pmZ(qi 1 li E I). Suppose that M is modular.

Let p be a prime such that plr. Then p ¢ {qi liE I}, and I= pmZ(qi 1 li E I). Hence every maximal ideal M in mZ(qi 1 1i E I) is of the form M= pmZ(qi 1 li E I). Suppose that M is modular. Then there 1 1 - mr E mZ(q: 1i E I) such that .!!!!. • m = m + pmf with s and s 1 s 1 s s 1 1 products of primes in {qili E 1}. Hence mrs - ss = pr 1s. Now (ss 1 , p) = 1, and so (m,p) = 1. ll i E I) with p a prime such that P ¢ {qi liE 1}, and (m,p) = 1. exists Theorem 4. 6: equivalent: 1) 2) 3) Let G be a rank one torsion free group.

Gp = Hp (f) Dp , Hp a reduced subgroup of Gp• 00 00 54 ) ). 10. For every positive integer n, pnGP ~ pn+lGP is an inclusion of ideals in R. Hence there exists a positive integer n, such that pnG = pn+lGP. Clearly pkG = pkH €) D for every positive integer p p p p k. Therefore pnHP = pn+lHP. , for every 1 -< i -< m, there exists a positive integer n1. n. such that H pi 4 (tj j=l a. Z(p~), cxJ. a cardinal number, j = 1 , ••• ,ni. J Clearly G is of the form of condition 2). 1. 8: Let G be a torsion group.