Additive Groups of Rings by S. Feigelstock

By S. Feigelstock

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0 0 beanassociativestrongly K principal ideal are ideal ring ringgroup. group. Then HH and K are cyclic. It suffices H and suffices totonegate negate that that H Proof: It 2 '! 0. ) be R2 ring satisfying R R R = (G,·) be aa ring K are both associative associative nil. nil. K are both Let = 2 K. There such that Suppose that that R R2 exist h0 1) Suppose There exist h0 E H, kk00 E K, such k0> •. Let hh E H. H. Since hh € R, there there exists an an integer integer n, R =

Group. Q+. Hence Hom(Q+@Q+, Z(p""J) '! 9. 9. G be torsion free Let G be aa rank rank one one torsion free group. group. 6. 6. Theorem Let G satisfy the theconditions conditionsofofthe thetheorem. theorem. 6isis that that satisfying tp(G) == =. 6 for which the set of J(R) = p1 ••• pkR' with {pl , ••. ,pk} the ofprimes primes pp for which = tp(G) = 0. NOW Now p1 ••• pkR+ ""G, so J(R) is aa radical radical ring ring with with G,, + G. [J(R)] ""G. 2]. 2]. For an proof,see see [44, [44,Theorem Theorem 4].

Hence V÷1 and so R(n+l) = 0, = n+ 1 ' k Claim: R2 +l For every integer k, For every positive positive integer c R(k+2). R(k+ 2 ). For k == 00 the k R(k+ 2 ) for some some positive positive integer integer k. 2k+l + 1 2k+l 2k+1 + . Then x == y·z with y Let xx E RR y E RR or zz E RR .. z E R(k+ 3). Therefore and so 2k+l R2 +1 true. Suppose R claim is claim is clearly true. c 1 2n-l ++ 11 R R2 c E ( 1) R(n+l) R n+ = = 0. 0. 6: Let GG be be aa torsion free free group group with following are are equivalent: equivalent: 1) Every Every ring RR with 2) 2) N(G) N(G) r(G) r(G) ==nn < ~.

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