By Shalom Feigelstock

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**Extra info for Additive Groups of Rings (Chapman & Hall CRC Research Notes in Mathematics Series) **

**Example text**

1 reduces the study of groups allowing only nilpotent (associative) ring structures to the torsion free case. 2: (associative) ring Let G be a torsion free group with r(G) < R with R+ = G is nilpotent if and only if m. Every 31 (v(G) < ~) N(G) < ~. J. A. S. Pierce [7], (8], and will be given in Chapter 5 section 4. 1 is nil potent, = l ••.. ,k. 1 is nilpotent. )<~. 2. v( Gi ) < ~. 4 is obviously necessary. 3. 5: (Wickless [74, pp. 253-254]): Let Gi be a rank one torsion free group with t(Gi) = {i,i, ••.

The descending chain v1 ~ v2 ~ ... terminates at 0, and has length at most n. Hence V = 0 ' and so R(n+l) = 0. n+ 1 k Claim: For every positive integer k, R2 +l c R(k+ 2 ). For k = 0 the claim is clearly true. Suppose R2k+l c R(k+ 2 ) for some positive integer k. 2k+l + 1 2k+l 2k+1 Let x E R Then x = y·z with y E R or z E R . z E R(k+ 3). Therefore R2n-l + 1 c R( n+ 1) = 0. 6: Let G be a torsion free group with following are equivalent: 1) Every ring R with 2) N(G) ~ 2n-l. r(G) = n < ~. The R+ = G is nilpotent.

1, either H and K are both cyclic, or both associative nil. If H and K are both associative nil, then they are both divisible, and so G is nil, Theorem 2. 1, a contradiction. Therefore G = (x 1) (f) (x 2 J, with lx; I = n1 , i = 1,2. If (n 1 , n2) = 1, then G is cyclic. Otherwise let p be a prime divisor of m. (n 1 ,n 2). 2. 1. m -1 The products Y;·Yj = p 2 y2 for i,j = 1,2, induce an associative ring structure R on G with R2 ~ 0. Therefore R = ~~, s 1 ,s 2 Every element x E R has the form m2-1 kxs 1y 1 + (kxs 2 + mxp )y2 , kx,mx integers. ~~