458 Algebra
( 1 −(xy)n)v =1; hencev(xy)n =(xy)nv =v−1. We claim that the inverse of
1 −(yx)nis 1+(yx)n−^1 yvx. Indeed, we compute
( 1 +(yx)n−^1 yvx)( 1 −(yx)n)= 1 −(yx)n+(yx)n−^1 yvx−(yx)n−^1 yvx(yx)n
= 1 −(yx)n+(yx)n−^1 yvx−(yx)n−^1 yv(xy)nx
= 1 −(yx)n+(yx)n−^1 yvx−(yx)n−^1 y(v− 1 )x= 1.Similarly,
( 1 −(yx)n)( 1 +(yx)n−^1 yvx)= 1 −(yx)n+(yx)n−^1 yvx−(yx)n(yx)n−^1 yvx
= 1 −(yx)n+(yx)n−^1 yvx−(yx)n−^1 y(xy)nvx
= 1 −(yx)n+(yx)n−^1 yvx−(yx)n−^1 y(v− 1 )x= 1.It follows that 1−(yx)nis invertible and its inverse is 1+(yx)n−^1 yvx.
295.(a) Letxandzbe as in the statement. We compute
(zxz−xz)^2 =(zxz−xz)(zxz−xz)
=(zxz)(zxz)−(zxz)(xz)−(xz)(zxz)+(xz)(xz)
=zxz^2 xz−zxzxz−xz^2 xz+xzxz
=zxzxz−zxzxz−xzxz−xzxz= 0.Therefore,(zxz−xz)^2 =0, and the property from the statement implies thatzxz−xz=0.
(b) We have seen in part (a) that ifzis an idempotent, thenxzx−xz=0. The same
argument works, mutatis mutandis, to prove thatzxz=zx. Hencexz=zxz=zx,
which shows thatzis in the center ofR, and we are done.
296.We will show that the elements
ac, a^2 c, a^3 c,...,anc,...are distinct. Let us argue by contradiction assuming that there existn>msuch that
anc=amc. Multiplying bycon the left, we obtainca(an−^1 c)=ca(am−^1 c), so by (iii),
ban−^1 c=bam−^1 c. Cancelbas allowed by hypothesis (ii) to obtainan−^1 c=am−^1 c.An
easy induction shows thatakc=c, wherek=n−m. Multiplying on the right byaand
usingca=b, we also obtainakb=b. The first condition shows thatbcommutes with
a, and sobak=b; cancelingbyieldsak=1. Henceais invertible anda−^1 =ak−^1.
The hypothesisca=bimplies
c=ba−^1 =bak−^1 =ak−^1 b=a−^1 b,
henceac=b, contradicting (iii). The contradiction proves that the elements listed in the
beginning of the solution are all distinct, and the problem is solved.
(Gazeta Matematica ̆(Mathematics Gazette, Bucharest), proposed by C. Gu ̧tan)