Bourbaki

Exercise 2, Appendix 0

Algebra, III, 654

Let FF be an alternative A-algebra, in which the relation 2x=02x = 0 implies x=0x = 0. Show that, for all x,y,zx, y, z in FF, x(yz)x=(xy)(zx)x(yz)x = (xy)(zx) (Moufang's identity). (In the identity

(xy)z+y(zx)=x(yz)+(yz)x (xy)z + y(zx) = x(yz) + (yz)x

successively replace yy by xyxy and zz by zxzx.)

Solution

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