Arturo Magidin wrote: >uasi wrote: >> William Elliot wrote: >> > >> >[In forum "Ask an Algebraist", user "Anu" asks] (edited): >> > >> >> If R is a finite commutative ring without multiplicative >> >> identity such that every nonzero element is a zero divisor, >> >> must there necessarily exist a nonzero element which >> >> annihilates all elements of R? >> >> Ok, I think I have it now. >> >> Consider a commutative ring R consisting of the following >> >> 8 distinct elements >> >> 0, x, y, z, x+y, y+z, z+x, x+y+z >> >> obeying the usual laws required for R to be a commutative >> ring (without identity), and also satisfying the following >> >> conditions: >> >> x^2 = x, y^2 = x, z^2 = x >> >> r+r = 0 for all r in R >> >> xy = yz = zx = 0 > >This is just (Z/2Z)^3 with its natural product structure; >the ring has a 1. > >Consider (x+y+z). We have > >(x+y+z)x = xx + yx + zx = x + 0 + 0 = x >(x+y+z)y = xy + yy + zy = 0 + y + 0 = y >(x+y+z)z = xz + yz + zz = 0 + 0 + z = z > >Hence, > >(x+y+z)(x+y) = x+y, >(x+y+z)(x+z) = x+z, >(x+y+z)(y+z)=y+z, and >(x+y+z)^2 = x+y+z. > >Thus, z+y+z is an identity.
No -- you misread some of my specified conditions.
Look more carefully at how I defined the products x^2, y^2, z^2.