```Date: Feb 4, 2013 12:12 PM
Author: quasi
Subject: Re: Finite Rings

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 productsx^2, y^2, z^2.quasi
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