```Date: Feb 4, 2013 12:01 PM
Author: magidin@math.berkeley.edu
Subject: Re: Finite Rings

On Monday, February 4, 2013 7:56:26 AM UTC-6, quasi wrote:> William Elliot <marsh@panix.com> 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 = 0This 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 = zHence, (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.-- Arturo Magidin
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