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 products

x^2, y^2, z^2.

quasi