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):

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> >

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> >> If R is a finite commutative ring without multiplicative

>

> >> identity such that every nonzero element is a zero divisor,

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> >> must there necessarily exist a nonzero element which

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> >> annihilates all elements of R?

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>

>

> Ok, I think I have it now.

>

>

>

> Consider a commutative ring R consisting of the following

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> 8 distinct elements

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>

>

> 0, x, y, z, x+y, y+z, z+x, x+y+z

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>

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> obeying the usual laws required for R to be a commutative

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> ring (without identity), and also satisfying the following

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> conditions:

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>

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> x^2 = x, y^2 = x, z^2 = x

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>

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> r+r = 0 for all r in R

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>

>

> 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.

--

Arturo Magidin