Date: Feb 4, 2013 12:33 AM
Author: quasi
Subject: Re: Finite Rings
Butch Malahide wrote:

>quasi wrote:

>> William Elliot wrote:

>>

>> >[In forum "Ask an Algebraist", user "Anu" asks:]

>>

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

>> >> identity and if every element is a zero divisor, then does

>> >> there exist a nonzero element which annihilates all elements

>> >> of the ring?

>>

>> >No - the trivial ring.

>> >So add the premise that R has a nonzero element.

>>

>> Even with that correction, the answer is still "no".

>>

>> Consider the commutative ring R consisting of the following

>> seven distinct elements:

>>

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

>

>Does this mean that the additive group of R is a group of order 7?

>

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

>> ring (without identity), we also require the following

>> relations:

>>

>> r^2 = r for all r in R

>>

>> r+r = 0 for all r in R

>

>If r is nonzero, then r is an element of order 2 in the additive

>group? What about Lagrange's theorem?

Oops. I missed the element x+y+z.

But then x+y+z is an annihilator -- which destroys my attempted

counteraxample.

>> xy = yz = zx = 0

>>

>> Note that the above relations imply

>>

>> (x+y)z = (y+z)x = (z+x)y = 0

>>

>> so every element of R is a zerodivisor.

>>

>> However, since all elements of R are idempotent, it follows

>> that no nonzero element of R annihilates all elements of R.

quasi