Date: Feb 4, 2013 1:09 AM
Author: quasi
Subject: Re: Finite Rings
quasi wrote:

>quasi wrote:

>>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.

>

>That's wrong too.

>

>x+y+z is not an annihilator.

>

>However it's not a zero-divisor -- in fact, it's an identity,

>

>Thus, my example still fails.

>

>>>> 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.

Ok, I think I now have a valid counterexample.

I'll post it shortly in a reply to the original post.

quasi