Date: Feb 4, 2013 12:50 AM
Author: quasi
Subject: Re: Finite Rings
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.

quasi