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