```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
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