```Date: Feb 4, 2013 2:38 AM
Author: quasi
Subject: Re: Finite Rings

quasi wrote:>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.Forget it.My new counterexample just vaporized.At this point, I'm not so sure there there _is_ a counterexample.quasi
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