quasi
Posts:
12,067
Registered:
7/15/05


Re: Finite Rings
Posted:
Feb 4, 2013 2:38 AM


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 zerodivisor  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

