quasi
Posts:
10,403
Registered:
7/15/05


Re: Finite Rings
Posted:
Feb 4, 2013 1:09 AM


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.
quasi

