quasi
Posts:
12,042
Registered:
7/15/05


Re: Finite Rings
Posted:
Feb 4, 2013 12:50 AM


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.
quasi

