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.
>> 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.