
Re: Finite Rings
Posted:
Feb 4, 2013 12:01 PM


On Monday, February 4, 2013 7:56:26 AM UTC6, quasi wrote: > William Elliot <marsh@panix.com> wrote: > > > > > >[In forum "Ask an Algebraist", user "Anu" asks] (edited): > > > > > >> If R is a finite commutative ring without multiplicative > > >> identity such that every nonzero element is a zero divisor, > > >> must there necessarily exist a nonzero element which > > >> annihilates all elements of R? > > > > Ok, I think I have it now. > > > > Consider a commutative ring R consisting of the following > > 8 distinct elements > > > > 0, x, y, z, x+y, y+z, z+x, x+y+z > > > > obeying the usual laws required for R to be a commutative > > ring (without identity), and also satisfying the following > > conditions: > > > > x^2 = x, y^2 = x, z^2 = x > > > > r+r = 0 for all r in R > > > > xy = yz = zx = 0
This is just (Z/2Z)^3 with its natural product structure; the ring has a 1.
Consider (x+y+z). We have
(x+y+z)x = xx + yx + zx = x + 0 + 0 = x (x+y+z)y = xy + yy + zy = 0 + y + 0 = y (x+y+z)z = xz + yz + zz = 0 + 0 + z = z
Hence, (x+y+z)(x+y) = x+y, (x+y+z)(x+z) = x+z, (x+y+z)(y+z)=y+z, and (x+y+z)^2 = x+y+z.
Thus, z+y+z is an identity.
 Arturo Magidin

