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Topic: Finite Rings
Replies: 28   Last Post: Feb 6, 2013 8:33 AM

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magidin@math.berkeley.edu

Posts: 11,155
Registered: 12/4/04
Re: Finite Rings
Posted: Feb 4, 2013 12:01 PM
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On Monday, February 4, 2013 7:56:26 AM UTC-6, 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



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