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

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quasi

Posts: 10,313
Registered: 7/15/05
Re: Finite Rings
Posted: Feb 4, 2013 12:12 PM
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Arturo Magidin wrote:
>uasi wrote:
>> William Elliot 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.


No -- you misread some of my specified conditions.

Look more carefully at how I defined the products
x^2, y^2, z^2.

quasi



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