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

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quasi

Posts: 10,451
Registered: 7/15/05
Re: Finite Rings
Posted: Feb 4, 2013 2:38 AM
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quasi wrote:
>quasi wrote:
>>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 zero-divisor -- 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.

>
>Ok, I think I now have a valid counterexample.
>
>I'll post it shortly in a reply to the original post.


Forget it.

My new counterexample just vaporized.

At this point, I'm not so sure there there _is_ a
counterexample.

quasi



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