quasi
Posts:
10,562
Registered:
7/15/05
|
|
Re: Finite Rings
Posted:
Feb 4, 2013 1:09 AM
|
|
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.
quasi
|
|
