quasi
Posts:
9,498
Registered:
7/15/05
|
|
Re: Finite Rings
Posted:
Feb 4, 2013 12:50 AM
|
|
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.
quasi
|
|
