Talk:Reduced ring

Thank you Darij for fixing my mistake. However, the criterion

every element $a\in A$ satisfying $a^{2}=0$ is zero itself

is also wrong: the ring Z/6Z satisfies this criterion but is not reduced. I have fixed this. Ninte (talk) 11:09, 20 October 2010 (UTC)[reply]

I did not notice this strange comment before: the condition is equivalent. The ring Z/6Z is clearly reduced. Even if one does not believe the squaring criterion, it is still obvious that 6 never divides any power of 1, 2, 3, 4, or 5 evenly. In fact, my main purpose today is to add the Z/6Z example because it is an example of a finite reduced ring, unlike the other two. Rschwieb (talk) 20:16, 29 December 2010 (UTC)[reply]

Oh yes, I agree now. Ninte (talk) 16:00, 28 February 2011 (UTC)[reply]

Proof

Suppose that $a^{n}=0$ . If n is even, divide it by 2. If n is odd, add 1 to it and divide by 2. Repeat this process until n=1, and then a=0. GeoffreyT2000 (talk) 01:56, 2 May 2015 (UTC)[reply]