An element , a ring, is called nilpotent if for some positive integer .
- Show that if for some integers , then is nilpotent in .
- If is an integer, show that the element is nilpotent if and only if every prime divisor of also divides . Find the nilpotent elements of explicitly.
- Let be a nonempty set and a field, and let be the ring of functions . Prove that contains no nonzero nilpotent elements.
- Suppose , where . Now mod .
- Suppose is nilpotent. Then for some and . Now if is a prime dividing , then divides , so that it divides . Thus every prime dividing divides . Let and , where for all and is some integer. Let . Then , where for each . Thus for some integer , and we have mod .
The nilpotent elements of , where , are as follows: 0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66.
- Suppose is nilpotent. If , then there exists such that . Let be minimal such that ; note that . Then , where are not zero. Thus contains zero divisors, a contradiction. Thus no nonzero element of is nilpotent.