Let A be a finite integral domain. Prove that if there is a nonzero a in A such that 256 * a = 0, then A has characteristic 2.
I'm not at all sure how to do this. Any advice is appreciated. Thanks for your time!
First we know that the characteristic of an ID is p, where p is prime (since ID is finite) .
Second notice that the additive order of every non zero element in the integral domain is equal to characteristic p.
Why?
Take $\displaystyle x \in D $ such that $\displaystyle x \not = 0 $ . Let the additive order of x be n. Then since $\displaystyle x \not = 0 $, n > 1. Since the characteristic is p, we know that $\displaystyle px = 0 $. Thus this means that n must divide p. Since the only things that divide p are 1 and p, since we said n > 1, it must be that n = p .
This means that p must divide 256 = 2^8.
Since 2 is the only prime which divides 256. So A has char 2.