Suppose that (R, + , •), is an integral domain, where + and • are the usual operations, addition and multiplication respectively, and the non zero element r, as considered as an element of the abelian group (R, +), has finite order k. Prove that k is prime.
So at this point I have myself quite confused on what seems like a very easy question. From my understanding I want to add r to itself until I reach the additive identity, 0. The question tells us r should be added k times. So,
(r + r + r + ... + r) k times = kr = 0.
However, since R is an integral domain kr cannot be equal to 0. This is where my problem is.
My first thought was my definition of order is incorrect, and I will need to multiply r, k times before it will equal the multiplicative identity. If this is the case I am still unsure how to proceed...
Any help would be great, Thanks.
This is related to the concept of the characteristic of a domain.
Note if $(r+r+\cdots+ r) = 0$, then $r(1+1+\cdots+1) = 0$. So since $r \neq 0$, $1+\cdots + 1 = 0$ as your ring is a domain. The characteristic of a ring is the order of $1$ in the additive group of the ring.
Now suppose that $1+1+\cdots + 1 = 0$ where the sum has $n$ terms. Then if $n$ is composite, say $n = pq$ so that $p,q<n$, we must have $(1+1+\cdots+1)(1+1+\cdots + 1) = 0$ where the former sum has $p$ terms and the latter $q$. In fact, we may call those terms $p$ and $q$ itself, but then we have $pq = 0$ in $R$, a domain. So one of $p$ or $q$ must be $0$. But by definition of order in a group, it must be minimal, and so we've contradicted our claim of $n$ as the order of $1$. So $n$ must be prime.