Let $R$ be an integral domain and suppose $R$ has characteristic $n > 0$.
Prove that $n$ must be prime.
I just proved this exercise, but I think it needs extra conditions. We can prove the statement if $R$ is ring without zero divisors. It's not needed $R$ to be commutative or to have an identity.
Here is the proof.
Let $n$ is characteristic which is not prime. So $n = mk$ and neither $m$ nor $k$ are characteristic so exists $a,b\in R$ for $ma\ne0$ and $kb\ne 0$. So we have $0 = n(ab) = mk(ab)=(ma)(kb)$ which contradicts that $R$ hasn't zero divisors.
What is wrong with this proof?
Definition 2.16 If $R$ is an arbitrary ring and there exists a positive
integer $n$ such that $nr = 0$ for every $r \in R$ (i.e. $r$ added to itself $n$
times is the zero element) then the least such positive integer $n$ is
called the characteristic of $R$, and $R$ is said to have positive
characteristic. If no such positive integer $n$ exists, $R$ is said to
have characteristic $0$.
Best Answer
Your proof does not work for the zero ring (fortunately, that is not an integral domain). It has non-prime characteristic $1$. Apart from that, your proof is ok.