So the question is simply.
If $D$ is an integral domain and has finite characteristic prove that the characteristic of $D$ is a prime number.
This is my proof.
Assume $p$ is the characteristic of $D$. Let $a$ be a non zero element of $D$. Seeking a contradiction assume $p$ is not prime. Then $p$ can be written as a factor: $rs=p$ for some $r$ and some $s$. By definition $pa=0$, so $(rs)a=0$. We know that $r,s$ are non-zero, so by definition of integral domain the only way this equation can equal zero is if $a=0$ however this is a contradiction as we chose a to be a non-zero element of $D$. Therefore $p$ is a prime.
Is this proof correct? The answer I have for this problem is slightly longer and I thought I might have missed something in my proof.
Hint $\ $ The finite characteristic $\,n\,$ is just the size of the natural image of $\Bbb Z$ in $D\,$, via $\,1_\Bbb Z \mapsto1_D.$ This image is a subring of $D$ isomorphic to $\,\Bbb Z/n,\,$ which is a domain $\iff n\,$ is prime.