If D is an Integral Domain and has finite characteristic p, prove p is prime. 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. 

Let $n$ be the characteristic of $D$ and suppose that $n=a\cdot b$, $a,b>1$.
Then
$$\underbrace{1+1+\cdots+1}_{n~~times} =0 \Rightarrow$$
$$ \underbrace{\left(1+\cdots+1\right)}_{a~~times}\cdot\underbrace{\left(1+\cdots+1\right)}_{b~~times} =0 $$
Now, since $D$ is an integral domain, then $a\cdot 1=0$ or $b\cdot 1=0$.
But the first case implies that characteristic of $D$ is $a<n$, and the second case implies that characteristic of $D$ is $b<n$. Contradiction!
Therefore, $n$ must be a prime number.
