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 the characteristic of $D$ be $p$, therefore $pa = 0$ and $p$ is
the smallest positive integer. 
Suppose that $p$ is not a prime, then we can write $p = rs$ for
some positive integers $r$ and $s$, with both not equal to $1$. Assume $a \in D$ but a is not zero, then $a^2 \in D$ too. 
So we have 
$pa^2 = 0$, which implies
 $(rs)(aa) = 0$ and 
 $(ra)(sa) = 0$
And since $D$ in an integral domain, it implies  that $ra = 0$ or $sa = 0$. When $ra = 0$, we have
 for all $x$ in $D$
$(ar)x = 0$, which implies that $(a+a+a+.....+a)x=0$; here a is added $r$ times 
also $(ax+ax+......+ax)=0$ implies that $ax$ is added $r$ times also. 
same for 
$a(x+x+.....+x)=0$, which implies that $a(rx)=0$ 
But $a$ is not zero and $D$ is integral domain then  $a(rx)=0$  implies that $rx=0$. 
Then we have
$rx = 0$ for all $x ∈ D$ with $1 < r < p$, which is a contradiction as $p$ is the smallest
such integer.
Similarly, when $sa = 0$ we have contradiction. Thus $p = rs$ is not
possible. Thus $p$ must be prime.
