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. 
 A: 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. 
A: 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.
A: 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.
A: Another (pretty similar) approach:
Let $k$ be the characteristic of $D$. Assume $k$ can be factored nontrivially as $k=mn$, where $\gcd(m,n)=1$. For any $a\in D$, we have $(ma)(na)=mna^2 = 0$, so either $ma=0$ or $na=0$, but not both (since $m$ and $n$ are coprime). Since $mn$ is the characteristic, there exists $0\neq x\in D$ with $mx=0$ and $0\neq y\in D$ with $ny=0$. But then $mxy=nxy=0$, contradiction.
So $k=p^j$ for some prime $p$. Suppose there is $0\neq a\in D$ with $pa\neq 0$. Then since $(pa)(p^{j-1} a) = p^n a^2 = 0$, we must have $p^{j-1}a = 0$. It follows $$0 = (p^{j-1} a)(a) = p^{j-1}a^2 = (p^{j-2} a)(pa)$$ so $p^{j-2} a = 0$. Continuing in this fashion, it follows $pa = 0$, contradiction. Thus $pa = 0$ for all $a\in D$ and $j=1$.
A: Here is another proof when the domain has unity.
Let $D$ be an integral domain and set $\mathbb{Z}[1]:=\{k \cdot 1 | k \in \mathbb{Z} , 1 \in D \}$  (we understand $k \cdot 1= \underbrace{1+1+ ..+1}_\text{k times}$)
We define $Char(D)=0$ if $\mathbb{Z}[1] \cong \mathbb{Z}$ and $Char(D)=n$ if $\mathbb{Z}[1] \cong \mathbb{Z_n}$.
If $Char(D)=n$ , then $n$ is the least integer such that $n \cdot 1 = 0$ . If $n$ is composite, then $n=sr$ with $1 < r<s<n $,  thus $(r \cdot 1)(s \cdot 1) =n \cdot 1= 0$  and $(r \cdot 1) \neq 0$ and $(s \cdot 1) \neq 0$ for the minimality of $n$, and contradicts that $D$ is an integral domain. 
In conclusion, n must be prime.
