Prove that the characteristic of a finite integral domain $A$ divides the order of $A$ I am a little confused as how to approach this problem. The title of this problem is the title of the section which it comes from. However, there is no information that the given integral domain is finite. The problem is as follows: 
Let $A$ be an integral domain. If $A$ has characteristic $q$, then $q$ is a divisor of the order of $A$.
I inferred that the $A$ a should be finite, since its characteristic is $q$. However, our class has not gone over any problems which state that this is the case. Also, some proofs I have seen make use of the fact that if the order of $A$ is $n$, then $n(x)=0$ for all $x \in A$. But I am wondering why this is. 
Also, the only definition of order of a group/ring that we have examined is that it is the number of elements. 

Let $I$ denote the integral domain since char ($I)=q$; $q$ is the least positive integer such that $q.a=0\forall a\in I;(0) $ denotes zero element of $I$
Assuming $o(I)=n$ we have $n.a=0$
To prove $q$ divides $n$; let $n=pq+r$ where $0\leq r<q$
So $na=p(qa)+ra\implies 0=0+ra\forall a\in I\implies ra=0\forall a$ 
But $q$ is the least positive integer such that $q.a=0\forall a\in I$ and $r<q$ so $r=0$
Thus $q$ divides $n$
