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. 
 A: Yes, $A$ should be finite to make such a conclusion.
For the proof of the statement, you should look at the underlying abelian group und use Lagrange's theorem. Characteristic $q$ implies that $A$ contains a copy of $\mathbb Z/q\mathbb Z$.
A: Moos is right: if you have a finite domain, it makes the most sense to solve is using Lagrange's theorem. Even if you haven't covered it, you certainly will at some point, and this is a great opportunity. You will find a proof in any group theory chapter, or in learnmore's post.
Since the characteristic of a ring with identity is the same as the additive order of $1$, Lagrange's theorem says right away that the order of the subgroup generated by 1 divides the order of the ring's entire additive group (assuming it is finite, of course.)
There is one thing that you said which needs clarification:

I inferred that the $A$ a should be finite, since its characteristic is $q$.

Such an inference would be false, though. Given a finite field $F$, the rational polynomial ring $F(x)$ is a field of finite characteristic with infinitely many elements.
You were still right in inferring the ring should have finite order, but for a different reason. You simply can't talk about divisibility between two orders unless they are both finite.
Speaking of finite integral domains, it's always worth mentioning that these are all fields. (You can see many questions and answers related to this fact in the Related column for this question.) Ultimately one can show that the ring generated by $1$ is the smallest subfield, and it has a prime number $p$ of elements. The entire ring, being a vector space over that subfield, must have $p^n$ elements for some $n$. Thus the characteristic of the ring ($p$) certainly divides the order of the ring ($p^n$).
A: 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 $n$
A: As its mentioned that q divides order of A. That means, order of A is a natural number(i.e. finite). You cannot deduce that domain of characteristic q(not equal to 0) is finite. There are fields of characteristic q(not equal to 0), which are infinite. Consider, for example, algebraic closure of Z/pZ, for p a prime number.
Now, that the integral domain given is finite, you can prove that any finite integral domain is a field.(You can find proof for this in algebra textbooks or on internet). If the cardinality of the domain is q. Then, q is a power of p, for some prime p. Thus, p is the characteristic of the domain. 
Hope this helps!  
