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. 

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!  
