How can a finite integral domain have a non-zero characteristic? I know that by virtue of being a finite integral domain, $n\cdot 1_{R}=m\cdot 1_{R}$. Hence, $(n-m)\cdot 1_{R}=0$. The smallest positive element in the set of all possible such $(n-m)$'s is the characteristic. 
But then again, in an integral domain, if $a\cdot b=0$, then either $a=0$ or $b=0$. Here, neither $n-m=0$ (by definiton, we have a non-zero characteristic), nor $1_{R}=0$ (we're assuming this is not a zero ring). Isn't this a contradiction?
 A: How can a finite integral domain have zero characteristic? It would contain a copy of the integers, which is a contradiction. So a finite integral domain must have non zero characteristic.
When you write $na$, for $n\in\mathbb{Z}$ and $a\in R$ you're not doing a product in $R$, but rather $\underbrace{a+a+\dots+a}_n$ (when $n>0$) or $-((-n)a)$ when $n<0$.
So there's no problem in having $3a=0$ in a domain. This just means that $a+a+a=0$, which happens in $\mathbb{Z}/3\mathbb{Z}$, for instance, which is a field.
I usually consider the unique ring homomorphism $\varepsilon\colon\mathbb{Z}\to R$, defined by $\varepsilon(n)=n1_R$.
In case the ring $R$ is finite, this homomorphism has a non zero kernel, say $k\mathbb{Z}$, so easily $k1_R=0$ and $k$ is the characteristic of $R$.
A: Consider what you mean when you write $na=0$ where $n$ is an integer:
$$\underbrace{a+\cdots+a}_{n\text{ times}}=0$$
It very well can be the case that this occurs for a non-zero $n$. This is not a contradiction because the expression $na$ does not mean a product of elements of the ring $R$; it is $\mathbb{Z}$ acting on the ring $R$.
Alternatively, you can think of "$n$" as being  a shorthand for
$$\underbrace{1_R+\cdots+1_R}_{n\text{ times}}$$
If you prefer this approach, you very well can have that $n=0$ in the ring $R$, and therefore it is not a contradiction to have $na=0$.
