$\operatorname{char}(F)=0$ or a prime number I have been taught that the characteristic of a field ($F$) is always $0$ or a prime number, and that this proves it:  
Assume $\operatorname{char}(F)=ab\space, a,b>1$
$\operatorname{char}(F)$ is the smallest number of F such that $ab \cdot 1=0$
$ab \cdot 1=(a\cdot 1)(b\cdot 1)\cdot 1=0$.
$F$ has no zero divisers therefor $(a\cdot 1)=0$ or $ (b\cdot 1)=0$
$a, b < ab$
$a$ or $b$ is a smaller number than $ab$ that acts as $\operatorname{char}(F)$   
MY QUESTION
Why is this statement and proof always concerning fields? As far as I can tell it is applicable to any integral domain - the only ring property being used is the lack of zero divisers. 
 A: Let $R$ be a ring with unit $1_R$. Then there exists unique ring homomorphism $\varphi\colon\mathbb Z\to R$ such that $\varphi(1) = 1_R$ and it has property $$\varphi(n) = \varphi(1+\ldots + 1) = \varphi(1) + \ldots +\varphi (1) = 1_R+\ldots + 1_R = n\cdot 1_R,\quad n\geq 0$$ and this can be easily extended to whole of $\mathbb Z$ since additive inverse must be sent to additive inverse.
Think about this statement: 

Ring characteristic is the nonnegative integer $n$ that generates
  ideal $\ker \varphi$, i.e. $\ker\varphi = n\mathbb Z$.

By the first isomorphism theorem, $\operatorname{im}\varphi \cong \mathbb Z/n\mathbb Z$. Finally, if $R$ is integral domain, so is $\operatorname{im}\varphi$, and $n\mathbb Z$ is prime ideal. Thus, $n$ is prime or $0$.
You are right, we don't need to work with fields only. However, to get a precise answer to "Why is this statement and proof always concerning fields?" you should include references to quantify "always".
For example, in Hungerford's Algebra, Ch. III, Theorem 1.9. proves this for integral domains.
