Proof that the characteristic of an integral domain is zero or prime

Proof

1. Let $R$ be an integral domain and let $n$ be its characteristic. For a positive integer $k$, write $k \cdot 1$ for the sum of $k$ copies of $1$ in $R$.
2. If $n = 0$, there is nothing to prove. So assume $n > 0$. Then $n \cdot 1 = 0$, and $k \cdot 1 \neq 0$ for $0 < k < n$.
3. Since $1 \cdot 1 = 1 \neq 0$ in an integral domain, $n \neq 1$. So $n \ge 2$.
4. We argue by contradiction. Suppose that $n$ is not a prime number. Then $n$ is composite, so it has a positive divisor $a$ with $a \neq 1$ and $a \neq n$.
5. Put $b = n / a$. Then $1 < a < n$ and $1 < b < n$.
6. For positive integers $a$ and $b$, $(a \cdot 1)(b \cdot 1) = (a b) \cdot 1$. Indeed, by the distributive laws of the ring $R$, the product is a sum of $a b$ terms, each equal to $1 \cdot 1 = 1$.
7. So $(a \cdot 1)(b \cdot 1) = n \cdot 1 = 0$.
8. Since $R$ is an integral domain, $a \cdot 1 = 0$ or $b \cdot 1 = 0$.
9. Both $a$ and $b$ are positive and less than $n$. So step 8 contradicts the minimality of $n$ in step 2.
10. Hence $n$ is a prime number. So the characteristic of $R$ is $0$ or a prime number.