Claim: the characteristic of an integral domain $D$ must be either 0 or prime.
Here is my attempt: Assume $D$ is an integral domain. Assume $k$ is the characteristic of $D$. Let $a \in D\setminus \{0\}$. Aiming for a contradiction, assume $k$ is neither prime nor $0$. Since $k$ is the smallest positive integer satisfying $k \cdotp a = 0$, $\exists m, n \in \mathbb{Z}^+$ s.t. \begin{equation} k = m \cdotp n \end{equation} Without loss of generality, assume that $m, n$ are the smallest positive integers satisfying $k = m \cdotp n$. Since $D$ is a ring with unity $1 \neq 0$, we have $k = (m \cdotp 1) \cdotp (n \cdotp 1)$. That is, $(m \cdotp 1) \cdotp (n \cdotp 1) \cdotp a= 0$. Since $D$ contains no divisors of $0$, either $(m \cdotp 1) = 0$ or $(n \cdotp 1) = 0$. If $(m \cdotp 1) = 0$, then by Theorem 19.15, $n$ is the characteristic of $D$ is $n$, which is a contradiction. If $(n \cdotp 1) = 0$, then by Theorem 19.15 again, the characteristic of $D$ is $m$, which is also a contradiction. $\square$
Theorem 19.15: Let $R$ be a ring with unity. If $n \cdotp 1 = 0$ for some $n \in \mathbb{Z}^+$, then the smallest such integer $n$ is the characteristic of $R$.
My question: I am not sure if my use of Theorem 19.15 is correct/ justified in my proof. I know that I have "Without loss of generality, assume that $m, n$ are the smallest positive integers satisfying $k = m \cdotp n$" in my proof but I am not sure if this is sufficient to use Theorem 19.15 the way I have in the last couple lines of my proof.
Can someone please verify if this proof is correct or if it needs any adjustments? Thanks!
Yes, it's all correct, though you don't need $n$ to be the smallest nontrivial divisor. (Note that saying the pair $n,m$ is 'smallest' makes no sense.)
Theorem 19.15 can be easily seen, as if $n\cdot 1=0$, then $n\cdot a=(n\cdot 1)\cdot a=0$ for every element $a$.