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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!

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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 $a$.

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  • $\begingroup$ Thanks! Can you please explain how Theorem 19.15 is "easily seen" using the equation you have in the answer? What is $a$ in your equation? How does it show that $n$ is the smallest integer satisfying $n \cdotp 1 = 0$. $\endgroup$ – Ricky_Nelson May 24 '20 at 0:29
  • $\begingroup$ @Ricky_Nelson We have $(\forall a:na=0)\iff (n1=0), and the definition of characteristic is the smallest positive $n$ satisfying the left side. $\endgroup$ – Berci May 24 '20 at 9:49

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