How do I show that the integral domain $Z/(p)$ is not an ordered integral domain? The properties of an ordered integral domain are the following: **1\. Closed under addition in $D^+$** **2\. Closed under multiplication $D^+$** **3\. Satisfy the trichotomy law which either $a=0,a\in D^+, or -a \in D^+$** I tried creating a multiplication and addition table for $Z/(3)$ and from the results, they're closed under $D^+$. So I figured it has something to do with not satisfying the trichotomy law. I know that $a=0$ is true since I have gotten a sum and product that equals to $0$. Any help please?

Suppose $1\in D^+$. If you keep adding $1$ to itself $p-1$ times, you arrive at $-1\in D^+$ as well. But by trichotomy, both $1$ and $-1$ cannot be in $D^+$ at the same time.