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?
On
Suppose that $\mathbb{Z}/3\mathbb{Z}$ were an ordered integral domain. By the trichotomy law, at least one of $1$ and $2$ must lie in $D^+$. Since $1+2=0$, it follows from closure under addition that $1$ and $2$ cannot both belong to $D^+$.
However, if $1\in D^+$ and $2\not\in D^+$, then $1+1=2\not\in D^+$, contrary to the closure property. Similarly, if $2\in D^+$ and $1\not\in D^+$, then $2+2=1\not\in D^+$, again contradicting the closure property.
Since both cases lead to a contradiction, it follows that $\mathbb{Z}/3\mathbb{Z}$ cannot be an ordered integral domain. A similar proof should work for general $p$.
On
Yet another argument:
The $p-1$ nonzero elements of $\Bbb Z/(p)$ fall into two subsets of equal size: $D^+$ and $-D^+$, which consists of the negatives of elements of $D^+$. Let $m$ be the largest integer in $[1,p-1]$ whose class is in $D^+$. This is certainly not $p-1$. Then $m+1\in-D^+$, a contradiction.
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.