I understand that an ordered ring is a ring $R$ with a total order such that for all $a,b$, and $c$ in $R$:
- if $a \leqslant b$ then $a+c \leqslant b+c$
- if $0 \leqslant a$ and $0 \leqslant b$, then $0 \leqslant ab$.
I understand that an integral domain is a nonzero commutative ring wherein $ab = 0$ implies $a= 0$ or $b = 0$. The set of real numbers, integers and rational numbers are all integral domains that are also ordered rings.
I don't understand how I can algebraically prove that an ordered ring is also an integral domain. I don't know what I can and cant assume.
It is not true that an ordered ring is always an integral domain.
An intuitive example is that of $R=\mathbb{R}[x]/(x^2)$. You can think of $x$ as a number which is strictly positive but so infinitesimally small that $x^2=0$. So the order on $R$ is given by: $a+bx\leqslant c+dx$ iff $a< c$, or $a=c$ and $b\leqslant d$.
You can check that all the axioms of an ordered ring are satisfied, but clearly $R$ is not integral since $x^2=0$.