# How can I prove an ordered ring is also an integral domain

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.

• Sure that's the definition you want? I see in some places people define an ordered ring using strict inequalities, and then the second condition just gives you that it is a domain automatically. – rschwieb Apr 22 '20 at 13:59

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 $$x^2=0$$.
• This definition of an ordered ring is apparently inequivalent to the one with strict inequalities. It seems like the "positive cone" idea suffers, or at least the positive and negative elements must share $0$ instead of partitioning the ring. – rschwieb Apr 22 '20 at 14:00