Investigating the Existence of totally ordered ring with zero divisors, I've got an interesting result:
If an Archimedean totally ordered ring (or rng) has a nonzero zero divisor, then $ab = 0$
for any elements $a$ and $b$ of the ring.
Is this correct?
My logic is below:
In a totally ordered ring $ac \le bc$ and $ca \le cb$ for any elements $a, b, c$ such that $a \le b$ and $0 \le c$.
Let $x$ be a nonzero zero divisor: $xy = 0$, $x \ne 0$, $y \ne 0$.
We can assume $0 < x$ and $0 < y$ since $-xy = xy = 0$.
For an element $v$ such that $0 < v \le x$: $0y \le vy \le xy$, $0 \le vy \le 0$, $vy = 0$.
Therefore, for any $v$, $0 < v \le x$:
- $vy = 0$;
- $v$ is a left nonzero zero divisor.
Similar, for any $w$, $0 < w \le y$:
- $xw = 0$;
- $w$ is a right nonzero zero divisor.
Let's note that if $xy = 0$, then $xy + xy + xy + \ldots = 0$, or $nxy = 0$ for a natural $n$.
Applying compatibility with addition ($0 < x \implies x \le x + x$): $0 < nx$ and $0 < ny$.
If $x$ is a left nonzero zero divisor, then $nx$ is also a left nonzero zero divisor for any natural $n$.
If $y$ is a right nonzero zero divisor, then $ny$ is also a right nonzero zero divisor for any natural $n$.
If the total order is Archimedean, then for any element $0 < z$ and a zero divisor $0 < x$
there is a natural $n$ such that $0 < z \le nx$, where $nx$ is a zero divisor.
Thus, in an Archimedean totally ordered ring with a nonzero zero divisor:
- all elements are zero divisors;
- $ab = 0$ for any elements $a$ and $b$.