Existence of totally ordered ring with zero divisors
Does there exist a totally ordered ring with zero divisors? I can't think of an example right now.
Too long for a comment.
It looks like there are two conflicting definitions of an ordered ring. The first one was given by @martin-brandenburg in his comment and it is also used by L. Fuchs in his book.
The second one is proposed by J.K. Hodge in his book Abstract Algebra: An Inquiry Based Approach and it goes like this:
An ordered ring is a commutative ring $R$ containing a subset $P$ such that:
- $P$ is not empty;
- if $a \in P$ and $b \in P$, then $a + b \in P$ and $ab \in P$; and
- for any $a \in R$, exactly one of following conditions holds: $a \in P$, $a = 0$, or $− a \in P$.
Let me reproduce the short proof of Theorem 18.3 pointed out by @victor-m.
Theorem 18.3. An ordered ring contains no zero divisors.
Proof. Suppose $R$ is an ordered ring, and let $a, b \in R$ with both $a$ and $b$ nonzero. If $a, b > 0$, then $ab > 0$ and so $ab \not= 0$. Suppose one of $a$ or $b$ is positive and the other negative. Without loss of generality, assume $a > 0$ and $b < 0$. Then $−b > 0$ and so $(a)(−b) > 0$. Thus, $−(ab) > 0$, which means that $ab \not= 0$. The final case is when $−a > 0$ and $−b > 0$. Then $ab = (−a)(−b) > 0$, and so $ab \not= 0$.
In conclusion, the answer depends on the definition of an ordered ring.
L. Fuchs, Partially Ordered Algebraic Systems. Pergamon Press, 1963
Chapt VIII, $\S 3$ "O-rings with divisors of zero".