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.
 A: 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.
A: L. Fuchs, Partially Ordered Algebraic Systems. Pergamon Press, 1963
Chapt VIII, $\S 3$ "O-rings with divisors of zero".
A: Assuming the definition used in Fuchs, Partially Ordered Algebraic Systems, take the set of integer numbers with the usual order relation and the usual addition, and define the multiplication to be trivial, so that every product be $0$.  Rings with trivial multiplication are one of the examples of ordered rings given in Fuchs's book.
