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Does there exist a totally ordered ring which is non-commutative? I have searched the web, but I have not found any examples.

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up vote 6 down vote accepted

Wikipedia's definition of "ordered ring" requires the ring to be commutative, but you could strip that condition and still have a perfectly sensible idea. I did a Google search and found that this is considered in A First Course in Noncommutative Rings, which, according to a Google Books preview, provides some constructions for examples in the Exercises for $\S17$.

Here is Ex. 17.3, reproduced as best I can (two typos are faithfully reproduced, any others are my own):

Let $R$ by the Weyl algebra generated over $\mathbb R$ by $x$ and $y$, with the relation $xy-yx=1$. Elements of $R$ have the canonical form $$r=r_0(x)+r_1(x)y+\cdots+r_n(x)y^n\text{,}$$where each $r_i(x)\in\mathbb R[x]$, $r_n(x)\ne0$ (if $r\ne0$). Let $P\subset R$ be the set of all nonzero elements $r\in R$ above for which $r_n(x)$ has a positive leading coefficient. Show that $P$ defines an ordering "$<$" on $R$ on which $$\mathbb R<x<x^2<\cdots<y<xy<x^2y<\cdots<y^2<xy^2<x^2y^2<\cdots\text{.}$$

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