Let $R$ be a ring. Is there at most one ordering $\leq$ such that $(R,\leq)$ is an ordered ring?

Question

I was wondering whether it is equivalent to consider an ordered ring or a ring that can be ordered. I assume that there are plenty of people who already thought about this issue, hence my question...I would also like to know if the existence of a multiplicative identity is relevant.

Answer 1

Even for fields, there are usually many (potentially infinitely many) non-equivalent orderings. For instance, $\mathbb{Q}(\sqrt{2})$ has the "obvious" ordering coming from its embedding into $\mathbb{R}$, but you can also embed it into $\mathbb{R}$ by sending $\sqrt{2}$ to $-\sqrt{2}$, and this embedding induces a different ordering on $\mathbb{Q}(\sqrt{2})$.

In general, for a field $K$, the orderings of $K$ form a compact topological space $X(K)$ (the so-called space of orderings, with its Harrison topology), which can also be described as the generic fiber of $Spec(W(K))\to Spec(\mathbb{Z})$ where $W(K)$ is the Witt ring of $K$ (this is called Artin-Schreier theory).