Uniquely ordered commutative rings

Question

I am wondering whether there are reasonable necessary and/or sufficient conditions to dedice whether a commutative ring can be uniquely ordered (like for instance $\mathbb{Z}$) or not. In the field case, for example, we know that every real closed field has a unique order (namely the sums of squares). I was hoping that the notion of "real closed rings" does the same, but this is not the case.

In a more specific context, if I have a valued ring $(R,v)$, I'd like to know under which conditions the residue ring $Rv := R_v/I_v$ is uniquely ordered, where $R_v := \{x \in R: v(x) \geq 0\}$ and $I_v := \{x \in R: v(x) > 0\}$

Answer 1

• In a field, an element can be made negative in some order iff it is not a sum of squares. Thus, $F$ is uniquely ordered iff for every $a\in F^\times$, $a$ or $-a$, but not both, is a sum of squares.

• An order on a domain $R$ extends uniquely to its fraction field, thus $R$ is uniquely ordered iff its fraction field is. Thus:

Proposition: The following are equivalent for any domain $R$:

1. $R$ is uniquely ordered.

2. $0$ is not a sum of nonzero squares, and for every $a\in R$, there is a nonzero $b\in R$ such that $ab^2$ or $-ab^2$ is a sum of squares.