# Is the class of linearly-orderable rings first order axiomatizable?

A linearly ordered ring is a commutative ring $$R$$ with unity equipped with a linear order $$\leq$$ that is compatible with addition, and such that the set of nonnegative elements are closed under multiplication. By writing down the axioms in the language $$L = \{0,1,+,-,\cdot,\leq\}$$, the class of linearly ordered rings is first order axiomatizable. Now, consider the algebraic reduct of linearly ordered rings, in the language $$L' = \{0,1,+,-,\cdot\}$$. Is that class first order axiomatizable?

• hmm. I think if you asked that the set of strictly positive elements be closed under multiplication, then the axiomatization would be given by asserting that $R$ is an integral domain and taking the axiom scheme $\forall x_1\dots\forall x_n\left[x_1^1+\dots+x_n^2=0\to\bigwedge_{i\in[n]}(x_i=0)\right]$. off the top of my head I'm not sure about the question as you pose it Commented Jul 19 at 0:11
• (just to elaborate quickly on the strictly ordered case, certainly any ordered ring in which the strictly positive elements are closed under multiplication satisfies those axioms. conversely, the axioms ensure that the fraction field of $R$, which makes sense since $R$ is an integral domain, is formally real and hence has a total ordering, the restriction of which to $R$ makes $R$ into an ordered ring) Commented Jul 19 at 0:19
• The nicest thing that could happen is that the proof of the Artin-Schreier theorem (that every formally real field has an order) just generalizes to formally real commutative rings. I can't tell at a glance whether division is really used in an essential way in the argument. A reference for the Artin-Schreier theorem is these notes by Pete Clark: alpha.math.uga.edu/~pete/realspectrum.pdf Commented Jul 19 at 0:52
• There are two common notions of "linearly ordered ring" in the literature: one requires that the positive elements are closed under multiplication, while the other only requires that the non-negative elements are closed under multiplication. The difference between the two is whether a non-zero ordered ring is required to be a domain. I believe @AtticusStonestrom is correct that the orderable domains are exactly the ones satisfying his axiom scheme - these are called "formally real" rings. (continued) Commented Jul 19 at 15:39
• Now it's an easy observation that if $R$ is a non-zero ordered ring which is not a domain, then $R$ contains non-zero elements which square to $0$. Indeed, if $ab = 0$ with $0\leq a\leq b$, then $0 \leq a^2 \leq ab = 0$, so $a^2 = 0$. This makes me wonder whether the weakening of "formally real" which only requires $x_1^2+\dots+x_n^2 = 0$ implies $x_i^2 = 0$ for all $i$ could be sufficient to axiomatize the orderable rings. But I'm not sure, and I haven't found a name for this condition by searching around. Commented Jul 19 at 15:43

Now, consider the algebraic reduct of linearly ordered rings, in the language $$𝐿′=\{0,1,+,−,\cdot\}$$. Is that class first-order axiomatizable?

Yes. This can be derived from Theorem 2.13 of

T. Frayne; A. Morel; D. Scott
Reduced direct products
Fundamenta Mathematicae (1962) Volume: 51, Issue: 3, page 195-228

and from the comment following the theorem. The theorem+comment state that a class of structures is first-order axiomatizable if and only if it is closed under the formation of isomorphic images, elementary substructures, and ultraproducts. Frayne, Morel, and Scott state this for relational languages only, but the theorem+comment are true for arbitrary languages.

Let's couple F-M-S with Theorem 1.6 of Keisler's doctoral dissertation, which asserts that any pseudo-elementary class is closed under ultraproducts. [A class $$K'$$ of $$L'$$-structures is pseudo-elementary if there is a language $$L\supseteq L'$$ and an elementary class (= first-order axiomatizable class) of $$L$$-structures $$K$$ such that $$K'$$ is the class of reducts of structures in $$K$$ to the language $$L'$$.]

The class $$K$$ of linearly-ordered rings in the language $$L=\{0,1,+,−,\cdot,\leq\}$$ is first-order axiomatizable, so the class $$K'$$ of reducts to $$L'=\{0,1,+,−,\cdot\}$$ is pseudo-elementary, hence is closed under ultraproducts. This class is easily seen to be closed under the formation of isomorphic images. It is also easy to see that any subring $$S$$ of a linearly-ordered ring $$R$$ is linearly-ordered by the restriction to $$S$$ of the order on $$R$$. Thus, the class of linearly-ordered rings in the language $$𝐿′=\{0,1,+,−,\cdot\}$$ has the necessary closure properties to apply the theorem of Frayne-Morel-Scott to deduce that it is a first-order axiomatizable class.

• More generally, don't we get the elementarity of any pseudoelementary class of the form "All structures expandable to a model of $\exists R(\theta)$" for $\theta$ a $\forall^*$-sentence? Commented Jul 19 at 3:27
• @NoahSchweber: Yes. Commented Jul 19 at 3:31
• That's neat, I didn't know that! (More generally, I wasn't familiar with this paper; I knew about Keisler/Shelah, but that's it.) Commented Jul 19 at 4:28