# Is every ordered abelian group the additive group of an ordered ring?

Let $$\Lambda$$ be an ordered abelian group, (there is a total order on $$\Lambda$$ which is compatible with addition). Is there a multiplication map on $$\Lambda$$ that turns it into an ordered ring?

I have tried using classification results, such as $$\Lambda$$ is a subgroup of $$\mathbb{R}^\Omega$$ for some set $$\Omega$$ (Hahn embedding theorem) and the classification of subgroups of $$\mathbb{Q}$$.

Partial results are also welcome.

No, and the classification of subgroups of $$\mathbb{Q}$$ can be used to find a counterexample. Consider the subgroup $$A$$ generated by $$\frac{1}{p}$$ as $$p$$ runs over all primes (or more generally infinitely many primes). I claim that $$A$$ cannot be the additive group of any ring (ordered or otherwise). To see this, first note that by partial fraction decomposition $$A$$ consists exactly of the set of fractions $$\frac{a}{b}$$ such that, when written in lowest terms, $$b$$ is squarefree.

Now suppose by contradiction that $$\star : A \times A \to A$$ is a multiplication for a ring structure on $$A$$, and let $$e = \frac{a}{b}$$ be its multiplicative identity. Let $$p$$ be a prime not dividing $$b$$; then $$\frac{e}{p} = \frac{a}{pb} \in A$$ satisfies

$$\underbrace{\frac{e}{p} + \dots + \frac{e}{p}}_{p \text{ times}} = e$$

(I am not writing this as $$p \frac{e}{p}$$ because we should carefully distinguish between multiplication of fractions and multiplication in $$A$$) and hence its square $$\frac{e}{p} \star \frac{e}{p}$$ satisfies

$$\underbrace{ \frac{e}{p} \star \frac{e}{p} + \dots + \frac{e}{p} \star \frac{e}{p} }_{p \text{ times}} = \frac{e}{p}.$$

It follows that $$\frac{e}{p} \star \frac{e}{p} = \frac{e}{p^2} = \frac{a}{p^2 b}$$, which is not in $$A$$; contradiction. So there is no such multiplication $$\star$$.

For instance, the lexicographically ordered group $$(\mathbb{Z}^2,+,<)$$ is the underlying ordered group of no ordered ring.

edit: in fact, what I show below is rather that there is no structure of ordered domain expanding $$(\mathbb{Z}^2,+,<)$$ - see Bart Michels' comment.

I pick the lexicographic ordering with prevalence on the first coordinate. So $$(1,0)$$ is larger than all $$(0,n)$$ for $$n \in \mathbb{Z}$$.

If $$(1,0)$$ is strictly larger than the unit in our ring, then this unit must be $$(0,k)$$ for some non-zero $$k \in \mathbb{N}$$. Then the element $$(1,0)^2$$ is larger than all $$(n,0) =(0,k n)\times(1,0)$$ for $$n \in \mathbb{N}$$, which cannot be.

So $$(1,0)$$ is smaller than the unit, whence $$(0,1)$$ is positive smaller than the unity, so $$(0,1)^2$$ is positive and smaller than $$(0,1)$$, which cannot be.

• It could still be that $(0,1)^2 = 0$. In fact, if I'm not mistaken this can be made into an ordered ring, by defining $(a,b) \times (c, d) = (ac, ad+bc)$. The unit is $(1,0)$ and all $(0,n)^2=0$. Oct 5, 2022 at 11:44
• @BartMichels Yes in fact I forgot that ordered ring need not be ordered domains. I'll edit accordingly. Oct 5, 2022 at 20:28