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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.

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2 Answers 2

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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 $\star$.

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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.

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    $\begingroup$ 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 $(0,n)^2=0$. $\endgroup$ Oct 5, 2022 at 11:44
  • $\begingroup$ @BartMichels Yes in fact I forgot that ordered ring need not be ordered domains. I'll edit accordingly. $\endgroup$
    – nombre
    Oct 5, 2022 at 20:28

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