Does sign-magnitude ordering admit infinite descending chains in any discretely ordered ring?

Question

I have a conjecture:

Let $R$ be an ordered ring $(S, 0, 1, +, ·, <)$.

Let $|x| = x$ if $0 \leq x$ and $|x| = -x$ if $x < 0$.

Define $x \preccurlyeq y$ to hold if either $|x| < |y|$ or both $|x| = |y|$ and $y < x$. Observe that $\preccurlyeq$ is an ordering, though not congruent with $+$ or $·$.

Assume $1$ is the smallest positive number.

Then $\preccurlyeq$ admits of no infinite descending chains.

Is this conjecture true?

Note: if $0 < q < 1$ then $0 = 0q < qq < 1q = q$ i.e. $0 < q^2 < q$, and all $q^{2^n}$ are distinct by transitivity and irreflexivity, and form an infinite descending chain. So I need 1 to be the smallest positive number.

But then if $n < q < n+1$ where $n = 1 + \ldots + 1$, then $0 < q - n < 1$ which we assumed to not be true. If $k < q < k+1 \leq 0$ then $-0 = 0 \leq -(k+1) < -q < -k$. So there cannot be any numbers "between the integers". Can there be numbers "beyond the integers"?

Otherwise, for any $x$ I think I can let $m$ be the smallest $1 + \ldots + 1$ such that $|x| \leq m$, and then there are only finitely many non-negative values smaller than $m$, and thus only finitely many values $\preccurlyeq$-smaller than $x$, and my conjecture holds.

I observe with curiosity that one can prove the Archimedean property for the reals using the least upper bound property. I'm not sure what to make of that, but it smells relevant.

Answer 1

For an explicit counterexample, consider the polynomial ring $\mathbb{Z}[t]$ ordered such that $t$ is infinitely large. Explicitly, a polynomial is positive with respect to this order iff its leading coefficient is positive. The smallest positive element is $1$, but $t,t-1,t-2,\dots$ is an infinite descending chain of positive elements.

Answer 2

If you know a little bit of model theory, then a counterexample to your hypothesis can be constructed via the compactness theorem and the fact that the theory of ordered rings is first-order. Let $\mathcal{L}=\{0,1,+,\cdot,<\}$ be the language of ordered rings, and define a new language $\mathcal{L}'=\mathcal{L}\cup\{c\}$ with a single new constant symbol $c$. Now, let $T$ be the $\mathcal{L}$-theory of ordered rings, and define a new $\mathcal{L}'$-theory $T'$ to be the union of

1. The $\mathcal{L}$-theory $T$.
2. The axiom $\forall v\left[(v\geqslant 1)\vee (v\leqslant 0)\right]$.
3. Axioms $c>\underbrace{1+\dots+1}_{n\text{ times}}$ for each $n\in\mathbb{N}$.

Any finite subset of $T'$ is satisfiable in $\mathbb{Z}$ by taking $c$ to be sufficiently large, and so, by the compactness theorem, $T'$ has a model $R$. Then $R$ is a model of $T$ and hence an ordered ring and, by the second axiom above, has $1$ as its smallest positive element. However, denoting by $r$ the realization of $c$ in $R$, the sequence $$r>r-1>r-(1+1)>r-(1+1+1)>\dots$$ is an infinite descending chain of positive elements, and hence a infinite descending chain under $\succ$, thus giving the desired counterexample.