Zeros of polynomials in discretely ordered rings

Question

Let's say that a discretely ordered ring has rank 1 if it has elements greater than any integer, and for any two such elements $x<y$ there is an integer $n$ such that $x^n>y$.

Question: Let $f(\bar{x})$ be a polynomial in any number of variables with integer coefficients that has a zero in at least one discretely ordered ring. Must $f$ have a zero in a discretely ordered ring of rank 1?

Answer 1

The answer is no. The argument below is essentially due to Kaye [1] (Lemmas 2.1 and 5.8). Put $$p(a,x,y)=x^2-2axy+y^2-1.$$

Lemma 1: In any discretely ordered ring (DOR), if $p(a,x,y)=0$, $0< x\le y$, and $a>0$, then $y\le 2ax\le x+y$ and $p(a,2ax-y,x)=0$.

Proof: Straightforward computation.

Lemma 2: Let $n\in\mathbb N$. In any DOR, if $p(a,x,y)=0$, $0\le x\le y$, $n\le b\le a-2$, $x\equiv b\pmod{a-1}$, and $y\equiv b+1\pmod{a-1}$, then $y\ge a^n$.

Proof: By induction on $n$. For $n=1$, $y< a$ together with the congruences implies $x=b$ and $y=b+1$, whence $-p(a,x,y)=2b^2(a-1)+2ab+1>0$. Assume the statement holds for $n$, we prove it for $n+1$. The assumption $p(a,x,y)=0$ together with Lemma 1 gives $p(a,2ax-y,x)=0$, where $0\le2ax-y\le x$ and $2ax-y\equiv b-1\pmod{a-1}$. By the induction hypothesis, $x\ge a^n$, hence by Lemma 1, $y\ge(2a-1)a^n\ge a^{n+1}$.

Now, let $q(\vec w)$ be an integer polynomial that has no roots in $\mathbb Z$, but has roots in some model of, say, PA (or rather the ring obtained from a model of PA by adding a negative part). Then the polynomial $r(a,\vec w)=\bigl(a-\sum_iw_i^2\bigr)^2+q^2(\vec w)$ is also solvable in a model $M$ of PA, but any its root in a DOR must have $a$ larger than every integer. Put \begin{multline}f(a,\vec w,u_0,u_1,u_2,u_3,v_0,v_1,v_2,v_3)=\\\\ \textstyle r^2(a,\vec w)+p^2\bigl(a,(1+\sum_iu_i^2)(a-1)-1,(1+\sum_iu_i^2+\sum_iv_i^2)(a-1)\bigr).\end{multline} If we take a root of $f$ in a DOR $R$, then $a>n$ for every $n\in\mathbb N$. Putting $b=a-2$, $x=(1+\sum_iu_i^2)(a-1)-1$, $y=(1+\sum_iu_i^2+\sum_iv_i^2)(a-1)$, Lemma 2 implies that $y\ge a^n$ for every $n\in\mathbb N$, hence $R$ does not have rank 1.

On the other hand, PA (or even much weaker theories, see Lemma 2.2, 2.3 in [1]) proves that for every $a\ge b+2$ there are $x,y$ satisfying the properties in Lemma 2, and it also proves Lagrange’s four-square theorem, hence $f$ has a root in $M$.

Reference:

[1] Richard Kaye, Diophantine induction, Annals of Pure and Applied Logic 46 (1990), no. 1, pp. 1–40.