Clarification needed on ordering $\mathbb{Z}[X]$ in the language of arithmetic.

Question

I have been reading the book by richard kaye, on peano arithmetic. In section 2.1, he gave the following statement.

"$\mathbb{Z}[X]$ is made into a $L_A$ structure by defining an order $<$ making $X$ 'infinitely large' "

($L_A$ here means language of arithmetic)

So exactly what does the author meant by that statement, I pondered about it for a way and cant figure out. Any help is appreciated.

Cheers

Answer 1

As you pointed out, it's not entirely clear what the author meant but I have a good feeling that I know what they had in mind:

$L_{A}$ - the language of arithmetic - includes two constant symbols $\dot{0},\dot{1}$, one (or two - depending on your convention) relation symbols $\dot{\prec}$ (,\dot{=}) and two function symbols $\dot{\circ}$ and $\dot{+}$. If we want to consider $\mathbb Z[X]$ as an $L_{A}$-structure, we must decide how to interpret all these symbols in this structure. The interpretation of $\dot{0}, \dot{1}$ should be clear (it's the zero polynomial and constant polynomial with value $1 \in \mathbb Z$ respectively). The interpretation of $\dot{\circ}, \dot{+}$ is given by polynomial multiplication and addition respectively. The interpretation of $\dot{=}$ is equality.

What remains now is the interpretation of $\dot{\prec}$. Call this interpretation $<$. We want that $(\mathbb Z[X]; 0,1, +, \circ, \prec)$ is an ordered ring and here is what we can do to achieve this.

1. Let $f, g \in \mathbb Z[X]$ be constant polynomials. We let $$f \prec g : \iff f(0) \lt g(0), $$ where $\lt$ is the natural strict order of $\mathbb Z$.
2. For any constant polynomial $f \in \mathbb Z[X]$, we let $f \prec X$,
3. We now expand $\prec$ to all polynomials via the rules for ordered rings, i.e. if $f \prec g$ and $0 \prec h$ then $f \circ h \prec g \circ h$ and the second rule: If $f \prec g$ then for all $h \in \mathbb Z[X] \colon f + h \prec g + h$.

Evaluating $\prec$ with only this information given can be quite painful so let me spare you some time and tell you that $$ f \prec g \iff g-f \text{ has non-negative leading coefficient}. $$ It's a good exercise to prove that this equivalence holds.

Answer 2

Here is an alternative approach (but it yields the same ordering as in Stefan Mesken's answer):

Let $A$ be a nonstandard model of the ring $\mathbb{Z}$ and let $a \in A$ be a nonstandard number $>0$. Then, the map $$\mathbb{Z}[X] \to A,\phantom{aaa}f(X) \mapsto f(a)$$ is an injective ring homomorphism. (Injectivity follows from the fact that the map has trivial kernel. This can be proved by using the overspill principle which is stated in Kaye's book.) Therefore, we can view $\mathbb{Z}[X]$ as a subring of $A$ and a subring of an ordered ring is also ordered (simply restrict $<$ from $A$ to $\mathbb{Z}[X]$).