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Every ordered field $\mathbb F$ contains $\mathbb Q$ in a canonical way. If the field is not Archimedean there exists an $x>n$ for all $n \in \mathbb N$. Since we are dealing with a field any polynomial expression $P(x)$ must lie in the field and be invertible unless it is the zero polynomial. So also $1/P(x)$ will lie in $\mathbb F$.

From the simple algebraic properties of inversion, addition and multiplication that hold in every field it appears that $\mathbb F$ must then contain the field of rational polynomials $\mathbb Q(x)$ as a subfield (although not necessarily in a canonical way).

My question whether the ordering on this subfield has to be the same ordering as the ordering of $\mathbb Q(x)$. I think this would be implied by $x^2>x$, which follows from $x>1$ and then multiplying with $x$. This should give $x^2 > \alpha x + \beta$, with $\alpha, \beta \in \mathbb Q$ and the ordering on the polynomial part is the same as the one you get on the polynomial ring $\mathbb Q[x]$.

But I don't see whether this implies that the embedding $\mathbb Q(x)$ is order preserving.

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  • $\begingroup$ (edited as an anwser) $\endgroup$ – nombre Apr 27 '16 at 12:34
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Basically your question is : for any $P\in \mathbb{Q}[X]$ of degree $n$, is it true that $x^{n+1}>P(x)$ ?

Now if $P=\sum a_k X^k$, then $P(x)\leqslant (\sum |a_k|)\cdot x^n$ since $x^k\leqslant x^n$ for all $k\leqslant n$.

Then since $x>(\sum |a_k|)$, you get $x^{n+1}>(\sum |a_k|)\cdot x^n$, so $x^{n+1}> P(x)$.

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  • $\begingroup$ Excellent, thanks. Can you sketch how the order extends to $\mathbb Q(x)$? Ie that $\frac{P(x)}{Q(x)} > \frac{\tilde P (x)}{\tilde Q(x)}$ if $\rm{deg}(P)-\rm{deg}(Q)>\rm{deg}(\tilde P)-\rm{deg}(\tilde Q)$ or when they are equal if the ratio of the leading coefficients of $P$ and $Q$ is larger than the that of $\tilde P$, $\tilde Q$ (if that is equal, take the next ratio). $\endgroup$ – s.harp Apr 26 '16 at 16:48
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Given an ordered ring $(A,<)$ (which is an integral domain) there is a unique order $<'$ on $Frac(A)$ which coïncides with $<$ on $A$, because given $a,c \in A, \ b,d > 0 \in A$, $\frac{a}{b} <' \frac{c}{d} \longleftrightarrow ad <' bc$.

Moreover, there is a universal property which is basically that of the fraction field along with order compatibility. So if an ordered ring embeds in an ordered field, so does its (unique) fraction ordered field.

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