Does every non-Archimedean ordered field contain $\mathbb Q (x)$ as an ordered subfield?

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.

• (edited as an anwser) – nombre Apr 27 '16 at 12:34

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 $x^{n+1}> P(x)$.
• 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). – s.harp Apr 26 '16 at 16:48
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 $a,c \in A, \ b,d > 0 \in A$, $\frac{a}{b} <' \frac{c}{d} \longleftrightarrow ad <' bc$.