Not every ordered field is Archimedean

Let $f=p/q\ne0$, where $p(x)=a_mx^m+\cdots+a_0$ and $q(x)=b_nx^n+\cdots+b_0$ have nonzero leading coefficients. For $x>1$, dividing by the leading terms gives $p(x)=a_mx^m(1+\sum_{i<m}(a_i/a_m)x^{i-m})$ and the analogous formula for $q$. If $x$ is larger than $1$ plus the sums of the absolute values of the lower coefficient ratios, then both lower-term sums have absolute value less than $1$. Thus $p(x)$ and $q(x)$ eventually have the signs of $a_m$ and $b_n$, respectively, and $f(x)$ has the constant nonzero eventual sign of $a_m/b_n$. Hence exactly one of $f\in P$ and $-f\in P$ holds.

If $f, g \in P$ then $f(x) > 0$ and $g(x) > 0$ for all large $x$, so $(f + g)(x) > 0$ and $(fg)(x) > 0$ for all large $x$, giving $f + g \in P$ and $fg \in P$.

For each natural number $n$ the rational function $t - n \cdot 1 = t - n$ satisfies $(t - n)(x) = x - n > 0$ for all $x > n$, so $t - n \in P$.

By the trichotomy of step 1.1 and the closure of step 1.2, $P$ is a positive cone, so $\mathbb{R}(t)$ is an ordered field.

By step 1.3, $t - n \cdot 1 \in P$ for every natural $n$, which by [L1] means $n \cdot 1 < t$ for every natural $n$.

In the ordered field $\mathbb{R}(t)$ the element $t$ satisfies $n \cdot 1 < t$ for every natural $n$ (step 2.2), so no natural $n$ has $t < n \cdot 1$; the canonical naturals are not cofinal and $\mathbb{R}(t)$ is not Archimedean, refuting the claim that every ordered field is Archimedean.

This is consistent with [L3], whose contrapositive states that a non-Archimedean ordered field cannot be complete: $\mathbb{R}(t)$ is an ordered field that is not complete.