$a^+$ Induced Order

Question

I am studying the different orders that can be induced from the usual polynomial order in $ \mathbb R[x]$: i.e: $$ p(x) >_{+\infty} 0 \iff a_n > 0 $$ Where $a_n$ is the leading coefficient. One can induce the following ordering: $$ p(x) >_{0^+} 0 \iff p\left(\frac{1}{x}\right) >_{+\infty} 0 $$ So we get that $ 0 <_{0^+} x <_{0^+} \varepsilon$ for every positive $ \varepsilon$. I am trying to generalize this order by taking any $a \in \mathbb R$ and letting: $$ p(x) >_{a^+} 0 \iff p\left(\frac{1}{x-a}\right) >_{+\infty} 0$$ However this doesn't seem to work properly, since I need that $ a <_{a^+} x <_{a^+} a+ \varepsilon$, but it is not true that $ a <_{a^+} x $ because the sign of $$ \frac {1}{x-a} - a = \frac{1-ax+a}{x-a} $$ depends on the sing of $a$. Is this the right way to define this? Am I making a silly mistake somewhere? Thanks!

Answer 1

I think this is easier if you formulate everything in terms of automorphisms of the field $\mathbb{R}(x)$. Your ordering $\leq_{+\infty}$ uniquely extends to an order on $\mathbb{R}(x)$, which I will write simply as $\leq$. Your other orders are then just obtained by taking some automorphism $f:\mathbb{R}(x)\to\mathbb{R}(x)$ and defining a new order $\preceq$ by $p\preceq q$ iff $f(p)\leq f(q)$. For instance, your order $\leq_{0^+}$ is given in this way by the automorphism $f$ which sends $x$ to $1/x$ and fixes $\mathbb{R}$. (So, explicitly, $f$ sends a rational function $p(x)$ to $p(1/x)$.)

To get your ordering "$\leq_{a^+}$", then, you just need to use an automorphism that sends $x$ to something which is infinitesimally larger than $a$, such as $a+1/x$. So you can use the automorphism which sends a rational function $p(x)$ to $p(a+1/x)$. If you use $p(1/(x-a))$ instead as you did, that will not work, since if $a\in\mathbb{R}$ then $x-a$ is infinitely large so $1/(x-a)$ is infinitesimal (so this in fact gives the same ordering as your $\leq_{0^+}$.