Are radicals dense in the real closure of an ordered field?

Question

Let $F$ be an ordered field and let $R$ denote its real closure.

It is well-known that $F$ is cofinal in $R$, but not necessarily dense. For example, consider $F=\mathbb{R}(\omega)$ with the order determined by $r<\omega$ for all $r\in \mathbb{R}$, and consider the open interval $(\sqrt{\omega}-1,\sqrt{\omega}+1)$ in $R$, which does not meet $F$.

Question 1. Is always the case that radicals of elements of $F$ in $R$ are dense in $R$? Equivalently, is it possible to find for all $0<\alpha<\beta$ in $R$ an element $\gamma\in F$ and some $n\in \mathbb{N}$ such that $\alpha^n<\gamma<\beta^n$?

It is not difficult to see that a positive answer would imply a positive answer to the next question, which is really what I am after.

Question 2. Let $a_0,a_1,\dots,a_t\in R$ be distinct elements. Is there always a polynomial $p\in F[x]$ satisfying $p(a_0)<0$ while $p(a_i)>0$ for all $i\in\{1,\dots,t\}$?

The most interesting case is when $a_0,\dots,a_t$ constitute the list of roots of an irreducible polynomial in $F[x]$.

Answer 1

Few years late, but the answer to your 2nd question is negative: let $F=\mathbb{Q}\left(\epsilon\right)$, ordered such that $\epsilon > 0$ is smaller than any positive rational. Let $a_1<a_2<a_3<a_4$ be the roots of the polynomial $f(x)=\left(x^2-2\right)^2-\epsilon$ in $R$, the real closure of $F$. I will show that, for any $p(x)\in F[x]$, we have $p(a_1)p(a_2)p(a_3)p(a_4) \geq 0$

First note that we can assume that $p(x)\in \mathbb{Q}[x]$: we first move from $F[x]=\mathbb{Q}(\epsilon)[x]$ to $\mathbb{Q}[\epsilon][x]$ by multiplying by the absolute value of any denominator; and then we move from $\mathbb{Q}[\epsilon][x]$ to $\mathbb{Q}[x]$ by replacing every $\epsilon$ with $\left(x^2-2\right)^2$.

It is easy to see that $a_1,a_2$ are infinitesimally close to $-\sqrt{2}$ and $a_3,a_4$ are infinitesimally close to $\sqrt{2}$. On the other hand, for any two rational numbers $u,v\in\mathbb{Q}$, not both zero, and for any $1\leq i \leq 4$, we have that $\left|a_i u+v\right|$ is bounded both from below and above by positive rationals.

Therefore, if we write $$ p(x) = \sum_{n\in S} \left(x^2-2\right)^n\left(u_nx+v_n\right) $$ where $\left(u_n,v_n\right)\neq (0,0)$ for any $n\in S$, we have, where $m=\min S$ $$ \operatorname{sign}(p(a_i)) = \operatorname{sign}\left(\left(a_i^2-2\right)^m\left(u_ma_i+v_m\right)\right) = \operatorname{sign}\left(a_i^2-2\right)^m \operatorname{sign}\left(u_ma_i+v_m\right) $$ Now we have \begin{align*} \operatorname{sign}\left(u_ma_1+v_m\right) = \operatorname{sign}\left(u_ma_2+v_m\right) \\ \operatorname{sign}\left(u_ma_3+v_m\right) = \operatorname{sign}\left(u_ma_4+v_m\right) \\ \operatorname{sign}\left(a_1^2-2\right)=\operatorname{sign}\left(a_4^2-2\right)=1 \\ \operatorname{sign}\left(a_2^2-2\right)=\operatorname{sign}\left(a_3^2-2\right)=-1 \end{align*} (First two equalities follow from the fact that each $a_i$ is infinitesimally close to an irrational number $\pm\sqrt2$)

From this it follows that $p(a_1)p(a_2)p(a_3)p(a_4) \geq 0$

Answer 2

The answer is negative, and your example of $F$ not being dense in $R$ gives an example of this as well. Let $\alpha=\sqrt{\omega}+1,\beta=\sqrt{\omega}+2$. Then $\alpha^n>\omega^{n/2}+n\omega^{(n-1)/2},\beta^n<\omega^{n/2}+(2n+1)\omega^{(n-1)/2}$. It's not hard to see there is no element of $\mathbb R(\omega)$ in $(\omega^{n/2}+n\omega^{(n-1)/2},\omega^{n/2}+(2n+1)\omega^{(n-1)/2})$ (it's easiest to check $n$ odd, even separately).

Answer 3

Update: the following doesn't answer the question, since the element $\omega^\pi$ is transcendental rather than algebraic.

Let $K$ be an ordered field, and let $$F = \left\{ \alpha = \sum_{k=k_0}^\infty a_k\omega^{-k/d}\Big\vert k_0\in\mathbb{Z},\,d\in\mathbb{Z}_{\geq 1},\, a_k\in K\right\}$$ be the field of Puiseaux series over $K$ ordered lexicographically (take the convention that $a_{k_0}\neq 0$ if $\alpha\neq 0$ and then $\alpha>0$ iff $a_{k_0}>0$).

Note that $F$ is complete with respect to the valuation $v(\alpha) = k_0/d$. We call $-k_0/d$ the order of $\alpha$ and $a_{k_0} \omega^{-k_0/d}$ the leading term of $\alpha$.

Let $R = F(\omega^\pi)$ where $\pi\in \mathbb{R}$ is any irrational. Then the valuation and order extend to $R$. In particular, elements of $R$ have leading terms and orders. Clearly if $\alpha,\beta\in R$ are positive of the same irrational order then no elements of $F$ separate them (the order of any positive element of $F$ is rational, hence either smaller or larger than that of $\alpha,\beta$). Now let $p(x) = \sum_{i=0}^n a_i x^i \in F[x]$ be a non-constant polynomial. Then for any $\alpha \in R$ of irrational order, the elements $a_i x^i$ (for $a_i\neq 0$) have distinct orders (because $1,\pi$ are linearly independent over $\mathbb{Q}$). It follows that the order of $p(\alpha)$ is irrational and depends only on the order of $\alpha$ (when that is irrational), and in particular that $p$ cannot separate elements of $R$ of the same irrational order.

Remark It is not hard to see that polynomials in $K[x]$ can be used to separate elements of $F$. More generally, if $K$ is an ordered field and $R$ is the field of multivariable Puiseaux series over $K$ in the variables $\omega_1,\ldots,\omega_r$ where we repeatedly order lexicographically then the same holds. In other words, the claim is true if $R=F(\omega_1,\ldots,\omega_r)$ where the extensions are successively non-archimedean in that $\omega_i$ is larger than every element of $F(\omega_1,\ldots,\omega_{i-1})$.