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

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 $$F[x]$$.

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 $$a_1 be the roots of the polynomial $$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 $$p(a_1)p(a_2)p(a_3)p(a_4) \geq 0$$

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).

• Thank you! Do you happen to know what might be the answer to my second question? Jul 29, 2019 at 18:33

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 $$F(\omega_1,\ldots,\omega_{i-1})$$.