17
$\begingroup$

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 $F[x]$.

$\endgroup$
11
$\begingroup$

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).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you! Do you happen to know what might be the answer to my second question? $\endgroup$ – Uriya First Jul 29 '19 at 18:33
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.