Polynomial inequalities in ordered fields

Let $p(x)$ be a polynomial over an ordered field. If $p'(x)\ge 0$ for all $x$ in an interval over that field, is it true that $p(x)$ is increasing over that interval?

• Don't we have arbitrary order integration formulae with positive rational coefficients and "rational" nodes? Or am I misunderstanding the question? – fedja Oct 12 '15 at 21:30
• I am asking if this is true in non-Archimedean fields, i.e. can this be proved purely algebraically. – Zbigniew Fiedorowicz Oct 12 '15 at 22:02
• That's what I'm saying. Isn't it true that $1>0$, so $m>0$ for all positive integer $m$, so $q>0$ for all positive rational $q$, so if $b>a$, then $p(b)-p(a)=(b-a)\sum_j q_jp'(a+r_j(b-a))>0$ for any rational positive coefficient quadrature formula of order greater than the degree of $p$? – fedja Oct 13 '15 at 1:46
• The answer to the OP is certainly 'yes' in the case of real closed fields, since these are elementarily equivalent to $\mathbb{R}$. Hopefully an affirmative answer can be obtained by exploiting the fact that each ordered field embeds in a real closed field. – Todd Trimble Oct 13 '15 at 13:58
• @fedja: I think it would be great if you could expand your comments into a proper answer. (And yes, a polynomial identity with rational coefficients like this holds automatically in all ordered fields if it holds in the reals.) – Emil Jeřábek Oct 13 '15 at 15:42

• @fedja I have to confess that I didn't understand your argument. What if the coefficients of $p$ aren't rational? – Todd Trimble Oct 13 '15 at 16:16
• @ToddTrimble: What fedja suggested would amount to this in the degree $2$ case (as an illustration): we have the identity $p(b)-p(a)=((b-a)/2)(p'(a)+p'(b))$ for all $p$ with $\deg p\le 2$. Obviously, this holds in all characteristic $0$ fields (just check it on monomials), and the OP's claim is immediate now. – Christian Remling Oct 13 '15 at 16:20