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?

$\begingroup$ Don't we have arbitrary order integration formulae with positive rational coefficients and "rational" nodes? Or am I misunderstanding the question? $\endgroup$ – fedja Oct 12 '15 at 21:30

$\begingroup$ I am asking if this is true in nonArchimedean fields, i.e. can this be proved purely algebraically. $\endgroup$ – Zbigniew Fiedorowicz Oct 12 '15 at 22:02

4$\begingroup$ 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 $p(b)p(a)=(ba)\sum_j q_jp'(a+r_j(ba))>0$ for any rational positive coefficient quadrature formula of order greater than the degree of $p$? $\endgroup$ – fedja Oct 13 '15 at 1:46

$\begingroup$ 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. $\endgroup$ – Todd Trimble♦ Oct 13 '15 at 13:58

$\begingroup$ @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.) $\endgroup$ – Emil Jeřábek Oct 13 '15 at 15:42
It is remarked in A Course in Model Theory: An Introduction to Contemporary Mathematical Logic by Bruno Poizat, page 101 (Google Books link), that the assertion is true for all ordered fields, but that its proof is difficult. He gives the proof for real closed fields just prior to that remark.

$\begingroup$ Todd, the remark is, indeed, there but I still wonder what subtlety I'm missing if any because the argument I proposed is totally trivial :( $\endgroup$ – fedja Oct 13 '15 at 15:31

$\begingroup$ @fedja I have to confess that I didn't understand your argument. What if the coefficients of $p$ aren't rational? $\endgroup$ – Todd Trimble♦ Oct 13 '15 at 16:16

1$\begingroup$ @ToddTrimble: What fedja suggested would amount to this in the degree $2$ case (as an illustration): we have the identity $p(b)p(a)=((ba)/2)(p'(a)+p'(b))$ 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. $\endgroup$ – Christian Remling Oct 13 '15 at 16:20

$\begingroup$ @ChristianRemling Thanks; that's very helpful indeed. That does seem like a promising idea (wonder what could go wrong?!). $\endgroup$ – Todd Trimble♦ Oct 13 '15 at 16:31