Let $F$ be an ordered field and $f\in F[X]$ be a polynomial such that $f(x)>0$ for all $x\in K$. Is it possible that there is an extension $L\supseteq K$ of ordered fields and $y\in L$ such that $f(y)\leq 0$? My conjecture is that this is not the case but I do not find a proof for it.



Positivity in extensions of ordered fields
Let $K=F=\mathbb Q,f=(x^2-2)^2,L=\mathbb Q(\sqrt{2})$, so that $f(\sqrt{2})=0$.


It is more interesting to ask for $f(y)<0$. This is also possible. Let $K=F=\mathbb Q(T)$, where $T$ is an indeterminate larger than all elements of $\mathbb Q$. Let $f=(x^2-T)^2-1$ and $L=\mathbb Q(\sqrt{T})$. Then $f$ is positive on $F$ but $f(\sqrt{T})=-1<0$.


