Let $E$ be a field and $F \subseteq E$ a subfield of $E$ and let $p(x), q(x) \in F[x]$ two polynomials with coefficients in $F$.
Show that if $p(x)$ divides $q(x)$ on $E$ $(= \exists h(x) \in E[x]$ t.q. $q(x) = p(x) · h(x)$), then $p(x)$ divides $q(x)$ on $F$ $(= \exists h(x) \in F[x]$ t.q. $q(x) = p(x)·h(x)$).
I tried to do the following : by an euclidian division on $F[x]$ $$q(x) =p(x) ·b(x) +r(x) $$ and by the given supposition : $$q(x) = p(x)·h(x)$$ Since, $p(x) ·b(x) +r(x) \in E[x]$, $$p(x)·h(x) = p(x) ·b(x) +r(x)$$ $$\iff p(x) [b(x)-h(x)]+r(x)=0$$ Then I don't know how to show $r(x) = 0$ to finish the proof. I thought of saying that since the remainder and the quotient are unique then $r(x)=0$ and $b(x)=h(x)$ but I feel there is something I am doing wrong...
Maybe, I could say that $r(x) = p(x)[h(x)-b(x)]$. If $r\ne0$, then $h-b\ne0$, and so $\deg(p(h-b))\ge \deg(p) > \deg(r)$. Thus, $r(x)=0$.
