Does the $\mathrm{Spec}(F)$ equal $\overline{F}$ for a field $F$?

Question

Let $F$ be a field and let $p\in F[X]$ be a monic irreducible polynomial. Let $P\in F[X]$ be an arbitrary polynomial. Does there exist a polynomial $Q\in F[X]$ such that $p(X)$ divides $Q(P(X))$, i.e., $$ Q(P(X))=p(X)R(X),\quad R\in F[X]? $$ Thank you.

Answer 1

Let $\alpha_i$ be the roots of $p$ in $\overline F$, with multiplicities $m_i$. Let $\overline Q \in \overline F[X]$ be any polynomial for which $P(\alpha_i)$ is a root with multiplicity at least $$\sum_{j : P(\alpha_j) = P(\alpha_i)} m_j$$ Then $p(X) \mid \overline Q(P(X))$. Now $\overline Q$ has coefficients in some finite extension $E/F$. Multiply $\overline Q$ by all its conjugates w.r.t. this extension to obtain a $Q \in F[X]$ with the same property.