What is the relation between $Irr(a, F)$ and $Irr(a, K)$?

Question

We have that $F \leq K \leq L$ and $a \in L$.
If $a$ is algebraic over $F$ then it is also algebraic over $K$.
What is the relation between $Irr(a, F)$ and $Irr(a, K)$?

Let $Irr(a, K)=p(x) \in K[x]$ and $Irr(a, F)=q(x) \in F[x]$.
Then $q(x) \in F[x] \subseteq K[x]$ with $q(a)=0$

How can I continue?

Answer 1

Hints

1. If $p=\operatorname{Irr}(a,K)$ is the minimal polynomial for $a$ over $K$ and if $q\in K[x]$ has the property that $q(a)=0$, then $p$ divides $q$ (remember that $K[x]$ is a principal ideal domain and the characterization of the minimal polynomial).

2. If $q=\operatorname{Irr}(a,F)$, then $q\in K[x]$ and $q(a)=0$.

3. So…