# Divisibility between the degree of two extension fields

I am working on this problem:

Let $K$ be an extension field of a field $F$, and let $\alpha \in K$ be algebraic over $F$, with minimal polynomial $p(x)$. Let $\beta \in F(\alpha)$ be algebraic over $F$, with minimal polynomial $q(x)$. Prove that $\deg(q)\mid \deg(p)$.

I'm stuck at starting this problem so I would appreciate some hints for it.

• $F \subseteq F(\beta) \subseteq F(\alpha)$ Aug 28, 2017 at 15:15

$$\dim_F F(\alpha)=\dim_{F(\beta)} F(\alpha)\cdot \dim_F F(\beta).$$
• do you mean this: $[F(\alpha):F] = [F(\alpha):F(\beta)] [F(\beta):F]$ Aug 28, 2017 at 15:28
• From the problem, I have $[F(\alpha):F]$ = deg $q$, and $[F(\alpha):F(\beta)] = 1$, hence $[F(\beta):F]$ = deg $q$. Do I understand it correctly? Aug 28, 2017 at 15:36
• Not quite: How do you know $F(\alpha)=F(\beta)$? Aug 28, 2017 at 15:45
• From $[F(\alpha):F] = [F(\alpha):F(\beta)] [F(\beta):F]$ one gets deg $p = [F(\alpha):F(\beta)]$ deg $q$. Aug 31, 2017 at 9:22