Prove that $[F(a, b) : F] ≤ \deg(f)\deg(g)$.

Question

Where $K/F$ is a field extension, and $a,b ∈ K$ is algebraic over $F$. $f(x)$, $g(x)$ are the minimal polynomials of $a,b$ respectively over $F$.

I know that equality holds if the two degrees are relatively prime. What happens if they aren't?

Answer 1

We have $$ [F(a,b):K] = [F(a,b): F(a)][F(a): K] = [F(a,b): F(a)] \deg{(f)} $$

where $f\in F[X]$ is the minimal polynomial of $a$. It suffices to show that $[F(a,b): F(a)] \leq \deg{(g)}$, where $g\in F[X]$ is the minimal polynomial of $b$ over $F$. But we have, $F[X] \subset (F(a))[X]$ and by assumption $g(b) = 0$, hence if $\tilde{g} \in (F(a))[X]$ is the minimal polynomial of $b$ over $F(a)$ then $\tilde{g}$ divides $g$ and $$[F(a,b): F(a)] = \deg{(\tilde{g})} \leq \deg{(g)},$$ as desired.