Field Theory: Degree of Composite equals degree of tower field extenstions (relatively prime)? Let $E_1,E_2$ be finite extensions of a field $F$, and assume $E_1,E_2$ contained in some field. If $[E_1:F]$ and $[E_2:F]$ are relatively prime, show that $[E_1E_2:F]=[E_1:F][E_2:F]$ and $[E_1E_2:E_2]=[E_1:F]$.
Would anyone know how to approach this problem/solve it?
 A: Two theorems apply here: first, if $k\subset E\subset\Omega$ and $k\subset F\subset\Omega$, where all the letters denote fields, then $[EF:E]\le[F:k]$, and in particular if both $[E:k]$ and $[F:k]$ are finite, so is $[EF:k]$. The other is that if $k\subset E\subset K$, then $[K:k]=[K:E][E:k]$.
The rest follows when you realize that the relatively prime numbers $[E_1:F]$ and $[E_2:F]$ both divide $[E_1E_2:F]$, and you then apply the inequality from the first quoted theorem.
A: (i) Suppose $F \subset E_2 \subset L$ and suppose $\alpha \in L$. Can you show that $[E_2(\alpha):E_2] \leq [F(\alpha):F]$?
[Hint: $[F(\alpha):F]$ is the degree of the minimal polynomial of $\alpha $ over $F$, and $[E_2(\alpha):E_2]$ is the degree of the minimal polynomial of $\alpha$ over $E_2$. How are the degrees of these minimal polynomials related?]
(ii) Let $E_1, E_2$ be finite extensions of $F$, all contained inside $L$. Convince yourself that $E_1 = F(\alpha_1, \dots, \alpha_n)$ for certain $\alpha_1, \dots, \alpha_n \in L$. Convince yourself that $E_1E_2 = E_2(\alpha_1, \dots, \alpha_n)$. Can you use the result of (i) inductively to show that $[E_1 E_2 : E_2] \leq [E_1 : F]$?
[Of course, a similar argument gives $[E_1 E_2 : E_1] \leq [E_2 : F]$.]
(iii) Apply the tower law to the inclusions $F \subset E_1 \subset E_1 E_2$ and $F \subset E_2 \subset E_1 E_2$. Use the fact that $[E_1 : F]$ and $[E_2 : F]$ are relatively prime, together with the result from (ii), to show that $[E_1 E_2 : E_2] = [E_1 : F]$ and $[E_1 E_2 : F] = [E_1 : F][E_2 : F]$.
