$L/k$ finite extension , $L_1,L_2 $ subfields of $L$ containing $k$ , $L_1/k$ separable and $L_2/k$ normal , then $[L_1L_2:L_2]=[L_1:L_1\cap L_2]$ ?

Question

Let $L/k$ be a finite extension . $L_1,L_2 $ subfields of $L$ containing $k$ such that $L_1/k$ is separable and $L_2/k$ is normal . Then it is easy to see $L_1L_2/L_2$ is separable . But how to show that $[L_1L_2:L_2]=[L_1:L_1\cap L_2]$ ?

Answer 1

$L_2/k$ is normal and $L_1/k$ is separable. Then we have the following: $$L_1\cap L_2= L_1\cap L_2 ^s$$ $$L_1L_2^s=(L_1L_2)^s$$ Now we have $L_2^s/k$ is a Galois extension and hence $$Gal(L_1L_2^s/L_1)\cong Gal(L_2^s/L_1\cap L_2 ^s)=Gal(L_2^s/L_1\cap L_2)$$ Thus we get $$[L_1L_2^s:L_1]=[L_2^s:L_1\cap L_2]$$ and $$[L_1L_2^s:L_2^s]=[L_1:L_1\cap L_2]$$ Note we haven't yet used the separability of $L_1$ explicitly. Using primitive element theorem we have $L_1=k(\alpha)$ where $\alpha$ is separable over $k$. Then we have $L_1L_2=L_2(\alpha)$ and $L_1L_2^s=L_2^s(\alpha)$. let $f$ be the min polynomial of $\alpha$ over $L_2$ and $f^s$ be that over $L_2^s$. Then $f$ and $f^s$ both are separable since $\alpha$ is separable. Thus we have the coefficients of $f$ and $f^s$ both are separable. This implies $f=f^s$. Thus again we have $$[L_2^s(\alpha):L_2^s]=[L_2(\alpha):L_2]$$ i.e. $$[L_1L_2^s:L_2^s]=[L_1L_2:L_2]$$ Combining the above equalities we have $$[L_1L_2:L_2]=[L_1:L_1\cap L_2]$$