Prove $[L(\alpha) : L]$ divides $[K(\alpha):K]$ for L finite normal extension of K

Question

We are also given $\alpha$ belongs to a finite extension of L.

With the tower law, I get:

$[L(\alpha):L][L:K] = [L(\alpha):K]$ and $[L(\alpha):K(\alpha)][K(\alpha):K] = [L(\alpha):K]$

But I'm stuck from here. Any hints?

Answer 1

Let $N/K$ be the normal closure of $L(\alpha)/K$ and $f(x)\in K[x]$ the minimal polynomial of $\alpha$. Factorize it over $L$ $$f(x)= \prod_j f_j(x)^{e_j}\in L[x]$$ Wlog $f_1$ is the $L$-minimal polynomial of $\alpha$. For each remaining $f_j(x)$ pick one of its root $\alpha_j\in N$. Then $f_j(x)$ is the $L$-minimal polynomial of $\alpha_j\in N$ which is a root of $f$. Thus, $\alpha_j$ is $K$-conjugate to $\alpha$ so there is some $\sigma_j\in Aut(N/K)$ such that $\sigma_j(\alpha_j)=\alpha$. Since $L/K$ is normal then $\sigma_j(L)=L$.

This implies that $f_j^{\sigma_j}(x),f_1(x)$ are both irreducible $\in L[x]$ and have a common root $\alpha$, therefore $f_j^{\sigma_j}(x)=f_1(x)$,$\deg(f_j)=\deg(f_1)$, $$[L(\alpha):L]=\deg(f_1)\ \ | \ \deg(f)=[K(\alpha):K]$$