I'm trying to prove the following:
Let $K$ be a field, $\alpha \in \overline{K}$ a separable element s.t. $K(\alpha)/K$ is normal, and let $L/K$ be some finite subextension of $\overline{K}$. Prove that $\deg_{\alpha,L}\mid\deg_{\alpha,K}$
My attempt:
We know that $\begin{array}{c} \left[L\left(\alpha\right):L\right]=\deg_{\alpha,L}\\ \left[K\left(\alpha\right):K\right]=\deg_{\alpha,K} \end{array}$ and therefore $\deg_{\alpha,L}\mid\deg_{\alpha,K}$ iff $\left[L\left(\alpha\right):L\right]\mid\left[K\left(\alpha\right):K\right]$ but since $$\begin{array}{c} \left[L\left(\alpha\right):K\right]=\left[L\left(\alpha\right):K\left(\alpha\right)\right]\left[K\left(\alpha\right):K\right]\\ \left[L\left(\alpha\right):K\right]=\left[L\left(\alpha\right):L\right]\left[L:K\right] \end{array}$$ we get that $$\left[L\left(\alpha\right):L\right]=\frac{\left[L\left(\alpha\right):K\right]}{\left[L:K\right]}\qquad\frac{\left[L\left(\alpha\right):K\right]}{\left[L\left(\alpha\right):K\left(\alpha\right)\right]}=\left[K\left(\alpha\right):K\right]$$ meaning in order to prove $\deg_{\alpha,L}\mid\deg_{\alpha,K}$ I need to just show (at least) one of the following holds:
- $\left[L\left(\alpha\right):L\right]\mid\left[K\left(\alpha\right):K\right]$
- $\left[L\left(\alpha\right):K\left(\alpha\right)\right]\mid\left[L:K\right]$
Now since $L/K$ is finite we can write $L=K\left(a_{1},\ldots,a_{k}\right)$ and since $\alpha$ is separable and $K(\alpha)/K$ is normal, it is also Galois.
From this point on I am stuck. I can't find a way to connect $K$ and $L$ in any way. I thought about looking at the normal closure of $L$ but didn't find this useful.
I also noticed that we can embbed $\mathrm{Aut}_L (L(\alpha))\hookrightarrow Gal(K(\alpha ))$ and get that $\left|\mathrm{Aut}_{L}\left(L\left(\alpha\right)\right)\right|\mid\left[K\left(\alpha\right):K\right]$ but again I can't do anything from here because I don't know that $L(\alpha)/L$ is Galois.