$L/K$ field extension, $\alpha\in\overline{K}$, does $\deg f_L^\alpha \mid \deg f_K^\alpha$? Let $L/K$ be a field extension.
Let $\alpha\in\overline{K}$.
We know that $f_L^\alpha \mid f_K^\alpha$, do we also have $\deg f_L^\alpha \mid \deg f_K^\alpha$?
Usually $f\mid g$ does not imply $\deg f\mid \deg g$, take for example $X^2 + 1 \mid (X^2+1)(X-1)$, than $2\nmid 3$.
The difference is that the minimal polynomials are irreducible.
We have the tower of fields as shown in the following diagram:

Notice that
$$
  \deg f_L^\alpha \mid \deg f_K^\alpha \iff [L(\alpha):L] \mid [K(\alpha):K].
$$
With the tower law this is equivalent to
$$
  \frac{[L(\alpha):K]}{[L:K]} \mid \frac{[L(\alpha):K]}{[L(\alpha):K(\alpha)]} \iff [L(\alpha):K(\alpha)] \mid [L:K].
$$
Is it true that $\deg f_L^\alpha \mid \deg f_K^\alpha$? Otherwise, do you have a counter example?
 A: In general, no. Take $K=\mathbb{Q}$, $L=\mathbb{Q}(2^{1/3})$ and $\alpha=j2^{1/3}$. Then $[L:K]=3$, but $[L(\alpha):K(\alpha)]=2$. 
However, if $L/K$ is Galois, and $\alpha$ is separable, the statement becomes true. 
Indeed, any $K(\alpha)$-automorphism of $\overline{K}$ will leave $\alpha$ invariant and $L$ stable, thus $L(\alpha)/K(\alpha)$ is Galois as well. 
But we have a natural restriction morphism $Gal(L(\alpha)/K(\alpha)) \rightarrow Gal(L/K)$, which is injective. One concludes using Lagrange’s theorem (on subgroups). 
Edit: just to add another (more ad hoc) sufficient condition. If $K(\alpha)$ and $L$ have coprime dimensions $a$ and $b$ over $K$, then $[L(\alpha):L]=[K(\alpha):K]$ and $[L(\alpha):K(\alpha)]=[L:K]$. 
Indeed, $d=[L(\alpha):K]$ is divisible both by $a=[K(\alpha):K]$ and $b=[L:K]$, so as $a$ and $b$ are coprime $d \geq ab$. But $d=b[L(\alpha):L]$. Now $\alpha$ vanishes a polynomial in $K \subset L$ of degree $a$, thus $[L(\alpha):L] \leq a$, thus $d \leq ba$ and $d=ba$, $a=[L(\alpha):L]$. Moreover, $ab=[L(\alpha):K]=[L(\alpha):K(\alpha)][K(\alpha):K]=a[L(\alpha):K(\alpha)]$, which concludes. 
