Question: Let $K \subset E \subset L = K(\alpha)$ be a tower of field extensions, with $\alpha$ algebraic over $K$. Prove that $E$ as a $K$-vectorspace is generated by the coefficients of the polynomial $f_{K}^{\alpha} / f_{E}^{\alpha} \in E[X]$.
Attempt: So far I have proven earlier that $E$ as a field extension of $K$ is generated by the coefficients of the polynomial $f_{E}^{\alpha} \in E[X]$. Furthermore, I have been given the hint to use the fact $f_{K}^{\alpha} / (X - \alpha) = (f_{K}^{\alpha} / f_{E}^{\alpha}) \cdot (f_{E}^{\alpha} / (X - \alpha))$. Lastly, I know that if $K \subset K(\alpha)$ is a primitive field extension of degree $n$, then we can define $c_{i} \in K(\alpha)$ by $\sum_{i=0}^{n-1} c_{i}X^{i} = f_{K}^{\alpha} / (X - \alpha)$. Then $\{c_{0}, \dots, c_{n-1}\}$ is a $K$-basis for $K(\alpha)$.
So far I have atempted: we know that $f_{K}^{\alpha} / (X - \alpha) = \sum_{i=0}^{m-1} c_{i}X^{i} \in K(\alpha)[X]$, then $\{c_{0}, \dots, c_{m-1}\}$ is a $K$-basis for $K(\alpha)$.
Also, we know that $f_{E}^{\alpha} / (X - \alpha) = \sum_{j=0}^{n-1} d_{j}X^{j} \in E(\alpha)[X]$, then $\{d_{0}, \dots, d_{n-1}\}$ is a $E$-basis for $E(\alpha)$.
I am however unsure how to proceed with this information. If anybody could help me out I would gladly appreciate that.
I'll expand on my comment.
As you wrote, if $f_{K}^{\alpha} / (X - \alpha) = \sum_{i=0}^{m-1} c_{i}X^{i} \in K(\alpha)[X]$, then $X = \{c_{0}, \dots, c_{m-1}\}$ is a $K$-basis for $K(\alpha)$, and if $f_{E}^{\alpha} / (X - \alpha) = \sum_{j=0}^{n-1} d_{j}X^{j} \in E(\alpha)[X]$, then $Y = \{d_{0}, \dots, d_{n-1}\}$ is a $E$-basis for $E(\alpha)$.
First note that in fact $E(\alpha) = K(\alpha)$, thus $X$ and $Y$ are in fact bases for $K(\alpha)$ as a K- and E-vectorspace respectively. Then, if $f_{K}^{\alpha} / f_{E}^{\alpha} = \sum_{i=0}^{m-n} e_i x^i$, we see that every element of $X$ can be written as a linear combination of elements of $Y$ with coefficients from $Z = \{ e_0, \ldots e_{m-n} \} \subseteq E$ (Since $f_K^{\alpha}/F_E^{\alpha} = (\sum_{i=0}^{m-n} e_i x^i) \cdot f_E^{\alpha}/(X-\alpha)$). In light of this, for arbitrary $k_0, \ldots, k_{m-1} \in K$ you can write:
$$\sum_{i=0}^{k-1} k_i c_i = \sum_{i=0}^{k-1} k_i \left( \sum_{j=0}^{i} e_{j} d_{i-j} \right) = \sum_{j=0}^{k-1} \left( \sum_{i=j}^{k-1} k_i e_{j} \right) d_{i-j} \quad (\dagger)$$
Thus, for arbitrary $e \in E$, consider the vector $v = e d_0 \in K(\alpha)$. Note that $v$ has this unique representation, or it would contradict the fact that $X$ was an basis for $E(\alpha)$. Then, since $X$ spans $E(\alpha)$ over $K$, exists $k_0, \ldots, k_{n-1}$ such that $\sum_{i=0}^{k-1} k_i c_i = e d_0$. In the light of the above, we conclude that $e$ is in the span of $Z$ over $K$. Since $e$ was an arbitrary element of $E$, we are done.