The question goes as follows: Let $K\subset E\subset K(\alpha)$ be a tower of field extensions, with $\alpha$ algebraic over $K$. Prove that as an extension of $K$, the field $E$ is generated by the coefficients of the minimum polynomial $f^\alpha_E\in E[X]$.
I have had the following idea, using a previous result: Let $K\subset K(\alpha)$ be a simple extension of degree $n$, and define $c_i\in L$ by
$$ \sum^{n-1}_{i=0} c_i X^i=\frac{f^\alpha_K}{X-\alpha}\in L[X]$$
Then $\{ c_0,c_1,…,c_{n-1}\}$ is a $K$-basis for $L$.
I think this result could be used to prove the statement above, but I am not sure how I could adapt it to the problem.
We have the extensions $K \subset E \subset K(\alpha)$, where $\alpha$ is algebraic over $K$. Therefore, $\alpha$ is algebraic over $E$. We write $f^{\alpha}_E := \sum^n_{k=0} a_k x^k \in E[x]$ and $L:= K(a_0,...,a_n)$. Because $E \subset K(\alpha)$, we see that $E(\alpha) \subset L(\alpha)=: K(a_0,...,a_n)(\alpha)$. However, because $L \subset E$, we also know that $L(\alpha) \subset E(\alpha)$. So, we get that $L(\alpha) = E(\alpha)$.
Let's look at the tower $L \subset E \subset E(\alpha) = L(\alpha)$. We denote by $f^{\alpha}_E$ resp. $f^{\alpha}_L$ the minimum polynomial of $\alpha$ over $E$ resp. $L$. Notice that $f^{\alpha}_L \in E[x]$, thus $f^{\alpha}_E \mid f^{\alpha}_L$.
Furthermore, $f^{\alpha}_E \in L[x]$, because $L$ is generated by the coefficients of $f^{\alpha}_E$. Thus, $f^{\alpha}_L \mid f^{\alpha}_E$ as well. Now, we can see that $f^{\alpha}_E = f^{\alpha}_L$, therefore $[ L(\alpha) :L] = [E(\alpha):E]$, and we can conclude that $[E:L] = 1$. In other words, $L = E$.