If $K$ and $k$ are fields, $K\supset k$ is a field extension and $\alpha \in K$ is algebraic over $k$, then we denote by $k[\alpha]$ the set of elements of $K$ which can be obtained as polynomial expressions of $\alpha$.
$$k[\alpha] = \left\{P(\alpha): \quad P\in k[X] \right\} $$
Also, we denote by $k(\alpha)$ the smallest subfield of $K$ containing both $k$ and $\{\alpha\}$. This is easily seen to be equal to the set of fractions of polynomial expressions in $\alpha$ (just thinking about the way the "smallest" subfield must be generated):
$$k(\alpha) = \left\{\dfrac{P(\alpha)}{Q(\alpha)}: \quad P,Q\in k[X] \text{ and } Q(\alpha)\neq 0 \right\} $$
If $M$ is the minimal polynomial of $\alpha$, it is easily shown that $k[\alpha]\simeq k[X]/\langle M\rangle$ is a field and therefore a subfield of $k(\alpha)$ containing both $k$ and $\{\alpha\}$. Therefore $k(\alpha)=k[\alpha]$, because $k(\alpha)$ is the smallest.
It follows that all quotients $\dfrac{P(\alpha)}{Q(\alpha)}$ are equal to $H(\alpha)$ for some polynomial $H\in k[X]$.
What I want to see is a more direct proof of this fact. Given $P$ and $Q$, how do you produce this $H$?