TITLE: Algebraic field extensions: Why $k(\alpha)=k[\alpha]$.
QUESTION [11 upvotes]: 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 $H$?

REPLY [6 votes]: It is clear that $k[\alpha]\subseteq k(\alpha)$, so all we need to show is that $k[\alpha]$ is a field. Indeed, $k(\alpha)$ is contained in every subfield of $K$ that contains $\alpha$.
Let $\beta\in k[\alpha]$, $\beta\notin k$. Then $\beta$ is algebraic over $k$; indeed, the powers of $\beta$ cannot be linearly independent over $k$, as $k[\alpha]$ is finitely generated as a vector space over $k$ (you need just polynomial division with remainder to show this).
Let $a_0+a_1\beta+\dots+a_n\beta^n=0$ with $a_0\ne0$ (this surely exists, why?). Then $n\ge1$ and so
$$
\beta\frac{a_1+a_2\beta+\dots+a_n\beta^{n-1}}{-a_0}=1
$$
so $\beta^{-1}\in k[\alpha]$.