Extension with algebraic element is finite I don't get how to proof this theorem:
Let $E/K$ be a field extension. If $\alpha$ is algebraic over K, then $K(\alpha ):K<\infty$.
I know that we can assume that there exists a nontrivial polynomial $f(X)$ with $f(\alpha)=0$. We didn't had the minimal polynomial in class yet. I would very much appreciate your help.
Best, KingDingeling
 A: The relation
$$\alpha^n + \sum_{i=0}^{n-1} a_i \alpha^i = 0$$
shows that you can write all powers of $\alpha$ as linear cobinations of $\{ \alpha^i\}_{0 \le i \le n-1}$.
Proof by induction.
CASE $k=0, \dots , n-1$ is obvious. $\alpha^k$ is a linear combinations of $\{ \alpha^i\}_{0 \le i \le n-1}$.
CASE $k=n$. Follows from
$\alpha^n =- \sum_{i=0}^{n-1} a_i \alpha^i $
INDUCTIVE STEP. For $k \ge n+1$ you have
$$\alpha^k = \alpha \cdot \alpha^{k-1}$$
since $\alpha^{k-1}$ is a linear combinations of $\{ \alpha^i\}_{0 \le i \le n-1}$, you can write
$$\alpha^{k-1} = \sum_{i=0}^{n-1} b_i \alpha^i$$
Thus
$$\alpha^k = \alpha \cdot \alpha^{k-1} = \alpha \cdot \sum_{i=0}^{n-1} b_i \alpha^i = \sum_{i=0}^{n-1} b_i \alpha^{i+1} = \sum_{i=0}^{n-2} b_i \alpha^{i+1} + b_{n-1} \alpha^n = \sum_{i=0}^{n-2} b_i \alpha^{i+1} - b_{n-1} \sum_{i=0}^{n-1} a_i \alpha^i$$
is a linear combination of $\{ \alpha^i\}_{0 \le i \le n-1}$.
This concludes the proof.
WHAT DOES THIS MEAN: Since
$$K( \alpha) = \{ \sum_j  a_j \alpha^j : a_j \in K \}$$
it shows that $K( \alpha)$ is generated by $\{ \alpha^i\}_{0 \le i \le n-1}$, i.e. it is finitely generated.
