# 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

## 1 Answer

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.

• Thank you taking the time and help with the problem! :) – KingDingeling Dec 30 '18 at 13:46