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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

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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.

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

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