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


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.