Minimum polynomial of generator of finite extension

Question

Let $K\subset L$ be an extension of finite fields and $G=Aut_K L$. Prove: for $\alpha\in L$ with $L=K(\alpha)$, we have

$$f_K^{\alpha}= \prod_{\sigma\in G} (X-\sigma(\alpha)) $$

What is the corresponding statement for an arbitrary $\alpha\in L$?

I know the proof must go along the lines of: I know that the zeros of the minimum polynomial over $\mathbb{F_p}$ of an element $\alpha\in\overline{\mathbb{F_p}}$ are exactly the elements $\sigma(\alpha)$. Now, I do not know how to generalize this to an arbitrary finite field, i.e., if $K$ is not of the form $\mathbb{F_p}$.

On the other hand, I am not sure what the second question means. From what I understand, it refers to an arbitrary element of an arbitrary finite extension of $K$, not necessarily generated by a single element. Am I right?

Answer 1

The equality is proved in my answer to the other question you posted. Let $g$ be the product in question. Basically, the idea is that:

1. $\sigma(\alpha)$ are roots of $f_K^\alpha$ for all $\sigma\in \text{Aut}_K(L)$, so $g$ divides $f_K^\alpha$.
2. $g$ is fixed by all elements in $\text{Aut}_K(L)$, so we can show that $g\in K[x]$ by comparing coefficients.

For the corresponding results when $\alpha\in L$ does not generate $L$ over $K$, note that $K(\alpha)$ is still a finite field containing $K$, so we can simply replace $G$ by $\text{Aut}_K(K(\alpha))$ and get a similar result.