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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?

cut
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  • It refers to an arbitrary element $a$ of $K$, which does not necessarily generate all of $L$; for example, say $K=\mathbb{F}3$, $L=\mathbb{F}{3^6}$. Maybe you pick an $a\in L$ such that $K(a)=\mathbb{F}_{3^3}$, rather than $K(a)=L$? – Arturo Magidin Mar 09 '23 at 20:18
  • As for the "corresponding statement", I can think of two things you can do: you can either say how to correct the product so that you actually do get the minimum polynomial of $\alpha$ over $K$; or you could correct the formula to say what it is that you get by taking the product in the general case. – Arturo Magidin Mar 09 '23 at 20:21
  • Understood. Any tips on how to prove the equality? – cut Mar 10 '23 at 10:12

1 Answers1

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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.

durianice
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