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Let $p$ be a prime.

Find how many $\alpha \in F_{p^6}$ such that $F_p(\alpha)=F_{p^6}$.

I read this post Number of elements $\alpha \in \mathbb F_{2^4}$ such that $\mathbb F_2(\alpha) = \mathbb F_{2^4}.$

I learn about field extension , for example $\mathbb Q(\sqrt2)$ is the rational numbers and $\sqrt(2)$

But I still don't understand the meaning $F_p(\alpha)=F_{p^6}$.

Thanks for your time.

Algo
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1 Answers1

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Ok, perhaps this portend an “Okay” explanation.

Consider F2(x) = F4 means:

F2 is the base field (with 2 elements) F4 is the degree-2 extension field (with 4 elements) x is an element of F4 The smallest subfield of F4 containing x is all of F4

As, x generates F4 as an extension of F2. Only one element of F4 has this property (can generate the whole extension).

Now for the general case:

Fp is the base prime field Fp6 is the degree-6 extension α is an element of Fp6 Fp(α) = Fp6 means: the smallest subfield of Fp6 containing α is all of Fp6 In other words, α generates Fp6 as an extension of Fp

I believe the question is asking how many such "generator" elements α there can be. As I showed in my other replies, this depends on the minimum polynomial of α over Fp, and can be 1, p, p2, p3, p4, p5, or p6.

Anon Imus
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  • It is not true that only one element of $F_4$ can generate $F_4$. Also, your work will be much easier to read if you study https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Gerry Myerson Mar 08 '23 at 12:46
  • Then provide an answer yourself instead of minimizing - it’s potentially engaging and important to Algo – Anon Imus Mar 11 '23 at 13:25
  • My attempts to engage with Algo have not borne much fruit. But why are you so stubborn about not correcting the error in your answer, and about not learning how to format mathematics on this website? – Gerry Myerson Mar 11 '23 at 21:06
  • Because we all have choices. – Anon Imus Mar 12 '23 at 03:03