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I have a question about finite field extensions over finite field $\mathbb{F}_p$ with characteristic $p > 0$:

It is well known from algebra that every finite field extension $\mathbb{F} \ | \ \mathbb{F}_p$ is exactly the field $\mathbb{F}_q$ where $q= p^n$ for $n=[\mathbb{F} \ | \ \mathbb{F}_p]$ or in other words $\mathbb{F}$ is the splitting field of the polynomial $X^q-X$.

Futhermore $\mathbb{F}$ has Galois Group isomorphic to cyclic group $\mathbb{Z}/n$ with generator $\sigma: a \to a^p$.

My questions:

  1. is what happens with the field extension $\mathbb{F}_p(\zeta_n) \ | \ \mathbb{F}_p$ where $p \nmid n$ and $\zeta_n$ is the n-th primitive root?

Obviously here $\mathbb{F}_p(\zeta_n)$ is the splitting field of $X^n-1$ but because $n \neq p^r$ by assumpting this contradicts the considerations above, doesn't it?

  1. What Galois Group $Gal(\mathbb{F}_p(\zeta_n) \ | \ \mathbb{F}_p)$ has $\mathbb{F}_p(\zeta_n)$ and why?
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    $\begingroup$ Do you actually mean $ X^n-1 $? $\endgroup$ – Arnaud Mortier Feb 6 '18 at 21:21
  • $\begingroup$ Yes sorry, you are right. $\endgroup$ – KarlPeter Feb 6 '18 at 21:34
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The multiplicative group of a field $\mathbb{F}_q$ is cyclic of order $q-1$. So for $p\nmid n$, there is a primitive $n$th root of unity in $\mathbb{F}_q$ iff $n$ divides $q-1$. This means $\mathbb{F}_p(\zeta_n)$ is just $\mathbb{F}_q$ where $q$ is the least power of $p$ such that $n$ divides $q-1$. If $q=p^r$, its Galois group is then cyclic of order $r$. (This $r$ can also be described as the least $r>0$ such that $p^r$ is $1$ mod $n$; that is, the multiplicative order of $p$ mod $n$.)

Obviously here $\mathbb{F}_p(\zeta_n)$ is the splitting field of $X^n-X$ but because $n \neq p^r$ by assumpting this contradicts the considerations above, doesn't it?

This is no contradiction. A field extension can be the splitting field of many different polynomials. Every finite extension of $\mathbb{F}_p$ is the splitting field of a polynomial of the form $X^{p^r}-X$, but they can also be described as splitting fields of polynomials that are of different forms.

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  • $\begingroup$ And this would imply $\mathbb{F}_p(\zeta_n) = \mathbb{F}_p(\zeta_{q-1}) = \mathbb{F}_q$? But then $\zeta_{q-1}$ and $\zeta_{n}$ are both primitive $n$-th AND $q-1$-th roots of unity, right? Especially there would be exist a $s$ coprime to $q-1$ with $\zeta_{q-1}^s =\zeta_{n}$. Then, if I consider $\zeta_{q-1} =e^{2 \pi i /{q-1}}$ and $\zeta_{n} =e^{2 \pi i /{n}}$, we get $\zeta_{q-1}^s =e^{2 \pi i s/{q-1}} = e^{2 \pi i /{n}}$, so $q-1/s = n$ in contradiction that $s$ and $q-1$ are coprime. What is my error in reasoning? $\endgroup$ – KarlPeter Feb 6 '18 at 23:53
  • $\begingroup$ Why would $s$ and $q-1$ be coprime? $\endgroup$ – Eric Wofsey Feb 7 '18 at 0:01
  • $\begingroup$ $\zeta_{q-1}$ is a primitive $(q-1)$st root of unity, and $\zeta_n$ is a primitive $n$th root of unity. There is absolutely no reason that $\zeta_{q-1}$ would also be a primitive $n$th root of unity, or that $\zeta_n$ would also be a primitive $(q-1)$st root of unity. $\endgroup$ – Eric Wofsey Feb 7 '18 at 0:03
  • $\begingroup$ Because $\zeta_{q-1}, \zeta_{n}$ must be $q-1$-th roots of unity and all primitive $q-1$-th roots of unity have the shape $\zeta_{q-1}^t$ with $t$ coprime to $q-1$. Or am I wrong? $\endgroup$ – KarlPeter Feb 7 '18 at 0:06
  • $\begingroup$ $\zeta_n$ is a $(q-1)$st root of unity, but probably not a primitive one. So it has the form $\zeta_{q-1}^t$ for some $t$, but $t$ is probably not coprime to $q-1$. $\endgroup$ – Eric Wofsey Feb 7 '18 at 0:07
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The polynomial $X^n-X$ is not irreducible over $\mathbb{F}_p$, you have to find the minimal polynomial of $\zeta_p$ to determine the degree of the extension.

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  • $\begingroup$ Do you mean $\zeta_n$? $\endgroup$ – KarlPeter Feb 6 '18 at 21:33
  • $\begingroup$ yes, I mean that. $\endgroup$ – Tsemo Aristide Feb 6 '18 at 21:36

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