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I have a question about the structure of this Galois group that I can't understand: suppose that $p>2$ is prime and $q$ is any power of p, and we have these two function fields: $$K=\mathbb{F}_{p}(t^{1/d^{'}},\mu_{d^{'}})$$ $$F=\mathbb{F}_{q}(t^{1/d})$$ where $\mu_{d^{'}}$ is the $d^{'}$-th root of unity and $d$ is some integer that divides $d^{'}=p^{f}+1$ (f is a non-negative integer).

I know that $K_{d^{'}}/F$ is a Galois extension with this Galois group: $$Gal(K/F)\cong(d\mathbb{Z}/d^{'}\mathbb{Z})\rtimes \langle Fr_{q}\rangle.$$ ($\langle Fr_{q}\rangle$ is a cyclic group with order $[\mathbb{F}_{q}(\mu_{d^{'}}):\mathbb{F}_{q}]$ generated by $q$-power Frobenius)

but I can't understand how?

Soroush Salehin
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    Is it not $\langle Fr_q \rangle$ but $\langle Fr_p \rangle / \langle F_p^{2f}\rangle$ the Galois group of $\Bbb{F}p(t^{1/d"},\mu{p^f+1})/\Bbb{F}p(t^{1/d"})$ isomorphic to that of $\Bbb{F}_p(\mu{p^f+1})/\Bbb{F}p=\Bbb{F}{p^{2f}}/\Bbb{F}_p$ – reuns Feb 12 '22 at 17:26
  • @reuns I read this from this paper https://arxiv.org/abs/1002.3313 page 7. writer used Frobenius q. – Soroush Salehin Feb 13 '22 at 05:16
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    Let $q=p^m$, $L=\Bbb{F}_p(t)$, $\zeta$ a primitive root of unity of order $d'$ and $u=t^{1/d'}$. Then $K=L(u,\zeta)$ is the splitting field of $x^{d'}-t\in L[x]$ over $L$. The Galois group $Gal(K/L)$ is generated by the cyclic permutation of roots $\sigma: u\mapsto \zeta u, \zeta\mapsto \zeta$, and the Frobenius $\phi:u\mapsto u, \zeta\mapsto \zeta^p$. We have the relation $\phi\sigma\phi^{-1}=\sigma^p$ implying that we have a semidirect product $$Gal(K/L)=\langle \sigma\rangle \rtimes \langle \phi\rangle.$$ The first factor has order $d'$ and the latter has order $2f$. – Jyrki Lahtonen Feb 13 '22 at 10:56
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    (cont'd) $F$ is a subfield of $K$ if and only if $m\mid 2f$. We see that $\sigma^{d'/d}$ fixes $F$ pointwise as does $\phi^{m}$. It seems clear to me that these two powers also generate the Galois group $Gal(K/F)$. Hence $$Gal(K/F)\simeq C_{d'/d}\rtimes C_{2f/m}.$$ – Jyrki Lahtonen Feb 13 '22 at 11:02
  • @JyrkiLahtonen Thank you so much, it was very helpful for me. – Soroush Salehin Feb 13 '22 at 19:07
  • Glad to hear that! If you want to post a detailed answer, filling in the holes I left in the comments, that would be welcome. You will then get more feedback on the details. – Jyrki Lahtonen Feb 13 '22 at 19:59
  • @JyrkiLahtonen sure. just two questions: why is the Frobenius map acting trivially on $u=t^{1/d^{'}}$ ? and can you refer me to a reference to complete the proof of semidirect product? – Soroush Salehin Feb 14 '22 at 07:03
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    Any element of the Galois group must map $u$ to some other element $z$ such that $z^{d'}=t$. The alternatives for $z$ are $u\zeta^j$ for some $j$. But the other automorphism $\sigma$ takes care of the possibilities $j\neq0$. So it is simply a natural choice to define $\phi(u)=u$ to get the other generator of the Galois group. – Jyrki Lahtonen Feb 22 '22 at 07:18
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    This and this older question try to explain where the semidirect product comes from. There $2$ plays the role of $t$ and $d$ happens to be a prime. But you do get the extension field by adjoining an appropriate root of $2$ and an appropriate root of unity, so it is very similar. – Jyrki Lahtonen Feb 22 '22 at 07:31
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    Here the order of the root is not a prime (does not change much). – Jyrki Lahtonen Feb 22 '22 at 07:38

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