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Suppose $K_0 = \mathbb{Q}(\xi_p)$ and $K_n = \mathbb{Q}(\xi_{p^{n+1}})$. Prove that $Gal(K_n/K) = \mathbb{Z}/p^n\mathbb{Z}$.

I know that $Gal(K_0/\mathbb{Q}) = \mathbb{Z}/(p-1)\mathbb{Z}$ and thus $[K_0 : \mathbb{Q}] = p-1$. I think that $[K_n : \mathbb{Q}] = \phi(p^{n+1}) = p^{n+1} - n - 1$, which is equal to the euler phi functions, that counts the amount of co-prime numbers to $p^{n+1}$.

However, I am not sure how to continue from here on.

CasMaster5000
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    We know that ${\rm Gal}(\Bbb Q(\zeta_n)/\Bbb Q)\cong U(n)$ of order $\phi(n)$. Of course, $U(p^{n+1})\cong C_{p^n}\times C_{p-1}$. We have $\phi(p^{n+1})=p^{n}(p-1)$, see here, and not $p^{n+1}-n-1$. Then $[K_n:K_0]=\frac{p^n(p-1)}{p-1}=p^n$ by the degree theorem: $$[K_n:\Bbb Q]=[K_n:K_0]\cdot [K_0:\Bbb Q].$$ Since ${\rm Gal}(K_n/K_0)$ is cyclic, we are done. – Dietrich Burde Jun 05 '25 at 10:55
  • @DietrichBurde how do we know that $Gal(K_n/K_0)$ is cyclic? Is it because it is completely determined by what an element does to $\xi_{p^{n+1}}$? – CasMaster5000 Jun 05 '25 at 11:41
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    By the Galois correspondence the group is a subgroup of the cyclic group $U(p^{n+1})$, and hence cyclic itself. Here $U(n)=(\Bbb Z/n\Bbb Z)^{\times}$ is the unit group of the ring $\Bbb Z/n\Bbb Z$. – Dietrich Burde Jun 05 '25 at 11:54

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