Galois group is isomorphic to the group of invertible affine transformations Let $p$ be a prime and suppose $f(x)=x^p-a$ is irreducible. Let $AGL(1,\mathbb{Z}_p)$ be the group of invertible affine transformations of $\mathbb{Z}_p$. Show that the Galois group of $f$ over $\mathbb{Q}$ is isomorphic to $AGL(1,\mathbb{Z}_p)$. Any advice on how to start to this would be wonderful.
 A: Hint. Show the splitting field of $f$ over $\mathbb{Q}$ is equal to $K=\mathbb{Q}(\sqrt[p]{a},\zeta_p)$ where $\zeta_p$ is a $p$th root of unity. 
Now suppose $\sigma\in\text{Gal}(K/\mathbb{Q})$. Then 


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*$\sigma(\sqrt[p]{a})$ must be one of the $p$ roots of $x^p-a$ (i.e., $\zeta_p^n\sqrt[p]{a}$ for some $0\leq n\leq {p-1}$)

*$\sigma(\zeta_p)$ must be one of the $p-1$ $p$th roots of unity in $K$ (i.e., $\zeta_p^r$ for some $1\leq r\leq p-1$) 
Any $\sigma$ is uniquely determined by a choice of where to send $\sqrt[p]{a}$,  and a choice of where to send $\zeta_p$.
Consider the affine transformations of $\mathbb{Z}_p$. There are the 


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*translations, which form a subgroup of $AGL(1,\mathbb{Z}_p)$ of order $p$ (the maps $f_n(a)=a+n$, one for each $n\in\mathbb{Z}_p$)

*dilations, which form a subgroup of $AGL(1,\mathbb{Z}_p)$ of order $p-1$ (the maps $g_r(a)=ra$, one for each nonzero $r\in\mathbb{Z}_p$)
Any affine transformation is uniquely determined as a combination of a translation and a dilation.
Do you see the correspondence?
(Note that, depending on what you've covered in class about these things, you may have to provide proofs that the above statements are in fact true to have a complete answer.)
