# 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.

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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

• $\sigma(\sqrt[p]{a})$ 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

• 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.)

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@ Zev..I kind of see the correspondence..I just don't know quite how to put everything together. Intuitively I feel like the galois group is going to be some kind of semi-direct product. Of G, where G is the galois group and some subgroup H. But I don't know quite how to wrap my fingers around this yet. – user8771 Apr 29 '11 at 1:02
For each $0\leq n\leq p-1$, define $\sigma_n\in\text{Gal}(K/\mathbb{Q})$ by $\sigma_n(\sqrt[p]{a})=\zeta_p^n\sqrt[p]{a}$ and $\sigma_n(\zeta_p)=\zeta_p$. The subgroup $\{\sigma_0=\text{id},\sigma_1,\ldots,\sigma_n\}$ of $\text{Gal}(K/\mathbb{Q})$ is of order $p$. For each $1\leq r\leq p-1$, define $\tau_r\in\text{Gal}(K/\mathbb{Q})$ by $\tau_r(\sqrt[p]{a})=\zeta_p^r\sqrt[p]{a}$ and $\tau_r(\zeta_p)=\zeta_p^r$. The subgroup $\{\tau_1=\text{id},\tau_2,\ldots,\tau_{p-1}\}$ of $\text{Gal}(K/\mathbb{Q})$ is of order $p-1$. – Zev Chonoles Apr 29 '11 at 1:09
@user8771: Also, you are correct that it is a semidirect product, but it would not make sense for a group to be isomorphic to a semidirect product of itself with something else. Here is something to work out that may help: What is $\#\text{Gal}(K/\mathbb{Q})$? – Zev Chonoles Apr 29 '11 at 1:11
@ Zev.. Yes..my mistake.I see the end result in my head, I know there will be an intersection between the two groups, which is 1. But now I just have to figure out how to piece everything together. Much thanks for your help. I hate when I have the right notion, but I cannot put it together right away. – user8771 Apr 29 '11 at 1:20
@user8771: So in $\text{Gal}(K/\mathbb{Q})$, we have a subgroup of order $p$ and a subgroup of order $p-1$, and together they generate all of $\text{Gal}(K/\mathbb{Q})$ (you would have to prove this; use the knowledge of what $\#\text{Gal}(K/\mathbb{Q})$ is). Also in $AGL(1,\mathbb{Z}_p)$ we have a subgroup of order $p$ and a subgroup of order $p-1$, and together they generate all of $AGL(1,\mathbb{Z}_p)$. Try figuring out how to map these two subgroups of $\text{Gal}(K/\mathbb{Q})$ to the corresponding subgroups of $AGL(1,\mathbb{Z}_p)$, and then extending this to a map to the whole thing. – Zev Chonoles Apr 29 '11 at 1:36