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Let $p$ be a prime number and $\alpha\in\mathbb{N}$ such that $\forall\beta\in\mathbb{Q} \space\alpha\neq \beta^p $ e.g. $\alpha$ is not a $p$-th power of any rational number. Let $E$ denote the splitting field of the polynomial $x^p-\alpha$.

Prove:

$Gal(E/\mathbb{Q}) \cong G $ where we define $G$ as the group:

$G =\{\begin{pmatrix} a & b\\ 0 & 1 \end{pmatrix} s.t. \space a,b\in \mathbb{F}_p \space \wedge a\neq0 \}$

My attempt: denote the relevant Galois group as $H$, so it is obvious that $H$ is of order $p(p-1)$ because we can write $E / \mathbb{Q}$ as $ \mathbb{Q}(\zeta_p, \sqrt[p]{\alpha})/ \mathbb{Q} $. This is not a proof but rather intuition why the isomorphism might hold. Then I tried to explicitly define the isomorphism. So, we have to map every $\sigma\in H$ to a matrix in $G$. The mapping that seemed natural to me is the following:

Denote the roots of $x^p-\alpha$ as $\{x_1,...,x_p\}$, then we map an autormphism to a matrix by:

if $\sigma(\zeta_p) = x_i$ and $\sigma(\sqrt[p]{\alpha}) = x_j$ then $\sigma \mapsto M$ where we have for $M: a=i \space\wedge\space b=j\space$ e.g. we map the automorphism (enough to define on the generators of the extension, I think) to a matrix in $G$ correlated to the way the automorphism acts on both generators. However, this does not work and I'm stuck. Any clues, ideas or insight would be greatly appreciated.

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I'll write $a$ instead of $\alpha$. The roots of $x^p-a$ are $\sqrt[p]{a}e^{\frac{2\pi ik}{p}}$ where $0\leq k\leq p-1$. Each element $\varphi\in Gal(E/\mathbb{Q})$ is uniquely determined by the images of $\sqrt[p]{a}$ and $e^{\frac{2\pi i}{p}}$ because they generate the field extension. What might the images be? $e^{\frac{2\pi i}{p}}$ is the root of $x^p-1$, hence $\varphi$ must send it to a root of this polynomial, though not to $1$. (since $\varphi$ is injective and $\varphi(1)=1$). So $\varphi(e^{\frac{2\pi i}{p}})=e^\frac{2\pi ij}{p}$ for some $1\leq j\leq p-1$. Now what is the image of $\sqrt[p]{a}$? It is the root of $x^p-a$, hence $\varphi(\sqrt[a]{p})=\sqrt[p]{a}e^{\frac{2\pi ik}{p}}$ for some $0\leq k\leq p-1$. And now define your isomorphism between $Gal(E/\mathbb{Q})$ and $G$ by sending such $\varphi$ to the matrix $\begin{pmatrix} j & k\\ 0 & 1 \end{pmatrix} $.

The map is injective because the images of $\sqrt[p]{a}$ and $e^{\frac{2\pi i}{p}}$ uniquely determine the element of $Gal(E/\mathbb{Q})$. It is onto because it is an injective map between finite sets of the same size. Finally, check that this map is indeed a group homomorphism. I'll leave it to you.

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  • $\begingroup$ Thanks! Though I still wonder if this can be proven without explicitly defining the isomorphism.. $\endgroup$ – Adar Gutman May 10 at 19:53

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