# Proving an isomorphism of Galois group

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.

## 1 Answer

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.

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