Prove that a Galois group is isomorphic to matrix group This was a problem in a test:
Prove that the Galois group of $x^p-2$ (where $p$ is a prime)  over $\mathbb Q$ isomorphic to the following matrix group under multiplication:
$$\begin{pmatrix}
a & b \\
0 & 1\\
\end{pmatrix}$$
where $a \in \mathbb F_{p}$ and $b \in \mathbb F_{p} \backslash \{0\}.$
I don't know why the Galois group is not simply the order $p$ cyclic group (since the roots of the polynomial over $\mathbb C$ are $\sqrt[p]{2}\epsilon_{i}$ where the $\epsilon_{i}$ are the $p$-th roots of unity),
Can someone help me?
 A: Let $\zeta = e^{\frac{2\pi i}p}$, a primitive $p$-th root of unity and let $K$ be the splitting field of $x^p-2$ over $\mathbb Q$. Then claim that $K = \mathbb Q(\sqrt[p]2, \zeta)$ and $[K:\mathbb Q] = p(p-1)$. Since $x^p -2$ is an separable polynomial over $\mathbb Q$ (why?) it follows that $K/\mathbb Q$ is Galois and so the Galois group has order $p(p-1)$. Following the comments one can show that $Gal(K/\mathbb Q)$ is isomorphic to the matrix group you state above, which is also isomorphic to the nonabelian group $C_p \rtimes C_{p-1}$
Proof of claim: As you said $\sqrt[p]2\zeta^k$ for $k = 0, 1, \dots, p-1$ are the roots of $x^p - 2$; so clearly $K \subset \mathbb Q(\sqrt[p]2, \zeta)$. Now note that $\sqrt[p]2 \in K$ and also $\zeta = \frac{(\sqrt[p]2\zeta)^{p+1}}{2\sqrt[p]2} \in K$; hence $\mathbb Q(\sqrt[p]2, \zeta) \subset K$. Now note that $K$ contains the subfields $\mathbb Q(\sqrt[p]2)$ and $\mathbb Q(\zeta)$, which have degree $p$ and $p-1$ resp. over $\mathbb Q$. Since $\gcd(p, p-1) = 1$, by a basic field theory result it follows that the compositum field $\mathbb Q(\sqrt[p]2)\mathbb Q(\zeta) = K$ has degree $p(p-1)$ over $\mathbb Q$. $\square$
