Splitting field of $f=X^p -a \in \mathbb{Q}[X]$.

Question

Let $p$ be a prime number, and $a \in \mathbb{Q}$, a number such that there is no integer $k$ satisfying $p^k=a$. Write $f= X^p -a \in \mathbb{Q}[X]$. I have to prove the following statements:

1. The degree of the splitting field $\Omega/\mathbb{Q}$ equals $p(p-1)$
2. Prove that the Galois group is isomorphic to the following:

$$ \{ \left( \begin{array}{ccc} a & b \\ 0 & 1 \end{array} \right) : \ a,b \in \mathbb{F}_p \ , \ a \neq 0\} $$ My own attempts

1. I should see the extension as a double extension I guess. I thought I had to add some root $^p\sqrt{a}$ and a primitive root of unity $\zeta$. The first degree would $p$, and the second one would be $p-1$ because $\sum_{k=0}^{p-1}X^k$ is the minimal polynomial of $\zeta$. This would give the degree $p(p-1)$, right?
2. Every element $\sigma \in G$ has to map $\zeta$ to $\zeta^k$ where $1\leq k \leq p-1$. The other root $\sqrt{a}$ has to be sent to some $^p\sqrt{a}^m$, where $1 \leq m \leq p$. So I took the map.

$$ \phi : \left( \begin{array}{ccc} a & b \\ 0 & 1 \end{array} \right) \longmapsto \left\{ \begin{array}{lr} \zeta \quad \mapsto \quad \zeta^a\\ ^p\sqrt{a} \quad \mapsto \quad ^p \sqrt{a} \cdot \zeta^b \end{array} \right.$$ If we multiply to matrices we get: $$ \left( \begin{array}{ccc} a & b \\ 0 & 1 \end{array} \right) \cdot \left( \begin{array}{ccc} x & y \\ 0 & 1 \end{array} \right) = \left( \begin{array}{ccc} ax & ay+b \\ 0 & 1 \end{array} \right) $$ The upper right corner troubles me. I don't see why it doesn't work componentwise. Could someone explain me?

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Answer 1

The definition of your map $\phi$ looks dubious to me. You have to map $\gamma=\sqrt[p]a$ to another root of $X^p-a$, and $\gamma^b$ just won't do. Since the set of roots of that polynomial is $\{\,\zeta^i\gamma\mid 0\leq i<p\,\}$, it would be more appropriate to choose $\zeta^b\gamma$ as its image. Then the image of an arbitrary root will be $\phi_{a,b}(\zeta^i\gamma)=\zeta^{ia+b}\gamma$. Moreover $\phi_{a,b}(\phi_{c,d}(\zeta^i\gamma))=\zeta^{(ic+d)a+b}\gamma = \zeta^{ica+da+b}\gamma$ which is beginning to look like matrix multiplication. Can you take it from here?