2
$\begingroup$

I tried to solve the following problem, but I couldn't; I hope can you help me.

Let $G$ the matrix group $G=\left.\left\{ \left(\begin{array}{cc} a & b \\ 0 & 1 \end{array}\right) \;\right|\; a,b \in \mathbb{F}_p, a \neq 0 \right\},$ with $p$ prime. Build a Galois extension $K$ of $\mathbb{Q}$ such that $\mathrm{Gal}(K/\mathbb{Q}) \cong G$. In addition, , determine the intermediate extension for the subgroups $H_1=\left.\left\{ \left(\begin{array}{cc} a & 0 \\ 0 & 1 \end{array}\right) \;\right|\; a \in \mathbb{F}_p, a \neq 0 \right\},$ and $H_2=\left.\left\{ \left(\begin{array}{cc} 1 & b \\ 0 & 1 \end{array}\right) \;\right|\; b \in \mathbb{F}_p, a \neq 0 \right\}.$

Thanks for you answer.

  • 1
    Dear @yoyo: Your comment looks like an answer...2012-01-03

1 Answers 1

5

I'm ripping this from Lang's "Algebra" ch 6 sec 9:

Let $p$ be an odd prime and consider the splitting field $K$ of the irreducible polynomial $x^p-a$, with $a$ not a $p$th power. Let $\alpha$ be a root and $\zeta$ a primitive $p$th root of unity. We see that $[\mathbb{Q}(\zeta):\mathbb{Q}]=p-1$ and $[\mathbb{Q}(\alpha):\mathbb{Q}]=p$ are coprime. Hence $K=\mathbb{Q}(\alpha,\zeta)$ has degree $p(p-1)$ over $\mathbb{Q}$.

For $\sigma\in G=G_{K/\mathbb{Q}}$, we have $\sigma(\alpha)=\alpha\zeta^{b(\sigma)}$ and $\sigma(\zeta)=\zeta^{d(\sigma)}$ for some $b\in\mathbb{F}_p, d\in\mathbb{F}_p^{\times}$. Composition gives $\sigma\tau(\alpha)=\alpha\zeta^{d(\sigma)b(\tau)+b(\sigma)}$ and $\sigma\tau(\zeta)=\zeta^{d(\sigma)d(\tau)}$. We can write this composition down as matrix multiplication under the correspondence you gave $ \sigma\mapsto \left( \begin{array}{cc} d(\sigma)&b(\sigma)\\ 0&1\\ \end{array} \right) $ whether or not it is obvious that this produces a bijection between $G$ and our matrix group, we have an injective homomorphism ($b$ and $d$ are unique) between groups of the same order $p(p-1)$, hence an isomorphism.

I'll leave the fixed fields of $H_1,H_2$ to you.