Generating Elements of Galois Group

Solution 1:

The crucial point is: It is not trivial at all to see that $\mathbb{Q}(\sqrt{p_1},\dotsc,\sqrt{p_n})$ has degree exactly $2^n$ (only $\leq 2^n$ is clear) over $\mathbb{Q}$, or equivalently, that $p_i \notin \mathbb{Q}(\sqrt{p_1},\dotsc,\sqrt{p_{i-1}})$ for all $i \leq n$.

Here is what you can do: Let $m_1,\dotsc,m_r$ be coprime and square free numbers $>1$ (for example distinct primes). Then I claim that $K_r := \mathbb{Q}(\sqrt{m_1},\dotsc,\sqrt{m_r})$ has degree $2^r$ over $\mathbb{Q}$.

It is enough to prove $\sqrt{m_{i+1}} \notin K_i$ for all $i < r$. But it turns out one should show something stronger: For $i<r$ and $i < j_1 < \dotsc < j_s \leq r$ we have $\sqrt{m_{j_1} \cdot \dotsc \cdot m_{j_s}} \notin K_i$. Now this can be shown by induction on $i$; I leave it for you as an exercise.

Now the rest is easy: $K_r$ is the splitting field of $(x^2-m_1) \cdot \cdots \cdot (x^2-m_r)$, hence normal, and everything in characteristic $0$ is separable. Thus $K_r$ is Galois over $\mathbb{Q}$; let $G$ be the Galois group. Since the conjugates of $\sqrt{m_i}$ are $\pm \sqrt{m_i}$, there is a homomorphism $\alpha : G \to \{\pm 1\}^r$ given by $\alpha(g) \sqrt{m_i} = g(\sqrt{m_i})$. Clearly it is injective. Since $G$ has order $2^r$, it is also surjective. Thus we have an isomorphism $G \cong \{\pm 1\}^r$. Explicitly, the Galois automorphisms are exactly as you've described them, they are given by $\sqrt{m_i} \mapsto \varepsilon_i \sqrt{m_i}$, where $\varepsilon_i = \pm 1$.

Remark: There are shorter proofs, using Kummer Theory.