$\text{Gal}(\Bbb Q(\sqrt{p_1},...,\sqrt{p_n})/\Bbb Q)=(\Bbb Z/2\Bbb Z)^n$

Question

Let $p_1,...,p_n\in\Bbb N$ be distinct primes. Compute $\text{Gal}(\Bbb Q(\sqrt{p_1},...,\sqrt{p_n})/\Bbb Q)$

Intuitively, $G:=\text{Gal}(\Bbb Q(\sqrt{p_1},...,\sqrt{p_n})/\Bbb Q) = (\Bbb Z/2\Bbb Z)^n$. My first idea is that if $\sigma\in G$, then $\sigma(\sqrt{p_i})=\{\sqrt{p_i},-\sqrt{p_i}\}$ for each $i$. Hence, from this, pick $\sigma_i\in G$ such that $\sigma_i(\sqrt{p_i})= -\sqrt{p_i}$ and $\sigma_i(\sqrt{p_j}) = \sqrt{p_j}$ for all $j\neq i$. Then, $\sigma_i$'s generates $G$ and each $\sigma_i$ does not generates $\sigma_j$ if $i\neq j$, $G = (\Bbb Z/2\Bbb Z)^n$. Is this valid?

Another way to do this is using the fact that 'if $K/F,L/F$ are both Galois and $K\cap L =F$ then $\text{Gal}(KL/F)\simeq\text{Gal}(K/F)\times\text{Gal}(L/F)$', we inductively apply this. The problem is whether $\sqrt{p_i}\in\Bbb Q(\sqrt{p_1},...,\sqrt{p_{i-1}})$. Any hint or comment will be appreciated.

Answer 1

Every element of that Galois group is $\pm\sqrt{p_i}$ for every $p_i$. There are $n$ $p_i$'s and we have choice of $2$ values on every of them. Hence we have $2^n$ automorphisms in that Galois group. because every of them is of order $2$, this gorup is ismorphic to product of $n$ copies of $\mathbb{Z}/2\mathbb{Z}$ (You should earlier prove that for every $j+1$, $\sqrt{p_{j+1}}$ is not contained in $\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_j})$.)