Hey I want to check if my solutions for this exercise are right.
Let $E = \mathbb{Q}( \sqrt{p_1}, . . . , \sqrt{p_n})$ with $n$ pairwise different prime numbers $p_1, . . . , p_n$. Show that $E/\mathbb{Q}$ is Galois and that $Gal(E/\mathbb{Q}) \cong (\mathbb{Z}/2\mathbb{Z})^n$.
Every minimal polynomial is normal and separable so we have that $E/\mathbb{Q}$ is Galois.
What I thought is to use this Theorem that we have done in class:
Let $p_1 < p_2 < p_3 < p_4 < . . .$ the sequence of prime numbers. For every $n ∈ \mathbb{N}$ we have $[\mathbb{Q}( \sqrt{p_1}, . . . , \sqrt{p_n}) : \mathbb{Q}] = 2^n$
So I have that $[E/\mathbb{Q}]=[\mathbb{Q}( \sqrt{p_1}, . . . , \sqrt{p_n}) : \mathbb{Q}( \sqrt{p_1}, . . . , \sqrt{p_{n-1}})]\cdot[\mathbb{Q}( \sqrt{p_1}, . . . , \sqrt{p_{n-1}}) : \mathbb{Q}( \sqrt{p_1}, . . . , \sqrt{p_{n-2}})]\cdot[\mathbb{Q}( \sqrt{p_1}) : \mathbb{Q}]=2^n$
Where every $[\mathbb{Q}( \sqrt{p_1}, . . . , \sqrt{p_{n-1}}) : \mathbb{Q}( \sqrt{p_1}, . . . , \sqrt{p_{n-2}})]=2$ and therefore every $|Gal(\mathbb{Q}( \sqrt{p_1}, . . . , \sqrt{p_{n-1}}) / \mathbb{Q}( \sqrt{p_1}, . . . , \sqrt{p_{n-2}}))|=2$ and $\cong \mathbb{Z}/2\mathbb{Z}$
we have $Gal(E/K)= Gal([\mathbb{Q}( \sqrt{p_1}, . . . , \sqrt{p_n})/ \mathbb{Q}( \sqrt{p_1}, . . . , \sqrt{p_{n-1}})])\cdot...\cdot Gal(\mathbb{Q}( \sqrt{p_1}) / \mathbb{Q}) \cong (\mathbb{Z}/2\mathbb{Z})^n$
Am I right or have I done any mistake?