Adding finite list of square roots of primes to $\mathbb{Q}$

Question

I have that idea which I'm pretty sure that is true- but I haven't succeded to prove it:
Given finite list of primes ${p_1,p_2,p_3,…,p_n}$ , and extension of the rational field with the square roots of these primes: $\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\sqrt{p_3},…,\sqrt{p_n})$, then given a prime which isn't in that list $q$,it follows that: $\sqrt{q}\notin \mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\sqrt{p_3},…,\sqrt{p_n})$.
Any hint will be great!

Answer 1

I will prove something slightly more general, by induction: if $q\ne1$ is any square-free integer coprime to $p_1,\dots,p_n$, then $\sqrt q\notin\mathbb Q(\sqrt{p_1},\dots,\sqrt{p_n})$. This is trivial when $n=0$.

Let $K=\mathbb Q(\sqrt{p_1},\dots,\sqrt{p_{n-1}})$ and suppose $\sqrt q\in K(\sqrt{p_n})=\mathbb Q(\sqrt{p_1},\dots,\sqrt{p_n})$, so $\sqrt q=a+b\sqrt{p_n}$ for $a,b\in K$.

Now, let $\sigma\in\mathrm{Gal}(K(\sqrt{p_n})/K)$ be the unique nontrivial automorphism (remember any degree-$2$ extension is Galois). Then, $\sigma(\sqrt q)=a-b\sqrt{p_n}$, but $\sigma(\sqrt q)^2=\sigma({\sqrt q}^2)=q$, so $\sigma(\sqrt q)=\pm\sqrt q$. Thus, we must have $a=0$ or $b=0$, so $\sqrt q$ or $\sqrt{qp_n}$ is in $K$. Either way, this contradicts our inductive hypothesis.