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.

Answer 2

I don’t know of a fully elementary solution, so here are two different solutions that use different kinds of nontrivial theory.

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Let $K$ be a number field and $p,q$ be two distinct primes with an odd $q$, $\mathfrak{q}$ any prime ideal of $\mathcal{O}_K$ above $q$, and let $L=K(\sqrt{p})$. Then $L/K$ is unramified at $\mathfrak{q}$, because it is generated by a root of the integer polynomial $X^2-p$ which is separable mod $\mathfrak{q}$.

Par induction, it follows that $\mathbb{Q}(\sqrt{p_1},\ldots,\sqrt{p_n})$ is unramified at $q$, so it cannot contain an extension ramified at $q$, such as $\mathbb{Q}(q^{1/2})$.

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Here’s another, that works in full generality:

Assume $n$ is a minimal integer that gives a counterexample. You can show by hand that $n \geq 2$. Then $K=\mathbb{Q}(\sqrt{p_1},\ldots,\sqrt{p_n})$ has degree $2^n$ over $\mathbb{Q}$ and has $\left(P_I=\prod_{i \in I}{\sqrt{p_i}}\right)_{I \subset \{1,\ldots,n\}}$ as a basis over $\mathbb{Q}$. $K$ is Galois over $\mathbb{Q}$, and by degree (or Kummer theory) its Galois group can change signs of arbitrary subsets of the $\sqrt{p_i}$.

If $\sqrt{q} \in K$, then $\sqrt{q}=\sum_I{a_IP_I}$ where the $a_I$ are rational numbers.

Clearly, there are at least some $I\neq J$ such that $a_Ia_J \neq 0$, and there is $s \in Gal(K/\mathbb{Q})$ such that $s(P_I)=uP_I$, $s(P_J)=vP_J$, where $u$ and $v$ are distinct signs.

Then the coordinates of $s(\sqrt{q})$ in the basis of the $(P_I)$ are neither those of $\sqrt{q}$, nor those of $-\sqrt{q}$, and yet $s(\sqrt{q})^2=q$ so $s(\sqrt{q})=\pm \sqrt{q}$. We get a contradiction.