$\sqrt{p_1}$ is not in $Q[\sqrt{p_2},...,\sqrt{p_n}]$

Question

How to show $\sqrt{p_1}$ is not in $Q[\sqrt{p_2},...,\sqrt{p_n}]$ if $p_1,...,p_n$ are distinct primes? Intuitively, this is pretty clear, but it makes me very uncomfortable to just believe. Any idea to prove this rigorously? I want this result because I am trying to compute the Galois group of $(X^2-p_1)...(X^2-p_n)$. If I know the statement is true, then the Galois group of this polynomial will be direct product of separate Galois group.

This question has several answers, including a few links that might help you : http://math.stackexchange.com/questions/30687/the-square-roots-of-different-primes-are-linearly-independent-over-the-field-of — Arnaud D., Aug 22 '16 at 19:29

score 0 · Accepted Answer · answered Aug 23 '16 at 00:03

You may also go through the following lines: by quadratic reciprocity and Dirichlet's theorem, there is some uber-huge prime $P$ for which $p_2,p_3,\ldots,p_n$ are quadratic residues, while $p_1$ is not. It follows that the algebraic numbers $\sqrt{p_1}$ and $\sqrt{p_2}+\ldots+\sqrt{p_n}$ have different degrees over $\mathbb{F}_P$ ($2$ and $1$, respectively), so they cannot be linearly dependent over $\mathbb{Q}$.

$\sqrt{p_1}$ is not in $Q[\sqrt{p_2},...,\sqrt{p_n}]$

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