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For distinct prime numbers $p_1,...,p_n$, what is the Galois group of $(x^2-p_1)\cdots(x^2-p_n)$ over $\mathbb{Q}$?

This problem appears to be quite common, however my understanding of Galois theory is quite poor, and I have no idea how to do this problem.

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  • $\begingroup$ Can you deal with the case where $n=1$? What about $n=2$. Once you've tried the first few cases, you may see a pattern emerging. $\endgroup$ – Mathmo123 Dec 10 '15 at 10:45
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    $\begingroup$ See this thread for IMO a good explanation as to why the dimension (and hence also the order of the Galois group) is $2^n$. With that out of the way it is easy to prove that the Galois group is an $n$-fold cartesian power of $C_2$. See also this and this. $\endgroup$ – Jyrki Lahtonen Dec 10 '15 at 11:14
  • $\begingroup$ Possible duplicate of Generating Elements of Galois Group $\endgroup$ – Watson Aug 17 '16 at 22:06
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This question has been asked many times, and been given many answers (see the links, especially those given by Jykri). I propose here a very simple proof based on Kummer theory. I recall the setting of this theory (which can be found in any course on Galois theory): let $m$ be a fixed integer, $K$ a field of characteristic not dividing $m$, containing the group $W_m$ of all $m$th roots of $1$; let $A$ be a subgroup of $K^\ast$ containing $K^{\ast m}$ , and let $L = K(A^{1/m})$, the field obtained by adding to $K$ all the $m$th roots of all the elements of $A$. Then $L/K$ is Galois, with abelian group $G$ of exponent $m$, isomorphic to $\operatorname{Hom}(A/K^{\ast m}, W_m)$.

Here we take $K = \Bbb Q$, $m = 2$, $A_n$ = the (multiplicative) subgroup generated by $p_1$ , ..., $p_n$ and $\Bbb Q^{\ast 2}$. We want to show that $\Bbb Q((A_n)^{1/2})/\Bbb Q$ has Galois group isomorphic to $\underbrace{\Bbb Z/2\Bbb Z \times \cdots \Bbb Z/2\Bbb Z}_{\text{$n$ times}}$. By Kummer theory, this amounts to show that $A_n \mod{\Bbb Q^{\ast 2}}$ has the same description. Here we shift perspectives and use elementary linear algebra: all the previous multiplicative groups are of exponent 2, hence can be viewed as vector spaces over the field $\Bbb Z/2\Bbb Z$. Let us show that $p_1 \mod {\Bbb Q^{\ast 2}}$, ..., $p_n \mod {\Bbb Q^{\ast 2}}$ form a basis of $A_n \mod {\Bbb Q^{\ast 2}}$. Any relation of linear dependence between them, written multiplicatively, would be of the form:

A product of distinct $p_i$'s is equal to an element of $\Bbb Q^{\ast 2}$.

This is impossible by the unicity of decomposition into primes in $\Bbb Z$. Thus we have shown that $A_n \mod{\Bbb Q^{\ast 2}}$, as a vector space over $\Bbb Z/2\Bbb Z$ (written multiplicatively), has dimension $n$. QED

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