# Galois Group of $(x^2-p_1)\cdots(x^2-p_n)$

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.

• 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. – Mathmo123 Dec 10 '15 at 10:45
• 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. – Jyrki Lahtonen Dec 10 '15 at 11:14
• Possible duplicate of Generating Elements of Galois Group – Watson Aug 17 '16 at 22:06

## 1 Answer

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