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hbghlyj
Post time 2024-4-10 21:08
$\text{Gal}(\Bbb Q(\sqrt{p_1},...,\sqrt{p_n})/\Bbb Q)=(\Bbb Z/2\Bbb Z)^n$ for distinct prime numbers $p_1,\dots,p_n$
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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青青子衿
Post time 2024-4-10 03:04
