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Let $\mathbb{E} = \mathbb{Q}[\sqrt{p_1}, \dots , \sqrt{p_n}]$, with $p_1, \dots, p_n$ distinct prime integers. Show that the group of Galois of the extension $G = Gal(\mathbb{E} | \mathbb{Q})$ has at least $2^n - 1$ subgroups of index $2$.

The extension $\mathbb{E} | \mathbb{Q}$ is Galois of degree $2^n$ which is equal to the order of $G$. I can see, for any $i = 1, \dots , n$, $Gal(\mathbb{E} | \mathbb{Q}[\sqrt{p_i}])$ being a subgroup of order $2^{n-1}$ of $G$, thus of index $2$ for Lagrange. However it's far away from the $2^n-1$ claimed in the text. Any hint about it?

Ale Dalba
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  • What about $\mathbb{Q}(\sqrt{p_1 p_2})$ etc etc etc – ancient mathematician Sep 18 '22 at 14:59
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    The Galois group is an elementary abelian group of order $2^n$. So the claim follows, see this post, or similar ones. – Dietrich Burde Sep 18 '22 at 15:03
  • Elaborating on @ancientmathematician's comment, Pick a nonempty subset $I={i_1,\dots,i_k}\subset{1,2,\dots,n}$ and consider $\mathbb Q[\sqrt{p_{i_1}\cdots p_{i_k}}]$. – Kenta S Sep 18 '22 at 15:25
  • I gave one argument of exactly this claim in my answer to the question Dietrich linked to. Is there something specific you want to ask about the argument? There I used the "inner product" on a vector space over $\Bbb{F}_2$ to tally the subgroups of index two. I'm sure there are other ways of seeing that. – Jyrki Lahtonen Sep 18 '22 at 15:59

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The Set of subgroups of index $2$ is in bijection with the set of degree-2 extensions $L/\Bbb Q$ such that $\Bbb Q \subset L \subset \Bbb Q[\sqrt p_1, \dots, \sqrt p_n]$. Any (non-empty) choice of a subset $I \subset \{1, \dots, n\}$ yields such a degree 2 extension via $L = \Bbb Q[\sqrt \pi]$, where $\pi = \prod_{i \in I} p_i$. On the other hand, given such an extension $L$, let $D$ be the discriminant of $L/\Bbb Q$. Now we can reconstruct $\pi$ via $$ \pi = \begin{cases} \prod_{p \mid D} p \quad & \text{if } 4 \nmid D \\ \prod_{4p \mid D} p \quad & \text{if } 4 \mid D.\end{cases} $$ By theory of quadratic number fields we find $L = \Bbb Q[\sqrt \pi]$, and as $L \subset \Bbb Q[\sqrt p_1, \dots, \sqrt p_n]$ we find that $\pi$ only has prime divisors among the $p_i$ (hence is of the form $\pi = \prod_{i \in I} p_i$ for some subset $I$).

LurchiDerLurch
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