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Consider the Galois extension $E=\mathbb{Q}(\sqrt{p_1}, \sqrt{p_2},\ldots)$ of $\mathbb{Q}$ where $p_1<p_2<\cdots$ are positive primes.

Then $\mbox{Gal}(E/\mathbb{Q})$ is $(\mathbb{Z}/2)\times (\mathbb{Z}/2)\times \cdots$ (countable infinite direct product).

This group has uncountable subgroups of order $2$.

Q. Can we explicitly find a subgroup of order $2$ which is not equal to $\mbox{Gal}(E/F)$ for some $\mathbb{Q}\subset F \subset E$ (and $[E:F]=2$)?

Any automorphism in $\mbox{Gal}(E/\mathbb{Q})$ will take $\sqrt{p}_i$ to $\pm \sqrt{p_i}$. I took the automorphism $\sigma$ which takes $\sqrt{p}_i$ to $-\sqrt{p_i}$ for all $i$. Then I thought that if $F=\mathbb{Q}(\sqrt{p_ip_{i+1}}:i=1,2,\ldots)$ then $[E:F]=2$ and this is fixed field of $\langle\sigma\rangle$. So $\langle\sigma\rangle$ corresponds to some intermediate field.

Maths Rahul
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1 Answers1

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The main theorem of infinite Galois theory states that if $L/K$ is a Galois extension, then $E \mapsto \mathrm{Gal}(L/E)$ provides an order-reversing bijection between the intermediate fields of $L/K$ and the closed subgroups of the topological group $\mathrm{Gal}(L/K)$. Moreover, $\mathrm{Gal}(L/K)$ is a Hausdorff topological group, so that in particular all finite subgroups $H$ are closed, and they have the form $\mathrm{Gal}(L/E)$ for some finite $L/E$ with $[L:E] = \mathrm{ord}(H)$. Namely, we can take $E = L^H$.

This answers your special case affirmatively.

Martin Brandenburg
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