For what $n$ is $$ \operatorname{Gal}(\mathbb{Q}(\sqrt{p_1},\dots,\sqrt{p_n})/\mathbb{Q}) $$ known, where the $p_i$ are primes?

By Kummer theory, I think that $$ \operatorname{Gal}(\mathbb{Q}(\sqrt{p_1},\dots,\sqrt{p_n}))/\mathbb{Q})\simeq A/(\mathbb{Q}^\times)^2 $$ where $A$ is the subgroup of $\mathbb{Q}^\times$ generated by $(\mathbb{Q}^\times)^n$ and the $p_i$. But it's not clear to me what $A/(\mathbb{Q}^\times)^n$ is. Intuitively, it should probably be $(\mathbb{Z}/2\mathbb{Z})^n$, since we should have an automorphism permuting $\pm\sqrt{p_i}$ for any choice of indices $i$, and those are the only possible ways to permute the roots. It is somehow obvious that $$ A/(\mathbb{Q}^\times)^2\simeq(\mathbb{Z}/2\mathbb{Z})^n $$ if one is thinking just in terms of groups?

  • $\begingroup$ Hint: Induct on the number of primes. $\endgroup$ – PVAL-inactive Mar 14 '14 at 23:54
  • $\begingroup$ You may find of interest the papers of Mordell and Siegel cited in my answer here. $\endgroup$ – Bill Dubuque Mar 15 '14 at 0:03

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