Let \[ L=\QQ(\sqrt{n_1},\ldots,\sqrt{n_m}), \qquad \alpha=\sqrt{n_1}+\cdots+\sqrt{n_m}. \] Describe \[ \Gal\bigl(L/\QQ(\alpha)\bigr). \] What happens when the square classes of the \(n_i\) are linearly independent in \(\QQ^\times/(\QQ^\times)^2\)?
Solution
The extension \(L/\QQ\) is multiquadratic. Let \(r\) be the dimension over \(\FF_2\) of the subgroup of \(\QQ^\times/(\QQ^\times)^2\) generated by the square classes of \(n_1,\ldots,n_m\). Then \[ \Gal(L/\QQ)\cong (C_2)^r. \] Every automorphism is determined by compatible sign changes of the square roots.
By the fundamental theorem of Galois theory, \[ \Gal(L/\QQ(\alpha)) = \{\sigma\in\Gal(L/\QQ):\sigma(\alpha)=\alpha\}. \] Thus the answer in complete generality is the stabilizer of the sum \[ \alpha=\sum_{i=1}^m\sqrt{n_i} \] under the sign-change action.
Suppose now that the square classes of \(n_1,\ldots,n_m\) are linearly independent in \[ \QQ^\times/(\QQ^\times)^2. \] Then \(r=m\), and \[ 1,\sqrt{n_1},\ldots,\sqrt{n_m} \] are linearly independent over \(\QQ\) as part of the standard multiquadratic basis. If \[ \sigma(\sqrt{n_i})=\varepsilon_i\sqrt{n_i}, \qquad \varepsilon_i\in\{\pm1\}, \] and \(\sigma(\alpha)=\alpha\), then \[ \sum_{i=1}^m(\varepsilon_i-1)\sqrt{n_i}=0. \] Linear independence forces every \(\varepsilon_i=1\). Hence the stabilizer is trivial: \[ \Gal(L/\QQ(\alpha))=1. \] Therefore \[ \QQ(\alpha)=L \] in the square-class-independent case.
When square-class relations are present, the stabilizer can be nontrivial; the displayed stabilizer formula is the exact general answer.