Mathematical Reflections 2014, Issue 1 - Problem U294

Problem:
Let $p_1, p_2, \ldots, p_n$ be pairwise distinct prime numbers. Prove that
$$\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_n})=\mathbb{Q}(\sqrt{p_1}+\sqrt{p_2}+\ldots+\sqrt{p_n}).$$

Proposed by Marius Cavachi.

Solution:
Clearly, $\mathbb{Q}(\sqrt{p_1}+\sqrt{p_2}+\ldots+\sqrt{p_n})$ is a subfield of $\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_n})$. Observe that $\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_n})$ is Galois over $\mathbb{Q}$, since it is the splitting field of the polynomial $(x^2-\sqrt{p_1})\cdots(x^2-\sqrt{p_n})$. Every automorphism $\sigma$ is completely determined by its action on $\sqrt{p_1},\ldots,\sqrt{p_n}$, which must be mapped to $\pm \sqrt{p_1},\ldots,\pm \sqrt{p_n}$, respectively. Therefore, $\textrm{Gal}(\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_n})/\mathbb{Q})$ is the group generated by $\{\sigma_1,\sigma_2,\ldots,\sigma_n\}$, where $\sigma_i$ is the automorphism defined by $$\sigma_i(\sqrt{p_j})=\begin{cases} -\sqrt{p_j} & \textrm{if } i=j \\ \sqrt{p_j} & \textrm{if } i \neq j. \end{cases}$$ Clearly, the only automorphism that fixes $\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_n})$ is the identity. Moreover, it's easy to see that the only automorphism that fixes the element $\sqrt{p_1}+\ldots+\sqrt{p_n}$ is the identity, which means that the only automorphism that fixes $\mathbb{Q}(\sqrt{p_1}+\ldots+\sqrt{p_n})$ is the identity. Hence, by the Fundamental Theorem of Galois Theory, $\mathbb{Q}(\sqrt{p_1}+\sqrt{p_2}+\ldots+\sqrt{p_n})=\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_n})$.