Sum of squareroots of primes under Galois action

Question

Let $p_1, ..., p_r$ be distinct prime numbers, $L = \mathbb{Q}(\sqrt{p_1}, ..., \sqrt{p_r})$ and $\alpha = \sqrt{p_1} + ... + \sqrt{p_r}$. Is there any way to proof, that $$ \{ \sigma(\alpha) \ | \ \sigma \in \text{Gal}(L/\mathbb{Q}) \}$$ is contained in $\mathbb{Q}(\alpha)$ without showing, that $L = \mathbb{Q}(\alpha)$?

Answer 1

If the set of $\alpha$'s conjugates in $\mathbb{C}$ is $A=\{\alpha_1= \alpha, \alpha_2, \dots, \alpha_n\}$, it suffices to show that $\mathbb{Q}(A) = \mathbb{Q}(\alpha)$. Since the former is a splitting field, it is Galois, with Galois group $G$.

Now note that if all of $G$'s subgroups are normal, then, in particular, $\mathbb{Q}(\alpha)$ is normal over $\mathbb{Q}$, therefore it contains all of $\alpha$'s conjugates and the result follows.

Since $L/\mathbb{Q}$ is normal $L$ and contains $\mathbb{Q}(A)$, $G$ is a subgroup of $\text{Gal}(L/\mathbb{Q}) \approx \mathbb{Z}_2^r$, and since $\mathbb{Z}_2^r$ is abelian, all of its subgroups are normal, including those of $G$.

In general, this argument works for any $\alpha \in L$, for example $\alpha = \sqrt{p_1}\sqrt{p_3}+\sqrt{p_2}$.