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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 $L = \mathbb{Q}(\alpha)$?

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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 $\alpha = \sqrt{p_1}\sqrt{p_3}+\sqrt{p_2}$.

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    Yeah thats a nice observation, in an Abelian extension, all elements are normal. One can go a step further in this case an note that $\sqrt{p_1}+\cdots +\sqrt{p_r}$ has $2^r$ conjugates and thus is equal to the whole extension. – 2017-01-27
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    Yes, normally that would be the standard solution, however the OP asked for a solution without showing that $\mathbb{Q}(\alpha)$ is equal to the whole extension. @ReneSchipperus – 2017-01-27
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    Granted. But what is the standard solution ? Is there some other way to see equality of the extensions ? – 2017-01-27
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    Yes. Since $\mathbb{Q}(\sqrt{p_1}, \dots, \sqrt{p_{r-1}})(\sqrt{p_r}):\mathbb{Q}(\sqrt{p_1}, \dots, \sqrt{p_{r-1}})=2$, there is $\sigma$ in Gal$(L/\mathbb{Q})$ such that $\sigma(\sqrt{p_r})=-\sqrt{p_r}$, and $\sigma$ fixes other $p_i$s. Then these $\sigma$s and their compositions, together, make $2^r$ distinct automorphisms. @ReneSchipperus – 2017-01-27
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    Ok, but that is the argument that I gave. – 2017-01-27
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    Then your argument is the standard solution :D. Also, without using Galois theory, it can be shown by induction that $\sqrt{p_1}+\dots+\sqrt{p_r}$ is a primitive element. @ReneSchipperus – 2017-01-27
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    Do you mean by squaring and doing algebra etc. That gets kind of complicated doesnt it ? In simple cases yeah its ok, but as a general method or is there some clever neat way of doing it ? – 2017-01-27
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    I just looked at your profile you say you are in high school, its amazing that you understand Galois theory so well. – 2017-01-27
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    @ReneSchipperus My brother is studying math in university, and I learn many things from him :D – 2017-01-27