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In this thread @Geoff Robinson gives a nice argument to show that if $p_1, \ldots, p_n$ are distinct primes then we have $\mathbf Q(\sqrt{p_1}, \ldots, \sqrt{p_n})=\mathbf Q(\sqrt{p_1}+\cdots+\sqrt{p_n})$.

In the comments @Paul Garret gives a sketch to show a more general statement that if $r>1$ is an integer then

$\mathbf Q(\sqrt[r]{p_1}, \ldots, \sqrt[r]{p_n})=\mathbf Q(\sqrt[r]{p_1}+\cdots+\sqrt[r]{p_n})$

@Paul Garret writes that "summing over the powers of a single Galois automorphism would annihilate the roots that are "moved" by it, entirely analogously to the effect your argument achieves, giving the analogue over $\mathbf Q(\zeta_r)$."

I am unable to see how @Geoff Robinson's argument can be adapted to prove the more general statement. One thing which Geoff uses in his argument is that the Galois group of $Q(\sqrt{p_1}, \ldots, \sqrt{p_n}):\mathbf Q$ is abelian. This is easy to see because the square of each automorphism is identity. It is not clear if the Galois group of $\mathbf Q(\sqrt[r]{p_1}, \ldots, \sqrt[r]{p_n}):\mathbf Q$ is also abelian.

Can somebody please elaborate on Paul's argument.

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You don't need to know the Galois group of the full extension, look at my answer here How can we prove $\mathbb{Q}(\sqrt 2, \sqrt 3, ..... , \sqrt n ) = \mathbb{Q}(\sqrt 2 + \sqrt 3 + .... + \sqrt n )$

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  • $\begingroup$ Can you please also see my comment to Geoff Robinson's answer here math.stackexchange.com/questions/93453/… and check if what I am saying is correct? $\endgroup$ – caffeinemachine Feb 23 '16 at 16:36
  • $\begingroup$ @caffeinemachine: Aha, yes, seems OK. However, do not need to focus too much on the Galois group. Just check this: " if an automorphism fixes $s$ then it fixes each term ". Now with Galois theory, this means $s$ generates each radical. This is the only result in Galois theory that you use. It just show you how powerful Galois theory really is. My feeling is that the proof given in that link was in a way suboptimal. $\endgroup$ – Orest Bucicovschi Feb 23 '16 at 22:15
  • $\begingroup$ I suppose I have made a mistake in my comment to Geoff's answer. I have said that of $L:M$ is separable then $[L:M]=|\text{Aut}(L:M)|$. This is not true. $\endgroup$ – caffeinemachine Feb 24 '16 at 5:09
  • $\begingroup$ @caffeinemachine: Indeed, needs to be Galois ( normal also). $\endgroup$ – Orest Bucicovschi Feb 24 '16 at 5:38
  • $\begingroup$ So I have finally added my answer in the thread I linked. I would be grateful if you can check that it is correct. I really worked hard these couple of days to understand Galois theory to the point so that I can solve this problem. Also, your proof which you linked of the much more general result is quite badass. I do not see why would you call it suboptimal. $\endgroup$ – caffeinemachine Feb 24 '16 at 8:25

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