field theory - To show that $mathbf Q(sqrt[r]{p_1}, ldots, sqrt[r]{p_n})=mathbf Q(sqrt[r]{p_1}+cdots+sqrt[r]{p_n})$

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.

Answer

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