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p-th root of rational numbers 2

Source: Iran 3rd round 2012-Algebra exam-P5

September 20, 2012
algebra proposedalgebra

Problem Statement

Let pp be an odd prime number and let a1,a2,...,an∈Q+a_1,a_2,...,a_n \in \mathbb Q^+ be rational numbers. Prove that Q(a1p+a2p+...+anp)=Q(a1p,a2p,...,anp).\mathbb Q(\sqrt[p]{a_1}+\sqrt[p]{a_2}+...+\sqrt[p]{a_n})=\mathbb Q(\sqrt[p]{a_1},\sqrt[p]{a_2},...,\sqrt[p]{a_n}).
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