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I know that the set of square roots of distinct square-free integers is linearly independent over $\mathbb{Q}$. To generalize this fact, define

$R_n = \{ \sqrt[n]{s} \mid s\text{ integer with prime factorization }s = p_1^{a_1} \ldots p_k^{a_k}, \text{ where } 0 \leq a_i < n \}$

For example, $R_2$ is the set of square roots of square-free integers.

Question: Is $R_n$ linearly independent over $\mathbb{Q}$ for all $n \geq 2$?

Harder (?) question: Is $\cup_{n\geq2}R_n$ linearly independent over $\mathbb{Q}$?

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1 Answer 1

This is true iff the radicands are multiplicatively independent. See the references to the work of Besicovitch, Mordell and Siegel in my answer to a similar question. Nowadays these results are special cases of the Galois theory of radical extensions (a.k.a. Kummer extensions).

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What does multiplicatively independent mean? –  Mikko Korhonen Jun 15 '12 at 16:00
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@m.k. Said simply it means that are no nontrivial multiplicative relations among the radicands, i.e. $\rm\:s_1^{e_1}\cdots s_k^{e_k}\in \mathbb Q\:\Rightarrow\: n\:|\:e_i.\:$ See the reviews in the linked posts for further detail. –  Bill Dubuque Jun 15 '12 at 16:17

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