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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 $\mathbb{Q}$?

Mikko Korhonen
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2 Answers2

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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).

Community
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Bill Dubuque
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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
  • @BillDubuque Hello~It seems to me that according to the review of Mordell and Siegel's work in your answer, $R_n$ in the question is algebraically independent iff the radicands are multiplicatively independent. However, if we only needs $R_n$ to be linearly independent (over $\mathbb{Q}$), I wonder if we need such restriction, as $R_2$ seems always linearly independent to me. – Wembley Inter Mar 27 '23 at 12:03
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We'll show that incommensurable real radicals are linearly independent.

Let $\alpha_i$ real numbers, $F$ subfield of $\mathbb{R}$ so that for any $i$ $\alpha_i^{n_i}\in F$ for some $n_i>1$. Assume moreover that $\frac{\alpha_i}{\alpha_j}\not \in F$ for all $i\ne j$. Then the $\alpha_i$ are linearly independent over $F$.

For the proof we use the Lemma: let $\beta$ a real number so that $\beta^m\in F$ for some $m>1$, and $\beta\not \in F$. Then $\operatorname{Trace}_F \beta = 0$. A proof is given below.

Let now $\sum a_i \alpha_i$ a linear relation. Take $i_0 \in I$. We get $$ a_{i_0} = \sum_{i \ne i_0} a_i \frac{\alpha_i}{\alpha_{i_0}}$$

Taking $\operatorname{Trace}^K_F$ on both sides ( $K$ is an arbitrary finite extension of $F$ containing all the $\beta_i =\frac{\alpha_i}{\alpha_{i_0}}$ we get, using lemma $d\cdot a_{i_0} = 0$, and so $a_{i_0} = 0$.

Proof of the lemma: May assume $\beta > 0$. Let $m>1$ minimal so that $\beta^m \in F$. The polynomial $X^m - \beta^m$ factors over $\mathbb{C}$ as $\prod_{j=0}^{m-1} ( X- \beta \omega^j)$. Assume that some factor $\prod_{j \in J} (X - \beta \omega^j)$ is in $F[X]$. The the free term $\prod_{ j \in J} ( - \beta \omega^j)$ is in $F$. Taking complex absolute values on both sides, we get $\beta^l \in F$ for some $1\le l < m$, contradiction. Now we know the basis $1$, $\beta$, $\ldots$, $\beta^{m-1}$ for $F(\beta)$ over $F$. It follows right away that the trace of $\beta$ is $0$.

Note: the condition $\beta$ real is necessary, as we see for $\beta = 1+i$, $\beta^4 = -4$, and $\operatorname{trace}^{\mathbb{Q}(i)}_{\mathbb{Q}} \beta = 2$.

orangeskid
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