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I've found several uses of the phrase "linearly independent over the rationals" that imply that any set of irrationals is linearly independent over the rationals if it is pairwise linearly independent over the rationals, but I can't find a reference to justify that claim. Can someone point me to such a reference?

What I want to show is that given some $k \notin ℚ$ for which $k^n$ is irrational when $n \in ℤ$ and $0<n<m$, the set $\lbrace k^n, 0<n<m \rbrace$ is linearly independent over the rationals. (For example, let $k=p^{1/q}$ for some prime $p$ and integer $q$, then $m=q$.) The reference I can't find would be sufficient to show that; I wonder if I'm failing to find one because it's put in some other terms that I'm failing to recognize as equivalent, and if that's the case I may need to reframe the question to ask for a translation.

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Say $\alpha$ and $\beta$ are linearly independent over the rationals. Then $\alpha$ and $\alpha+\beta$ are linearly independent over the rationals, and so are $\alpha+\beta$ and $\beta$. But clearly, $\alpha$, $\beta$, and $\alpha+\beta$ are not linearly independent over $\mathbb{Q}$. So pairwise linear independence does not imply linear independence. –  Arturo Magidin Jun 8 '12 at 18:51
    
Ahh, so I've misunderstood the criteria for independence in some things my search turned up. The sum case will not occur in $k^n$, so it doesn't answer the underlying question. I expect the $k^n$'s to be independent because ${k^{n_1} \over k^{n_2}} = k^{n_1-n_2} \notin ℚ$, which is a separate premise from pairwise independence. I'll ask a new question about that. Thanks @ArturoMagidin –  Polyergic Jun 8 '12 at 20:46
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1 Answer

up vote 3 down vote accepted

Here is a counterexample. Let $\alpha=1+\sqrt{2}$. The powers of $\alpha$ are all irrational, but the set $\{\alpha, \alpha^2, \alpha^3\}$ is linearly dependent over the rationals.

If it is required that some positive power of $\alpha$ is rational, let $\alpha$ be a primitive $6$-th root of unity. Then $\alpha$ is a root of $x^2-x+1$, so again $\{\alpha, \alpha^2, \alpha^3\}$ is linearly dependent over the rationals.

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I think $\alpha$ is required to be real. –  Zander Jun 9 '12 at 1:11
    
I actually asked a far to general question, which this answer covers well. I got closer to what I meant to ask in math.stackexchange.com/questions/155812 –  Polyergic Jun 9 '12 at 1:43
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