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I expected it to be basic, but seem unable to find a proof of the following:

Let $p_0, p_1, .., p_m$ be distinct primes. Then the $m+1$ terms $\dfrac{\log p_0}{\log p_j}$, are rationally independent.

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Perhaps you've already observed this, but it follows from Schanuel's conjecture that in fact the logarithms of the primes are algebraically independent. –  Qiaochu Yuan Sep 25 '11 at 23:50
Sorry, I voted to close because I misread the question as asking about $\log p_j / \log p_0$ which is trivial. –  Felipe Voloch Sep 26 '11 at 10:41
@Felipe, no problem. Although, now that I think about it, I would have preferred it to be closed because of triviality, because I need this result :D @Qiaochu, thanks. Actually, I don't really mind if the part of the paper I'm writing in which I use this result, was based on a conjecture like Schanuel's. But if that should happen, should I then use the 'easiest' conjecture from which this follows, or the most well-known, or some other criterium? –  Woett Sep 26 '11 at 12:45
@Qiaochu, I probably misread your observation. I now believe that what you mean is that IF my question has an affirmative answer, then Schanuel's conjecture shows that the terms are algebraically independent, right? –  Woett Sep 26 '11 at 13:18
I think that Qiaochu is saying that because $\log p_0, \dots, \log p_m$ are linearly independent over $\mathbb{Q}$, Schanuel's conjecture would imply that they are algebraically independent over $\overline{\mathbb{Q}}$. This then would imply the linear independence result you want. See Chapter 1 of Waldschmidt's book "Diophantine approximation on linear algebraic groups" for more information on algebraic independence of logarithms. Conjecture 1.15 there is all that you would need, but it's just a special case of Schaneul's conjecture which is more widely known. –  Matt Papanikolas Sep 30 '11 at 4:20

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