For each odd prime $p$, choose an arbitrary $x_p \in \mathbb Q(\zeta_p) \smallsetminus \mathbb Q$, where $\zeta_p = e^{\frac{2\pi i}{p}}$. Is it always true that the set $\{x_p : p \in \mathbb P\}$ is linearly independent over $\mathbb Q$?
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8$\begingroup$ Yes, it is true. $\mathbb{Q}(\zeta_n)\cap\mathbb{Q}(\zeta_m)=\mathbb{Q}$ if $(m,n)=1$. Look at ramification or use Galois theory in $\mathbb{Q}(\zeta_{mn})$. $\endgroup$– Felipe VolochApr 9, 2011 at 9:26
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8$\begingroup$ $\mathbb{Q}(\zeta_m)\cap\mathbb{Q}(\zeta_n)=\mathbb{Q}$ also follows from $[\mathbb{Q}[\zeta_m]=\varphi(n)$ and multiplicativity of Euler's totient function $\varphi$. $\endgroup$– George LowtherApr 9, 2011 at 14:25
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I think everything is clear now by George Lowther's comment, reproduced here with two minor typos corrected:
$\mathbb Q(\zeta_m) \cap \mathbb Q(\zeta_n) = \mathbb Q$ also follows from $[\mathbb Q[\zeta_m] : \mathbb Q] = \varphi(m)$ and multiplicativity of Euler's totient function $\varphi$.