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If $\zeta$ is a primitive nth root of unity, it appears that the cyclotomic field $\mathbb{Q}(\zeta )$can be partitioned into ${\mathbb{Q}}(\zeta +{{\zeta }^{-1}})$ and ${\mathbb{Q}}(\zeta -{{\zeta }^{-1}})$ where ${{\mathbb{Q}}^{+}}$= ${\mathbb{Q}}(\zeta +{{\zeta }^{-1}})$is the well-known maximal totally real subfield of $\mathbb{Q}(\zeta )$. This subfield has dimension $\varphi (n)/2$ over $\mathbb{Q}$, so ${{\mathbb{Q}}^{-}}$= ${\mathbb{Q}}(\zeta -{{\zeta }^{-1}})$ should also have dimension $\varphi (n)/2$. Some authors refer to ${{\mathbb{Q}}^{-}}$as the maximal totally complex subfield of $\mathbb{Q}(\zeta )$but I have never seen a proof of this.

Since ${{\mathbb{Q}}^{+}}$= $\mathbb{Q}(\zeta )$$\cap \mathbb{R}$, the complement as a vector space is $\mathbb{Q}(\zeta )\cap \mathbb{R}i$ and each of these must have dimension $\varphi (n)/2$. ${{\mathbb{Q}}^{+}}$ is always ${\mathbb{Q}}(\zeta +{{\zeta }^{-1}})$, and the question is how to characterize $\mathbb{Q}(\zeta )\cap \mathbb{R}i$. The symmetric ‘odd’ representation would be ${\mathbb{Q}}(\zeta -{{\zeta }^{-1}})$but clearly that does not always hold.

  • In what sense do you intend this subfield to be a complement? – Jyrki Lahtonen Aug 28 '15 at 17:48
  • I don't think that dimension formula holds. For example with $n=3$ we have $\zeta-\zeta^{-1}=i\sqrt3$, so $\Bbb{Q}^-=\Bbb{Q}(\zeta)$. – Jyrki Lahtonen Aug 28 '15 at 17:50
  • Your comment is well-taken Jyrki –what I am looking for is an algebraic characterization of $\mathbb{Q}(\zeta )$$\cap \mathbb{R}i$, so the complement is with respect to $\mathbb{Q}(\zeta )$$\cap \mathbb{R}$ as a vector space, and of course for n = 3, these are both dimension 1. In all cases $\mathbb{Q}(\zeta )$is a quadratic extension of ${{\mathbb{Q}}^{+}}$and for n = 3, ${{\mathbb{Q}}^{+}}$= -1 and $\mathbb{Q}(\zeta )$ = $\mathbb{Q}(i\sqrt{3})$. – sunshineghh Aug 29 '15 at 00:02
  • Jyrki, I could only edit the last comment for 5 minutes and I just realized that there is a misprint: For n = 3, $\zeta +{{\zeta }^{-1}}$ is -1, not ${{\mathbb{Q}}^{+}}$ – sunshineghh Aug 29 '15 at 00:17

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