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- Thread starter oblixps
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- Feb 7, 2012

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Sorry for the long delay in responding – I didn't see this thread when it was first posted.

I believe that the key to this problem is to use the Vandermonde matrix formed by the $p$th roots of unity. If $\zeta = e^{2\pi i/p}$, let $V_p$ be the $p\times p$ matrix whose $(i,j)$-element is $\zeta^{\,ij}$ for $0\leqslant i,j\leqslant p-1$ (notice that the rows and columns are labelled from 0 to $p-1$ rather than from 1 to $p$). The first thing to check is that $V_pV_p^* = pI_p$, where the star denotes the hermitian transpose and $I_p$ is the $p\times p$ identity matrix. It follows that $|\det(V_p)| = p^{p/2}.$

Next, the matrix $V_p$ is unchanged if for $1\leqslant i\leqslant (p-1)/2$ we interchange row $i$ with row $p-i$ and then take the complex conjugate of each element of the resulting matrix. Interchanging two rows of a matrix changes the sign of its determinant, so if $(p-1)/2$ is even then the number of sign changes is even and so $\det(V_p) = \overline{\det(V_p)}$ (the bar denoting the complex conjugate). Therefore $\det(V_p)$ is real and it follows from the previous paragraph that $p^{-(p-1)/2}\det(V_p) = \pm\sqrt p$. But the left side of that equation is in $\mathbb{Q}(\zeta)$, so $\sqrt p \in\mathbb{Q}(\zeta)$ and hence $\mathbb{Q}(\sqrt p)\subseteq \mathbb{Q}(\zeta)$.

If $(p-1)/2$ is odd, then a similar argument shows that $\det(V_p)$ is purely imaginary and therefore $p^{-(p-1)/2}\det(V_p) = \pm\sqrt{-p}$, from which $\mathbb{Q}(\sqrt {-p})\subseteq \mathbb{Q}(\zeta)$.