MHB Showing that Q(sqrt(p)) is in Q adjoined the pth root of unity

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The discussion revolves around proving that Q(sqrt(p*)) is contained in Q(zeta_p), where p* is defined based on the congruence of p modulo 4. The key insight involves using the Vandermonde matrix formed by the pth roots of unity, which leads to the conclusion that the determinant of this matrix can be either real or purely imaginary depending on the parity of (p-1)/2. If (p-1)/2 is even, it shows that sqrt(p) is in Q(zeta), while if it is odd, it demonstrates that sqrt(-p) is in Q(zeta). This establishes the desired inclusion for both cases of p.
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i am having trouble showing that \mathbb{Q}(\sqrt{p*}) \subset \mathbb{Q}(\zeta_{p}) where p* = (-1)^{\frac{p-1}{2}}p. in other words, if p = 1 (mod 4) then p* = p and if p = 3 (mod 4) then p* = -p. i encountered this in the context of galois theory and i have no idea how to start. it seems that i need to know what \zeta_{p} looks like before i decide if \sqrt{p*} \in \mathbb{Q}(\zeta_{p}) but for arbitrary p that is hard to figure out. i also can't figure out why we have the 1 mod 4 and 3 mod 4. can someone give me some hints on this question?
 
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oblixps said:
i am having trouble showing that \mathbb{Q}(\sqrt{p*}) \subset \mathbb{Q}(\zeta_{p}) where p* = (-1)^{\frac{p-1}{2}}p. in other words, if p = 1 (mod 4) then p* = p and if p = 3 (mod 4) then p* = -p. i encountered this in the context of galois theory and i have no idea how to start. it seems that i need to know what \zeta_{p} looks like before i decide if \sqrt{p*} \in \mathbb{Q}(\zeta_{p}) but for arbitrary p that is hard to figure out. i also can't figure out why we have the 1 mod 4 and 3 mod 4. can someone give me some hints on this question?
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 $\mathbb{Q}(\sqrt {-p})\subseteq \mathbb{Q}(\zeta)$.
 
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