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

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In summary: But $\sqrt{-p} = \sqrt{p*}$, so in this case $\mathbb{Q}(\sqrt{p*})\subseteq \mathbb{Q}(\zeta)$.Note that the condition $p = 1 \pmod 4$ guarantees that $(p-1)/2$ is even, so this argument covers both cases. I hope this helps!In summary, the problem at hand is to show that $\mathbb{Q}(\sqrt{p*}) \subset \mathbb{Q}(\zeta_{p})$, where $p* = (-1)^{\frac{p-1}{2}}p$. This is equivalent to showing that $\sqrt{p
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oblixps
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i am having trouble showing that [tex] \mathbb{Q}(\sqrt{p*}) \subset \mathbb{Q}(\zeta_{p}) [/tex] where [tex] p* = (-1)^{\frac{p-1}{2}}p [/tex]. 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 [tex] \zeta_{p} [/tex] looks like before i decide if [tex] \sqrt{p*} \in \mathbb{Q}(\zeta_{p}) [/tex] 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 [tex] \mathbb{Q}(\sqrt{p*}) \subset \mathbb{Q}(\zeta_{p}) [/tex] where [tex] p* = (-1)^{\frac{p-1}{2}}p [/tex]. 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 [tex] \zeta_{p} [/tex] looks like before i decide if [tex] \sqrt{p*} \in \mathbb{Q}(\zeta_{p}) [/tex] 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 $\mathbb{Q}(\sqrt {-p})\subseteq \mathbb{Q}(\zeta)$.
 

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

What is Q(sqrt(p))?

Q(sqrt(p)) is the set of all rational numbers multiplied by the square root of a prime number p. In other words, it is the set of numbers of the form a + b*sqrt(p), where a and b are rational numbers.

What does it mean for Q(sqrt(p)) to be in Q adjoined the pth root of unity?

Q adjoined the pth root of unity is a field extension of Q, which means it includes all elements of Q as well as the pth root of unity. Therefore, showing that Q(sqrt(p)) is in Q adjoined the pth root of unity means that all elements of Q(sqrt(p)) can also be written as a polynomial involving the pth root of unity.

Why is it important to show that Q(sqrt(p)) is in Q adjoined the pth root of unity?

This proof is important because it helps to understand the relationship between rational numbers and roots of unity. It also has applications in fields such as number theory and algebraic geometry.

How is this proof typically approached?

This proof is typically approached using algebraic and number theory techniques. It often involves showing that any element of Q(sqrt(p)) can be written as a polynomial involving the pth root of unity, and then using properties of polynomial rings and field extensions to show that Q(sqrt(p)) is indeed a part of Q adjoined the pth root of unity.

What are some real-world examples of where this proof can be applied?

This proof has applications in areas such as cryptography, where the security of some algorithms relies on the properties of field extensions. It can also be used in the study of modular forms and in understanding the behavior of certain equations involving roots of unity.

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