Determine element in a field extension

Question

Let $F=\mathbb{Q}(\xi)$ with $\xi$ satisfying $\xi^4+\xi^3+\xi^2+\xi+1=0$. Find an element $z\in F$, $z=a+b\cdot\xi+c\cdot\xi^2+d\cdot\xi^3$ satisfying $z^2=5$.

Is there any way to solve this problem without having to solve a nonlinear system of equations?

Answer 1

You know that $z=\sqrt{5}$ or $z=-\sqrt{5}$, while $\xi$ is a primitive fifth root of $1$; so $$ \xi=\cos\frac{2\pi}{5}+i\sin\frac{2\pi}{5} $$ or a power thereof. Now $$ \cos\frac{2\pi}{5}=\frac{\sqrt{5}-1}{4} $$ Can you go on from here?

Further hint: $\xi+\xi^{-1}=2\cos(2\pi/5)=(\sqrt{5}-1)/2$, so $$ \sqrt{5}=1+2\xi+2\xi^{-1} $$ Now $\xi^{-1}=\dots$

Answer 2

$F$ has an automorphism $\phi$ induced by $\xi\mapsto \xi^2$. If $z$ is a solution of $z^2=5$, then $\phi(z)$ must also be a solution. Hence either $\phi(z)=z$ or $\phi(z)=-z$. At any rate $\phi(\phi(z))=z$. But $$\begin{align}\phi(\phi(a+b\xi+c\xi^2+d\xi^3))&=a+b\xi^4+c\xi^8+d\xi^{12}\\&=(a-b)-b\xi+(c-b)\xi^3+(d-b)\xi^2\end{align}$$ so that $a-b=a$, $-b=b$, $c-b=d$, $d-b=c$ and hence $b=0$, $c=d$. So we need $$ z=a+c(\xi^2+\xi^3)$$ for some $a,b\in\mathbb Q$. With this we have $$\phi(z)=a+c(\xi^4+\xi^6)=a+c(-1-\xi^2-\xi^3)=(a-c)-c(\xi^2+\xi^3).$$ If we assume $\phi(z)=z$, this leads to $c=0$, so $z=a\in\mathbb Q$, which is impossible as $\sqrt 5$ is irrational. Hence $\phi(z)=-z$, i.e. $a-c=-a$, i.e. $$ z=a(1+2\xi^2+2\xi^3)$$ for some $a\in\mathbb Q$. So we are looking for $\in\mathbb Q$ with $$5=a^2(1+2\xi^2+2\xi^3)^2.$$ Expand and simplify the RHS to see how easy such $a$ can be found.

Answer 3

This is actually a nice property of Gauss sums. Given any odd prime $p$ and a primitive $p$th root of unity $\zeta_p$ the element:

$G = \sum_{a=1}^{p-1} \left(\frac{a}{p}\right)\zeta_p^a$

is such that:

$G^2 = (-1)^{\frac{p-1}{2}}p$.

This shows you that the unique quadratic subfield of $\mathbb{Q}(\zeta_p)$ is $\mathbb{Q}\left(\sqrt{(-1)^{\frac{p-1}{2}}p}\right)$. This is the starting point for more a modern proof of Quadratic Reciprocity.

In the case where $p=5$ you find that $G = \zeta_5 - \zeta_5^2 - \zeta_5^3 + \zeta_5^4$ is such that $G^2 = 5$.