$ ℚ(\zeta + \zeta^{-1}) = ℚ(\sqrt5)$

Question

Show that $ℚ(\zeta + \zeta^{-1}) = ℚ(\sqrt5)$ where $\zeta$ is the primitive $5^{th}$ root of unity.

I have seen a similar question asked, from this I gather that the quadratic polynomial is $ ^2+−1$, however I don't really understand how to conclude that it is equal to $ℚ(\sqrt5)$.

Answer 1

Let $L/K$ be a quadratic extension given as the splitting field of a degree 2 polynomial $f \in K[x]$. When $\operatorname{char} K \neq 2$ the extension is determined by the discriminant.

Why? Well the roots of $f = ax^2 + bx + c$ are given by the quadratic formula (not that this is where we use the assumption on the characteristic of $K$), which only requires us to adjoin $\sqrt{\Delta(f)} = \sqrt{b^2 - 4ac}$ - i.e., the splitting field of $f$ is obtained by adjoining $\sqrt{\Delta(f)}$. In your case the discriminant is $5$.

Answer 2

To answer the question in the title, here is a roadmap:

• $\zeta$ is a root of $x^4 + x^3 + x^2 + x + 1$

• $\zeta^{-1}=\zeta^4$

• $z=\zeta + \zeta^{-1}$ implies $z^2=\zeta^2+\zeta^3+2$

• $z$ is a root of $x^2+x−1$ and so $z=(-1\pm \sqrt 5)/2$; no need to know the actual sign in $\pm$

• $\mathbb Q(z) = \mathbb Q(\sqrt 5)$