# Thread: Field extension

1. ## Field extension

I am having trouble with this problem. It would be great if someone can help.
I want to prove that $Q(\zeta)=Q(\delta)$ where $Q(\zeta)$ is the cyclotomic field of 5th roots of unity, and $\delta=\frac{i}{2}\sqrt{10+2\sqrt{5}}$

2. Originally Posted by namelessguy

I am having trouble with this problem. It would be great if someone can help.
I want to prove that $Q(\zeta)=Q(\delta)$ where $Q(\zeta)$ is the cyclotomic field of 5th roots of unity, and $\delta=\frac{i}{2}\sqrt{10+2\sqrt{5}}$
first see that $\delta$ is a root of the polynomial $f(x)=x^4 + 5 x^2 + 5,$ which is irreducible over $\mathbb{Q}$ by Eisenstein criterion. thus $[\mathbb{Q}(\delta): \mathbb{Q}]=4.$ since $[\mathbb{Q}(\zeta): \mathbb{Q}]=4,$ we'll be done if we prove that

$\mathbb{Q}(\delta) \subseteq \mathbb{Q}(\zeta).$ now suppose $\zeta=\cos a + i \sin a.$ then from $1=\zeta^5=(\cos a + i \sin a)^5,$ we'll get: $5 \cos^4 a \sin a - 10 \cos^2a \sin^3 a + \sin^5 a = \text{Im} \ (\cos a + i \sin a)^5 = 0,$ which after simplifying

gives us: $16\sin^4a - 20 \sin^2 a + 5 = 0.$ thus: $(2i \sin a)^4 + 5(2i \sin a)^2 + 5 = 0.$ therefore $2i \sin a=\zeta - \zeta^{-1}$ is a root of $f(x)=0.$ we actually proved that the roots of $f(x)=0$ are exactly

$x_k=\zeta^k - \zeta^{-k}, \ k=1,2,3,4.$ thus there exists $1 \leq k \leq 4$ such that $\delta = x_k \in \mathbb{Q}(\zeta). \ \ \Box$

3. Thank you very much for your help, NonCommAlg. So I understand that you let $x=\frac{i}{2}\sqrt{10+2\sqrt{5}}$. Then squaring this equation twice and manipulate to get the polynomial. It is also the minimal polynomial correct? But I don't understand the notation you use below. Does it mean degree of $Q(\delta)/Q=4$

$[\mathbb{Q}(\delta): \mathbb{Q}]=4.$ since $[\mathbb{Q}(\zeta): \mathbb{Q}]=4,$

4. Originally Posted by namelessguy

Does it mean degree of $Q(\delta)/Q=4$

$[\mathbb{Q}(\delta): \mathbb{Q}]=4.$ since $[\mathbb{Q}(\zeta): \mathbb{Q}]=4,$
yes, that's what it means. it's a standard notation for degree.