$\mathbb{Q}(\zeta_{10}) $ degree of field extension over $\mathbb{Q}$

Question

I wish to calculate the degree of $\mathbb{Q}(\zeta_{10}) $ over $\mathbb{Q}$.

Using the dimensions theorem: $[\mathbb{Q}(\zeta_{10}) : \mathbb{Q}]=[\mathbb{Q}(\zeta_{10}) : \mathbb{Q(\zeta_5)}] \cdot[\mathbb{Q}(\zeta_{5}) : \mathbb{Q}] = 2\cdot4=8$. Due to the fact that $x^4+x^3+x^2+x+1$ is irreducible and $x^2-\zeta_5$ is the minimal polynomial for $\zeta_{10}$ over $\mathbb{Q}(\zeta_5)$.

However: $x^5+1|_{\zeta_{10}} = {(e^{\frac{2\pi i}{10}})}^5 + 1=e^{\pi i}+1=0$.

I am confused, because the extension should be $8$, yet it seems like $\zeta_{10}$ is a root of $x^5+1$.

Answer 1

$\zeta_{10}$ is a root of $x^4-x^3+x^2-x+1=0$. This is the tenth cyclotomic polynomial, and is irreducible over $\Bbb Q$.

The flaw in your argument is that $|\Bbb Q(\zeta_{10}):\Bbb Q(\zeta_5)|=1$. Note that $\zeta_{10}=\exp(\pi i/5)=-\exp(6\pi i/5)=-\zeta^3_5\in\Bbb Q(\zeta_5)$.

Answer 2

If $\zeta$ is a primitive $n$-th root of unity for an odd number $n$, then it is easily checked that $-\zeta$ is a primitive $2n$-th root of unity.

It follows that the field extension of the rational numbers corresponding to these numbers are one and the same.

In this case if you know the $n$-th cyclotomic polynomial, then easy to see that simply by changing the signs of odd powers of the variable we get the $2n$-th cyclotomic polynomial. (In particular they both have the same degree).