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

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

$$\zeta_{10}$$ 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 $$\zeta_{10}=\exp(\pi i/5)=-\exp(6\pi i/5)=-\zeta^3_5\in\Bbb Q(\zeta_5)$$.
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.
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).