Why is $[\mathbb{Q}(\zeta):\mathbb{Q}] = 8$ and not $14$? (Where $\zeta$ is a primitive $15^{th}$ root of unity)

Question

I have a field extension $\mathbb{Q}(\zeta)/\mathbb{Q}$, where $\zeta$ is a primitive $15^{th}$ root of unity.

So, since $x^{15}-1 = \phi_{1}(x)\phi_{3}(x)\phi_{5}(x)\phi_{15}(x)$, where $\phi_{n}(x)$ is the $n^{th}$ cyclotomic polynomial, and I have that $\mathbb{Q}(\zeta) =$ splitting field of $\phi_{3}(x)\phi_{5}(x)\phi_{15}(x)$ over $\mathbb{Q}$, and has degree $14$ (since $\deg(\phi_3) = 2, \deg(\phi_5) = 4, \deg(\phi_{15}) = 8$).

But the answer is supposed to be $8$, and I do not see how I can get that. If anyone can point out my mistake, I would be grateful.

Answer 1

The degree of a field extension of the form $\mathbb{Q}(\alpha)$ is the degree of the minimal polynomial of $\alpha$, i.e. the degree of a monic irreducible polynomial of which $\alpha$ is a root. The polynomial $x^{15} - 1$ is not irreducible (you factored it, after all). You should be taking the degree of the irreducible factor that you wrote as $\phi_{15}$, which as you say correctly is $8$.