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

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.

• $\mathbb Q(\zeta)$ is the splitting field of $\phi_{15}(x)$; the roots of $\phi_3(x)$ and $\phi_5(x)$ are in subfields thereof Aug 26, 2019 at 13:06
• The polynomial $\phi_3\phi_5\phi_{15}$ is not irreducible (somehow), so the dimension if its splitting field cannot be lower-bounded by its degree. Aug 26, 2019 at 13:07
• You could prove the following theorem that has as consequence what you are looking for : Let $\zeta = e^{2\pi i/n}$ where $n$ is a positive integer and let $F = \mathbb{Q}(\zeta)$. Then the extension $F\supset \mathbb{Q}$ is Galois and the Galois group $\textrm{Gal}(F/\mathbb{Q})$ is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^{\times}$ the group of reduced residue classes modulo $n$ which cardinality is $\phi(n)$. Aug 26, 2019 at 13:20

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 $$8$$.
• How come is it that we have that the splitting field is $\phi_{15}$ and not one of the other cyclotomic polynomials? Because if it is supposed to be the minimal irreducible polynomial, with $\zeta$ as a root? Aug 26, 2019 at 13:09
• @Ponky, if $\zeta$ were a root of $\phi_3$ or $\phi_5$, would it be a primitive 15th root of unity? Aug 26, 2019 at 13:13
• @PeterTaylor, no, since those are made up of the primitive 3rd and 5th roots of unity, which are not the same as the primitive 15th roots of unity. Okay, so I see, the only cyclotomic polynomial of those three that has $\zeta$ as a root is $\phi_{15}$ and so since it is irreducible over $\mathbb{Q}$, it is the minimal irreducible polynomial. Aug 26, 2019 at 13:17