Cyclotomic polynomials being irreducible over Q

Question

So, task is to, using algebra, write polynomial $X^n-1$ as a product of irreducible polynomials over $Q$. Our prof told us that the solution is

$$X^n-1 = \prod_{d|n} \Phi_d(x),$$

where $\Phi_d(x)$ is $d$-th cyclotomic polynomial over $Q$. Now, this means that every $\Phi_d(x)$ is irreducible over $Q$. Every textbook I have looked in, there are just some basic and "in general case", "in general field", nothing to say about $Q$. I am curious how to prove this ($d$-th cyclotomic polynomial over $Q$ is irreducible) can not find proof and has absolutely not a point how to even begin. Thanks for any help!

Answer 1

The definition I've seen of the $n^{th}$ cyclotomic polynomial (Galois Theory by David Cox pg 75) is that $\Phi_n(x)$ is the minimal polynomial of the $n^{th}$ root of unity $e^{\frac{2\pi i}{n}}$.

here's a quick proof that minimal polynomials are irreducible.

Suppose by way of contradiction that $h(x)$ is the minimal polynomial for an algebraic number $\alpha$ and $h(x) = g(x) \cdot f(x)$ for some $g(x)$ and $f(x)$ of degree less then that of $h(x)$. Then evaluation at $\alpha$ gives $$0 = h(\alpha) = g(\alpha)f(\alpha)$$ thus one of either $g(\alpha)$ or $f(\alpha)$ is zero, contradicting that $h(x)$ is the minimal polynomial of $\alpha$.

Thus since the cyclotomic polynomials are minimal polynomials, they are irreducible. Note that we didn't use anywhere a specific field, so if we define our cyclotomic polynomial in a similar way for a field other than $\mathbb{Q}$ we get the irreducibility of $\Phi_n(x)$ using the same argument.

Hope that helps!