For each $n\in \mathbb{N},\, \,$ $\phi_{n}(x):=\displaystyle \prod_{\substack{1\leq k\leq n\\ (k,n)=1}}(x-e^{\frac{2\pi i}{n}k})$
Show that the irreducible factors of $\phi_n$ in $\mathbb{Q}[x]$ have the same degree.
I have a confusion with this exercise. I understand that the cyclotomic polynomial $\phi_n$ is irreducible in $\mathbb{Q}[x]$ so $\phi_n$ has a unique irreducible factor, the same. Is this correct or am I confusing something?
Showing that the irreducible factors of the cyclotomic polynomial in $\mathbb{Q}[x]$ have the same degree.
cyclotomic-polynomials
Best Answer
The argument you gave is correct, but proving irreducibility of the cyclotomic polynomials over $\mathbb Q$ takes some work. You can solve the question at hand without citing this result, which I'm guessing is the intent of the exercise. I will, however, be using some field theory.
Let $\phi_n = \prod f_i$ with each $f_i$ monic and irreducible. The $f_i$ are distinct as $\phi_n$ has no repeated roots. Let $\alpha_i$ be a root of $f_i$. These are therefore roots of $\phi_n$ so they are primitive $n^{th}$ roots of unity. Let's consider the field $\mathbb Q(\alpha_i)$. The degree of this field is the degree of the minimal polynomial of $\alpha_i$, which is $f_i$. Furthermore, as $\alpha_i$ is a primitive $n^{th}$ root of unity, the set $\{\alpha_i^k : k \in \mathbb Z\}$ is precisely the set of $n^{th}$ roots of unity. But since $\mathbb Q(\alpha_i)$ is a field, each of these powers lies in $\mathbb Q(\alpha_i)$. Hence, $\mathbb Q(\alpha_i)$ contains all the $n^{th}$ roots of unity. Furthermore, because the above result held for all $i$, we have $\alpha_j \in \mathbb Q(\alpha_i)$ for all pairs $i,j$. The definition of $\mathbb Q(\alpha_j)$ is the smallest field containing $\mathbb Q$ and $\alpha_j$, so $\mathbb Q(\alpha_j) \subseteq \mathbb Q(\alpha_i)$ for all pairs $i,j$. Thus, all of these fields $\mathbb Q(\alpha_i)$ are the same. In particular, they have the same degree over $\mathbb Q$. We have that $[\mathbb Q(\alpha_i) : \mathbb Q] = \deg f_i$, so for all $i,j$ $\deg f_i = \deg f_j$ as desired.