when is the $n$-th cyclotomic polynomial irreducible over $\mathbb{R}$ It is well known that the cyclotomic polynomials
$\Phi_n(x)$
are irreducible over the field of rationals $\mathbb{Q}$.
I am curious about their reducibility over the real numbers $\mathbb R$.
We have $$\Phi_n(x)=\prod_{\substack{1\leqslant k\leqslant n \\\gcd(k,n)=1}}(x-\zeta_n^k)$$
where $\zeta_n:=e^{2\pi i/n}$.
So $\Phi_n$ is reducible when $2 k/n\in\mathbb Z$ for all $\gcd(k,n)=1$ which is clearly impossible. 
So am I safe to conclude that $\Phi_n(x)$ are irreducible also over $\mathbb R$?
EDIT
As is mentioned by @Wojowu, every polynomial of degree $\geqslant 3$ is reducible in $\mathbb R$, so my question becomes... How to write down the irreducible factors of the cyclotomic polynomials over $\mathbb R$.
 A: If $n \neq 1,2,3,4,6$ then $\phi(n) >2$ and
$$Q(x)=(x-\zeta_n)(x-\zeta_n^{n-1})$$
is a divisor of your polynomial with real coefficients.
As for the edited question For each $\gcd(k,n)=1$ the polynomial
$$Q(x ) = ( x-\zeta_n^k) (x-\zeta_n^{n-k}) $$
is irreducible over $\mathbb R$.
Extra Here is the proof that, for $n \neq 1,2$, and all $(k,n)=1, Q(x)$ is irreducible over $\mathbb R$.
First, it is easy to see that $\zeta_n^{n-k}=\overline{\zeta_n^{k}}$. You can see this either using the polar/trig form or by observing that
$$\zeta_n^{n-k}=\frac{\zeta_n^{n}}{\zeta_n^{k}}=\frac{1}{\zeta_n^{k}}=\frac{\overline{\zeta_n^{k}}}{\zeta_n^{k}\cdot \overline{\zeta_n^{k}}}=\overline{\zeta_n^{k}}$$
From here it follows immediately that the coefficients are real.
If $Q(x)$ were reducible, it would be a product of 2 linear terms, and hence its roots would be real. Therefore
$$\zeta_n^{k} \in \mathbb R$$
But then $\zeta_n^{k}$ would be a real root to $x^n-1$, and hence $\zeta_n^{k}  \in \pm 1$.
Since $(k,n)=1$ this is also a primitive $n$th root of unity, and hence $n \in \{1,2\}$, meaning the cyclotmic polynomial is linear.
