# 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 $$\mathbb R$$.

• Every polynomial of degree greater than $2$ is reducible over $\mathbb R$. – Wojowu May 8 '19 at 20:58
• @Wojowu, Thanks for the comment... – qinr May 8 '19 at 21:04

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 $$Q(x)$$ would be 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.

• Excellent answer. Any hint on how to prove that $Q(x)$ are irreducible? – qinr May 8 '19 at 21:15
• Check that the coefficients of each $Q(x)$ are real, but the roots are not. – Sameer Kailasa May 8 '19 at 21:52
• @qinr See the edit. – N. S. May 8 '19 at 22:25
• @N.S. Thanks a lot! – qinr May 8 '19 at 22:42