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 $\mathbb R$?


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 $\mathbb R$.

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

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

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