# Are the prime cyclotomic polynomials irreducible over any field where they're not obviously reducible ?

My question is the following : if $$p$$ is a prime number, $$\Phi_p = \frac{X^p-1}{X-1}$$, is $$\Phi_p$$ irreducible over any field $$K$$ where it has no root ?

Phrased differently, if $$K$$ is of characteristic $$\neq p$$ and has no nontrivial $$p$$th root of unity, is $$\Phi_p$$ irreducible over $$K$$ ?

Note that any $$p$$th root of unity is primitive as $$p$$ is prime; so let $$K$$ be a field of char $$\neq p$$ with no nontrivial roots of unity: if $$\zeta$$ is such a root and $$L=K(\zeta)$$, then $$L$$ is the decomposition field of $$\Phi_p$$ over $$K$$.

Its Galois group is generated by $$\zeta \mapsto \zeta^k$$ for some $$k$$, so the question is linked to subgroups of $$(\mathbb{Z/pZ})^\times$$. The question reduces to : is there a proper subgroup $$H$$ of $$(\mathbb{Z/pZ})^\times$$, and a field $$K$$ of char $$\neq p$$ with no nontrivial roots of unity such that $$\displaystyle\prod_{l\in H}(X-\zeta^l) \in K[X]$$ ?

I have tried to inspect the roots/coefficients relations to see what it would yield but I don't seem to get anywhere.

• Wikipedia has a section on the cyclotomic polynomials over finite fields. – Arthur Dec 14 '18 at 23:03
• @Arthur indeed, thank you ! – Max Dec 14 '18 at 23:29
• To give the possibly best known characteristic zero example: $$\Phi_5(x)=(x^2+\frac{1+\sqrt5}2x+1)(x^2+\frac{1-\sqrt5}2x+1)$$ over $\Bbb{Q}(\sqrt5)$. All the zeros of $\Phi_5$ are complex and $\Bbb{Q}(\sqrt5)$ is real, so there are no zeros in that field. – Jyrki Lahtonen Dec 16 '18 at 6:55
• May be $$\phi_8(x)=x^4+1=(x^2+1)^2-(\sqrt2x)^2=(x^2+\sqrt2x+1)(x^2-\sqrt2x+1)$$ is actually even better known for many of us, because it is needed when integrating $1/(x^4+1)$ with the partial fractons method? – Jyrki Lahtonen Dec 16 '18 at 7:05

If you choose $$K=\mathbb{Q}[\zeta]^H$$, it should work, shouldn’t it?

• Right, I definitely should have thought of that, thank you ! – Max Dec 14 '18 at 23:30