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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.

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    $\begingroup$ Wikipedia has a section on the cyclotomic polynomials over finite fields. $\endgroup$ – Arthur Dec 14 '18 at 23:03
  • $\begingroup$ @Arthur indeed, thank you ! $\endgroup$ – Max Dec 14 '18 at 23:29
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    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Dec 16 '18 at 6:55
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    $\begingroup$ 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? $\endgroup$ – Jyrki Lahtonen Dec 16 '18 at 7:05
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If you choose $K=\mathbb{Q}[\zeta]^H$, it should work, shouldn’t it?

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

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