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Let $F$ be any field with $\operatorname{Char} F=q$. Let $p$ be a prime such that $p\neq q$. Suppose $F$ has no $p$-th root of unity except $1$. Is it true that the cyclotomic polynomial $X^{p-1}+\dotsb+X+1$ is irreducible over $F$?

I could not find any reference for this result in the literature.

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    $\begingroup$ You could not find a proof because it is not true. For example, when $q = 2$ and $F$ is the field with $2$ elements, when $p = 7,$ the cyclotomic polynomial $x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1$ is not irreducible over $F$, but factors as a product of two irreducible polynomials of degree $3$. $\endgroup$
    – Geoff Robinson
    Commented Oct 4 at 18:13
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    $\begingroup$ Let $o$ be the multiplicative order of $q$ modulo $p$. In general, $\phi_p(X)$ factors over $\mathbb{F}_q$ into a product of $(p-1)/o$ irreducible polynomials of degree $o$. (See MO question 252835.) In particular, $\Phi_{p}(X)$ remains irreducible if and only if $q$ generates $\mathbb{F}_p^{\times}$. In Geoff Robinson's comment and Dave Benson's answer, $o=3$ since $2^3 \equiv 1 \bmod 7$ (equivalently, $2$ is a square as $2=4^2 \bmod 7$, and so $2$ generates a subgroup of index $2$). $\endgroup$
    – Ofir Gorodetsky
    Commented Oct 4 at 18:35

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No, it's not true. Take for example the cyclotomic polynomial $X^{6}+\dots+X+1$ over $\mathbb{F}_2$. This factorises as $(X^3+X^2+1)(X^3+X+1)$.

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    $\begingroup$ In fact for every prime $p$ and every $d$ dividing $p-1$, there exists a field of characteristic not $p$ over which $(x^p-1)/(x-1)$ factors into a product of irreducible polynomials of degree $d$ (and in particular, if $d\neq 1,p-1$, has no roots but isn't irreducible). $\endgroup$
    – Will Sawin
    Commented Oct 4 at 18:34

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