Let $f_p(x)=x^{p-1} +x^{p-2} +......+x+1$ where $p$ is any prime then I know that $f_p(x)$ is irreducible over $\mathbb{Q}$.
Is $f_p(x^n)$ irreducible over $\mathbb{Q}?$ Is there any condition on $n$ or $f_p(x^n)$ is also irreducible over $\mathbb{Q}$ for every $n\in \mathbb{N}$ where $\mathbb{N}$ is set of positive integers.
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dezdichado
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ASHWINI SANKHE
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The roots of $h = f_p(x^n)$ are all $pn$th roots of unity that are not $n$th roots of unity.
So $\Phi_{pn}$ divides $h$. Therefore $h$ is irreducible if and only if it is itself a cyclotomic polynomial, that is if $\phi(pn) = n(p-1)$. This occurs if and only if $n = p^k$ for some $k \geq 0$.
user49640
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Here k can be any real no? – ASHWINI SANKHE Jun 14 '17 at 08:13
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Any nonnegative integer. – user49640 Jun 14 '17 at 08:15
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Here I am using the fact that all cyclotomic polynomials $\Phi_{p^k}$ are irreducible. – user49640 Jun 14 '17 at 08:16
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Ok but if I take k as any real no. then will your result work? – ASHWINI SANKHE Jun 14 '17 at 08:30
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If $k$ can be any real number, then $p^k$ can be any positive real number, so none of what I said will make any sense. – user49640 Jun 14 '17 at 08:32
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I think it will work because ultimately we are taking mod p to $f_p(xn)$ by substituting x as x+1. – ASHWINI SANKHE Jun 14 '17 at 08:36
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1I'm sorry, but I don't think I can say any more about this. – user49640 Jun 14 '17 at 08:40
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Consider the polynomial $$f_p(x^p)=1+x^p+x^{2p}+\dots +x^{p(p-1)},$$ that is put $n=p$. I claim that this polynomial is irreducible over $\mathbb{Q}$. You can see the answer to this question for a proof of that fact. (Also, notice that the proof is more or less the same of the proof of the irreducibility of the $p$-th cyclotomic polynomial. We only have to take a detour over $\mathbb{Z}_p$).
Can you generalize the same approach to powers of $p$? The same proof as in the linked question will no longer work since $\mathbb{Z}_{p^k}$ is not a field if $k>1$.
Mathematician 42
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