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Is it true that the polynomial $\frac{x^p- 1}{x-1}$ ($p$ is prime) is irreducible in $\mathbb{F}_2[x]$ iff $p$ is prime?

I know it will be true in $\mathbb Q[x]$. Can anyone please help me to understand what happens in $\mathbb{F}_2[x]$?

Batominovski
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cmi
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  • Are you sure that "($p$ is prime)" is supposed to be there? Should it, perhaps, be "($p$ is a natural number)"? – Arthur Dec 14 '18 at 15:30
  • https://en.wikipedia.org/wiki/Cyclotomic_polynomial – vadim123 Dec 14 '18 at 15:33
  • Is the statement true for any natural number?@Arthur – cmi Dec 14 '18 at 15:35
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    This is false when $p=7$. For prime $p> 2$, it is true if and only if $2$ is a generator mod $p$. – KCd Dec 14 '18 at 15:43
  • "2 is a generator mod p" - means?@KCd – cmi Dec 14 '18 at 15:45
  • It is never irreducible when $p$ is not a prime (characteristic zero factors survive reduction modulo two). Over $\Bbb{F}_2$ it is irreducible, as KCd said, if and only if $2$ is a primitive root modulo $p$. So it is irreducible for example, when $p=3,5,11,13,19$ and not irreducible when $p=7,17,23$. – Jyrki Lahtonen Dec 14 '18 at 16:18
  • Artin's conjecture suggests that we get irreducibility for roughly $40$ per cent of the primes. Observe that $2$ is a quadratic residue modulo $p$ whenever $p\equiv\pm1\pmod8$. In those case the polynomial will automatically be reducible. This already proves that the polynomial is irreducible for at most one half of the primes. – Jyrki Lahtonen Dec 14 '18 at 16:21
  • primitive root modulo p means?@JyrkiLahtonen – cmi Dec 14 '18 at 16:23
  • An integer $a$ is a primitive root modulo $p$ if $n=p-1$ is the smallest positive integer such that $a^n\equiv1\pmod p$. In other words, here we want the order of $2$ in the group $\Bbb{Z}_p^*$ to be $p-1$. See here for a bit more. – Jyrki Lahtonen Dec 14 '18 at 16:26
  • Yet in other words, $2^0,2^1,2^2,\ldots,2^{p-2}$ are all pairwise noncongruent modulo $p$. – Jyrki Lahtonen Dec 14 '18 at 16:35

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