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I am reading the group theory text of Eugene Dickson. Theorem 33 shows this polynomial is irreducible

$$ \frac{x^p - 1}{x-1} = 1 + x + \dots + x^{p-1} \in \mathbb{Z}[x]$$

He shows this polynomial is irreducible in $\mathbb{F}_q[x]$ whenever $p$ is a primitive root mod $q$.

By Dirichlet's theorem there are infinitely many primes $q = a + ke$, so this polynomial is "algebraiclly irreducible", I guess in $\mathbb{Q}[x]$.

Do you really need a strong result such as the infinitude of primes in arithmetic sequences in order to prove this result? Alternative ways of demonstrating this is irreducible for $p$ prime?

COMMENTS Dirichlet's theorem comes straight out of Dickson's book. I am trying to understand why he did it. Perhaps he did not know Eisenstein's criterion. It's always good to have a few proofs on hand.

Another thing is that Eisenstein's criterion is no free lunch since it relays on Gauss lemma and ultimately on extending unique factorization from $\mathbb{Z}$ to $\mathbb{Z}[x]$.

Martin Sleziak
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cactus314
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    You can use a simple change of variables to show it's irreducible over $\Bbb Q$ (via Eisenstein). I am not sure what you are using Dirichlet's theorem to conclude anyway - is irreducible mod infinitely many primes supposed to imply it's irreducible over $\Bbb Q$ or something? – anon May 11 '15 at 18:50
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    Do you want to show it is irreducible in $\mathbb Z[x]$ or in $\mathbb F_q[x]$? Because those are two different things. Over $\mathbb Z[x]$ I believe there's a cute trick using Eisenstein's criteria. – Gregory Grant May 11 '15 at 18:51
  • Look at this page: http://en.wikipedia.org/wiki/Cyclotomic_polynomial

    Look at the section on cyclotomic polynomials.

     – Gregory Grant May 11 '15 at 18:52
  • Actually the exact proof is on this page: http://en.wikipedia.org/wiki/Eisenstein%27s_criterion#Examples – Gregory Grant May 11 '15 at 18:53
  • Look at that last wiki page for the "cyclotomic polynomial" example – Gregory Grant May 11 '15 at 18:54
  • @GregoryGrant Eisenstein's criterion is not a total free lunch as you have to invoke Gauss' Lemma (also here) or detour into notions of total ramification of primes. – cactus314 May 11 '15 at 19:49
  • @johnmangual Sure but it's a whole lot easier than Dirichlet's theorem – Gregory Grant May 11 '15 at 21:48
  • If you just want a proof of Gauss' lemma, then go to Theorem 17.14 in this online textbook. – Noble Mushtak Apr 16 '16 at 19:43
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    To me this proof is completely absurd. Dirichlet's theorem is a thousand times more complicated than Eisenstein criterion (which can be proved without too much trouble by a good first year undergrad with a few hints), and using it to prove such an easy thing makes absolutely no sense to me. – Captain Lama Apr 16 '16 at 19:55
  • Related posts (for the case of $\mathbb Z[x]$ or $\mathbb Q[x]$): http://math.stackexchange.com/questions/82876/given-a-prime-p-in-mathbbn-is-a-fracxp2-1xp-1-irreducible-i http://math.stackexchange.com/questions/215042/irreducibility-of-xp-1-cdots-x1 http://math.stackexchange.com/questions/linked/215042 – Martin Sleziak Apr 18 '16 at 07:02

1 Answers1

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To clarify an issue that came up in the comments, by Gauss's lemma a monic polynomial is irreducible over $\mathbb{Q}[x]$ iff it's irreducible over $\mathbb{Z}[x]$, and by reduction $\bmod p$, if a polynomial is irreducible $\bmod p$ for any particular prime $p$, then it's irreducible over $\mathbb{Z}[x]$. So once you know that the polynomial is irreducible $\bmod q$ for any prime $q$ which is a primitive root $\bmod p$, it suffices to exhibit one such prime.

Unfortunately, as far as I know this is no easier than Dirichlet's theorem. Among other things, it's a straightforward exercise to show that Dirichlet's theorem is equivalent to the assertion that for any $a, n, n \ge 2$ such that $\gcd(a, n) = 1$ there is at least one prime (rather than infinitely many primes) congruent to $a \bmod n$.

Regarding your last comment, Eisenstein's criterion is still much, much, much easier to prove than Dirichlet's theorem, even with all of the details filled in.

Qiaochu Yuan
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