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This question already has an answer here:

Eisenstein's criterion says that for a prime $p$ if the following conditions are satisfied for a primitive polynomial, $f(x)$, then that polynomial is irreducible in $\mathbb{Z}[x]$

$p \mid a_0, a_0, ..., a_{n-1}$

$p \not\mid a_n$

$p^2 \not\mid a_0^2$

Now I am almost certain I read somewhere that this lemma can be used to prove that the polynomial $x^{p-1} + x^{p-2} + ... + x^2 + x + 1$ is irreducible.

All coefficients are $1$ though and we have have an awkward amount of terms, $p - 1$ terms as opposed to $p$ terms - so how can Eisenstein's criterion be applied here?

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marked as duplicate by Git Gud, Martin Sleziak, Zev Chonoles, Bruno Joyal, user61527 Dec 1 '13 at 17:45

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  • $\begingroup$ Do you mean $x^{p-1}+x^{p-2}+....+1$? $\endgroup$ – LASV Dec 1 '13 at 17:09
  • $\begingroup$ Fixed the mistakes. $\endgroup$ – AlgebraGuy Dec 1 '13 at 17:13
  • $\begingroup$ There is another issue. EC tells you $p^2$ does not divide $a_0$. The way you have it, that is, $p$ divides $a_0$ but $p^2$ does not divide $a_0^2$, Eisenstein' would be impossible to use =] $\endgroup$ – LASV Dec 1 '13 at 17:17
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You can't directly apply Eisenstein to $f$. But you can combine it with the substitution criterion: Look at the substituted polynomial $g(x) = f(x+1)$.

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