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I'm studying for an upcoming exam, and have been given the following question:

For which primes p is the polynomial $f(x) = x^2 + x + 1$ irreducible in $\mathbb{Z}_p[x]$?

I think I can use Eisenstein's Criteria here, but I am confusing myself as I try to apply it. Any help is appreciated!

Sil
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GabeT
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1 Answers1

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Hint:

For $p=2$ the polynomial is easily seen to be irreducible.

For odd prime $p$, the polynomial has roots in $\mathbb Z_p$ (and hence reducible) if and only if $4x^2+4x+4=(2x+1)^2+3$ has roots. Do you know under what circumstances $x^2\equiv-3\pmod p$ has integer solutions?

awllower
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    You should look at $\frac{1+\sqrt{-3}}{2}$ not $\sqrt{-3}$ – reuns Dec 11 '19 at 16:58
  • That is why I exclude the even case at the beginning. – awllower Dec 11 '19 at 16:59
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    This is, of course, an argument that leads to a solution. But I, too, would go via the route that $x^2+x+1$ has solutions iff $x^3-1$ has solutions other than $x=1$, and use cyclicity of $\Bbb{Z}_p^*$. – Jyrki Lahtonen Dec 11 '19 at 17:02
  • Indeed. I didn't notice the relation with $x^3+1$ when I wrote this. – awllower Dec 11 '19 at 17:05
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    This is how we prove quadratic reciprocity : the splitting of $x^2-d\bmod p$ is complicated, the splitting of $x^{D}-1\bmod p$ is much easier, we can relate the two because $\Bbb{Q}(\sqrt{d})/\Bbb{Q}$ is an abelian extension thus contained in a cyclotomic field (which is shown to be $\Bbb{Q}(\zeta_d)$ or $\Bbb{Q}(\zeta_{4d})$) – reuns Dec 11 '19 at 17:12
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    No harm done. Likely this will get closed as a duplicate, but I don't have the time to look for one now. – Jyrki Lahtonen Dec 11 '19 at 18:15