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Find all finite fields making $f (x)=x^2+x+1$ irreducible.

Obviously, we have only two cases:

  1. For each prime $p $, $f $ is reducible over $\mathbb {F}_p $. Then, for any $n\ge 1$, it's reducible over $\mathbb {F}_{p^n}. $

  2. $f $ is irreducible over $\mathbb {F}_p $. Then, since the deg of $f $ is 2, $f $ is reducible over $\mathbb {F}_{p^{2n}} $ if $n\ge 1$.

So it suffices to classify the cases $\mathbb {F}_p. $ From a direct observation, $f $ is reducible over $\mathbb {F}_p $ iff $p $ is a factor of $n^2+n+1$ for some integer $n $. Can I reduce it more? I wonder if $f $ is reducible over $\mathbb {F}_p $ iff $p$ has the form $n^2+n+1$.

user253816
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  • Because $f(x)(x-1)=x^3-1$ we see that $f(x)$ has a zero in $\Bbb{F}{p^n}$, iff there is a non-trivial cubic root of unity in $\Bbb{F}{p^n}$. Because the multiplicative group is cyclic, this is the case, iff $3\mid (p^n-1)$. (Actually Cauchy's theorem is enough, we don't need cyclicity). – Jyrki Lahtonen Jul 11 '15 at 05:58
  • But I think this has been asked before. That's why I posted the argument as a comment rather than as an answer. Gimme a minute. My search-fu is weak. – Jyrki Lahtonen Jul 11 '15 at 05:59
  • The first related hit. Anyway the zeros of $x^2+x+1$ are either in the prime field $\Bbb{F}p$ or in its quadratic extension $\Bbb{F}{p^2}$. The former (resp. latter) case happens when $p\equiv1\pmod3$ (resp. $p\equiv-1\pmod3$). The field $\Bbb{F}{p^n}$ has $\Bbb{F}{p^2}$ as a subfield, iff $2\mid n$. If $p=3$, then $x=1$ is a double root. – Jyrki Lahtonen Jul 11 '15 at 06:02

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