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Possible Duplicate:
How do I prove that $x^p-x+a$ is irreducible in a field of $p$ elements when $a\neq 0$?

How to prove that $x^p - x - 1$ is irreducible over $F_{p}[x]$. I thought proving that this polynomial has no roots over $F_p$ but that is not enough to irreducibility.

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In a splitting field $K$, if $\alpha\in K$ is a root, then $(\alpha+1)^p= \alpha^p+1$ implies that $\alpha+1$ is a root as well. We conclude that the cyclic group of order $p$ operates on the roots (by addition of multiples of $1$), and necessarily does so transitively (as there are $p$ roots). A non-trivial factor of the polynomial would correspond to a non-trivial subgroup, which does not exist.

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  • $\begingroup$ Can you explain the last line "A non-trivial factor of the polynomial would correspond to a non-trivial subgroup, which does not exist." $\endgroup$ – Satvik Sep 2 '12 at 18:30
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    $\begingroup$ Maybe a different view helps): If we factor $X^p-X-1=f(X)g(X)$ non-trivially, then the Galois group leaves $f$ and $g$ fixed, hence the cyclic group of order $p$ operates on less than $p$ roots of $f$ and $g$ respectively. Since $p$ is prime, this operation must be trivial. But $\alpha\mapsto\alpha+1$ is definitely not the trivial operation. $\endgroup$ – Hagen von Eitzen Sep 2 '12 at 18:41
  • $\begingroup$ @Mystic: See zyx's or Greg Martin's answer to the duplicate question for an argument. (+1 to Hagen) $\endgroup$ – Jyrki Lahtonen Sep 2 '12 at 18:43
  • $\begingroup$ +1 Simple and beautiful. Yet, perhaps the following more expanded argument (which already exists in Hagen's answer) could prevent doubt: it's been proved $\,\alpha\,$ is a root $\,\Longrightarrow \alpha+1\,$ is a root. Now, the same argument shows that $$\alpha +1\,\,\text{is a root}\Longrightarrow \,\alpha +1+1=\alpha +2$$ is a root , and etc. One gets that $\,\alpha, \alpha+1,...,\alpha+p-1$ are all different roots of the polynomial... $\endgroup$ – DonAntonio Sep 3 '12 at 3:04

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