It is possible, for $p\in\mathbb{N}$ prime, that the polynomial $x^p-x+1$ is irreducible in $\mathbb{Z}_p$?
By the identity $a^p\equiv a$ mod $p$ for any $a\in \mathbb{Z}_p$ surely there is not a root for this polinomial.
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Bart Michels
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Federico Fallucca
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Is $\mathbb Z_p = \mathbb Z/p$ or the $p$-adic integers? – Bart Michels Jun 15 '18 at 15:36
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Sorry, $Z_p=Z/pZ$ – Federico Fallucca Jun 15 '18 at 15:37
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1This is an Artin-Schreier polynomial. Google this. If it had a root then it completely splits (“one out, all out”). – Nicky Hekster Jun 15 '18 at 15:38
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Presumably the intended question was is this polynomial irreducible for all primes $p$? – Derek Holt Jun 15 '18 at 15:44
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Yes I want to prove what you said – Federico Fallucca Jun 15 '18 at 15:50
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1Several approaches to proving this are given here. But, barto's method is nice (+1) even though variants of it are also in that "mother" thread. – Jyrki Lahtonen Jun 23 '18 at 06:30
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It is always irreducible: by Little Fermat, if $f(x)$ is a monic irreducible factor, then so is $f(x+n)$ for all $n$. Let $k>0$ be minimal for which $f(x) = f(x+k)$. Then $k \mid p$, hence $k=1$ or $k=p$.
If $k=p$, we conclude that all $f(x+n)$ are distinct mod $p$ for $n=0,1,\ldots, p-1$. Hence $f$ has degree $1$. But $x^p-x+1$ has no roots, contradiction.
If $k=1$, $f(x)=f(x+1)$ hence $f(x)-f(0)$ has $p$ roots, so $f$ has degree $p$, and $x^p-x+1$ is irreducible.
By the exact same argument, we see that $x^p-x-\alpha$ is irreducible for $\alpha \neq 0$.
Bart Michels
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