For any prime p, prove that $x^p-x-1$ is irreducible over $\mathbb{Q}$[x].

(In a field of characteristic p this is true).

I asummed exist root in $\mathbb{Q}$, let's call $\frac{\alpha}{\beta} \in \mathbb{Q}$. Then following that $\frac{\alpha ^p}{\beta ^p} - \frac{\alpha}{\beta}-1 = 0$ and then $\alpha ^p - \alpha \beta ^{p-1} - \beta ^p =0$. So $\alpha ^p = -\beta^p (1 + \alpha \beta ^{-1})$. But this only proves that $x^p -x -1$ don't have rational roots.

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    $\begingroup$ A polynomial can be reducible without having a root. For example, the polynomial $x^4+4$ is reducible in $\mathbb{Q}[x]$ since $$x^4+4=(x^2+2x+2)(x^2-2x+2)$$ but it has no roots in $\mathbb{Q}$. $\endgroup$ – Zev Chonoles Apr 19 '15 at 0:53
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    $\begingroup$ The polynomial does not have any rational root, your argument will work because you can add an hypothesis on $\gcd(\alpha,\beta)=1$ and you finish by the same equation but it's not pssible as $\alpha$ will divide $\beta^p$ and hence $\alpha=1$, But this is far to be a proof of irreductibility $\endgroup$ – Elaqqad Apr 19 '15 at 0:56
  • $\begingroup$ How about (Frank) Eisenstein's criterionhttp://en.wikipedia.org/wiki/Eisenstein's_criterion ? $\endgroup$ – gary Apr 19 '15 at 1:12
  • $\begingroup$ What @Zev said holds obviously for degree 3 and higher. $\endgroup$ – gary Apr 19 '15 at 1:14
  • $\begingroup$ What Zev says. And this polynomial is not irreducible in all fields of characteristic $p$. For example it has $p$ zeros in the field of $p^p$ elements and factors into linear factors over that field. $\endgroup$ – Jyrki Lahtonen Apr 19 '15 at 3:47

For instance we have $x^p-x-1$ is irreducible in $F_p$ which implies that the same polynomial is irreducible over the field of rationals, the converse is not correct we can find polynomial which are irreducible in the field of rationals and reducible for any prime $p$.

Assume that $P=x^p-x-1$ is reducible over rational numbers and write: $$P=QR$$ so in $F_p$ we have $\overline{P}=\overline Q\overline R$ which is a factorization if $F_p$ and this is a contradiction.

  • $\begingroup$ I don't know why are you changing the polynomial but note that the same thing holds for every polynomial of the form $x^p-x+a$ where $a\neq 0$ $\endgroup$ – Elaqqad Apr 19 '15 at 11:10
  • $\begingroup$ You should explain why $x^p-x-1$ is irreducible in $\Bbb F_p$, most readers will not find this obvious. $\endgroup$ – Alex M. Mar 5 '17 at 16:03

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