TITLE: $x^p-x-1$ is irreducible over $\mathbb{Q}$[x]
QUESTION [6 upvotes]: 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.

REPLY [4 votes]: 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.