$X^3-X-1$ irreducible in $\mathbb{Q}[X]$

Question

I need to show that $f=X^3-X-1$ irreducible in $\mathbb{Q}[X]$. But reduction modulo $3$ gives $f_3=X^3+X+1$ and $f_3(1)=0$. Where is my mistake?

Answer 1

If a cubic polynomial is reducible, at least one of the factors must be linear. But the only possible rational roots are integer divisors of $-1$, i.e. $\pm 1$. Substituting into the given polynomial, we see that neither $1$ nor $-1$ is a root. So the polynomial has no linear factor and thus is irreducible.