Prove that $x^4-x-1$ is irreducible over $\mathbb{Q}$

Question

Prove that $f(x)=x^4-x-1$ is irreducible in $\mathbb{Q}[x]$.

All methods I know failed. I can only exclude that $f$ admits a factorization with a factor of degree 3, because in this case $f$ would have a root in $\mathbb{Q}$, and I can prove that this is not the case. But I can't exclude $f=gh$ with $g,h$ both of degree $2$. I also know that $f$ has two real roots and a pair of conjugate complex roots, but don't know how to use this. I know that if $f$ were reducible over $\mathbb{Q}$ then it would be reducible over $\mathbb{Z}$, but again I don't know how to deduce irreducibility. What can be done in this case to prove that $f$ is irreducible?

Answer 1

In this case you can just look at $f$ in $\mathbb F_2[x]$. The only irreducible quadratic polynomial is $x^2+x+1$ and it doesn't divide $x^4+x+1=x^2(x^2+1)+(x^2+x+1)$.

Answer 2

You can use the Rational Root Theorem to solve this problem. The possible roots under consideration are $\pm 1$ (because the roots under consideration are $\frac{p}{q}$, (where $p$ is the coefficient of the last term and $q$ is the coefficient of $x^4$ in $f(x)$), and neither of them are roots of $f(x)$.

Answer 3

The basic idea behind the rational root theorem can be used to show that this can only factor into two quadratics if the factorization is of the form:

$$ x^4 - x - 1 = (x^2 + ax + 1) (x^2 + bx - 1) $$