Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Proving irreducibility over $\mathbb{Z}$ can be done in a straightforward crude way. – André Nicolas Oct 1 '13 at 18:31
If you had rational constants $a,b,c,d,$ what could you conclude from $$ (x^2 + a x + b)(x^2 + c x + d) = x^4 - x - 1? $$ – Will Jagy Oct 1 '13 at 18:31
up vote 5 down vote accepted

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 $x^4+x+1=x^2(x^2+1)+(x^2+x+1)$.

share|cite|improve this answer
You beat me to it. +1 and thanks for saving me the trouble :-) – Jyrki Lahtonen Oct 1 '13 at 18:34

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:

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

share|cite|improve this answer

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 $f(x)$.

share|cite|improve this answer
this excludes that $f$ has a rational root, not that $f$ is reducible – Danae Kissinger Oct 1 '13 at 18:35
What I forgot to mention is that because $f$ does not have a rational root, it is not factorizable over $\mathbb Q$. – Anonymous Oct 1 '13 at 18:36
Anonymous, what about $x^4 + 4 = (x^2 - 2 x + 2)(x^2 + 2 x + 2)?$ No rational roots. – Will Jagy Oct 1 '13 at 18:38
$$\begin{eqnarray}x^4+4&=&(x^2+2i)\cdot (x^2-2i)\\ &=& (x-(1-i))\cdot (x+(1-i))\cdot (x-(1+i))\cdot(x+(1+i)) \\ &=& ((x-1)+i)\cdot ((x-1)-i)\cdot((x+1)-i)\cdot((x+1)+i) \\ &=& ((x-1)^2+1)\cdot((x+1)^2+1).\end{eqnarray}$$ Reducible. – Anonymous Oct 1 '13 at 18:40
Oh I see. No... – Anonymous Oct 1 '13 at 18:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.