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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?

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Proving irreducibility over $\mathbb{Z}$ can be done in a straightforward crude way. –  André Nicolas Oct 1 '13 at 18:31
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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

3 Answers 3

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

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

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 $f(x)$.

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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
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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
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Oh I see. No... –  Anonymous Oct 1 '13 at 18:42

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:

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

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