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This is an exercise for the book Abstract Algebra by Dummit and Foote (pg. 519): show $x^{5}-ax-1\in\mathbb{Z}[x]$ is irreducible for $a\neq-1,0,2$

I need help with this exercise, I don't have an idea on how to prove there are no quadratic factors for $a\neq-1,0,2$ (for $a=-1$ there is a quadratic factor).

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Irreducible over $\mathbb{Q}$ or just the integers? –  Bombyx mori Sep 9 '12 at 15:43
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Actually, in this case the irreductibility over $\mathbb{Q}$ is equivalent to the irreductibility over $\mathbb{Z}$, by Gauss' theorem. –  Radu Titiu Sep 9 '12 at 15:51
    
@user32240 - by gauss lemma a factorization over $\mathbb{Q}$ yeild a factorization over $\mathbb{Z}$ –  Belgi Sep 9 '12 at 15:52
    
@Belgi: I see. Thanks for reminding. –  Bombyx mori Sep 9 '12 at 15:57
    
It might not be pretty, and it might not lead to an answer, but I would just assume there was a factorization $(xp(x) + 1)(xq(x) -1)$, write out the four different possibilities of the degree of the monic polynomials $p$ and $q$, assume the quadratic, cubic and quartic terms vanish, and see what the linear term must be. –  Arthur Sep 9 '12 at 16:07
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3 Answers

up vote 2 down vote accepted

Let $P=X^5-aX-1$. If $P$ had a rational root $\frac{p}{q}$ with $p$ and $q$ coprime, we would deduce $p^5-apq^4-q^5=0$. Then $p$ divides $p^5-apq^4$, we see that $p$ divides $q^5$. By iterating Gauss' lemma, $p$ divides $q^4,q^3,q^2$ etc and finally $p$ divides $1$, so $p=\pm 1$. Similarly $q$ divides $p^5$ so $q=\pm 1$. So the only rational roots possible are $-1$ (corresponding to $a=2$) and $1$ (corresponding to $a=0$). So $P$ has no degree $1$ factors.

So all we've left to show is that there is no factor of degree $2$. So assume $P=UV$, with $U,V \in {\mathbb Z}[X]$ and ${\sf deg}(U)=3, {\sf deg}(V)=2$. Since $P$ is unitary, $U$ and $V$ are unitary also (we replace them by their opposites if necessary).

Write $U=X^3+u_2X^2+u_1X+u_0$ and $V=X^2+v_1X+v_0$. Then,

$$ P=UV=X^5+(u_2+v_1)X^4+(u_2v_1+u_1+v_0)X^3+(u_2v_0+u_1v_1+u_0)X^2+(u_1v_0+u_0v_1)X+u_0v_0 $$ Identifying the coefficients in $X^4,X^3,X^2$, we express $u_0,u_1,u_2$ in terms of the other coefficients : $$ u_2=-v_1,u_1=v_1^2-v_0,u_0=2v_0v_1-v_1^3 $$ Then the product $UV$ becomes $$ UV=X^5-(v_1^4-3v_0v_1^2+v_0^2)X+(2v_0^2v_1-v_0v_1^3) $$ The constant coefficient can be factorized as $v_0v_1(2v_0-v_1^2)$. So $v_0,v_1$ and $2v_0-v_1^2$ must all be equal to $1$ or $-1$. So necessarily $v_0=1,v_1=-1$, and hence $U=X^3+X^2-1,V=X^2-X+1,a=-1$.

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This exercice is corrected here (example [8]): http://mathbyjames.files.wordpress.com/2011/06/ch13sec1part2.pdf

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Here's a messy answer; they probably wanted something nicer. Any putative factorization looks like $$ x^5 - ax - 1 = (x^2 + Bx \pm 1)(x^3 + Dx^2 + Ex \mp 1) $$ In the case that the first sign is $+$ and the second is $-$, we get the four equations $$ B + D = 0 $$ $$ E + BD + 1 = 0 $$ $$ -1 + BE + D = 0 $$ $$ -B + E = a $$ Adding the second two equations together gives $(B+1)(D+E) = 0$. If $B = -1$ then $D = 1$ and $E = 0$, so $a = -1$. Else $D = -E$, and then the first and last equation together give $a = 0$.

In the case that the first sign is $-$, the equations are

$$ B + D = 0 $$ $$ E + BD - 1 = 0 $$ $$ 1 + BE - D = 0 $$ $$ B - E = a $$ Suppose $D$ is positive. Then $B$ is negative (first equation) and so $E$ is positive and greater than $D$. Then equation three is contradicted. So $D$ is negative and $B$ is positive, so $E$ is positive by equation 2; this also contradicts equation 3.

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(there's roughly a 70% chance of a mistake in there as I was free-writing without editing, please feel free to edit and fix the argument.) –  user29743 Sep 9 '12 at 16:15
    
I think you missed two possibilities, namely $(x^4 + Bx^3 + Cx^2 + Dx \pm 1)(x \mp 1)$. –  Arthur Sep 9 '12 at 16:20
    
The OP has already dealt with the case where there is a linear factor (otherwise, I agree, those would need to be treated separately). –  user29743 Sep 9 '12 at 16:26
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