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).

  • $\begingroup$ Irreducible over $\mathbb{Q}$ or just the integers? $\endgroup$ Sep 9 '12 at 15:43
  • 2
    $\begingroup$ Actually, in this case the irreductibility over $\mathbb{Q}$ is equivalent to the irreductibility over $\mathbb{Z}$, by Gauss' theorem. $\endgroup$
    – Radu Titiu
    Sep 9 '12 at 15:51
  • $\begingroup$ @user32240 - by gauss lemma a factorization over $\mathbb{Q}$ yeild a factorization over $\mathbb{Z}$ $\endgroup$
    – Belgi
    Sep 9 '12 at 15:52
  • $\begingroup$ @Belgi: I see. Thanks for reminding. $\endgroup$ Sep 9 '12 at 15:57
  • $\begingroup$ 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. $\endgroup$
    – Arthur
    Sep 9 '12 at 16:07

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 monic, $U$ and $V$ are monic 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 $U=X^3+X^2-1,V=X^2-X+1,a=-1$.


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

  • $\begingroup$ By "corrected" do you mean "solved"? $\endgroup$
    – Junglemath
    Aug 31 '20 at 2:47

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.

  • $\begingroup$ (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.) $\endgroup$
    – user29743
    Sep 9 '12 at 16:15
  • $\begingroup$ I think you missed two possibilities, namely $(x^4 + Bx^3 + Cx^2 + Dx \pm 1)(x \mp 1)$. $\endgroup$
    – Arthur
    Sep 9 '12 at 16:20
  • $\begingroup$ The OP has already dealt with the case where there is a linear factor (otherwise, I agree, those would need to be treated separately). $\endgroup$
    – user29743
    Sep 9 '12 at 16:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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