# Showing $x^{5}-ax-1\in\mathbb{Z}[x]$ is irreducible

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).

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

• By "corrected" do you mean "solved"? 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.

• (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 $(x^4 + Bx^3 + Cx^2 + Dx \pm 1)(x \mp 1)$. 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