2

Is the following polynomial irreducible in $\mathbb{Z}[x]$ or in $\mathbb{Q}[x]$?

$x^5 - 5x^4 + 7x^3 + x^2 + x - 1$

If it's reducible, there should be a linear factor with degree $1, 2$ or $3.$

I try $\mathbb{Z}_2[x]$ and we get $x^5 +x^4 + x^3 + x^2 + x + 1$ looking for roots $f(0) = 1$ and $f(1) = 6 = 6 \mod 2 = 0$. So it's reducible in $\Bbb Z_2[x]$?

Does this help me? I see the polynomial is also primitive, so if I prove that it's irreducible in $\mathbb{Q}[x]$ or $\mathbb{Z}[x]$, then it's also irreducible in $\mathbb{Z}[x]$ or $\mathbb{Q}[x]$.

If I use rational root system, there is no root for $x^5 - 5x^4 + 7x^3 + x^2 + x - 1$, so there can only be a decomposition in degree $2+3$.

I don't know how to show now, that it's irreducible.

Aryaman Maithani
  • 15,771
Vek
  • 303
  • The statement "a polynomial $f(x)$ is irreducible iff $f(x)$ (mod $p$) is irreducible" holds if degree($f) \leq 3$. – MAS Jul 10 '21 at 08:07
  • That is not correct @Why. For example $f(x)=x^2+1$ is irreducible over $\Bbb{Q}$, but modulo $p=5$ we have $f(x)=x^2-4=(x-2)(x+2)$. – Jyrki Lahtonen Jul 10 '21 at 08:30
  • 1
    (cont'd) If you meant that "When $\deg f(x)\le 3$ then $f(x)$ is irreducible over $\Bbb{Q}$ if and only if it is irreducible modulo $p$ for some prime $p$", then that is correct. – Jyrki Lahtonen Jul 10 '21 at 08:38
  • Another way to try: translate and then Eisenstein? – xbh Jul 10 '21 at 08:49
  • 1
    @JyrkiLahtonen, thanks. Yes, my statement was not in clear but I indeed meant your 2nd comment – MAS Jul 10 '21 at 08:55
  • 2
    Over ${\bf Z}_2$, it's $x^4(x+1)+x^2(x+1)+(x+1)=(x^4+x^2+1)(x+1)$. But that doesn't prove it's reducible over the integers. Also, linear factor means degree one. – Gerry Myerson Jul 10 '21 at 09:02
  • Modulo $2$ this is $(x+1)(x^2+x+1)^2$ and modulo $3$ we get $(x^2+2x+2)(x^3+2x^2+x+1)$. It is irreducible modulo five, but proving that by paper & pencil calculations is unpleasant. Basically you would need to show that it has no common factors with $x^{24}-1$ modulo five. – Jyrki Lahtonen Jul 10 '21 at 09:27
  • For $a \in \Bbb Z$ with $|a| \le 1000$: the non-leading coefficients of $g(x - a)$ have gcd $1$. So this should convince one that we must do something apart from Eisenstein. – Aryaman Maithani Jul 10 '21 at 09:57
  • We have $$f(2x+1)=2(16x^5-12x^3+4x^2+9x+2 )=2g(x)$$ and we can apply Murty's criterion on $g(x)$ with $g(5)$ being prime. Thus the given polynomial is irreducible. – Paramanand Singh Aug 03 '21 at 13:23

1 Answers1

8

By the Rational Root Theorem, the only possible linear factors are $(x\pm1)$, but evaluation at $\mp1$ shows that this is not the case.

Modulo $2$, we have the factorization $$x^5+x^4+x^3+x^2+x+1\equiv (x^2+x+1)(x^3+1)\equiv (x^2+x+1)^2(x+1)$$ Hence any quadratic factor in $\Bbb Z[x]$ must $\equiv x^2+x+1\pmod 2$, i.e, all coefficients odd, moreover leadeing $1$, and of course constant term $\pm1$, in other words: $ x^2+ax\pm1$ with odd $a$. Similarly, the corresponding cubic factor must be $x^3+bx^2+cx\mp1$ with $b,c$ even. We compute $$(x^2+ax\pm1)(x^3+bx^2+cx\mp1) =x^5+(a+b)x^4+(ab+c\pm1)x^3+(ac\pm b\mp1)x^2+(\pm c\mp a)x-1 $$ so from the linear term $$ c=a\pm1$$ and from the fourth power term $$ b= -a-5.$$ Plug this into the cubic term $$ ab+c=7\mp1$$ to arrive at $$ a(a+4)=\pm2-7=-5\text{ or }-9,$$ which has no integer solution (two factors of $-5$ or $-9$ must differ by at least $6$).

Hagen von Eitzen
  • 374,180