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Why is $x^6-x^5 + x^3 - x^2 + 1$ irreducible over $\mathbb{Z}[x]$?

It clearly has no integer roots, and in fact no real roots. Every polynomial with real coefficients can be written as a product of quadratic and linear terms. But I'm not sure how to factor the above polynomial. Maybe some sort of theorem involving irreducible integer polynomials might be useful?

N. F. Taussig
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user33096
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  • With respect to which set are you interested in factoring the polynomial? – N. F. Taussig Apr 01 '23 at 13:26
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    With respect to $\mathbb{Z}[x]$. Was that not clear in the problem statement? – user33096 Apr 01 '23 at 13:57
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    Observe that the equation $x^6 + x^3 + 1 = x^5 + x^2$ has no integer solutions due to opposite parity, (comparing LHS with RHS) – Dstarred Apr 01 '23 at 14:07
  • I did not find your goal clear since you cited the theorem that every polynomial with real coefficients can be expressed as a product of linear factors and irreducible quadratic factors with respect to the real numbers. – N. F. Taussig Apr 01 '23 at 15:21
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    It is irreducible modulo $2$, because in $\Bbb{Z}_2[x]$ it isn't divisible by any of the irreducible polynomials of degree $\le 3$: $x$, $x+1$, $x^2+x+1$, $x^3+x+1$, $x^3+x^2+1$ (a quick paper & pencil check gives this, ask if you need help). The claim follows from this because reduction modulo a prime is a homomorphism of rings. – Jyrki Lahtonen Apr 01 '23 at 15:52
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    @JyrkiLahtonen why is your last statement sufficient to solve the problem? Also, for finding irreducible polynomials, isn’t there a sieve of Eratosthenes method similar to the analogous one for prime number checking? – user33096 Apr 01 '23 at 15:57
  • Sufficiency is explained for example here. Yes, we can run an analogue of sieve of Eratosthenes to find irreducible monic polynomials in $\Bbb{Z}_p[x]$. But I think there are even more efficient methods to do the same. Of course, in this small case any method will do. The catch is that there are polynomials irreducible in $\Bbb{Z}[x]$ but reducible modulo every prime $p$. So this is not a universal tool, just one of many. – Jyrki Lahtonen Apr 01 '23 at 16:04
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    @user33096 For this problem Jyrki doesn't need a general method for finding irreducible polynomials over $\Bbb Z_2$, He just needs to know what they are up through degree 3. There are only 4 of these, and he already had them memorized. – MJD Apr 01 '23 at 16:34

3 Answers3

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Let $f(x) = x^{6} - x^{5} + x^{3} - x^{2} + 1 \in \mathbb{Z}[x]$. Consider the polynomial $g(x) = f(x+2) = 37 + 120x + 165x^2 + 121x^3 + 50x^4 + 11x^5 +x^6$. We note that the coefficients are all positive and bounded by $165$. We have that $g(172) = 27592587407701$ and $g(172)$ is prime. By Cohn's Criterion , $g(x)$ is irreducible in $\mathbb{Z}[x]$, and it follows $f(x)$ is irreducible in $\mathbb{Z}[x]$. A bit overkill, but it gets the job done.

conan
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We have a monic degree-$6$ polynomial with no real roots and constant term $1$, so if it is reducible it has a monic quadratic factor with constant term dividing $1$. There are only three such quadratic polynomials with nonreal roots: $x^2-x+1$, $x^2 + 1$, and $x^2+x+1$. None of these divide your polynomial, so it is irreducible.

Another method: Any monic factor $g(x)$ satisfies $g(x) > 0$ for all real $x$. Combined with the fact that $f(-1) = f(0) = f(1) = 1$, we have $g(-1) = g(0) = g(1) = 1$. Thus if $g(x)$ has degree $\le 2$, $g(x)$ must be a constant. So $f(x)$ has no degree-$2$ factors and is therefore irreducible.

Benjamin Wright
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    Interesting. I'm not sure I have ever seen the combination of lack of real roots and constant term $+1$ used in this way :-) – Jyrki Lahtonen Apr 05 '23 at 05:27
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Here is a similar proof as one in @conan's answer, with a smaller prime. We use Murty's irreducibility criterion:

Let $f(x)=a_mx^m+a_{m-1}x^{m-1}+\dots+a_1x+a_0$ be a polynomial of degree $m$ in $\mathbb{Z}[x]$ and set $$H=\max_{0\leq i\leq m-1} |a_i/a_m|.$$ If $f(n)$ is prime for some integer $n\geq H+2$, then $f(x)$ is irreducible in $\mathbb{Z}[x]$.

We can use this directly on our polynomial $f(x)=x^6-x^5 + x^3 - x^2 + 1$ where $H=1$ and for $n=4$ we get $f(n)=3121$ a prime.

(Note that using it on the reciprocal $\overline{f}(x)=x^6-x^4+x^3-x+1$ for $n=3$ gives slightly smaller prime $\overline{f}(n)=673$).

Sil
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