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

Question

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?

Answer 1

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.

Answer 2

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.

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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.

Answer 3

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 $\overline{f}(n)=673$).