Irreducibility of $x^5-10x^4+55x^3-110x^2+184x-60$ in $\mathbb{Q}[x]$

Question

Let $g(x)$ be the above polynomial, I know that this polynomial is indeed irreducible, but I'm not entirely sure how to show it since the reduction mod $p$ test gives $g(k) \neq 0 \enspace \forall \enspace k \in \mathbb{Z_{11}}$ and mostly $g(k) = 0$ for some $k \in \mathbb{Z_{p}}$ for primes $p=\{2,3,5,7,11\}$. Is it enough to say that it's reducible in $\mathbb{Z_{11}}$ that it is irreducible in $\mathbb{Q}[x]$? Is there any way to get to 11 without brute force calculations?

Answer 1

The polynomial is irreducible over $\Bbb F_{11}$ by Berlekamp, or by a direct computation, i.e., $f=x^5+x^4+8x+6$ cannot be written as a quadratic polynomial times a cubic one, by considering the linear equations over $\Bbb F_{11}$ arising by comparing coefficients. This is easy and not really a "brute force" calculation.

Hence $x^5-10x^4+55x^3-110x^2+184x-60$ is irreducible over $\Bbb Q$.

Edit: From $$ (x^3+ax^2+bx+c)(x^2+dx+e)=x^5-10x^4+55x^3-110x^2+184x-60 $$ we obtain the equations \begin{align*} ce & = -60,\\ a+d & -10,\\ ad + b + e & =55,\\ ae + bd + c & = - 110,\\ be + cd & = 184. \end{align*} Over $\Bbb F_{11}$ we can easily solve these equations, i.e., we obtain $$ d= 10(a + 10),\; e= a^2 + 10a + 10b, c= -a^3 + a^2 + 2ab - b, $$ which gives two equations in $a$ and $b$, which are contradictory.

Answer 2

If the polynomial is reducible in $\Bbb{Z}[x]$ then $g=fh$ for some nonconstant polynomials $f,h\in\Bbb{Z}[x]$ and so $\overline{g}\equiv\overline{f}\overline{h}\pmod{p}$, meaning that $\overline{g}$ is reducible in $\Bbb{F}_p[x]$. It follows that if $\overline{g}$ is irreducible in $\Bbb{F}_p[x]$ for some $p$, then it is irreducible in $\Bbb{Z}[x]$.

However, to show that $\overline{g}$ is irreducible in $\Bbb{F}_p[x]$ it does not suffice to show that $\overline{g}(k)\neq0$ for all $k\in\Bbb{F}_p$. This only shows that $\overline{g}$ has no linear factor; it might still be the product of an irreducible quadratic and cubic.

You have already checked that $g$ is reducible mod $2$, $3$, $5$ and $7$, so the smallest possible prime that can show $g$ is irreducible is $11$. You have already checked that $g$ has no linear factors, so by the argument above it remains to check that it has no quadratic factors. You can check by brute force that $g$ is not the product of an irreducible quadratic and cubic, either showing that $$g=(x^3+ax^2+bx+c)(x^2+dx+e),$$ has no solution $a,b,c,d,e\in\Bbb{F}_{11}$, or alternatively, by showing that $$g(a+bi)=0,$$ has no solution $a,b\in\Bbb{F}_{11}$, where $i^2=-1$.