Procedure for showing a polynomial is irreducible in $\mathbb{Q[x]}$

Question

Let $f(x) = x^5 − 6x^2 + 21x + 13$

What is the procedure for showing $f(x)$ is irreducible in $\mathbb{Q}[x]$?

Answer 1

Suppose $f(x) = p(x) q(x)$ over $\mathbb{Z}$, where we can take $p$ and $q$ monic and non-constant. Use subscript $l$ to indicate taking coefficients mod $l$. Note that $\deg p_l(x) = \deg p(x)$, etc., and $f_l(x) = p_l(x) q_l(x)$.

One computes $$ f_2(x) = x^5+x+1 = (x^2+x+1)(x^3+x^2+1) $$ $$ f_3(x) = x^5 + 1 = (x+1)(x^4-x^3+x^2-x+1) $$

The factors of $f_2(x)$ have no roots, so are irreducible. The quartic factor of $f_3(x)$ also has no root, so if it factors, it's the product of monic irreducible quadratics. The quadratics must have discriminant $-1$ since otherwise they would factor with the quadratic equation, hence we get factors of the form $x^2+bx+c$ where $b^2-c = -1$. The possible quadratics are $x^2+1$, $x^2+x-1$, $x^2-x-1$, and no two of these give the quartic as product, so the quartic is irreducible

Hence the degrees of $p$ and $q$ are ($2$ and $3$ from $l=2$) and ($1$ and $4$ from $l=3$), a contradiction, so $f$ is irreducible over $\mathbb{Z}$, hence over $\mathbb{Q}$.

Maybe you can find a less computational way to show the quartic is irreducible....

Answer 2

You can use the following arguments $f$ does not have an integer root so there is no linear factor of f and alsothere is no factore of degree. If f is not irreducible then $$x^5-6\,x^2+21\,x+13=\\ \left(x^2+a\,x+b\right)\,\left(x^3+c\,x^2+d\,x+e\right)=\\ x^5+\left(c+a\right)\,x^4+\left(d+a\,c+b\right)\,x^3+\left(e+a\,d+b \,c\right)\,x^2+\left(a\,e+b\,d\right)\,x+b\,e$$ We have $c=-a$ because the coefficient of $x^4$ vanishes.

So we get $$x^5-6\,x^2+21\,x+13=\\ x^5+\left(d+b-a^2\right)\,x^3+\left(e+a\,d-a\,b\right)\,x^2+\left(a \,e+b\,d\right)\,x+b\,e $$

$b$ must be a factor of $13$ so we have only four possible pairs for $(b,e)$. We can enlist all possible values for a,b,d,e. None of these quadruples will fit for the coefficient of power 3 and 1

$$ \begin{array}{ r | r | r | r | r | r | r } b\mid 13 & e=\frac{13}{6} & -6-e & a \mid (-6-e) & d = \frac{-6-e}{a}+b & d+b-a^2 & ae+bd \\ \hline{} -13 & -1 & -5 & -5 & -12 & -50 & 161 \\ -13 & -1 & -5 & -1 & -8 & -22 & 105 \\ -13 & -1 & -5 & 1 & -18 & -32 & 233 \\ -13 & -1 & -5 & 5 & -14 & -52 & 177 \\ -1 & -13 & 7 & -7 & -2 & -52 & 93 \\ -1 & -13 & 7 & -1 & -8 & -10 & 21 \\ -1 & -13 & 7 & 1 & 6 & 4 & -19 \\ -1 & -13 & 7 & 7 & 0 & -50 & -91 \\ 1 & 13 & -19 & -19 & 2 & -358 & -245 \\ 1 & 13 & -19 & -1 & 20 & 20 & 7 \\ 1 & 13 & -19 & 1 & -18 & -18 & -5 \\ 1 & 13 & -19 & 19 & 0 & -360 & 247 \\ 13 & 1 & -7 & -7 & 14 & -22 & 175 \\ 13 & 1 & -7 & -1 & 20 & 32 & 259 \\ 13 & 1 & -7 & 1 & 6 & 18 & 79 \\ 13 & 1 & -7 & 7 & 12 & -24 & 163 \end{array} $$

A simpler way is to notice that $f(27)=14345113$ is prime. An explanation is here: Showing the polynomial $x^4 + x^3 + 4x + 1$ is irreducible in $\mathbb{Q[x]}$.

In school mathematic exercises often Eisenstein's criterion can be used, but I don't see how it can be used here. If one changes the constant term from $13$ to $12$ one gets $$x^5-6\,x^2+21\,x+12$$ and Eisensteins's criterion can be applied immediately: Modulo the prime $3$ all the coefficients vanish except the coefficient of $x^5$, but the constant term $12$ does not vanish modulo $3^2$.