Irreducibilty of polynomial $x^9-6x^6+282x^3-8$ over $\mathbb {Q} $

Solution 1:

The $2$-adic Newton polygon for $f(x) = x^9-6x^6+282x^3-8$ immediately tells us that if $f(x)$ is reducible, its factorization must look like $$f(x) = (x^6 + \ldots \pm 2)(x^3 + \ldots \mp4).$$ In particular, $f(x)$ must have three roots whose product is $\pm4$. Using any root estimation technique (see below), we find that $f(x)$ has six roots satisfying $2 < |x| < 3$ and three roots satisfying $1/4 < |x| < 1/3$. But no combination of three values from those ranges has product with absolute value $4$, so $f(x)$ is irreducible.

To estimate the roots of $f(x)$, we can use an (unfortunately) obscure technique which is essentially a version of Newton polygons for $\mathbb{C}$ instead of $\mathbb{C_p}$:

For this problem, the upper convex hull of the points $(i, \log|a_i|)$ has vertices at $(0, \log 8)$, $(3, \log 282)$, and $(9, 0)$. As with $p$-adic Newton polygons, an estimation for the logarithm of the absolute value of the roots, $\log|x_i|$, is given by the negative slopes of the convex hull. In this case we have six roots with $\log|x_i| \approx -\dfrac{0-\log 282}{9-3} = \log \sqrt[6]{282}$ and three roots with $\log|x_i| \approx -\dfrac{\log 282 - \log 8}{3-0} = \log\sqrt[3]{\dfrac{8}{282}}$, which gives us the estimates above, noting that $2 < \sqrt[6]{282} < 3$ and $\dfrac{1}{4} < \sqrt[3]{\dfrac{8}{282}} < \dfrac{1}{3}$.

To formalize this into a proof, we would use Rouché's Theorem in the two annuli $2 < |x| < 3$ and $1/4 < |x| < 1/3$.

This method is essentially what is used in some implementations of the Aberth root-finding algorithm as a first approximation of the roots; see the paper Numerical computation of polynomial zeros by means of Aberth's method by Bini (1996). One word of caution: this technique is very accurate when the polynomial has roots that are well-separated, but gives inaccurate results when there are several roots that are close together on a log scale.

Solution 2:

The polynomial $P(x)$ factors $\!\!\!\mod \!\!13$ as $(x^3+6) ( x^6+ x^3 + 3) $ (link).

Assume that $P(x)$ is reducible. We conclude that $P(x)$ is a product of two polynomials $P_1(x)$, $P_2(x)\in \mathbb{Z}[x]$. with $P_1(x)\equiv x^3 + 6 \!\!\!\mod \!\!13$, $P_2(x)\equiv x^6 + x^3 + 3 \!\!\!\mod\!\! 13$.

Now, $P(1) =269$, a prime number. We conclude $P_1(1)=\pm 1$, or $P_2(1)=\pm 1$. But $P_1(1)\equiv 7 \!\!\!\mod \!\!13$, while $P_2(1)\equiv 5 \!\!\!\mod \!\!13$, contradiction.

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$\bf{Added:}$ About the factorization $\mod 13$. Our polynomial is $P(x) = Q(x^3)$, with $Q(x)= x^3 - 6 x^2 + 282 x - 8$. Now $Q(x)$ has root $7$ $\mod 13$, so we have $Q(x) = (x-7)(x^2 + x + 3)$, and so $P(x) = (x^3 - 7)(x^6+x^3 + 3)$. Now $x^3-7$ has no root $\mod 13$, because $7^{\frac{12}{3}}= 7^4 \not \equiv 1 \mod 13$. Also, $x^2 + x + 3 = x^2 + 2 \cdot 7 x + 7^2 + 3 - 7^2= (x+7)^2 - 7$, and $7$ is not a quadratic residue $\mod 13$, so $x^2 + x + 3$ is irreducible.

Based on Jyrki's idea: Note that since $13\equiv 1 \mod 3$, there exists $\omega \in \mathbb{F}_{13}$, such that $\omega^3 = 1$, and $\omega\ne 1$ ( a primitive root of $1). \mod 13$). Now, consider the action of the group of roots of $1$ on monic polynomials $f(x) \mapsto \frac{1}{\omega^{\deg f}} f(\omega x)$. The polynomials in $x^3$ are invariated by the action ( the fixed points are $x^m \cdot $ pol in $x^3$). Therefore, the a polynomial in $x^3$, its irreducible factors will be either polynomials in $x^3$, or group in orbits of size $3$ under this action. Now, if the polynomial $x^6 + x^3 + 3$ were reducible, it would equal $R(x) R(\omega x) R(\omega^2 x)$, where $R$ is a polynomial of degree $2$. But that would imply that the free term $3$ is a cube. However, we have $3^{\frac{12}{3}} = 3^4 =3 \mod 13$. Therefore, $x^6 + x^3 + 3$ is irreducible $\mod 13$.