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

Question

While trying to deal with the final parts of this answer I found that one needs to establish $$a=\sqrt[3]{1+\sqrt{11}}\notin\mathbb{Q} (\sqrt{11},\sqrt[3]{10})=F$$ Since both $a, F$ are of degree $6$ over $\mathbb {Q} $ the tower theorem does not help much.

Then I reasoned via contradiction. If $a\in F$ then $b=\sqrt[3]{1-\sqrt{11}}=-\sqrt[3]{10}/a\in F$ and hence $c=a+b\in F$.

One can check that $$c^3=a^3+b^3+3abc=2-3\sqrt[3]{10}c$$ or $$(c^3-2)^3=-270c^3$$ or $c$ is a root of $$f(x) =x^9-6x^6+282x^3-8$$ I checked using pari/gp that the polynomial $f(x) $ is irreducible over $\mathbb {Q} $. Further it can be observed that $f(289)$ is prime (also checked via pari/gp) so that the polynomial $f(x) $ is irreducible by Murty's criterion.

This shows that $c$ is of degree $9$ over $ \mathbb {Q} $ and hence $c\notin F$.

It can be observed that $f(x) $ can not be handled by Eisenstein as $2$ is the only prime which divides all non-leading coefficients and $4\mid 8$. I also checked the reducibility mod $3$ and the polynomial is reducible mod 3. So even that approach does not work. I don't know if using reducibility modulo other primes would help.

Is there is any other simpler way to prove the irreducibilty of $f(x) $ using hand computation?

Update: We have $$f(2x)=8(64x^9-48x^6+282x^3-1)=8g(x)$$ and it is somewhat easier to apply Murty's criterion on $g(x) $ with $g(8)=8577496063$ being prime. But I still find this unsuitable for hand calculation.

Answer 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.

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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.

Answer 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 $\mod 13$.

Answer 3

The following is the most elementary solution that I have.

Let us first consider the problem of factoring $f(x)$ modulo $7$. In the field $\Bbb{F}_7$ we have $11=4$, so $\sqrt{11}=2$. This gives us the counterparts $\tilde{a}=\root3\of{1+\sqrt{11}}=\root3\of{3}$, $\tilde{b}=-\root3\of{10}/a=-\root3\of{3}/a=-1$ and hence $\tilde{c}=\tilde{a}+\tilde{b}=\root3\of3-1$. It is easy to check that $3$ is not a cubic residue modulo $7$, so the minimal polynomial of $\tilde{c}$ over $\Bbb{F}_7$ is thus $$h(x)=(x+1)^3-3=x^3+3x^2+3x-2\in\Bbb{F}_7[x].$$ Paramanand Singh's calculations survive to the extent that we can conclude that $h(x)$ must be a factor of $f(x)$ modulo $7$.

Another ingredient is that $f(x)=x^9-6x^3+282x^3-8$ has the property $f(\omega x)=f(x)$ for any cubic root of unity $\omega$. This follows from the fact that all the terms of $f(x)$ have degrees divisible by three. In $\Bbb{F}_7$ we have the primitive third roots of unity: $\mu_3=\{1,2,4\}$. Therefore $h(2x)=x^3+5x^2+6x-2$ and $h(4x)=x^3+6x^2+5x-2$ must also be factors of $f(x)$ in $\Bbb{F}_7[x]$. Indeed, it is easy to verify that over $\Bbb{F}_7$ the factorization of $f(x)$ into irreducibles is $$f(x)=h(x)h(2x)h(4x).$$

This implies that any factorization of $f(x)$ over $\Bbb{Q}$ can only have factors of degrees $3$ or $6$, so it suffices to exclude the possibility of a cubic factor. We make the observation that the same applies over the field $\Bbb{Q}(\omega)$, $\omega=(-1+i\sqrt3)/2$. This is because the ring $E=\Bbb{Z}[\omega]$ of Eisensteinian integers is known to be a Euclidean domain (w.r.t. the complex norm) and also because the prime ideal $\mathfrak{p}=\langle 2+i\sqrt3\rangle$ has index seven in $E$. The argument is familiar. By Gauss's lemma a polynomial is irreducible over $\Bbb{Q}(\omega)$ iff it is irreducible over $E$, and then reduction modulo $\mathfrak{p}$ works the same way as reduction modulo $7$ does over $\Bbb{Z}$. This is because $E/\mathfrak{p}\simeq\Bbb{F}_7$.

Assume that $g(x)=x^3+Ax^2+Bx+C\in\Bbb{Z}[x]$ is a cubic factor of $f(x)$. By the above observations then $g(\omega x)$ and $g(\omega^2x)$ are also factors of $f(x)$, but possibly they reside in $E[x]$. Should we have $g(x)=g(\omega x)$ (when also $g(x)=g(\omega^2x)$ by complex conjugation), then $g(x)=x^3+C$. But this is impossible because $f(x)=F(x^3)$ with $F(x)=x^3-6x^2+282x-8$. The rational root test then shows that $F(x)$ is has no rational zeros, so $F(-C)\neq0$.

So we can conclude that $g(x)$, $g(\omega x)$ and $g(\omega^2x)$ are all distinct and irreducible over $E$. We are left with the possibility that $$f(x)=g(x)g(\omega x)g(\omega^2x).$$ Expanding everything gives $$g(x)g(\omega x)g(\omega^2x)=x^9+(A^3-3AB+3C)x^6+(B^3-3ABC+3C^2)x^3+C^3.$$ A comparison of the constant terms immediately reveals $C=-2$. Plugging that into the $x^6$-term gives the equation $A^3-3AB=0$. So either $A=0$ or $A^2=3B$.

• If $A=0$ a comparison of $x^3$ terms yields $B^3=270$ which is impossible given that $B$ is an integer.
• If $A^2=3B$ and $B^3+3AB=270$, then $$A^6+27A^3=27(B^3+3AB)=7290.$$ But it is easy to check that this has no integer solutions.

The irreducibility of $f(x)$ over $\Bbb{Q}$ follows from this.