Is $x^6 + 108$ irreducible over $\mathbb{Q}$?

I'm trying to determine whether or not $$x^6 + 108$$ is irreducible over $$\mathbb{Q}$$. Is there an easy way to determine this ? I tried Eisenstein's Criterion directly, and with the substitutions $$x \longmapsto x + 1, x \longmapsto x-1, x \longmapsto x+2, x \longmapsto x-2$$ to try and show it is irreducible, but that didn't work out. Clearly, there's no linear factor of $$x^6 + 108$$ in $$\mathbb{Q}$$, since $$x^6 + 108$$ doesn't have a root in $$\mathbb{Q}$$. Showing there's no two cubic polynomials in $$\mathbb{Q}[x]$$ that multiply to $$x^6 + 108$$, or no cubic and quadratic polynomial in $$\mathbb{Q}[x]$$ that multiply to $$x^6 + 108$$ also quickly becomes a hassle. Is there an easier way to determine this more quickly?

Thanks!

• Since the polynomial is strictly positive it has no real roots, and so its roots are conjugate pairs. In particular, the polynomial is irreducible iff it factors as a product of a quadratic and a quartic. No need to check the cubic-cubic case. Nov 19, 2019 at 3:53
• I think $p$-adic Newton polygons (so Eisenstein on steroids) settle this. Only two terms in the polynomial, so its polygon is a line. Furthermore, $108=2^2\cdot3^3$. So any $2$-adic zero must have exponential valuation $1/3$, implying that the $2$-adic factors must be of degree three. Similarly any $3$-adic zero has valuation $1/2$, implying $3$-adic factors of even degree only. An eventual rational factor would also be both a $2$-adic and a $3$-adic factor (not necessarily irreducible), but we now know that its degree must be a multiple of both two and three so... Nov 19, 2019 at 4:01

Hint Since the polynomial $$x^6 + 108$$ is monic, if the field factors over $$\Bbb Q$$, it factors over $$\Bbb Z$$. Reducing modulo $$7$$, we have $$x^6 + 108 \pmod 7 \equiv x^6 - 4 \pmod 7 \equiv (x^3 + 2) (x^3 - 2) \pmod 7 .$$

Thus, if $$x^6 + 108$$ factors over $$\Bbb Q$$, it factors as a product of two cubics.

• Good job! ${}{}$ Nov 19, 2019 at 4:11
• Cheers, @JyrkiLahtonen ! I cut out the additional hint, as it suggested an unnecessary detour: As Anton points out in his first paragraph, having a cubic factor means that the polynomial must have a (real) root, which $x^6 + 108$ does not. Nov 19, 2019 at 4:15

Let $$x^6 + 108 = p(x)q(x)$$ for non-constant polynomials $$p,q$$ with coefficients in $$\mathbb Z$$. Then, note that $$p,q$$ cannot have any real roots, since $$x^6 + 108$$ does not. This rules out any of $$p,q$$ having odd degree. Thus, it must happen that $$\deg p = 2$$ and $$\deg q = 4$$, without loss of generality.

Therefore, going modulo $$p$$ for a prime $$p$$ we get $$x^6 +108 = \overline{p(x)}\overline{q(x)}$$ modulo $$p$$. However, note that $$\overline{p}$$ and $$\overline{q}$$ must be of degree at most $$2$$ and at most $$4$$, since the degree cannot increase while going modulo $$p$$.

In particular, suppose that $$x^6 + 108 = p_1(x)p_2(x)$$ for some cubic $$p_1,p_2$$ which are irreducible modulo $$p$$. Then, we have $$\bar p \bar q = p_1p_2$$ as two distinct factorizations of $$x^6 + 108$$. By irreducibility in $$\mathbb Z/p \mathbb Z$$ we get that $$\bar p$$ is either a multiple of either $$p_1$$ or $$p_2$$, a contradiction by degree.

Now, we look at $$x^6 + 108$$ modulo $$7$$. This gives $$x^6-4$$ modulo $$7$$, which becomes $$(x^3 + 2)(x^3 - 2)$$ , both of which are irreducible modulo $$7$$ since the only cubic residues modulo $$7$$ are $$0,\pm 1$$. Thus, the statement follows.

Also, I am inclined to think that $$x^6 + 108$$ is in fact reducible mod $$\mathbb Z/p \mathbb Z$$ for every $$p$$, but I can't see it immediately.

EDIT : As Jyrki points out below, the splitting field of the polynomial $$x^6 + 108$$ is the same as the splitting field of $$x^3- 2$$ (this is fairly easy to see from the factorization $$108 =2^23^3$$). Therefore the Galois group of $$x^6 + 108$$ is $$S_3$$, since that is true of $$x^3 - 2$$.

However, the brilliant Dedekind lemma, as stated by Yuan in the link provided in Jyrki's comment, has as a corollary the following : if $$f$$ is irreducible modulo $$p$$ for any $$p$$ not dividing the discriminant of $$f$$, then the Galois group of $$f$$ must contain an element of order $$\deg f$$.

One can calculuate the discriminant of $$x^6 - 2$$/ use other ways to see that it only has $$2$$ and $$3$$ as prime factors. Since the Galois group contains no elements of order $$6$$, it immediately follows that $$x^6 + 108$$ is reducible modulo $$p$$ for every $$p > 3$$. Along with reducibility for $$p=2,3$$ this proves the assertion I had made earlier.

• Nice! ${}{}{}{}$ Nov 19, 2019 at 4:17
• @JyrkiLahtonen Thank you. I would like you to think about the reducibility of this polynomial modulo every $\mathbb Z/p\mathbb Z$, it feels like it can be done. Nov 19, 2019 at 4:18
• Yes. The splitting field of this polynomial is $K=\Bbb{Q}(\root3\of2,\sqrt{-3})$. For $x_1=i\root3\of2\sqrt{3}$ is a zero, and the others are gotten by multiplying $x_1$ by sixth roots of unity, all in $K$. $K$ is also the splitting field of $x^3-2$, so we know that the Galois group is $S_3$. This has no elements of order six, so it splits modulo any prime $>3$ by Dedekind. Nov 19, 2019 at 4:26
• Qiaochu Yuan explains the theory here. I want to add that because this Galois group is not abelian, the factorization behavior modulo $p$ cannot be described in terms of the residue class of $p$ modulo some $m$. At least, if I have correctly understood a relevant piece of class field theory. When the Galois group is abelian, like here, then the remainder of $p$ modulo a conductor $m$ ($m=8$ in that example) will tell everything. Nov 19, 2019 at 4:39
• That's fairly brilliant, I will edit my answer and add it. Nov 19, 2019 at 4:45