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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!

Travis Willse
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mayflowers46
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    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. – Travis Willse Nov 19 '19 at 03:53
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    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... – Jyrki Lahtonen Nov 19 '19 at 04:01

2 Answers2

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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.

Travis Willse
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  • Good job! ${}{}$ – Jyrki Lahtonen Nov 19 '19 at 04:11
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    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. – Travis Willse Nov 19 '19 at 04:15
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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.

Sarvesh Ravichandran Iyer
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    Nice! ${}{}{}{}$ – Jyrki Lahtonen Nov 19 '19 at 04:17
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    @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. – Sarvesh Ravichandran Iyer Nov 19 '19 at 04:18
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    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. – Jyrki Lahtonen Nov 19 '19 at 04:26
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    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. – Jyrki Lahtonen Nov 19 '19 at 04:39
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    That's fairly brilliant, I will edit my answer and add it. – Sarvesh Ravichandran Iyer Nov 19 '19 at 04:45
  • Even if the full factorization cannot be determined, computing the factorizations for the first few dozen primes suggests that $x^6 + 108$ factors modulo $p$ (at least) into a product of two cubics for $p \equiv 1 \pmod 6$ and (at least) into a product of three quadratics for $p \equiv -1 \pmod 6$. – Travis Willse Nov 19 '19 at 05:12
  • @TravisWillse That is a good point, I will see if anything fits with it. – Sarvesh Ravichandran Iyer Nov 19 '19 at 05:15
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    @TravisWillse Chebotaryev's density theorem tells us that this factors into three quadratics for one half of the primes (the conjugacy class of 2-cycles in $S_3$ covers half the elements), into two cubic for one third of the primes (the conjugacy class of 3-cycles in $S_3$ has one third of the elements), and splits completely into six linear factors modulo one sixth of the primes (the identity element of $S_3$ is one sixth of the elements). – Jyrki Lahtonen Nov 19 '19 at 18:12
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    Frankly I was a bit surprised to see the division according to residue classes modulo six. I think I once calculated these for $x^3-2$, and I don't recall seeing that. Must come from the abelian intermediate field $\Bbb{Q}(\sqrt{-3})$ somehow, but I never truly grokked class field theory. – Jyrki Lahtonen Nov 19 '19 at 18:12