# Splitting Field of $x^6-6x^3+7$

Given the polynomial $p(x)=x^6-6x^3+7 \in \mathbb Q$, find its splitting field $\mathbb F\subset \mathbb C$ and the Galois group of the extension $\mathbb F /\mathbb Q$. In fact, the exercise asked to show that $p$ is irreducible, then the degree and a basis of the splitting field as a vector space over $\mathbb Q$, and finally the Galois group of the extension.

The six complex roots are $\zeta_3^k \cdot \sqrt[3]{3\pm \sqrt2}$ with $\zeta_3$ a primitive third root of 1 and $k=0,1,2$.

I've worked out a clumsy solution, so I was thinking to myself:

• Is there a tricky/easy/obvious(!) way to show that $p$ is irreducible over $\mathbb Q$?
• Is there an easy way to show that $[\mathbb Q(\sqrt 2, \sqrt[3]{3\pm\sqrt2}):\mathbb Q(\sqrt 2, \sqrt[3]{3+\sqrt2})]=3$? It seems to me that there's no way to avoid this step or a similar one, like $[\mathbb Q(\sqrt 2, \sqrt[3]{3-\sqrt2}),\mathbb Q(\sqrt 2, \sqrt[3]7)]=3$ or else (you could add $\zeta_3$ just after $\sqrt 2$..). I'm quite sure that $[\mathbb F:\mathbb Q]=36$, but the only non trivial fact in the proof is the one I've just mentioned. Anyway, I could be missing something..
• Claim: the Galois group is $S_3\times S_3$, that is $D_3\times D_3$ (dihedral). Am I right? How can I prove it?
• Hint: Since $\mathbb{Q}$ is a field, $\mathbb{Q}[x]$ is a UFD, so by Gauss' lemma factorizations over $\mathbb{Q}$ are equivalent to factorizations over $\mathbb{Z}$. Then use the prime factorizations of the leading and trailing coefficients to not find factors over $\mathbb{Z}$. – Eric Towers Jan 26 '14 at 10:12
• This way you can say there can't be linear factors in a hypotetic factorization. But that doesn't mean the polynomial is irreducible. I also used Gauss to show that, but then had to prove that there can't be quadratic or cubic factors in another way (tried all the products). How can you show it only with Gauss' lemma? – Johan Jan 26 '14 at 10:23
• @Johann: You know that roots of $p$ over the integers divide the constant term, 7, a prime. After finding that all four of -7,-1,1,7 aren't roots, use the converse of Gauss' lemma: If $p$ does not factor over $\mathbb{Z}$ then $p$ does not factor over $\mathbb{Q}$. – Eric Towers Jan 27 '14 at 6:24
• Ok: take $x^5+x^3-2x^2-2$. Clearly $\pm 1$ and $\pm 2$ aren't roots, but the polynomial is reducible, because it factors in $(x^2+1)(x^3-2)$. That's what I meant. – Johan Jan 28 '14 at 8:17

There is no "big hammer" for proving irreducibility. Eisenstein's criterion sometimes works, but does not do so for the problem polynomial nor for the additional polynomial in your comment.

Tricky/obvious solution: $x$ only appears as powers of its cube, so consider the substitution $x^3 \rightarrow y$, yielding $y^2-6y+7$. This has discriminant 36-28 = 8 which is not a square, so $y^2-6y+7$ does not split over $\mathbb{Q}$. Let $r \in \mathbb{C}$ be (either) root of this equation. Then $y^2-6y+7$ does split over $\mathbb{Q}(r)$.

Index = 3: This is easy since you know how to split the cube, yielding three roots in $x$ per root in $y$.

Solved this way, you get a sequence of two extensions from which you should be able to write down your Galois group.

Partial answer - only considering irreducibility at this time.

Let us first reduce the coefficients modulo $5$, and show that the polynomial is irreducible in the ring $\Bbb{F}_5[x]$. I first show that the splitting field over $\Bbb{F}_5$ is of degree six. Obviously the element $3+\sqrt{2}\in\Bbb{F}_{25}$ needs to be in the splitting field as do its cube roots. Note that $2$ is not a quadratic residue modulo five, so $\Bbb{F}_5[\sqrt2]=\Bbb{F}_{25}$.

Claim. $3+\sqrt2$ is not a cube in $\Bbb{F}_{25}$.

Proof. Every non-zero element $\beta$ of $\Bbb{F}_{25}$ satisfies the equation $\beta^{24}=1$. So assuming contrariwise that $\beta^3=3+\sqrt2$, we would get $$1\equiv(\beta^3)^8=((3+\sqrt2)^2)^4\equiv(1+\sqrt2)^4=(3+2\sqrt2)^2\equiv2+2\sqrt2\pmod 5,$$ which is a contradiction. QED

This shows that the order of $3+\sqrt2$ in $\Bbb{F}_{25}^*$ is a multiple of three. Therefore the order of its cube root (in some extension field of $\Bbb{F}_5$ is a multiple of nine. The smallest extension field of $\Bbb{F}_5$ that has elements of order nine is $\Bbb{E}=\Bbb{F}_{5^6}$. The irreducibility of $p(x)$ over $\Bbb{F}_5$ follows from this.

If $p(x)$ were reducible over $\Bbb{Q}$ it would factor in $\Bbb{Z}[x]$ and, by reducing that factorization modulo five, also in $\Bbb{F}_5[x]$. Hence we can conclude that $p(x)$ is irreducible over the rationals.

• The modulo five consideration here also implies that the Galois group, when viewed as a group of permutations of the six roots, contains a 6-cycle. See this very nice answer by Qiaochu Yuan for a detailed explanation of that bit of theory. – Jyrki Lahtonen Jan 31 '14 at 11:08