Help with Galois Groups

Question

Suppose $f(x) = x^5 + 2x^4 + 4x^3 + 5x^2 + 4x + 4 \in \mathbb{Q}[x]$. Let $K$ be the splitting field of $f(x)$. I wish to determine $[K: \mathbb{Q}]$ and to show that the Galois Group $Gal(K/\mathbb{Q})$ is non - commutative. Any suggestions or solutions? My initial idea was that $f(x)$ is probably not irreducible in $\mathbb{Q[x]}$; I tried using Eisenstein's Criterion and that did not work so that's why I am guessing that $f(x)$ is reducible. But is this true? How do I determine this, and how does this answer my other questions on $[K: \mathbb{Q}]$ and $Gal(K/\mathbb{Q})$? Any suggestions or hints will be appreciated.

Answer 1

Thanks to Dan_Fulea answer and your computation it's easy to describe all $\text{Gal}(K/\mathbb Q)$.

Denoted with $L$ the splitting field of $x^3+x^2+x+2$ and with $F$ the splitting field of $x^2+x+2$, then the splitting field of $f(x)=(x^3+x^2+x+2)(x^2+x+2)$ is $LF =K$.

Thanks to your computations $\text{Gal}(L/ \mathbb Q)\cong S_3$ and $\text{Gal}(F/\mathbb Q)\cong \mathbb Z/(2)$.

Moreover $L\cap F=\mathbb Q$. Infact, this intersection has degree $2$ or $1$ over $\mathbb Q$, and if its grade is $2$ then it coincides with $F$. Now $F=\mathbb Q(\sqrt{-7})$, and the unique subfield of $L$ of degree $2$ (here I'm using the Galois corrispondance) is $\mathbb Q(\sqrt{-83})$ (thanks to the computation of the discriminant). Since $(-7)\cdot(-83)$ is not a square in $\mathbb Q$, then $F\neq \mathbb Q(\sqrt{-83})$, and $F\cap L = \mathbb Q$.

Now we are in the correct hypotesis to use the fact that $$\text{Gal}(LF/ \mathbb Q) \cong \text{Gal}(L/\mathbb Q)\times \text{Gal}(F/\mathbb Q)\cong S_3\times \mathbb Z/(2)$$ So $\text{Gal}(K/\mathbb Q)$ is not abelian.

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You can show that $\text{Gal}(K/\mathbb Q)$ is not abelian also in a faster way: thanks to the Galois corrispondance $\text{Gal}(L/\mathbb Q)\cong S_3$ is a quotient of $\text{Gal}(K/\mathbb Q)$ and we have a contraddiction if $\text{Gal}(K/\mathbb Q)$ is abelian because every quotient of an abelian group is abelian.

Answer 2

The given polynomial can be rewritten as $$ f=(x^2 + x + 2) (x^3 + x^2 + x + 2)\ . $$ The splitting field $K=K(f)$ is generated by

• the root $a$ of the first factor, which determines a quadratic subfield $k$ of $K$,
• and for the second factor we have a splitting field $L$, which already has a non-commutative Galois group over $\Bbb Q$.

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Later EDIT: Let us try to figure out explicitly the Galois group of $K:\Bbb Q$.

The factor $(x^2+x+2)$ has discriminant $-7$, so it splits over the quadratic field $K_2=\Bbb Q(\sqrt{-7})$. This is a subfield of $K$.

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If $b$ is a root of $x^3+x^2+x+2$, then $(3b+1)$ is a root of $x^3 + 6x + 47$. So we replace $x^3+x^2+x+2$ by this polynomial, which has discriminant $3^6$ times bigger.

Let $s$ be a primitive third root of unity, so $s^3=1$, $s\ne 1$. (It is simpler to type here as $\varepsilon$.)

For this polynomial we search roots of the shape $$ \begin{aligned} c_k &= s^kA+s^{2k}B\ , \qquad k=0,1,2\ .\text{ Explicitly:}\\[3mm] c_0 &= A+B\ ,\\ c_1 &= sA+s^2B\ ,\\ c_2 &= s^2A+sB\ .\\ \end{aligned} $$ Note that this Ansatz already insures $c_0+c_1+c_2=0$. The other two Vieta relation will determine $A,B$. Taking the product (recall the formula for the factorization of $u^3+v^3+w^3-3uvw$) $$ (X-A-B)(X-sA-s^2B)(X-s^2A-sB)=X^3-3ABX-(A^3+B^3)\ , $$ and comparing with $X^3+6X+47$, we search for $A,B$ with $A^3+B^3=-47$, $AB=-2$, so $A^3B^3=-8$. We get an equation of second degree for $A^3,B^3$, which is $T^2 +47T-8$, having the roots $\frac 12(-47\pm 3\sqrt{249})$. We fix a choice, $A$ for a third root of $\frac 12(-47+ 3\sqrt{249})$. Let $B$ be such that $A=-2$. The roots are then of the shape: $$ c_k= s^k \underbrace{\sqrt[3]{\frac 12(-47+ 3\sqrt{249})}}_{=A} + s^{2k} \underbrace{\sqrt[3]{\frac 12(-47- 3\sqrt{249})}}_{=B} \ ,\qquad AB=-2\ . $$ Quick check:

sage: A = ( (-47 + 3*sqrt(3*83))/2 )^(1/3)
sage: B = -2/A
sage: (A+B).minpoly()
x^3 + 6*x + 47
sage: ((A+B-1)/3).minpoly()
x^3 + x^2 + x + 2

The Galois group of the splitting field of $(x^3+6x+47)$ acts by permutation of the elements of the set $$ \{ \ A+B\ ,\ \ sA+s^2B\ ,\ \ s^2A+sB\ ,\ \}\ . $$ One permutation is obtained by replacing the choice of $A$ by an other cubic root, so we replace for instance $A$ by $sA$ and in the same time $B$ by $s^2B$. This implements the cyclic permutation $(0,1,2)$ of the three roots $c_0,c_1,c_2$ above. The tranposition $(12)$ (i.e. $c_1\leftrightarrow c_2$, $c_0$ invariated) is formally implemented by changing $s$ into $\bar s=s^2$.

The product (invariated by the cycle $(012)$) $$ (c_0-c_1)(c_1-c_2)(c_2-c_3) $$ is the square root of the discriminant $-83\cdot 3^6$, so the quadratic field $\Bbb Q(\sqrt{-83})$ is a subfield of $K$. (And it has "nothing to do with $\Bbb Q(\sqrt{-7})$".)

Now consider also $c_3,c_4$ to be $\pm\sqrt{-7}$.

Note that each permutation of $\{0,1,2;3,4\}$ that invariates the subsets $\{0,1,2\}$ and $\{3,4\}$ induces a Galois automorphism of $K$ mapping correspondingly the generators $c_0,c_1,c_2;c_3,c_4$. This implies $$ \operatorname{Gal}(K:\Bbb Q)\cong S_3\times S_2\ . $$ The subfields of $K$ correspond (in inverse inclusion order) to the subgroups of the above Galois group.

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sage code supporting the above:

sage: R.<x> = PolynomialRing(QQ)
sage: f = x^5 + 2*x^4 + 4*x^3 + 5*x^2 + 4*x + 4
sage: f.factor()
(x^2 + x + 2) * (x^3 + x^2 + x + 2)
sage: (g, gmul),  (h, hmul) = f.factor()
sage: h
x^3 + x^2 + x + 2
sage: L.<b> = h.splitting_field()
sage: GL.<t> = L.galois_group()
sage: L
Number Field in b with defining polynomial x^6 + 5*x^5 + 74*x^4 + 227*x^3 + 1573*x^2 + 2096*x + 9124
sage: GL
Galois group of Number Field in b with defining polynomial x^6 + 5*x^5 + 74*x^4 + 227*x^3 + 1573*x^2 + 2096*x + 9124
sage: GL.order()
6
sage: GL.is_commutative()
False


sage: K.<c> = f.splitting_field()
sage: GK = K.galois_group()
sage: GK.structure_description()
'D6'

The above is not a solution, but a hint for the fact that computer algebra systems may be very useful when trying to understand and experiment with structural mathematical objects.