Is this Galois Group is isomorphic to $S_5$

Question

We want $\text{Gal}_\mathbb{Q}(f) \cong S_5$ where $f = x^5 + 2x + 2$. This polynomial has 5 roots, one real and 4 imaginary, call them $a,b_1,b_2,c_1,c_2$ where $b_1, b_2$ and $c_1, c_2$ are conjugates. Then $|\text{Gal}_\mathbb{Q}(f)| = [\mathbb{Q}(a,b_1,c_1): \mathbb{Q}(a, b_1)][\mathbb{Q}(a, b_1): \mathbb{Q}(a)][\mathbb{Q}(a): \mathbb{Q}]$. I think this equals $2 \times 2 \times 5$, but I'm not sure and I'm not sure how to show it. In this case $|\text{Gal}_\mathbb{Q}(f)|=20$ and therefore is isomorphic to a subgroup of $S_5$ with 20 elements, which I guess would be $S_3 \times \mathbb{Z}_2$. Is this correct and how would I show that those are the degrees of the extensions?

Answer 1

The discriminant of $f=x^5+2x+2$ is $2^4\times 3637$, a non-square. I present several method to compute the Galois group, all of them involved some computation.

1. Reduction modulo $p$

Modulo $5$ shows it contains a $4$-cycle, modulo $13$ shows it contains a $3$-cycle, hence $60$ divides the order of Galois group, it is $S_5$ as you conjectured.

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2. Resolvent

Denote $x_1,\cdots,x_5$ be roots of the equation $x^5+2x+2$, let $$\theta_1 = x_1^2 x_2 x_5 + x_1^2 x_3 x_4 + x_2^2 x_1 x_3 + x_2^2 x_4 x_5 + x_3^2 x_1 x_5 + x_3^2 x_2 x_4 + x_4^2 x_1 x_2 + x_4^2 x_3 x_5 + x_5^2 x_1 x_4 + x_5^2 x_2 x_3$$ The stabilizer of $\theta_1$ under the action of $S_5$ is a group $M$, which has order $20$ (isomorphic to $C_5\rtimes C_4$). The orbit consists of six elements, denote them by $\{\theta_1, \cdots, \theta_6\}$. Any conjugates of $M$ in $S_5$ is the stabilizer of some $\theta_i$ (because $M$ is self-normalizing in $S_5$).

The sextic $g=(x-\theta_1)\cdots (x-\theta_6)$ has coefficients in $\mathbb{Q}$, one computes that $$g = x^6+16x^5+160x^4+1280x^3+6400x^2-33616x-283616$$

If the Galois group $G$ is a conjugate subgroup of $M$, then some $\theta_i$ will be in $\mathbb{Q}$. However, it is easily checked that $g$ has no rational root. Thus $G$ cannot be $C_5, D_5$ or $M$. The discriminant tells $G$ is $S_5$.

The following Mathematica code is used for computing $f$:

theta1=x1^2*x2*x5+x1^2*x3*x4+x2^2*x1*x3+x2^2*x4*x5+x3^2*x1*x5+x3^2*x2*x4+x4^2*x1*x2+x4^2*x3*x5+x5^2*x1*x4+x5^2*x2*x3;theta2=theta1/.{x1->x2,x2->x3,x3->x1};theta3=theta1/.{x1->x3,x3->x2,x2->x1};theta4=theta1/.{x1->x2,x2->x1};theta5=theta1/.{x2->x3,x3->x2};theta6=theta1/.{x1->x3,x3->x1};SymmetricReduction[Expand[(x-theta1)(x-theta2)(x-theta3)(x-theta4)(x-theta5)(x-theta6)],{x1,x2,x3,x4,x5},{0,0,0,2,-2}]

Alternatively, you can compute $\theta_i$ to enough accuracy to discern whether they are integers.

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3. Splitting of prime ideals

This method is reminiscent to the first. Let $K$ be the normal closure, $\alpha$ be a root of $f$, it is easily seen that $\mathcal{O}_{\mathbb{Q}(\alpha)} = \mathbb{Z}[\alpha]$. Thus, decomposition of $p$ in $\mathbb{Q}(\alpha)$ can be computed modulo $p$.

In $\mathbb{F}_5$, $f$ factors as $$(4 + x) (3 + x + x^2 + x^3 + x^4)$$ hence prime ideals of $K$ lying above $5$ has inertial degree divisible by $4$.

In $\mathbb{F}_{13}$, $f$ factors as $$(3 + x) (5 + x) (1 + 10 x + 5 x^2 + x^3)$$ hence prime ideals of $K$ lying above $13$ has inertial degree divisible by $3$.

Therefore $3$, $4$ and $5$ divides $[K:\mathbb{Q}]=|G|$, the non-square discriminant says $G\cong S_5$.