# Is this Galois Group is isomorphic to $S_5$

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?

• When you adjoin $b_1$ there is no guarantee that you would also get $b_2$ free of charge. This time you don't. Also, the minimal polynomial of $b_1$ over $\Bbb{Q}(a)$ may have degree four (this time that happens). – Jyrki Lahtonen May 17 '18 at 4:44
• Some irreducible quintics do have a Galois group of order $20$. That group is isomorphic to $C_5\rtimes C_4$. As a subgroup of $S_5$ a copy of it is generated by $(12345)$ and $(2354)$. – Jyrki Lahtonen May 17 '18 at 4:48
• @Jyrki How do you know that this time you get degree 4? Could you please elaborate? I tried to do something like this in a similar question but lost points for just saying that if you factor $f = (x-a)g(x)$ then $g$ is some irreducible polynomial of degree 4 that has $b_1$ as a root so $[\mathbb{Q}[a, b_1): \mathbb{Q}(a)] = 4$. Also, why isn't it guaranteed that $F(b_1)$ contains $b_2$? I thought it would be since they are complex congujates – Vinny Chase May 17 '18 at 4:53
• The splitting field is an extension of degree $120$ (see pisco's answer), so when you adjoin the roots one by one, the relative extensions must have the maximal degrees 5,4,3,2,1. – Jyrki Lahtonen May 17 '18 at 5:00
• @JyrkiLahtonen Ok makes sense. Thanks. But how you know the splitting field has degree 120 without using that thing you linked? I've never seen it before and I'd really prefer to use only theorems and whatnot that were covered in the class. – Vinny Chase May 17 '18 at 5:13

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.

## 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.

## 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 $G\cong S_5$.

• +1 for I have used the same technique so many time on this site. Because Dedekind's theorem is not included in all the courses on Galois theory, I refer the OP to the statement as well as a proof. – Jyrki Lahtonen May 17 '18 at 4:51
• How could I do this without needing to use that statement that was linked? I've never seen it before and I would like to stay with things that were covered in the class – Vinny Chase May 17 '18 at 4:54
• @VinnyChase I added a new method. Note that computation of Galois group is in generally very complicated, except some special cases (e.g. irreducible quintic over $\mathbb{Q}$ with exactly two complex roots), heavy computation is required. – pisco May 17 '18 at 12:55
• @pisco What is $\mathcal{O}_{\mathbb{Q}(\alpha)}$? – Vinny Chase May 22 '18 at 2:17
• @VinnyChase $\mathcal{O}_{\mathbb{Q}(\alpha)}$ is the ring of integer of $\mathbb{Q}(\alpha)$. If $x^5+2x+2 = 0$ had three real and two complex roots, then complex conjugation would be a transposition. However, it has four complex roots, so it is not clear whether it has a transposition. For example, $x^5-5x+12$ has four complex roots but its Galois group is $D_5$. – pisco May 22 '18 at 5:38