Exercise 13.2.19

Proof. We use the Exercise 14 resolvent() procedure to compute a sextic resolvent, and the additional procedure to verify that this resolvent has a rational root (the polynomials are supposed, as in this exercise, monic with integer coefficients, i.e., $f\in \mathbb{Z}[x]$, hence resolvent rational root is an integer):

def rational_root(theta):
    n = Integer(theta.subs(y=0))
    if n == 0:
        return True, 0
    for d in n.divisors():
        if theta.subs(y = d) == 0:
            return True, d
        if theta.subs(y = -d)== 0:
            return True, -d
    return False, None

(a)

     R.<x> = QQ[]
     f = x^5 + x + 1;f

$$x^5+x+1$$

     f.is_irreducible()

$$\text{False}$$

     f.factor()

$$f=(x^2+x+1)\cdot (x^3-x^2+1)$$

The roots of $x^2+x+1$ are $\omega ,{\omega }^2$. Write $x_1,x_2,x_3$ the roots of $x^3-x^2+1$. Then $L=\mathbb{Q}(\omega ,x_1,x_2,x_3)$ is the splitting field of $f$ over $\mathbb{Q}$. As $L=\mathbb{Q}(\omega )(x_1,x_2,x_3)$, $L$ is also the splitting field of $g=x^3-x^2+1$ over $\mathbb{Q}(\omega )$. The discriminant of $g$ is $\mathrm{\Delta }(g)=-23$. We show that $-23\notin \mathbb{Q}{(\omega )}^2$.

If $-23={(a+\mathit{b\omega })}^2,a,b\in \mathbb{Q}$, then $-23=a^2-b^2-\mathit{\omega b}(b-2a)$. Therefore

$$\{\begin{array}{ccc}\hfill -23& \hfill =\hfill & a^2-b^2,\hfill \\ \hfill 0& \hfill =\hfill & b(b-2a).\hfill \end{array}$$

Thus $b=0$ or $b=2a$. If $b=0$, then $-23=a^2$, which is impossible since $a\in \mathbb{Q}$, and $b=2a$ gives $a^2=23∕3$, which is also impossible. Therefore $\mathrm{\Delta }(g)$ is not a square in $\mathbb{Q}(\omega )$, so the Galois group $G_1$ of $g$ over $\mathbb{Q}(\omega )$ is $S_3$. This implies that $[L:\mathbb{Q}(\omega )]=|G_1|=6$. Since $[\mathbb{Q}(\omega ):\mathbb{Q}]=2$, $[L:\mathbb{Q}]=12$, so the Galois group $G$ of $f=x^5+x+1$ has order $12$:

$$|G|=12.$$

Since $-23$ is not a square of $\mathbb{Q}$, $\mathrm{Gal}(\mathbb{Q}(x_1,x_2,x_3)∕\mathbb{Q})\simeq S_3$.

Let

$$\phi \{\begin{array}{ccc}\hfill \mathrm{Gal}(L∕\mathbb{Q})\hfill & \hfill \to \hfill & \hfill \mathrm{Gal}(\mathbb{Q}(x_1,x_2,x_3)∕\mathbb{Q})\times \mathrm{Gal}(\mathbb{Q}(\omega )∕\mathbb{Q})\hfill \\ \hfill \sigma \hfill & \hfill \mapsto \hfill & \hfill \left(\sigma {\text{|}}_{\mathbb{Q}(x_1,x_2,x_3)},\sigma {\text{|}}_{\mathbb{Q}(\omega )}\right).\hfill \end{array}$$

Then $\phi$ is a group homomorphism, and the kernel of $\phi$ is $\{\mathrm{id}\}$, since every $\mathbb{Q}$-automorphism of $L$ which fixes $\omega ,x_1,x_2,x_3$ is the identity of $L$. So $\phi$ is injective, and $|\mathrm{Gal}(L∕\mathbb{Q})|=|\mathrm{Gal}(\mathbb{Q}(\omega )∕\mathbb{Q})\times \mathrm{Gal}(\mathbb{Q}(x_1,x_2,x_3)∕\mathbb{Q})|=12$, thefore $\phi$ is a group isomorphism.

$$\mathrm{Gal}(L∕\mathbb{Q})\simeq C_2\times S_3.$$

If we choose the numbering $x_1,x_2,x_3,x_4=\omega ,x_5={\omega }^2$ of the roots of $f$, then the Galois group $G$ of $f$ is

$$G={\mathrm{Gal}}_{\mathbb{Q}}(f)=⟨(12),(123),(45)⟩\simeq S_3\times C_2.$$

(b)

     R.<x> = QQ[]
     f = x^5 + 20 *x + 16;f

$$x^5+20x+16$$

     f.is_irreducible()

$$\text{True}$$

     theta = resolvent(f)[0]; theta.subs(Delta = f.discriminant()).expand()

$$y^6-800y^5+352000y^4-71680000y^3+7168000000y^2-557056000000y+6553600000000$$

     res = rational_root(theta); res

$$\text{(False, None)}$$

     f.discriminant().factor(),f.discriminant().is_square()

$$(2^{16}\cdot 5^6,\text{True})$$

Thus the Galois group of $f=x^5+20x+16$ over $\mathbb{Q}$ is $A_5$.

Verification:

     f.galois_group().gens()

$$⟨(3,4,5),(1,2,3,4,5)⟩$$

(c)

     R.<x> = QQ[]
     f = x^5 + 2;f

$$x^5+2$$

     f.is_irreducible()

$$\text{True}$$

     theta = resolvent(f)[0]; theta.subs(Delta = f.discriminant()).expand()

$$y^6-51200000y$$

     res = rational_root(theta); res

$$\text{(True, 0)}$$

     f.discriminant().factor(),f.discriminant().is_square()

$$(2^4\cdot 5^5,\text{False})$$

Thus the Galois group of $f=x^5+2$ over $\mathbb{Q}$ is $\mathrm{AGL}(1,{𝔽}_5)$, up to conjugacy.

Verification:

     f.galois_group().gens()

$$⟨(1,2,3,4,5),(1,2,4,3)⟩$$

(d)

     R.<x> = QQ[]
     f = x^5 -5*x + 12;f

$$x^5-5x+12$$

     f.is_irreducible()

$$\text{True}$$

     theta = resolvent(f)[0]; theta.subs(Delta = f.discriminant()).expand()

$$y^6+200y^5+22000y^4+1120000y^3+28000000y^2-66016000000y+1600000000$$

     res = rational_root(theta); res

$$\text{(True, 100)}$$

     f.discriminant().factor(),f.discriminant().is_square()

$$(2^{12}\cdot 5^6,\text{True})$$

     K.<alpha> = NumberField(f)
     S.<X> = K[]
     g = f.change_ring(S)
     g.factor()

$$\begin{array}{llll}\hfill & (x-\alpha )\cdot \left(x^2+\left(\frac{1}{4}{\alpha }^4+\frac{1}{4}{\alpha }^3+\frac{1}{4}{\alpha }^2+\frac{1}{4}\alpha -1\right)x-\frac{1}{2}{\alpha }^3-\frac{1}{2}\alpha -1\right)& \hfill & \\ \hfill & \cdot \left(x^2+\left(-\frac{1}{4}{\alpha }^4-\frac{1}{4}{\alpha }^3-\frac{1}{4}{\alpha }^2+\frac{3}{4}\alpha +1\right)x-\frac{1}{4}{\alpha }^4-\frac{1}{4}{\alpha }^3-\frac{1}{4}{\alpha }^2-\frac{5}{4}\alpha +2\right)& \hfill & \end{array}$$

Thus the Galois group of $f=x^5+20x+16$ over $\mathbb{Q}$ is $\mathrm{AGL}(1,{𝔽}_5)\cap A_5$, up to conjugacy.

Verification:

     f.galois_group().gens()

$$⟨(1,2,3,4,5),(1,4)(2,3)⟩$$

(e)

     R.<x> = QQ[]
     f = x^5 + x^4 - 4*x^3 - 3*x^2 + 3*x + 1
     f

$$x^5+x^4-4x^3-3x^2+3x+1$$

     f.is_irreducible()

$$\text{True}$$

     theta = resolvent(f)[0]; theta.subs(Delta = f.discriminant()).expand()

$$y^6-264y^5+25168y^4-1022208y^3+14992384y^2-14992384y$$

     res = rational_root(theta); res

$$\text{(True, 0)}$$

     f.discriminant().factor(),f.discriminant().is_square()

$$(11^4,\text{True})$$

     K.<alpha> = NumberField(f)
     S.<X> = K[]
     g = f.change_ring(S)
     g.factor()

$$\begin{array}{lll}\hfill (x-\alpha )\cdot (x-{\alpha }^2+2)\cdot (x+{\alpha }^4+{\alpha }^3-3{\alpha }^2-2\alpha +1)\cdot (x-{\alpha }^3+3\alpha )\cdot (x-{\alpha }^4+4{\alpha }^2-2)& & \hfill \end{array}$$

Thus the Galois group of $f=x^5+20x+16$ over $\mathbb{Q}$ is $⟨(12345)⟩$, up to conjugacy.

Verification:

     f.galois_group().gens()

$$⟨(1,2,3,4,5)⟩$$