Exercise 13.2.14

Proof.

(a)

We obtain the discriminant with Sage:

     S.<p,q,x> = QQ[]
     f = x^5 + p*x^3 +(1/5)*p^2*x + q
     Delta = f.discriminant(x);Delta.factor()

$$\left(\frac{1}{3125}\right)\cdot {(4p^5+3125q^2)}^2$$

So

$$\mathrm{\Delta }(f)=\frac{1}{5^5}\cdot {(4p^5+3125q^2)}^2.$$

We use the following procedure to compute a sextic resolvent with the same method as in Exercise 6:

     def resolvent(f):
         l = f.coefficients(sparse = False)
         R.<Delta,x1,x2,x3,x4,x5,y1,y2,y3,y4,y5,y,P,Q,e> = PolynomialRing(QQ, order = ’degrevlex’)
         elt = SymmetricFunctions(QQ).e()
         e = [elt([i]).expand(5).subs(x0=x1, x1=x2, x2=x3, x3 = x4, x4 = x5)
                        for i in range(6)]
         J = R.ideal(e[1]-y1, e[2]-y2, e[3]-y3,e[4]-y4,e[5]-y5)
         G = J.groebner_basis()
         u1 =x1*x2 + x2*x3 + x3*x4 + x4*x5 + x5*x1 - x1*x3 - x3*x5 -x5*x2 - x2*x4 -x4*x1
         u2 = u1.subs(x1 = x2, x2 = x3, x3 = x1)
         u3 = u1.subs(x2 = x3, x3 = x4, x4 = x2)
         u4 = u1.subs(x3 = x4, x4 = x5, x5 = x3)
         u5 = u1.subs(x1 = x4, x4 = x5, x5 = x1)
         u6 = u1.subs(x1 = x2, x2 = x5, x5 = x1)
         f1 = (y-u1) * (y-u2) * (y-u3) * (y-u4) * (y-u5) * (y-u6)
         var(’sigma_1,sigma_2,sigma_3,sigma_4,sigma_5’)
         g = f1.reduce(G).subs(y1=sigma_1, y2=sigma_2, y3=sigma_3, y4=sigma_4, y5= sigma_5)
         h = g.collect(y);
         B2 = h.coefficient(y,4)
         B4 = h.coefficient(y,2)
         B6 = h.coefficient(y,0)
         b2 = B2.subs(sigma_1 = -l[4], sigma_2= l[3],sigma_3 = -l[2], sigma_4 = l[1], sigma_5 = -l[0])
         b4 = B4.subs(sigma_1 = -l[4], sigma_2= l[3],sigma_3 = -l[2], sigma_4 = l[1], sigma_5 = -l[0])
         b6 = B6.subs(sigma_1 = -l[4], sigma_2= l[3],sigma_3 = -l[2],
                     sigma_4 = l[1], sigma_5 = -l[0])
         theta_f = [(y^3+b2*y^2+b4*y+b6)^2 - 2^10*Delta*y,b2,b4,b6]
         return theta_f

Then we obtain $b_2,b_4,b_6$ and ${𝜃}_f(y)$:

     K.<p,q> = QQ[]
     S.<x> =PolynomialRing(K, order = ’degrevlex’)
     f = x^5 + p*x^3 +(1/5)*p^2*x + q
     resolvent(f)[1:4]

$$\left(-7p^2,11p^4,\frac{3}{25}{p}^6+4000p{q}^2\right)$$

     theta_f=resolvent(f)[0];theta_f

$$\begin{array}{llll}\hfill & {𝜃}_f(y)=\frac{1}{625}{\left(3p^6+275p^{4}y-175p^2{y}^2+100000p{q}^2+25y^3\right)}^2-1024\mathrm{\Delta }(f)y& \hfill & \\ \hfill & ={\left(y^3-7p^2{y}^2+11p^{4}y+\frac{3}{25}{p}^6+4000p{q}^2\right)}^2-2^{10}\mathrm{\Delta }(f)y& \hfill & \end{array}$$

We obtained the results given in the text.

(b)

To find the rational root of $f$ we write

     theta_f.subs(Delta = (1/5)^5*(4*p^5+3125*q^2)^2)).factor()

$$\begin{array}{llll}\hfill & \frac{1}{3125}(5p^2-y)(9p^{10}-1625p^{8}y+74250p^6{y}^2+600000p^5{q}^2-81250p^4{y}^3& \hfill & \\ \hfill & +50000000p^3{q}^{2}y+28125p^2{y}^4-25000000p{q}^2{y}^2-3125y^5+10000000000q^4)& \hfill & \end{array}$$

Thus

$${𝜃}_f(5p^2)=0.$$

By Corollary 13.2.11, $f$ is solvable by radicals over $\mathbb{Q}$.

(c)

The substitution $x=z-\frac{p}{5z}$ is obtained by

     z = var(’z’)
     g = f.subs(x = z - p/(5*z))
     g.expand()

$$z^5+q-\frac{p^5}{3125z^5}$$

Thus

$$g(z)=f\left(z-\frac{p}{5z}\right)=z^5-\frac{p^5}{5^5{z}^5}+q.$$

(d)

Let $\beta \in ℂ$. $$\begin{array}{llll}\hfill g(\beta )=0& ⟺{\beta }^{10}+q{\beta }^5-{\left(\frac{p}{5}\right)}^5=0& \hfill & \\ \hfill & ⟺{\left({\beta }^5+\frac{q}{2}\right)}^2-\left[{\left(\frac{q}{2}\right)}^2+{\left(\frac{p}{5}\right)}^5\right]=0& \hfill & \\ \hfill & ⟺\left[{\beta }^5+\frac{q}{2}-\sqrt{{\left(\frac{q}{2}\right)}^2+{\left(\frac{p}{5}\right)}^5}\right]\left[{\beta }^5+\frac{q}{2}-\sqrt{{\left(\frac{q}{2}\right)}^2+{\left(\frac{p}{5}\right)}^5}\right]=0.& \hfill & \end{array}$$

So the $10$ roots of $g$ are

$${\beta }_{k,𝜀}={\zeta }^k\sqrt[5]{-\frac{q}{2}+𝜀\sqrt{{\left(\frac{q}{2}\right)}^2+{\left(\frac{p}{5}\right)}^5}},𝜀=\pm 1,k=0,1,2,3,4.$$

where $\zeta ={\zeta }_5=e^{2\mathit{i\pi }∕5}$.

Let $\alpha$ be a root of $f$ in $ℂ$. There exists $\beta \in ℂ$ such that $\alpha =\beta -\frac{p}{5\beta }$, so

$$0=f(\alpha )=f\left(\beta -\frac{p}{5\beta }\right)=g(\beta ).$$

Since $g(\beta )=0$, $\beta ={\beta }_{k,𝜀}$ for some $k\in \{0,\dots ,4\},𝜀\in \{-1,1\}$. If $L$ is the splitting field of $f$ in $ℂ$, then

$$L\subset \mathbb{Q}({\beta }_{0,1},\dots ,{\beta }_{4,1},{\beta }_{0,-1},\dots ,{\beta }_{4,-1}).$$

Write $\delta ={\left(\frac{q}{2}\right)}^2+{\left(\frac{p}{5}\right)}^5\in \mathbb{Q}$. Since ${\beta }_{k,𝜀}\in \mathbb{Q}\left(\sqrt{\delta },\zeta ,\sqrt[5]{-\frac{q}{2}+𝜀\sqrt{\delta }}\right)$,

$$L\subset \mathbb{Q}\left({\zeta }_5,\sqrt{\delta },\sqrt[5]{-\frac{q}{2}+\sqrt{\delta }},\sqrt[5]{-\frac{q}{2}-\sqrt{\delta }}\right),$$

where $\delta ,q\in \mathbb{Q}$, so $L$ is included in some radical extension of $\mathbb{Q}$.

Therefore $f$ is solvable by radicals over $\mathbb{Q}$.

Note: We can choose $\sqrt[5]{-\frac{q}{2}-\sqrt{\delta }}$ so that $\sqrt[5]{-\frac{q}{2}+\sqrt{\delta }}\sqrt[5]{-\frac{q}{2}-\sqrt{\delta }}=-\frac{p}{5}\in \mathbb{Q}$. Therefore

$$L\subset \mathbb{Q}\left({\zeta }_5,\sqrt{\delta },\sqrt[5]{-\frac{q}{2}+\sqrt{\delta }}\right),$$

where the chain of inclusions

$$\mathbb{Q}\subset \mathbb{Q}({\zeta }_5)\subset \mathbb{Q}\left({\zeta }_5,\sqrt{\delta }\right)\subset \mathbb{Q}\left({\zeta }_5,\sqrt{\delta },\sqrt[5]{-\frac{q}{2}+\sqrt{\delta }}\right)$$

proves that this last field is a radical extension.