Techniques for Solving Solvable Equations via Galois Theory

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[3-15] Computing the roots of $f(x)$

At last we have obtained the minimal polynomial $g_4(x)$ of $v$. Since $g_4(x)$ is linear in $x$, solving $g_4(x)=0$ is immediate, and its root gives the desired value of $v$.

\begin{align} &g_4(x)= x+\frac{{a_3}}{2}+{a_4} \\ &g_4(x)=0 \quad \Longrightarrow \quad \therefore \ v=-\frac{{a_3}}{2}-{a_4} \\ \end{align}

Substituting this value of $v$ into (4.6–9) yields the four roots of $f(x)$. We now illustrate the concrete computation. In (15.3), $\alpha$ is written as a polynomial in $v$.
Using the computer algebra system Maxima, $\alpha$ can be computed by the commands below. In (15.4), we reduce $\alpha$ modulo $g_4(v)$. Because $g_4(v)$ is linear in $v$, taking the remainder is equivalent to substituting the value of $v$ from (15.2) into (15.3).
From (15.4) onward, we reduce by the four binomials defining the cyclic tower of extensions. Finally, reducing modulo $\Omega$ as in (15.5) gives the final result (15.6).

\begin{align} &\alpha=\frac{801167701943012874015343807 }{10126546386824616812436636833146824818688}{{v}^{23}}\notag \\ &\quad +\frac{51207699710669004924125}{199474971178044691573821786887815168} {{v}^{22}} \notag \\ \notag \\ &\qquad .............................. \notag \\ \notag \\ &\quad -\frac{1279063375083309131586881879157101 }{11886064249859873624873982160300416}v \notag \\ &\quad -\frac{2718803338720300088760700554765}{3043746508454051079922817793088} \\ \notag \\ \end{align}

\begin{align} & \qquad \Downarrow \notag \\ \notag \\ &\alpha:remainder(\alpha,g_4(v),v)$ & &\leftarrow \quad ( \ mod \ g_4(v) \ ) \\ &\alpha:remainder(\alpha,B_4,a_4)$ & &\leftarrow \quad ( \ mod \ B_4(a_4) \ ) \notag \\ &\alpha:remainder(\alpha,B_3,a_3)$ & &\leftarrow \quad ( \ mod \ B_3(a_3) \ ) \notag \\ &\alpha:remainder(\alpha,B_2,a_2)$ & &\leftarrow \quad ( \ mod \ B_2(a_2) \ ) \notag \\ &\alpha:remainder(\alpha,B_1,a_1)$ & &\leftarrow \quad ( \ mod \ B_1(a_1) \ ) \notag \\ &\alpha:remainder(\alpha,Ω,ω); & &\leftarrow \quad ( \ mod \ \Omega \ ) \\ \end{align}

\begin{align} &\Rightarrow \ \frac{1}{166133760}\Bigl[\left( \left( {a_1}+129600\right) \, {{a}_{2}^{2}} {a_3} {a_4}+599040 {a_2} {a_3} {a_4}\right) \omega \notag \\ &\quad +\left( 69120-22 {a_1}\right) \, {{a}_{2}^{2}} {a_3} {a_4}-199680 {a_2} {a_3} {a_4}+83066880 {a_4}+20766720 {a_3} \Bigr] \\ \end{align}

The remaining three roots are obtained in exactly the same way. The final expressions of the four roots of $f(x)$ are as follows.

\begin{align} &f(x)=x^4+4x+2 =(x-\alpha)(x-\beta)(x \ -\gamma)(x \ -\delta) \\ \end{align}

\begin{align} \notag \\ &\alpha=\frac{1}{166133760}\Bigl[\left( \left( {a_1}+129600\right) \, {{a}_{2}^{2}} {a_3} {a_4}+599040 {a_2} {a_3} {a_4}\right) \omega \notag \\ &\quad +\left( 69120-22 {a_1}\right) \, {{a}_{2}^{2}} {a_3} {a_4}-199680 {a_2} {a_3} {a_4}+83066880 {a_4}+20766720 {a_3} \Bigr] \\ \notag \\ &\beta=-\frac{1}{166133760}\Bigl[ \left( \left( {a_1}+129600\right) \, {{a}_{2}^{2}} {a_3} {a_4}+599040 {a_2} {a_3} {a_4}\right) \omega \notag \\ &\quad +\left( 69120-22 {a_1}\right) \, {{a}_{2}^{2}} {a_3} {a_4}-199680 {a_2} {a_3} {a_4}+83066880 {a_4}-20766720 {a_3} \Bigr] \\ \notag \\ &\gamma=-\frac{1}{166133760}\Bigl[\left( \left( {a_1}+129600\right) \, {{a}_{2}^{2}} {a_3} {a_4}+599040 {a_2} {a_3} {a_4}\right) \omega \notag \\ &\quad +\left( 69120-22 {a_1}\right) \, {{a}_{2}^{2}} {a_3} {a_4}-199680 {a_2} {a_3} {a_4}-83066880 {a_4}+20766720 {a_3} \Bigr] \\ \notag \\ &\delta=\frac{1}{166133760}\Bigl[ \left( \left( {a_1}+129600\right) \, {{a}_{2}^{2}} {a_3} {a_4}+599040 {a_2} {a_3} {a_4}\right) \omega \notag \\ &\quad +\left( 69120-22 {a_1}\right) \, {{a}_{2}^{2}} {a_3} {a_4}-199680 {a_2} {a_3} {a_4}-83066880 {a_4}-20766720 {a_3} \Bigr] \\ \end{align}

We conclude by reviewing the cyclic tower computed so far in Fig. 3-6 and listing the associated binomials for each extension.

\begin{align} &\left\{ \begin{array}{l} \varOmega = \omega^2+ \omega +1 =0 \\ B_1(x)=x^2-A_1=0 \qquad a_1=\sqrt{A_1} \\ B_2(x)=x^3-A_2=0 \qquad a_2=\sqrt[3]{A_2}\\ B_3(x)=x^2-A_3=0 \qquad a_3=\sqrt{A_3} \\ B_4(x)=x^2-A_4=0 \qquad a_4=\sqrt{A_4} \\ \end{array} \right. \\ \end{align}

\begin{align} & A_1=-17510400 \\ & A_2= \frac{14 {a_1} \omega }{27}+2304 \omega +\frac{89 {a_1}}{135}-1088 \notag \\ & A_3=\frac{23 {a_1} {{a}_{2}^{2}} \omega }{324480}+\frac{63 {{a}_{2}^{2}} \omega }{338} -\frac{8 {a_2} \omega }{13}+\frac{{a_1} {{a}_{2}^{2}}}{324480} +\frac{135 {{a}_{2}^{2}}}{338}-\frac{32 {a_2}}{13} \notag \\ & A_4= -\frac{11 {a_1} {{a}_{2}^{2}} \omega }{2595840} +\frac{9 {{a}_{2}^{2}} \omega }{676}+\frac{2 {a_2} \omega }{13} -\frac{23 {a_1} {{a}_{2}^{2}}}{5191680}-\frac{63 {{a}_{2}^{2}}}{5408}+\frac{3 {a_2}}{26} \notag \\ \end{align}

\begin{align} g_0(x)=& \ x^{24}-160x^{20}++5440x^{18}+30080x^{16}+739840x^{14} \notag \\ & +25400832x^{12} -29593600x^{10}+1520414720x^8 \notag \\ &+35532554240x^6 +411134296064x^4+700091596800x^2 \notag \\ & +4691625312256\\ \notag \\ &\qquad \Downarrow \notag \\ \notag \\ g_4(x)=&\ x+\frac{{a_3}}{2}+{a_4} \qquad \therefore \ v=-\frac{{a_3}}{2}-{a_4} \\ \end{align}

This completes the solution procedure via Galois theory. (It was a long computation.)

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[3-15] Computing the roots of \(f(x)\)