Techniques for Solving Solvable Equations via Galois Theory

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[8-1] Problem Setting

We want to find the roots of the following quintic equation \(f(x)\). Even though it is a quintic, the Galois group of this equation is a solvable group called the Frobenius group, so it is possible to determine its roots.

\begin{align} &f(x)=x^5+x^4+2x^3+4x^2+x+1 \\ &\quad \{\alpha,\beta,\gamma,\delta,\epsilon\}: \ roots \ of \ f(x) \notag \\ \end{align}

Using the roots \(\{\alpha,\beta,\gamma,\delta,\epsilon\}\) of \(f(x)\), we define the following primitive element \(\boldsymbol{v}\):

\begin{align} v=1 \cdot \alpha+2 \cdot \beta+3 \cdot \gamma+4 \cdot \delta+5 \cdot \epsilon \\ \end{align}

From here, we will use the minimal polynomial of this primitive element \(v\) to define the simple extension \(F_0(v)\).
(Note): The coefficients \([1,2,3,4,5]\) attached to \(\{\alpha,\beta,\gamma,\delta,\epsilon\}\) may generally be replaced by many other choices of distinct integers.

[8-2] Derivation of the Galois Resolvent \(V(x)\)

The computational method described below was suggested by Keita Ikumi.
Using the five factors of \(f(x)\), such as \((x-\alpha)\), we divide step by step as follows:

\begin{align} f(x)&=x^5+x^4+2x^3+4x^2+x+1 \notag \\ &=(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)(x-\epsilon) \\ \notag \\ \end{align}

\begin{align} &\left\{ \begin{array}{l} f(x)=(x-\alpha)q_1(x)+r_1 \\ q_1(x)=x^4+(\alpha+1)x^3+(\alpha^2+\alpha+2)x^2+(\alpha^3+\alpha^2+2\alpha+4)x \\ \qquad +\alpha^4+\alpha^3+2\alpha^2+4\alpha+1 \\ r_1=\alpha^5+\alpha^4+2\alpha^3+4\alpha^2+\alpha+1 \\ \end{array} \right.\\ \notag \\ &\left\{ \begin{array}{l} q_1(x)=(x-\beta)q_2(x)+r_2 \\ q_2(x)=x^3+(\beta+\alpha+1)x^2+(\beta^2+(\alpha+1)\beta+\alpha^2+\alpha+2)x+\beta^3+(\alpha+1)\beta^2 \\ \qquad +(\alpha^2+\alpha+2)\beta+\alpha^3+\alpha^2+2\alpha+4 \\ r_2=\beta^4+(\alpha+1)\beta^3+(\alpha^2+\alpha+2)\beta^2+(\alpha^3+\alpha^2+2\alpha+4)\beta \\ \qquad +\alpha^4+\alpha^3+2\alpha^2+4\alpha+1\\ \end{array} \right.\\ \notag \\ &\left\{ \begin{array}{l} q_2(x)=(x-\gamma)q_3(x)+r_3 \\ q_3(x)=x^2+(\gamma+\beta+\alpha+1)x+\gamma^2+(\beta+\alpha+1)\gamma+\beta^2+(\alpha+1)\beta \\ \qquad +\alpha^2+\alpha+2 \\ r_3=\gamma^3+(\beta+\alpha+1)\gamma^2+(\beta^2+(\alpha+1)\beta+\alpha^2+\alpha+2)\gamma \\ \qquad +\beta^3+(\alpha+1)\beta^2 \\ \qquad +(\alpha^2+\alpha+2)\beta+\alpha^3+\alpha^2+2\alpha+4\\ \end{array} \right.\\ \notag \\ &\left\{ \begin{array}{l} q_3(x)=(x-\delta)q_4(x)+r_4 \\ q_4(x)=x+\delta+\gamma+\beta+\alpha+1 \\ r_4=\delta^2+(\gamma+\beta+\alpha+1)\delta+\gamma^2+(\beta+\alpha+1)\gamma+\beta^2+(\alpha+1)\beta+\alpha^2+\alpha+2 \\ \end{array} \right.\\ \notag \\ &\left\{ \begin{array}{l} q_4(x)=(x-\epsilon)q_5(x)+r_5 \\ q_5(x)=1 \\ r_5=\alpha+\beta+\gamma+\delta+\epsilon +1\\ \end{array} \right.\\ \notag \\ \end{align}

From the above, \(\{f(x),q_1(x),q_2(x),q_3(x),q_4(x)\}\) have roots \(\{\alpha,\beta,\gamma,\delta,\epsilon\}\), respectively. Thus the following relations hold. Then, by moving the right-hand side of (1.2) and defining it as \(r_6\), we obtain:

\begin{align} &\left\{ \begin{array}{l} &f(\alpha)=0 \ \Rightarrow \ r_1=0 \quad &q_1(\beta)=0 \ \Rightarrow \ r_2=0 \\ &q_2(\gamma)=0 \ \Rightarrow \ r_3=0\quad &q_3(\delta)=0 \ \Rightarrow \ r_4=0 \\ &q_4(\epsilon)=0\ \Rightarrow \ r_5=0 \quad &eq(1.2) \quad \Rightarrow \quad r_6=v-(\alpha+2\beta+3\gamma+4\delta+5\epsilon)=0\\ \end{array} \right.\\ \end{align}

By (2.7), the system \(\{r_1=0, r_2=0, r_3=0, r_4=0, r_5=0, r_6=0\}\) can be viewed as a system of simultaneous polynomial equations in the unknowns \(\{\alpha,\beta,\gamma,\delta,\epsilon,v\}\). We aim to solve it.
In such cases, one reduces the number of variables and simplifies the problem by using the resultant. Ultimately, it turns out we can reduce to a single-variable polynomial of degree 120 in \(v\).
Below we show the results computed with the software Maxima. Except for the first expression, intermediate steps are omitted since the formulas become very large. (We write Res(x,y,z) for resultant(x,y,z).)

\begin{align} &s_1:Res(r_6,r_5,\epsilon); \quad s_1=-\delta -2 \gamma -3 \beta -4 \alpha -v =0 \notag \\ & \qquad \qquad \Downarrow \notag \\ &s_2:Res(s_1,r_4,\delta); \quad s_2=v^2+(3\gamma+5\beta+7\alpha+9)v+....+22 =0\notag \\ & \qquad \qquad \Downarrow \notag \\ &s_3:Res(s_2,r_3,\gamma); \quad s_3=v^6+(12\beta+18\alpha+24)v^5+...+27620\alpha+7388=0\notag \\ & \qquad \qquad \Downarrow \notag \\ &s_4:Res(s_3,r_2,\beta); \quad s_4=v^{24}+(60\alpha+84)v^{23}+ .....+1177129867156681=0\notag \\ & \qquad \qquad \Downarrow \notag \\ &s_5:Res(s_4,r_1,\alpha); \quad s_5=v^{120}+360v^{119}+64740v^{118}+....=0 \\ \notag \\ &V(v) \equiv s_5=v^{120}+360v^{119}+64740v^{118}+....=0 \\ \end{align}

Therefore, the primitive element \(v\) must satisfy (2.8).
Accordingly, as in (2.9), we define \(s_5\) as \(V(v)\). Incidentally, \(V(v)\) is called the “Galois resolvent”. Furthermore, when \(V(v)\) is factorized in Maxima, it splits into six degree-20 polynomials as shown in (2.10).

\begin{align} V(v) &= v^{120}+360v^{119}+64740v^{118}+....+31146731373284877082690427466971286244056290471141012169870457904759580625\notag \\ \end{align}

\begin{align} V(v) &=\displaystyle \prod_{i=1}^{6}V_{i}(v) \\ \end{align}

We denote the six factors obtained in this factorization by \(\{V_1,V_2,\dots,V_6\}\), shown below.

\begin{align} V_{1}&={{v}^{20}}+60 {{v}^{19}}+1790 {{v}^{18}}+35100 {{v}^{17}}+505261 {{v}^{16}}+5652840 {{v}^{15}}+50799180 {{v}^{14}}+373971600 {{v}^{13}} \notag \\ &+2281089966 {{v}^{12}}+11590327440 {{v}^{11}}+49084357780 {{v}^{10}}+172620188400 {{v}^{9}}+500267581306 {{v}^{8}} \notag \\ &+1181164237800 {{v}^{7}}+2242276888380{{v}^{6}}+3401909638560 {{v}^{5}}+4254933143241 {{v}^{4}} \notag \\ &+4933817387460 {{v}^{3}}+6084439227750 {{v}^{2}}+7164705190500 v+5919446204113 \\ \notag \\ V_{2}&={{v}^{20}}+60 {{v}^{19}}+1790 {{v}^{18}}+35100 {{v}^{17}}+505261 {{v}^{16}}+5655960 {{v}^{15}}+50923980 {{v}^{14}}+376545600 {{v}^{13}} \notag \\ &+2317001166 {{v}^{12}}+11964256320 {{v}^{11}}+52107272740 {{v}^{10}}+191853693600 {{v}^{9}}+596962688906 {{v}^{8}} \notag \\ &+1566271269720 {{v}^{7}}+3458738963260 {{v}^{6}}+6423623108400 {{v}^{5}}+9958500810441 {{v}^{4}} \notag \\ &+12461335397940 {{v}^{3}}+11642922808790 {{v}^{2}}+7043489742900 v+2039290237153 \\ \notag \\ V_{3}&={{v}^{20}}+60 {{v}^{19}}+1790 {{v}^{18}}+35100 {{v}^{17}}+505261 {{v}^{16}}+5657520 {{v}^{15}}+51001980 {{v}^{14}}+378300600 {{v}^{13}} \notag \\ &+2341243566 {{v}^{12}}+12196523160 {{v}^{11}}+53761666780 {{v}^{10}}+201106997400 {{v}^{9}}+639709411106 {{v}^{8}} \notag \\ &+1735342823280 {{v}^{7}}+4035147522220 {{v}^{6}}+8090223429240 {{v}^{5}}+13945832028441 {{v}^{4}} \notag \\ &+20035467005580 {{v}^{3}}+22125472711550 {{v}^{2}}+16079157042780 v+6837132314233 \\ \notag \\ V_{4}&={{v}^{20}}+60 {{v}^{19}}+1790 {{v}^{18}}+35100 {{v}^{17}}+505261 {{v}^{16}}+5657520 {{v}^{15}}+51033180 {{v}^{14}}+379611000 {{v}^{13}} \notag \\ &+2367283086 {{v}^{12}}+12520678680 {{v}^{11}}+56578689820 {{v}^{10}}+219122002200 {{v}^{9}}+726542093666 {{v}^{8}} \notag \\ &+2051322345840 {{v}^{7}}+4880224987660 {{v}^{6}}+9607234534200 {{v}^{5}}+15166159082361 {{v}^{4}} \notag \\ &+ 18160354917900 {{v}^{3}}+14980013767550 {{v}^{2}}+7296354097500 v+1761063045625 \\ \notag \\ V_{5}&={{v}^{20}}+60 {{v}^{19}}+1790 {{v}^{18}}+35100 {{v}^{17}}+505261 {{v}^{16}}+5659080 {{v}^{15}}+51064380 {{v}^{14}}+379400400 {{v}^{13}} \notag \\ &+2351679966 {{v}^{12}}+12238407600 {{v}^{11}}+53492805460 {{v}^{10}}+195251014800 {{v}^{9}}+587807064506 {{v}^{8}} \notag \\ &+1431313551240 {{v}^{7}}+2742285017260 {{v}^{6}}+3974293793760 {{v}^{5}}+4088970374841 {{v}^{4}} \notag \\ &+2626911907620 {{v}^{3}}+675718360070 {{v}^{2}}-213666065340 v+15823731313 \\ \notag \\ V_{6}&={{v}^{20}}+60 {{v}^{19}}+1790 {{v}^{18}}+35100 {{v}^{17}}+505261 {{v}^{16}}+5662200 {{v}^{15}}+51220380 {{v}^{14}}+383284800 {{v}^{13}} \notag \\ &+2414329566 {{v}^{12}}+12961651680{{v}^{11}}+59789211940 {{v}^{10}}+237890884800 {{v}^{9}}+816378602506 {{v}^{8}} \notag \\ &+2407592691960 {{v}^{7}}+6059842042380 {{v}^{6}}+12880967649840 {{v}^{5}}+22758667394841 {{v}^{4}} \notag \\ &+32586900570900 {{v}^{3}}+36272321795190 {{v}^{2}}+28893115997460 v+13542366738433 \\ \end{align}

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