Techniques of Solving Equations à la Galois
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1st upload: 2023/06/17
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[8-3] Determining the Minimal Polynomial \(g_0(x)\)
As computed in the previous section, the Galois resolvent \(V(x)\) is a polynomial of degree \(120\); in the end it factors into six polynomials \(\{V_1,V_2,\dots,V_6\}\).\begin{align} V(x) &= x^{120}+360x^{119}+64740x^{118}+....=\displaystyle \prod_{i=1}^{6}V_{i}(x) \\ \end{align}
\begin{align} g_0(x) \equiv V_1(x) \\ \end{align}
\begin{align} \notag \\ V_{1}(x)&={{x}^{20}}+60 {{x}^{19}}+1790 {{x}^{18}}+35100 {{x}^{17}}+505261 {{x}^{16}}+5652840 {{x}^{15}}+50799180 {{x}^{14}}+373971600 {{x}^{13}} \notag \\ &+2281089966 {{x}^{12}}+11590327440 {{x}^{11}}+49084357780 {{x}^{10}}+172620188400 {{x}^{9}}+500267581306 {{x}^{8}} \notag \\ &+1181164237800 {{x}^{7}}+2242276888380{{x}^{6}}+3401909638560 {{x}^{5}}+4254933143241 {{x}^{4}} \notag \\ &+4933817387460 {{x}^{3}}+6084439227750 {{x}^{2}}+7164705190500 x+5919446204113 \notag \\ \end{align}
Next we factor the polynomial \(f(x)\) over the simple extension \(F_0(v)\) determined by the minimal polynomial \([g_0(v)=0]\).
Maxima has a command \(factor(p,q)\) that factors a polynomial \(p\) over the algebraic field generated by \(q\), so we use this. Equation (3.3) is the factoring command, and (3.4) shows the result. Since the factored expressions are extremely complicated, (3.5) records a heavily abbreviated form. Each of \(\{x_1(v),x_2(v),x_3(v),x_4(v),x_5(v)\}\) is a polynomial in \(v\) of degree \(19\).
\begin{align} f_{g_0} &:factor(f(x),g_0(v)); \\ \notag \\ \Rightarrow \quad f_{g_0}&=(x-x_1(v))(x-x_2(v))(x-x_3(v))(x-x_4(v))(x-x_5(v)) \\ \end{align}
\begin{align} \notag \\ x_1&=x_1(v) \notag \\ &=\frac{1}{1805610907325788018215199866227491256974007107113418338878394352169956056029101346198715864} \notag \\ &\times \bigl(20399914128957102114355730051252429667468197886785684817206258694119405575139320v^{19}+.... \bigr. \notag \\ &+\bigl. 16081798561393987072650538574298238544699323139325350416481161239277565867115418404308931185 \bigr) \\ \end{align}
The set \(\{x_1(v),x_2(v),x_3(v),x_4(v),x_5(v)\}\) coincides with \(\{\alpha,\beta,\gamma,\delta,\epsilon\}\) in some order, but the correspondence is not yet identified. We now compute to determine this correspondence.
Let \([ \ w = x_{1}+2 x_{2}+3 x_{3}+4x_{4}+5x_{5} \ ]\). Let \(\tau_i\) range over the elements of the symmetric group \(S_5\) acting by permuting the ordering of \(\{x_1,x_2,x_3,x_4,x_5\}\). The permutations produce \(\{w_1,w_2,\dots,w_{120}\}\) as in (3.7). Since these are permutations of the \(\{x_i\}\), we purposely use \(\{\tau_i\}\) rather than \(\{\sigma_i\}\).
\begin{align} &w = x_{1}+2 x_{2}+3 x_{3}+4x_{4}+5x_{5} \\ \end{align}
\begin{align} \notag \\ &\left\{ \begin{array}{l} \tau_1 (w)=w_1=x_1+2x_2+3x_3+4x_{4}+5x_{5} & &\tau_2 (w)=w_2=x_1+2x_2+3x_3+4x_{5}+5x_{4} \\ \tau_3 (w)=w_3=x_1+2x_2+3x_4+4x_{3}+5x_{5} & &\tau_4 (w)=w_4=x_1+2x_2+3x_4+4x_{5}+5x_{3} \\ \qquad ....... & & \\ \tau_{119} (w)=w_{119}=x_5+2x_4+3x_3+4x_{1}+5x_{2} & &\tau_{120} (w)=w_{120}=x_5+2x_4+3x_3+4x_{2}+5x_{1} \\ \end{array} \right. \\ \end{align}
Substituting (3.5) into \(\{w_i\}\) in terms of \(\{x_1,x_2,x_3,x_4,x_5\}\), we search for a permutation \(\tau_i\) for which the value becomes exactly \(v\). As shown in (3.8), we find that \(\tau_{46}\) is the desired permutation. The correspondence is given in (3.9).
Although the coefficients are very complicated, we record abbreviated forms indicating how the actual five roots \(\{\alpha,\beta,\gamma,\delta,\epsilon\}\) are expressed. Thus the five roots of \(f(x)\) are represented by polynomials in \(v\) of degree \(19\).
From this we find that the simple extension \(F_0(v)\) generated by the minimal polynomial \([g_0(v)=0]\) is the splitting field of \(f(x)\).
\begin{align} &\tau_{46} (w)=w_{46}=x_2+2x_5+3x_3+4x_{4}+5x_{1} =v \\ \notag \\ &\therefore \ \ \alpha=x_2(v), \quad \beta=x_5(v), \quad \gamma=x_3(v), \quad \delta=x_4(v), \quad \epsilon=x_1(v) \end{align}
\begin{align} \alpha&=\frac{1}{28741663893023098374894074591802026018471908286940284736663131550284264654482411768768824} \notag \\ &\times \bigl( 210567501608336257455406877608503070848764997115834429441727871265227830902505v^{19}+........ \bigr. \notag \\ &+\bigl. 57442210272809374169345337580868722435898720939368040111653691808072392787022194552213299 \bigr)\notag \\ \notag \\ \beta&=\frac{-1}{857252457835009000213110196109220524468524761248558374280911860562745030594448} \notag \\ &\times \bigl( 3489672752300150933392195695645112172288494139509663115078892274970v^{19}+........ \bigr. \notag \\ &+\bigl. 3160686259198568119139434052803948615255814775832360709381422206175471892235787 \bigr)\notag \\ \notag \\ \gamma&=\frac{-1}{13214374102432605525586821997382112143731339298639171446156053662313051022867808487112} \notag \\ &\times \bigl( 33551790541675810392282267393932986968662189984953840556354972077255294275v^{19}+........ \bigr. \notag \\ &+\bigl. 83050394964553626522006732647984223980120065803477113655047195713571787342880161878127 \bigr)\notag \\ \notag \\ \delta&=\frac{-1}{6634428894032683488687127171128885141617384968585523393306641160158162726923810672} \notag \\ &\times \bigl( 79709193917302725524707317987542422011469313081618774778762485686856770v^{19}+........ \bigr. \notag \\ &+\bigl. 96219092449220493069543091774396449857033300915454544410229243200192229204492103293 \bigr)\notag \\ \notag \\ \epsilon&=\frac{1}{1805610907325788018215199866227491256974007107113418338878394352169956056029101346198715864} \notag \\ &\times \bigl( 20399914128957102114355730051252429667468197886785684817206258694119405575139320v^{19}+........ \bigr. \notag \\ &+\bigl. 16081798561393987072650538574298238544699323139325350416481161239277565867115418404308931185 \bigr)\notag \\ \end{align}
Since we now have polynomial expressions in \(v\) for the five roots \(\{\alpha,\beta,\gamma,\delta,\epsilon\}\), we use them to determine the roots of the minimal polynomial \(g_0(x)\). First, act on \(v\) in (3.10) by elements \(\sigma_i\) of the symmetric group \(S_5\) (which induce permutations of \(\{\alpha,\beta,\gamma,\delta,\epsilon\}\)) to produce \([\ \sigma_i(v)=v_i\ ]\).
\begin{align} &v = \alpha+2 \beta+3 \gamma+4\delta+5\epsilon \\ \notag \\ &\left\{ \begin{array}{l} \sigma_1 (v)=v_1=\alpha+2\beta+3\gamma+4\delta+5\epsilon & &\sigma_2 (v)=v_2=\alpha+2\beta+3\gamma+4\epsilon+5\delta \\ \sigma_3 (v)=v_3=\alpha+2\beta+3\delta+4\gamma+5\epsilon & &\sigma_4 (v)=v_4=\alpha+2\beta+3\delta+4\epsilon+5\gamma \\ \qquad ....... & & \\ \sigma_{119} (v)=v_{119}=\epsilon+2\delta+3\gamma+4\alpha+5\beta & &\sigma_{120} (v)=v_{120}=\epsilon+2\delta+3\gamma+4\beta+5\alpha \\ \end{array} \right. \\ \notag \\ \end{align}
\begin{align} \sigma_i(v)&=v_i \quad i=[1,2,3,4,5,.........117,118,119,120] \\ \end{align}
\begin{align} \notag \\ g_0(v_i)&=0 \quad i=[1,8,18,23,30,33,40,43,52,59,61,70,73,80,90,95,99,108,110,117] \\ \end{align}
\begin{align} &Galois \ Group \quad Gal(F_0(v)/F_0) \equiv F_{20}:Frobenius Group \notag \\ \notag \\ \end{align}
\begin{align} &F_{20}=\{\rho_i\} \quad i=[1,8,18,23,30,33,40,43,52,59, \notag \\ &\qquad \qquad \qquad \qquad 61,70,73,80,90,95,99,108,110,117] \\ \notag \\ &\rho_i(v) \equiv \sigma_i(v)=v_i \\ \end{align}