Techniques of Solving Equations à la Galois
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1st upload: 2023/06/17
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[2-3] Introducing the minimal polynomial \(g_0(x) \equiv V(x)\)
The \(V(x)\) introduced in the previous section will in the end be what we call the "minimal polynomial."First, substitute (2.3) into the \(v_i\) in (3.1) and expand.
\begin{align} V(x)&\equiv \displaystyle \prod_{i=1}^6\sigma_i(x-v) =(x-v_{1})(x-v_{2})(x-v_{3})(x-v_{4})(x-v_{5})(x-v_{6}) \\ \notag \\ &= 36 {{\gamma }^{6}}+288 \beta {{\gamma }^{5}}+288 \alpha {{\gamma }^{5}}-132 x {{\gamma }^{5}}+863 {{\beta }^{2}} {{\gamma }^{4}}+1802 \alpha \beta {{\gamma }^{4}} \notag \\ &-828 x \beta {{\gamma }^{4}}+863 {{\alpha }^{2}} {{\gamma }^{4}}-828 x \alpha {{\gamma }^{4}}+193 {{x}^{2}} {{\gamma }^{4}}+1226 {{\beta }^{3}}{{\gamma }^{3}} \notag \\ & \qquad ......... \notag \\ &-828 x {{\alpha }^{4}} \beta +910 {{x}^{2}} {{\alpha }^{3}} \beta -480 {{x}^{3}} {{\alpha }^{2}} \beta +122 {{x}^{4}} \alpha \beta -12 {{x}^{5}} \beta \notag \\ &+36 {{\alpha }^{6}}-132 x {{\alpha }^{5}}+193 {{x}^{2}} {{\alpha }^{4}}-144 {{x}^{3}} {{\alpha }^{3}} +58 {{x}^{4}} {{\alpha }^{2}}-12 {{x}^{5}} \alpha +{{x}^{6}} \end{align}
[step1] Reduce \(V(x,\alpha,\beta,\gamma)\) modulo \(r_3(\gamma)\)
\begin{align} &V(x,\alpha,\beta,\gamma)= r_3(\gamma) \cdot Q_1(x,\alpha,\beta,\gamma)+V_1(x,\alpha,\beta) \equiv V_1(x,\alpha,\beta) \qquad( \ \because \ r_3=0 \ ) \notag \\ \notag \\ &=-4 {{\beta }^{6}}-12 \alpha {{\beta }^{5}}+\left( 3 {{\alpha }^{2}}+9 {{x}^{2}}\right) {{\beta }^{4}}+\left( 26 {{\alpha }^{3}}+18 {{x}^{2}} \alpha \right) {{\beta }^{3}} +\left( 3 {{\alpha }^{4}} +27 {{x}^{2}} {{\alpha }^{2}}-6 {{x}^{4}}\right) {{\beta }^{2}} \notag \\ & +\left( -12 {{\alpha }^{5}}+18 {{x}^{2}} {{\alpha }^{3}}-6 {{x}^{4}} \alpha \right) \beta -4 {{\alpha }^{6}}+9 {{x}^{2}} {{\alpha }^{4}}-6 {{x}^{4}} {{\alpha }^{2}}+{{x}^{6}} \\ \end{align}
[step2] Reduce \(V_1(x,\alpha,\beta)\) modulo \(r_2(\beta)\)
\begin{align} &V_1(x,\alpha,\beta)= \ r_2(\beta) \cdot Q_2(x,\alpha,\beta)+V_2(x,\alpha) \equiv V_2(x,\alpha) \qquad( \ \because \ r_2=0 \ ) \notag \\ \notag \\ &V_2(x,\alpha) =27 {{\alpha }^{6}}+162 {{\alpha }^{4}}+243 {{\alpha }^{2}}+{{x}^{6}}+18 {{x}^{4}}+81 {{x}^{2}}+108\\ \end{align}
[step3] Reduce \(V_2(x,\alpha)\) modulo \(r_1(\alpha)\)
\begin{align} &V_2(x,\alpha)= \ r_1(\alpha) \cdot Q_3(x,\alpha)+V_3(x) \equiv V_3(x) \qquad( \ \because \ r_1=0 \ ) \notag \\ \notag \\ & V_3(x)={{x}^{6}}+18 {{x}^{4}}+81 {{x}^{2}}+135 \equiv V(x)\\ \end{align}
\begin{align} V(x,\alpha,\beta,\gamma) &\equiv V(x) \qquad (mod \ r_3(\gamma) ), (mod \ r_2(\beta)), (mod \ r_1(\alpha)) \notag \\ \notag \\ \ V(x)&={{x}^{6}}+18 {{x}^{4}}+81 {{x}^{2}}+135 \qquad \in \ F_0[x]\\ \end{align}
It is striking that, simply by reducing in the order \([ r_3, r_2, r_1 ]\), the polynomial \(V(x,\alpha,\beta,\gamma)\) immediately becomes a polynomial over \(F_0\). Incidentally, the \(V(x)\) in (3.1)(3.6) is called the "Galois resolven".
Since \(V(x)\) has \(v=v_1\) as a root and is irreducible over \(F_0\), we may define it to be the minimal polynomial \(g_0(x)\) of \(v\).
\begin{align} g_0(x)&:minimal \ polynomial \ of \ v \ on \ F_0 \notag \\ \notag \\ &V(x) \equiv g_0(x)= x^6+18x^4+81x^2+135 \\ \notag \\ &g_0(v)=v^6+18v^4+81x^2+135=0 \\ \end{align}
Exactly as in Chapter 1, to determine the three roots \(\{\alpha,\beta,\gamma\}\) we will need the reciprocal \(V^{'}(x)^{-1}\) of \(V^{'}(x)\), so we compute it in advance.
Because \(V^{'}(v)^{-1}\) is an element of the simple extension \(F_0(v)\), it should be expressible as a linear combination of the basis \(\{1,v,v^2,v^3,v^4,v^5\}\) of \(F_0(v)\).
We therefore assume that \(V^{'}(v)^{-1}\) can be written as in (3.10). Then, from the condition \(V(v)^{'} \, V^{'}(v)^{-1}=1\), it suffices to solve the linear system (3.12) satisfied by the coefficients \(\{c_0,c_1,c_2,c_3,c_4,c_5\}\).
\begin{align} &V(x)=x^6+18x^4+81x^2+135 \quad \therefore V^{'}(v)=6v^5+72v^3+162v\\ \notag \\ &V^{'}(v)^{-1}=c_5v^5+c_4v^4+c_3v^3+c_2v^2+c_1v+c_0\\ &\qquad \qquad \Downarrow \notag \\ \notag \\ \end{align}
\begin{align} V(v)^{'} V^{'}(v)^{-1}&=(324c_4-36c_2+6c_0)v^5+(-3726c_5+324c_3-36c_1)v^4 \notag \\ &+(2106c_4-324c_2+72c_0)v^3+(-21384c_5+2106c_3-324c_1)v^2 \notag \\ &+(4860c_4-810c_2+162c_0)v-43740c_5+4860c_3-810c_1=1 \\ \notag \\ \end{align}
\begin{align} &\left\{ \begin{array}{l} &v^5:324c_4-36c_2+6c_0=0 & &v^4:-3726c_5+324c_3-36c_1=0 \\ &v^3:2106c_4-324c_2+72c_0=0 & &v^2:-21384c_5+2106c_3-324c_1=0 \\ &v^1:4860c_4-810c_2+162c_0=0 & &v^0:-43740c_5+4860c_3-810c_1=1 \\ \end{array} \right.\\ \notag \\ &\qquad \qquad \Downarrow \notag \\ \notag \\ &\biggl[ \ c_5=\frac{1}{3645}, \ c_4=0, \ c_3=\frac{1}{243}, \ c_2=0, \ c_1=\frac{7}{810}, \ c_0=0 \ \biggr] \\ \notag \\ &\therefore \quad V^{'}(v)^{-1}=\frac{{{v}^{5}}}{3645}+\frac{{{v}^{3}}}{243}+\frac{7 v}{810}\\ \end{align}
We have thus obtained the reciprocal \(V^{'}(x)^{-1}\).
(Note) In the computation of (3.11), remember to reduce modulo \(g_0(v)\).