【例題3】の解法手順
EX3-1\begin{align*} &f(x)=x^5-10x^3+5x^2+10x+13 \\ \\ &\qquad \{\alpha,\beta,\gamma,\delta,\epsilon\}: \ roots \ of \ f(x)\\ &\qquad v: \ Primitive \ element \\ \\ & v=1\cdot\alpha+2\cdot \beta+3\cdot\gamma+4 \cdot \delta+5 \cdot \epsilon \end{align*}
\[ \qquad The \ system \ of \ equations \]
\[ \left\{ \begin{array}{l} r_1=\alpha^5-10\alpha^3+5\alpha^2+10 \alpha +1=0\\ r_2=\beta^4+\alpha\beta^3+(\alpha^2-10)\beta^2+(\alpha^3-10 \alpha +5)\beta \\ \qquad +\alpha^4-10\alpha^2+5 \alpha +10=0\\ r_3=\gamma^3+( \beta +\alpha )\gamma^2+(\beta^2+\alpha \beta +\alpha^2-10)\gamma +\beta^3 \\ \qquad +\alpha\beta^2+(\alpha^2-10) \beta +\alpha^3-10 \alpha +5=0\\ r_4=\delta^2+( \gamma +\beta +\alpha )\delta +\gamma^2+( \beta +\alpha )\gamma \\ \qquad +\beta^2+\alpha \beta +\alpha^2-10=0 \\ r_5=\alpha+\beta+\gamma+\delta+\epsilon=0 \\ r_6=v-(\alpha+2\beta+3\gamma+4\delta+5\epsilon )=0 \\ \end{array} \right.\\ \quad \\ \qquad \qquad \qquad \Downarrow \]
\[ \qquad Elimination \ Theory \]
\[ V(v)= {{v}^{120}}-3000 {{v}^{118}}+4350000 {{v}^{116}}.... \\ \qquad \qquad ..........\\ \quad \\ \qquad =( v^5-125v^3+2500v-4375 )\times .....\\ \qquad \times (v^5-125v^3+800v^2-1750 v+1225 ) \\ \]
\[g_{0}(x)=x^5-125x^3+2500x-4375 \]
\[ g_0(x):minimal \ polynomial \ of \ v \ on \ F_0=Q(\zeta) \]
\[Factorization \ of \ f(x) \ on \ F_0(v)\] \[\quad "maxima's \ function \ "\] \[\qquad factor(f(x),g_0(v))\]
\begin{align*} &\alpha=\alpha(v)=\frac{22 {{v}^{4}}+25 {{v}^{3}}-2575 {{v}^{2}}-5125 v+35250}{5375}\\ &\beta=\beta(v), \ \gamma=\gamma(v), \ \delta=\delta(v), \ \epsilon=\epsilon(v)\\ \quad \\ \end{align*}
\begin{align*} &roots \ of \ V(x)\\ &\quad [ \ v_1=v_1(v), \ ....\ , \ v_{120}=v_{120}(v) \ ] \\ \end{align*}
\begin{align*} &g_0(v_i)=0 \quad for \ (i=1,34,65,91,97) \\ &\qquad \qquad \Downarrow\ \\ &C_5: Galois \ group \ of \ F(x) \\ &\qquad composition \ series \quad C_5 \rhd \{e\} \end{align*}
\[g_1(x)\ : \ minimal \ polynomial \ of \ v\\ \qquad g_1(x) \ \in \ F_0(a_1)[x] \\ \quad \\ B_1=a_1^5+ 500\zeta^3+1000\zeta^2+875 \zeta +125 =0 \]
\begin{align*}
&v=v(a_1,\zeta) \\
\quad \\
&\left\{
\begin{array}{l}
\alpha=\alpha(a_1,\zeta), \quad
\beta=\beta(a_1,\zeta) \\
\gamma=\gamma(a_1,\zeta), \quad
\delta=\delta(a_1,\zeta) \\
\epsilon=\epsilon(a_1,\zeta) \\
\end{array}
\right.\\
\end{align*}
（覚書：\(t_1=0 \ \Rightarrow \ A_1=0\) の時どうしますか？
EX3-8 Roots \( \{\alpha,\beta,\gamma,\delta,\epsilon\} \ of \ f(x)\)
前節までで \(v\) の最小多項式 \(g_1(x)\) の次数を1次式まで下げる事が出来ました。
従って \(g_1(x)=0\) の根が、\(v\) の値となります。
\begin{align} \setCounter{60} &g_1(x) = x+a_1-\frac{{{a}_{1}^{2}}\, \left( {{\zeta }^{3}}+2 {{\zeta }^{2}}+2\right) }{5} \notag \\ &\quad -\frac{{{a}_{1}^{3}}\, \left( \zeta +1\right) \, \left( 4 {{\zeta }^{2}}-3 \zeta +4\right) }{25} +\frac{{{a}_{1}^{4}}\, \left( 4 {{\zeta }^{3}}+3 {{\zeta }^{2}}+2 \zeta +6\right) }{125} \\ \notag \\ \therefore &\quad v=-a_1+ \frac{{{a}_{1}^{2}}\, \left( {{\zeta }^{3}}+2 {{\zeta }^{2}}+2\right) }{5} \notag \\ &\quad +\frac{{{a}_{1}^{3}}\, \left( \zeta +1\right) \, \left( 4 {{\zeta }^{2}}-3 \zeta +4\right) }{25} -\frac{{{a}_{1}^{4}}\, \left( 4 {{\zeta }^{3}}+3 {{\zeta }^{2}}+2 \zeta +6\right) }{125} \\ \end{align}
この \(v\) の値を式(22)に代入したものが、下式の様に最終的に求めたい \(f(x)\) の根となります。
\begin{align} \alpha=&\frac{1}{125} \Bigl[ {a_1} \left( 25 {{\zeta }^{3}}+25 {{\zeta }^{2}}+25 \zeta +50\right) +{{a}_{1}^{2}}\, \left( 5 {{\zeta }^{3}}-10 {{\zeta }^{2}}+5 \zeta \right) \Bigr.\notag \\ &\qquad \qquad \Bigl. +{{a}_{1}^{3}}\, \left( -4 {{\zeta }^{3}}+2 {{\zeta }^{2}}-2 \zeta -1\right)+{{a}_{1}^{4}}\, \left( {{\zeta }^{3}}+{{\zeta }^{2}}+2\right) \Bigr] \\ \beta=& -\frac{1}{125} \Bigl[ {a_1} \left( 50 {{\zeta }^{3}}+25 {{\zeta }^{2}}+25 \zeta +25\right) +{{a}_{1}^{2}}\, \left( 5 {{\zeta }^{3}}+5 {{\zeta }^{2}}+15\right) \Bigr. \notag \\ &\qquad \qquad \Bigl. +{{a}_{1}^{3}}\, \left( 4 {{\zeta }^{3}}+3 {{\zeta }^{2}}+2 \zeta +6\right) +{{a}_{1}^{4}}\, \left( {{\zeta }^{2}}-\zeta +1\right) \Bigr] \\ \gamma=&-\frac{1}{125} \Bigl[ {a_1} \left( 25 {{\zeta }^{2}}-25 {{\zeta }^{3}}\right)+ {{a}_{1}^{2}}\, \left( 15 {{\zeta }^{3}}+5 \zeta +5\right) \Bigr. \notag \\ &\qquad \qquad \Bigl. +{{a}_{1}^{3}}\, \left( -{{\zeta }^{3}}+3 {{\zeta }^{2}}-3 \zeta +1\right) +{{a}_{1}^{4}}\, \left( {{\zeta }^{3}}-{{\zeta }^{2}}+\zeta \right) \Bigr] \\ \delta=&\frac{1}{125} \Bigl[ {a_1} \left( 25 {{\zeta }^{2}}-25 \zeta \right) +{{a}_{1}^{2}}\, \left( 5 {{\zeta }^{2}}-10 \zeta +5\right) \Bigr. \notag \\ &\qquad \qquad +{{a}_{1}^{3}}\, \left( 6 {{\zeta }^{3}}+2 {{\zeta }^{2}}+3 \zeta +4\right) +{{a}_{1}^{4}}\, \left( 2 {{\zeta }^{3}}+\zeta +1\right)\Bigl. \Bigr] \\ \epsilon=&-\frac{1}{125} \Bigl[ {a_1} \left( 25-25 \zeta \right) +{{a}_{1}^{2}}\, \left( -15 {{\zeta }^{3}}-10 {{\zeta }^{2}}-10 \zeta -15\right) \Bigr. \notag \\ & \qquad \qquad \Bigl. +{{a}_{1}^{3}}\, \left( -{{\zeta }^{3}}-2 {{\zeta }^{2}}+2 \zeta -4\right)+{{a}_{1}^{4}}\, \left( 2 {{\zeta }^{3}}+{{\zeta }^{2}}+\zeta +2\right) \Bigr] \\ \notag \\ &但し上の計算は \ ( \ mod \ B_1 \ ) \ \rightarrow \ ( \ mod \ Z \ ) \ で剰余を取る事！ \notag \\ \end{align}
\begin{align} &\left\{ \begin{array}{l} Z=\zeta^4+\zeta^3+\zeta^2+\zeta+1=0 \\ B_1=a_1^5-A_1\qquad a_1\equiv \sqrt[5]{-500 {{\zeta }^{3}}-1000 {{\zeta }^{2}}-875 \zeta -125} \end{array} \right. \\ \notag \\ &g_0(x)=x^5-125x^3+2500x-4375 \ \in \ F_0[x] \\ &g_1(x)=g_1(x,a_1) \ \in \ F_0(a_1)[x]=F_1[x] \\ \end{align}
以上がガロア理論を使って5次方程式 \( f(x)=x^5-10x^3+5x^2+10x+13 \) を解く計算手順でした。
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