Techniques of Solving Equations à la Galois
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1st upload: 2023/06/17
revision2 : 2023/07/27
revision3 : 2024/12/22
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[9-3] A second failure example \(f_2(x)=x^5-5x^3+5x+6 \)
In the previous section we used the Trager Algorithm twice.There we skipped the details; in this section we return to those computations with a quintic, and explain the steps again. The example is a bit involved, but it makes the procedure of the Trager Algorithm clear, so we go through the calculation process. We consider the following equation.
\begin{align} f(x)&=x^5-5x^3+5x+5 =(x-\alpha)(x-\beta)(x \ -\gamma)(x \ -\delta)(x \ -\epsilon) \\ \notag \\ v&=1 \cdot \alpha+2 \cdot \beta+3 \cdot \gamma+4 \cdot \delta+5 \cdot \epsilon \\ \end{align}
\begin{align} &\left\{ \begin{array}{l} f(x)=(x-\alpha)q_1(x)+r_1 \\ q_1(x)=(x-\beta)q_2(x)+r_2 \\ q_2(x)=(x-\gamma)q_3(x)+r_3 \\ q_3(x)=(x-\delta)q_4(x)+r_4 \\ q_4(x)=(x-\epsilon)q_5(x)+r_5 \\ r_5=\alpha+\beta+\gamma+\delta+\epsilon\\ \end{array} \right.\\ \end{align}
\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(3.2) \quad \Rightarrow \quad r_6 \equiv v-(\alpha+2\beta+3\gamma+4\delta+5\epsilon)=0\\ \end{array} \right.\\ \end{align}
Moreover, using Maxima’s factor command, \(V(x)\) factors over the field of rational numbers \(Q\) into six polynomials of degree \(20\). We take the first factor \(V_1(x)\) as a candidate for the minimal polynomial of \(v\).
\begin{align} V(v) &= v^{120}-1500v^{118}+1085250v^{116}-504625000v^{114}+....... \\ \end{align}
\begin{align} V(v) &=\displaystyle \prod_{i=1}^{6}V_{i}(v) \\ \end{align}
\begin{align} V_1(v) &= v^{20}-250v^{18}+22025v^{16}-......+12500000v^2+3200000 \\ \end{align}
So we move to the field \(Q(\zeta)\) obtained by adjoining a primitive 5th root of unity \(\zeta\) and check whether “\(V_1(x)\) is irreducible (i.e., whether it factors)”. Writing the minimal polynomial of \(\zeta\) as \([ \ Z=\zeta^4+\zeta^3+\zeta^2+\zeta+1 \ ]\), the computer algebra system Maxima has a command \(factor(V_1,Z)\) that factors \(V_1\) over the number field generated by \(Z\). Nevertheless, we prefer to carry out the computation ourselves using the Trager Algorithm.
[9-4] Factor \(V_1(x)\) over the number field \(Q(\zeta)\) by hand
To factor \(V_1(x)\), we change variables as in (4.1), \([ \ x \ \rightarrow \ x+s \cdot \zeta \ ]\).In this run of the Trager Algorithm we take \(s=1\). We then compute the resultant of \(V_1(x+s \cdot \zeta)\) and \(Z\).
\begin{align} &V_1(x+s \cdot \zeta) =x^{20}+ 20\zeta x^{19}+(190\zeta^{2}-250)x^{18}+.... +406559376 \\ \notag \\ &Resultant(V_1(x+s \cdot \zeta),Z,\zeta) = R_1(x) \cdot R_2(x)\\ \notag \\ &\left\{ \begin{array}{l} R_1(x)={{x}^{40}}-10 {{x}^{39}}-445 {{x}^{38}}+3980 {{x}^{37}}+....+1591615199615670401 \\ R_2(x)={{x}^{40}}-10 {{x}^{39}}-445 {{x}^{38}}+5080 {{x}^{37}}+...+81117808067017201 \\ \end{array} \right. \\ \end{align}
\begin{align} &r_{i-1}=q_{i} \cdot r_{i}+c_{i+1} \cdot r_{i+1} \quad [ \ i=1,2,3,...,11 \ ] \\ \end{align}
\begin{align} &\left\{ \begin{array}{l} r_0 \equiv R_1(x), \quad r_1 \equiv V_1(x+s \cdot \zeta) \quad \rightarrow \quad r_{12}=0 \quad \therefore \ r_{11}=Y_1 \\ r_0 \equiv R_2(x), \quad r_1 \equiv V_1(x+s \cdot \zeta) \quad \rightarrow \quad r_{12}=0 \quad \therefore \ r_{11}=Y_2 \\ \end{array} \right. \\ \notag \\ &\qquad \left\{ \begin{array}{l} GCD(R_1(x),V_1(x+s \cdot \zeta) )=Y_1(x)=x^{10}+10\zeta x^9+...-4011535\zeta+217696416\\ GCD(R_2(x),V_1(x+s \cdot \zeta) )=Y_2(x)=x^{10}+10\zeta x^9+...+4795560\zeta+83166086\\ \end{array} \right. \\ \end{align}
\begin{align} \therefore \ V_1(x)&=Y_1(x-s \cdot \zeta) \cdot Y_2(x-s\cdot \zeta) \\ \notag \\ g_0(x) &\equiv Y_1(x-s \cdot \zeta) \notag \\ &=x^{10}-(110\zeta^3+110\zeta^2+180)x^8+(9625\zeta^3+9625\zeta^2+15575)x^6 \notag \\ &-(341000\zeta^3+341000\zeta^2+551750)x^4+(4228125\zeta^3+4228125\zeta^2+6841250)x^2 \notag \\ &+134208800\zeta^3+134208800\zeta^2+217154400 \\ \end{align}
[9-5] Factor \(f(x)\) over the number field \(Q(\zeta,v)\) by hand
With the minimal polynomial \(g_0(x)\) determined, we now factor the equation \(f(x)\) inside the number field generated by this polynomial. Since the field generated by \(g_0(x)\) is the minimal splitting field for \(v\), \(f(x)\) should factor all the way down to linear polynomials in \(x\).To factor \(f(x)\), we change variables as in (4.1), \([ \ x \ \rightarrow \ x+s \cdot \zeta \ ]\).
In this application of the Trager Algorithm we again take \(s=1\). We then compute the resultant of \(f(x+s \cdot v)\) and \(g_0(v)\).
\begin{align} &f(x+s \cdot v)=(x+v)^5-5(x+v)^3+5(x+v)+6 \\ \notag \\ \end{align} \begin{align} &Resultant(f(x+s \cdot v),g_0(v),v)=R_1(x) \cdot R_2(x) \cdot R_3(x) \cdot R_4(x) \cdot R_5(x) \\ \notag \\ &\left\{ \begin{array}{l} R_1(x)=x^{10}-(150\zeta^3+150\zeta^2+260)x^8+....+787597200\zeta^2+1274359536 \\ \qquad ..... \\ R_5(x)=x^{10}-(70\zeta^3+70\zeta^2+120)x^8+....+16888960\zeta^2+27326916 \\ \end{array} \right. \\ \end{align}
\begin{align} &\left\{ \begin{array}{l} r_{i-1}=q_{i} \cdot r_{i}+c_{i+1} \cdot r_{i+1} \quad [ \ i=1,...,5 \ ] \\ \qquad r_0 \equiv R_k(x), \quad r_1 \equiv f(x+s \cdot v) \quad \rightarrow \quad r_{6}=0 \quad \therefore \ r_{5}=Y_k \quad [ \ k=1,..,5 \ ] \\ \end{array} \right. \\ \notag \\ &\left\{ \begin{array}{l} GCD(R_1(x),f(x+s \cdot v))=Y_1(x)=x-\frac{2207 {{v}^{6}} {{\zeta }^{3}}}{300}+\frac{623 {{v}^{4}} {{\zeta }^{3}}}{60}+... +\frac{143 {{v}^{2}}}{60}+\frac{3 v}{2} \\ \qquad ......\\ GCD(R_5(x),f(x+s \cdot v))=Y_5(x)=x-\frac{2207 {{v}^{6}} {{\zeta }^{3}}}{300}+\frac{623 {{v}^{4}} {{\zeta }^{3}}}{60}+... +\frac{143 {{v}^{2}}}{60}+\frac{v}{2} \\ \end{array} \right. \\ \end{align}
\begin{align} &\therefore \ f(x)=Y_1(x-s \cdot v) \cdot Y_2(x-s \cdot v) \cdot Y_3(x-s \cdot v) \cdot Y_4(x-s \cdot v) \cdot Y_5(x-s \cdot v) \\ \notag \\ &Y_i(x-s \cdot v)=0 \qquad [\ i=1,...,5 \ ] \notag \\ &\qquad \Downarrow \notag \\ &\left\{ \begin{array}{l} x_1=\frac{1}{300}\bigl((2207v^6-3115v^4+1170v^2+150v-100)ζ^3+...-715v^2-150v \bigr) \\ x_2=-\frac{1}{300}\bigl((843v^6-1190v^4+455v^2+300v-100)ζ^3+...-260v^2-150v-100 \bigr) \\ x_3=-\frac{1}{150}\bigl((1364v^6-1925v^4+715v^2)ζ^3+... -455v^2+100 \bigr) \\ x_4=-\frac{1}{300}\bigl( (843v^6-1190v^4+455v^2-300v-100)ζ^3+...-260v^2+150v-100 \bigr) \\ x_5=\frac{1}{300}\bigl( (2207v^6-3115v^4+1170v^2-150v-100)ζ^3+...-715v^2+150v \bigr) \\ \end{array} \right. \\ \end{align}
The set \(\{ \ x_1,...,x_5 \ \}\) is some permutation of \(\{ \ \alpha, \ \beta, \ \gamma, \ \delta, \ \epsilon \}\), but we do not yet know the one-to-one correspondence. In the next section we compute that correspondence.