Techniques for Solving Solvable Equations via Galois Theory

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[2-13] Computing \(F_2/F_1\): finding the minimal polynomial \(g_2(x)\) (2)

Since we have already obtained \(\{t_0, \tilde{t_1},\tilde{t_2}\}\) in the previous section, we now move to the final stage of the computation.

Step 3 ILRT (Inverse Lagrange Resolvent Transformation) \begin{align} &\begin{bmatrix} \tilde{h_0} \\ \tilde{h_1} \\ \tilde{h_2} \end{bmatrix} = \begin{bmatrix} 1&1&1 \\ 1&\omega^2&(\omega^2)^2\\ 1&\omega&(\omega^2)\\ \end{bmatrix} \cdot \begin{bmatrix} t_0 \\ \tilde{t_1} \\ \tilde{t_2} \end{bmatrix} \ \Longrightarrow \ \left\{ \begin{array}{l} g_1(x)=\tilde{h_0}\cdot \tilde{h_1} \cdot \tilde{h_2} \ \in \ F_1[x] \\ g_2(x) \equiv \tilde{h_0} \ \in \ F_2[x] \end{array} \right. \\ \end{align}

We now compute the inverse transform ILRT in (13.1). The result of the inverse transform is the triple \(\{ \ \tilde{h_0}, \tilde{h_1},\tilde{h_2} \ \}\) (with tildes).

\begin{align} \begin{bmatrix} \tilde{h_0} \\ \tilde{h_1} \\ \tilde{h_2} \end{bmatrix} &= \begin{bmatrix} 1&1&1 \\ 1&\omega^2&\omega\\ 1&\omega&\omega^2 \end{bmatrix} \cdot \begin{bmatrix} t_0 \\ \tilde{t_1} \\ \tilde{t_2} \end{bmatrix} = \begin{bmatrix} t_0+\tilde{t_1}+\tilde{t_2}\\ t_0+\omega^2 \tilde{t_1}+\omega \tilde{t_2}\\ t_0+\omega \tilde{t_1}+\omega-2 \tilde{t_2} \end{bmatrix} \notag \\ \notag \\ &= \begin{bmatrix} x+a_2+a_2^2\left(-\frac{\omega }{3}+\frac{{a_1}}{18}-\frac{1}{6}\right) \\ x+a_2\omega^2+a_2^2\omega\left(-\frac{\omega }{3}+\frac{{a_1}}{18}-\frac{1}{6}\right) \\ x+a_2\omega+a_2^2\omega^2\left(-\frac{\omega}{3}+\frac{{a_1}}{18}-\frac{1}{6}\right) \end{bmatrix} \quad \in \ F_2[x] \\ \end{align}

All of \(\{ \ \tilde{h_0},\tilde{h_1},\tilde{h_2} \ \}\) have been expressed as polynomials in \(F_2[x]\) that do not involve \(v\).
Moreover, since \(\tilde{h_0}\) originally had \((x-v)\) as a factor, the minimal polynomial of \(v\) over the extension field \(F_2\) is \(g_2(x)=\tilde{h_0}\). Because the degree of the minimal polynomial \(g_2(x)\) is finally linear, the value of \(v\) is the root of \(g_2(x)\), giving (13.4).

\begin{align} &\tilde{h_0} \equiv g_2(x)=x+a_2+a_2^2\left(- \ \frac{\omega }{3}+\frac{{a_1}}{18}-\frac{1}{6}\right) \ \in F_2[x] \\ \notag \\ &g_2(x)=0 \quad \Rightarrow \quad v=-a_2+a_2^2\left(\ \frac{\omega }{3}-\frac{{a_1}}{18}+\frac{1}{6}\right) \ \in F_2 \\ \end{align}

[2-14] Solutions of the equation

Substituting \(v\) from (13.4) into (14.1) yields the roots of \(f(x)\). Below we list the results of that computation together with the formulas obtained so far.

\begin{align} \alpha&=\frac{v^4+15v^2-9v+36}{18} & \beta&=-\frac{{{v}^{4}}+15 {{v}^{2}}+36}{9} & \gamma&=\frac{{{v}^{4}}+15 {{v}^{2}}+9 v+36}{18} \\ \end{align}

\begin{align} &\alpha=\omega\left(-\frac{a_1a_2^2}{54}-\frac{a_2^2 }{6}+\frac{a_2}{3}\right)+\frac{{a_1} {{a}_{2}^{2}}}{54}-\frac{a_2^2}{6}+\frac{2a_2}{3}\\ &\beta=\omega\left(\frac{a_1a_2^2}{27}-\frac{2a_2}{3}\right)+\frac{a_1a_2^2}{54}+\frac{a_2^2}{6}-\frac{{a_2}}{3} \\ &\gamma=\omega\left(-\frac{a_1a_2^2}{54}+\frac{a_2^2}{6}+\frac{a_2}{3}\right)-\frac{a_1a_2^2}{27}-\frac{a_2}{3} \\ \end{align}

Note that in the above calculation it is important to reduce in the order \( (mod \ B_2) \rightarrow (mod \ B_1) \rightarrow (mod \ \varOmega) \). Finally, we summarize the minimal polynomials, binomial equations, etc., obtained thus far.

\begin{align} &\left\{ \begin{array}{l} B_1=a_1^2-A_1=0 \qquad A_1=-135 \qquad \quad a_1=\sqrt{A_1} \\ B_2=a_2^3-A_2=0 \qquad A_2=3\omega+\frac{a_1+3}{2} \quad a_2=\sqrt[3]{A_2}\\ \varOmega = \omega^2+ \omega +1 =0 \\ \end{array} \right. \\ \notag \\ &g_0(x)=(x-v_1)....(x-v_6)=x^6+18x^4+81x^2+135\\ \notag \\ &g_1(x)=(x-v_1)(x-v_4)(x-v_5)=x^3+9x+a_1\\ \notag \\ &g_2(x)=(x-v_1)=x+a_2^2\left(-\frac{\omega}{3}+\frac{a_1}{18}-\frac{1}{6}\right)+a_2 \\ \end{align}

The above constitutes the computational procedure for solving the equation using Galois theory.

[2-15] Numerical verification of the computed roots

Although (14.2), (14.3), and (14.4) give the roots of \(f(x)\), let us check numerically that they are correct. The computer algebra system "Maxima" provides the command allroots(). We will use it for numerical computation, proceeding in the order \(\omega \rightarrow a_1 \rightarrow a_2\).

Enter    \(rw:allroots(\varOmega);\)    and you obtain
\(rw=\left[ \ \omega =0.8660254037844386 \% i-0.5, \ \omega =-0.8660254037844386 \% i-0.5 \ \right]\)
as a list output, so we take \(\omega=0.8660254037844386 \% i-0.5\).

Enter    \(ra_1:allroots(B_1);\)    and you obtain
\(ra_1= \left[ a_1=11.61895003862225 \% i, \ a_1=-11.61895003862225 \% i\right] \)
as a list output, so we take \(a_1=11.61895003862225 \% i\).

With \(\omega\) and \(a_1\) determined, substitute these into \(B_2\) in (14.5). Then
\(B_2=a_2^3-5.809475019311125 \% i-3 \left( 0.8660254037844386 \% i-0.5\right) -1.5\).
Using this \(B_2\),

enter    \(ra_2:allroots(B_2);\)    and you obtain
\(ra_2=[ \ a_2=1.016700830808605\%i+1.760977495057993,\)
\(\qquad\) \(a_2=1.016700830808605\%i-1.760977495057993,\)
\(\qquad\) \(a_2=-2.03340166161721\%i \ ]\).
Hence we take \(a_2=1.016700830808605\%i+1.760977495057993\).

We now have numerical values for \(\{ \ \omega, \ a_1, \ a_2 \ \}\). Substituting these into (14.2)-(14.4) yields the three roots below.
\(\alpha=0.1610926773130429+1.754380959783721\%i\)
\(\beta=0.1610926773130429-1.754380959783722\%i\)
\(\gamma=-0.3221853546260854+6.579099405186113 \times 10^{-17}\%i \)

Independently, let us find the roots of \(f(x)\) numerically as well. Enter    \(abc:allroots(f(x));\)    to obtain
\(abc=[ \ x=-0.3221853546260856,\)
\(\qquad\) \( x=0.1610926773130428+1.754380959783722\%i,\)
\(\qquad\) \(x=0.1610926773130428-1.754380959783722\%i \ ]\).

Comparing the final results (order disregarded), we see agreement to at least 15 decimal places. Therefore, the lengthy computation carried out so far is also confirmed numerically to be correct.

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