Techniques of Solving Equations à la Galois
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1st upload: 2023/06/17
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[7-7] Roots of the cyclotomic polynomial \(\Phi_{17}=0\)
At the end of the previous section we obtained the value of \(v\). Substituting it into (1.4), we now compute all roots of \(\Phi_{17}=0\).When doing so, be sure to perform remainder calculations in the order \( (\mathrm{mod}\,g_4(v)) \rightarrow (\mathrm{mod}\,B_4(a_4)) \rightarrow (\mathrm{mod}\,B_3(a_3)) \rightarrow (\mathrm{mod}\,B_2(a_2)) \rightarrow (\mathrm{mod}\,B_1(a_1)) \). Also, the formulas for the \(v_i\) are rearranged so that similar polynomial shapes are paired, e.g., \(\{ \ [ v_{1},v_{16} ], \ [v_{2},v_{15} ], \ldots \}\).
\begin{align} v_{1}&=-{a_4}-\frac{{a_3}}{2}-\frac{{a_2}}{4}-\frac{{a_1}}{8}-\frac{1}{16} \notag \\ v_{16}&= {a_4}-\frac{{a_3}}{2}-\frac{{a_2}}{4}-\frac{{a_1}}{8}-\frac{1}{16} \notag \\ \notag \\ \notag \\ v_{2}&={a_3} {a_4}+\frac{{a_2} {a_4}}{2}+\frac{{a_1} {a_4}}{4}+\frac{{a_4}}{8}+\frac{{a_2} {a_3}}{2}+\frac{{a_1} {a_3}}{4}+\frac{{a_3}}{8}+\frac{{a_2}}{4}-\frac{{a_1}}{8}-\frac{1}{16} \notag \\ v_{15}&=-{a_3} {a_4}-\frac{{a_2} {a_4}}{2}-\frac{{a_1} {a_4}}{4}-\frac{{a_4}}{8}+\frac{{a_2} {a_3}}{2}+\frac{{a_1} {a_3}}{4}+\frac{{a_3}}{8}+\frac{{a_2}}{4}-\frac{{a_1}}{8}-\frac{1}{16} \notag \\ \notag\\ \notag \\ v_{3}&=-{a_2} {a_3} {a_4}-\frac{{a_1} {a_3} {a_4}}{2}-\frac{{a_3} {a_4}}{4}-\frac{{a_2} {a_4}}{2}+\frac{{a_1} {a_4}}{4}-\frac{7 {a_4}}{8}-\frac{{a_1} {a_2} {a_3}}{4}-\frac{3 {a_2} {a_3}}{8} \notag \\ &\qquad -\frac{{a_3}}{4}+\frac{{a_1} {a_2}}{8}-\frac{{a_2}}{16}+\frac{{a_1}}{8}-\frac{1}{16} \notag \\ v_{14}&={a_2} {a_3} {a_4}+\frac{{a_1} {a_3} {a_4}}{2}+\frac{{a_3} {a_4}}{4}+\frac{{a_2} {a_4}}{2}-\frac{{a_1} {a_4}}{4}+\frac{7 {a_4}}{8}-\frac{{a_1} {a_2} {a_3}}{4}-\frac{3 {a_2} {a_3}}{8} \notag \\ &\qquad -\frac{{a_3}}{4}+\frac{{a_1} {a_2}}{8}-\frac{{a_2}}{16}+\frac{{a_1}}{8}-\frac{1}{16} \notag \\ \notag \\ \notag \\ v_{4}&=\frac{{a_1} {a_2} {a_3} {a_4}}{2}+\frac{3 {a_2} {a_3} {a_4}}{4}+\frac{3 {a_3} {a_4}}{2}-\frac{{a_1} {a_2} {a_4}}{4}+\frac{5 {a_2} {a_4}}{8}+\frac{{a_4}}{4}+\frac{{a_3}}{2}-\frac{{a_2}}{4} \notag \\ &\qquad -\frac{{a_1}}{8}-\frac{1}{16} \notag \\ v_{13}&=-\frac{{a_1} {a_2} {a_3} {a_4}}{2}-\frac{3 {a_2} {a_3} {a_4}}{4}-\frac{3 {a_3} {a_4}}{2}+\frac{{a_1} {a_2} {a_4}}{4}-\frac{5 {a_2} {a_4}}{8}-\frac{{a_4}}{4}+\frac{{a_3}}{2}-\frac{{a_2}}{4} \notag \\ &\qquad -\frac{{a_1}}{8}-\frac{1}{16} \notag \\ \notag \\ \notag \\ v_{5}&=-{a_2} {a_3} {a_4}-\frac{{a_1} {a_3} {a_4}}{2}-\frac{5 {a_3} {a_4}}{4}+\frac{{a_1} {a_4}}{2}-\frac{3 {a_4}}{4}+\frac{{a_1} {a_2} {a_3}}{4}+\frac{3 {a_2} {a_3}}{8}+\frac{{a_3}}{4} \notag \\ &\qquad +\frac{{a_1} {a_2}}{8}-\frac{{a_2}}{16}+\frac{{a_1}}{8}-\frac{1}{16} \notag \\ v_{12}&= {a_2} {a_3} {a_4}+\frac{{a_1} {a_3} {a_4}}{2}+\frac{5 {a_3} {a_4}}{4}-\frac{{a_1} {a_4}}{2}+\frac{3 {a_4}}{4}+\frac{{a_1} {a_2} {a_3}}{4}+\frac{3 {a_2} {a_3}}{8}+\frac{{a_3}}{4} \notag \\ &\qquad +\frac{{a_1} {a_2}}{8}-\frac{{a_2}}{16}+\frac{{a_1}}{8}-\frac{1}{16} \notag \\ \notag \\ \notag \\ v_{6}&={a_3} {a_4}-\frac{{a_1} {a_2} {a_4}}{2}+\frac{3 {a_2} {a_4}}{4}-\frac{{a_1} {a_4}}{4}+\frac{3 {a_4}}{8}-\frac{{a_1} {a_2} {a_3}}{4}-\frac{3 {a_2} {a_3}}{8}-\frac{{a_1} {a_3}}{2} \notag \\ &\qquad -{a_3}-\frac{{a_1} {a_2}}{8}+\frac{{a_2}}{16}+\frac{{a_1}}{8}-\frac{1}{16} \notag \\ v_{11}&=-{a_3} {a_4}+\frac{{a_1} {a_2} {a_4}}{2}-\frac{3 {a_2} {a_4}}{4}+\frac{{a_1} {a_4}}{4}-\frac{3 {a_4}}{8}-\frac{{a_1} {a_2} {a_3}}{4}-\frac{3 {a_2} {a_3}}{8}-\frac{{a_1} {a_3}}{2} \notag \\ &\qquad -{a_3}-\frac{{a_1} {a_2}}{8}+\frac{{a_2}}{16}+\frac{{a_1}}{8}-\frac{1}{16} \notag \\ \notag \\ \notag \\ v_{7}&=\frac{{a_1} {a_2} {a_3} {a_4}}{2}-\frac{{a_2} {a_3} {a_4}}{4}+\frac{{a_1} {a_3} {a_4}}{2}+\frac{3 {a_3} {a_4}}{4}+\frac{{a_1} {a_2} {a_4}}{4}-\frac{{a_2} {a_4}}{8}+\frac{{a_1} {a_4}}{4} \notag \\ &\qquad -\frac{5 {a_4}}{8}+\frac{{a_1} {a_2} {a_3}}{4}+\frac{3 {a_2} {a_3}}{8}+\frac{{a_1} {a_3}}{2}+{a_3}-\frac{{a_1} {a_2}}{8}+\frac{{a_2}}{16}+\frac{{a_1}}{8}-\frac{1}{16} \notag \\ v_{10}&= -\frac{{a_1} {a_2} {a_3} {a_4}}{2}+\frac{{a_2} {a_3} {a_4}}{4}-\frac{{a_1} {a_3} {a_4}}{2}-\frac{3 {a_3} {a_4}}{4}-\frac{{a_1} {a_2} {a_4}}{4}+\frac{{a_2} {a_4}}{8}-\frac{{a_1} {a_4}}{4} \notag \\ &\qquad +\frac{5 {a_4}}{8}+\frac{{a_1} {a_2} {a_3}}{4}+\frac{3 {a_2} {a_3}}{8}+\frac{{a_1} {a_3}}{2}+{a_3}-\frac{{a_1} {a_2}}{8}+\frac{{a_2}}{16}+\frac{{a_1}}{8}-\frac{1}{16} \notag \\ \notag \\ \notag \\ v_{8}&= -\frac{{a_1} {a_2} {a_3} {a_4}}{2}-\frac{3 {a_2} {a_3} {a_4}}{4}-{a_1} {a_3} {a_4}-{a_3} {a_4}-\frac{{a_1} {a_2} {a_4}}{4}+\frac{5 {a_2} {a_4}}{8}-\frac{{a_1} {a_4}}{2} \notag \\ &\qquad +\frac{{a_4}}{2}-\frac{{a_2} {a_3}}{2}-\frac{{a_1} {a_3}}{4}-\frac{{a_3}}{8}+\frac{{a_2}}{4}-\frac{{a_1}}{8}-\frac{1}{16} \notag \\ v_{9}&= \frac{{a_1} {a_2} {a_3} {a_4}}{2}+\frac{3 {a_2} {a_3} {a_4}}{4}+{a_1} {a_3} {a_4}+{a_3} {a_4}+\frac{{a_1} {a_2} {a_4}}{4}-\frac{5 {a_2} {a_4}}{8}+\frac{{a_1} {a_4}}{2} \notag \\ &\qquad -\frac{{a_4}}{2}-\frac{{a_2} {a_3}}{2}-\frac{{a_1} {a_3}}{4}-\frac{{a_3}}{8}+\frac{{a_2}}{4}-\frac{{a_1}}{8}-\frac{1}{16} \notag \\ \end{align}
We also list the minimal polynomials and binomial equations computed at each step:
\begin{align} g_1(x)&=x^8+\frac{x^7}{2}+\frac{5x^6}{2}+\frac{7x^5}{2}+2x^4+\frac{7x^3}{2} \notag \\ &\qquad \quad +a_1x(x+1)^2(x^2+1)(x^2-x+1)+\frac{5x^2}{2}+\frac{x}{2}+1 \notag \\ g_2(x)&=x^4+\frac{(4a_2+2a_1+1)}{4}x^3+\frac{ \bigl( (2a_1-1)a_2+2a_1+7 \bigr) }{4}x^2 \notag \\ &\qquad \quad +\frac{(4a_2+2a_1+1)}{4}x+1 \notag \\ g_3(x)&=x^2+\biggl(a_3+\frac{a_2}{2}+\frac{a_1}{4}+\frac{1}{8}\biggr)x+1 \notag \\ g_4(x)&=x+{a_4}+\frac{{a_3}}{2}+\frac{{a_2}}{4}+\frac{{a_1}}{8}+\frac{1}{16} \notag \\ \end{align} \begin{align} \notag \\ B_1(x)&=a_1^2-A_1=0 \quad a_1=\sqrt{A_1} \qquad A_1=\frac{17}{4} \notag \\ B_2(x)&=a_2^2-A_2=0 \quad a_2=\sqrt{A_2} \qquad A_2=\frac{2 {a_1}+17}{8} \notag \\ B_3(x)&=a_3^2-A_3=0 \quad a_3=\sqrt{A_3} \qquad A_3=-\frac{{a_1} \left( 4 {a_2}+6\right) -6 {a_2}-17}{16} \notag \\ B_4(x)&=a_4^2-A_4=0 \quad a_4=\sqrt{A_4} \notag \\ &\qquad \qquad \qquad A_4=\frac{{a_1} \left( 4 {a_3}-2\right) +\left( 8 {a_2}+2\right) {a_3}+4 {a_2}-17}{32} \notag \\ \end{align}
That is all.