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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[7-2] Solving the cyclotomic equation \(\Phi_{17}(x)=0\)
Since we have determined the composition series, we can now finally use Galois theory to compute the minimal polynomials of the extension fields of \(g_0(x)\).
[step0]
The tower of fields
\(\{F_0,F_1,F_2,F_3,F_4\}\)
corresponds to
the composition series of groups
\(\{C_{16},C_8,C_4,C_2,e\}\)
[step1]
At the stage \([F_1,C_8]\), split the yellow part.
Green part: \(Gal(F_1/F_0)=C_{16}/C_8\)
Thus \(F_1/F_0\) is a quadratic cyclic extension.
[step2]
At the stage \([F_2,C_4]\), split the yellow part.
Green part: \(Gal(F_2/F_1)=C_{8}/C_4\)
Thus \(F_2/F_1\) is a quadratic cyclic extension.
[step3]
At the stage \([F_3,C_2]\), split the yellow part.
Green part: \(Gal(F_3/F_2)=C_{4}/C_2\)
Thus \(F_3/F_2\) is a quadratic cyclic extension.
[step4]
The remainder split off in step3
turns out to be itself a cyclic extension!
Green part: \(Gal(F_4/F_3)=C_{2}\)
Thus \(F_4/F_3\) is a quadratic cyclic extension.
[summary]
The yellow series in step0 is ultimately decomposed
into four cyclic extensions.
[7-3] Cyclic extension \(F_1/F_0\): computation of the minimal polynomial \(g_1(x)\)
Following the calculation procedure described in earlier chapters, we can compute \(g_1(x)\) straightforwardly.\begin{align} h_0&=\prod_{\rho_i \in \ C_8}\rho_i(x-v)=(x-v_1)(x-v_2)...(x-v_{15})(x-v_{16}) \\ h_1&=\prod_{\rho_i \in \ (C_{16}-C_8)}\rho_i(x-v)=(x-v_3)(x-v_5)...(x-v_{12})(x-v_{14}) \\ \end{align} \begin{align} \notag \\ & \begin{bmatrix} t_0 \\ t_1 \end{bmatrix} =\frac{1}{2} \begin{bmatrix} 1&1 \\ 1&-1 \end{bmatrix} \cdot \begin{bmatrix} h_0 \\ h_1 \end{bmatrix} \\ \end{align}
Substituting the polynomial expressions of \(v\) from (1.4) into the \(\{h_0,h_1\}\) of (3.1)(3.2) and carrying out the calculation, we obtain the following. From here on, all computations must be done \( (\mathrm{mod}\,g_0(v)) \).
\begin{align} h_0&=(x-v_1)(x-v_2)(x-v_4)(x-v_8)(x-v_9)(x-v_{13})(x-v_{15})(x-v_{16}) \notag \\ &=x^8+(v^{14}+v^{12}+v^{11}+v^{10}+v^7+v^6+v^5+v^3+1)x^7+ .... \notag \\ &\quad +(v^{14}+v^{12}+v^{11}+v^{10}+v^7+v^6+v^5+v^3+1)x+1 \\ h_1&=(x-v_3)(x-v_5)(x-v_6)(x-v_7)(x-v_{10})(x-v_{11})(x-v_{12})(x-v_{14}) \notag \\ &=x^8-(v^{14}+v^{12}+v^{11}+v^{10}+v^7+v^6+v^5+v^3)x^7+ ....\notag \\ &\quad -(v^{14}+v^{12}+v^{11}+v^{10}+v^7+v^6+v^5+v^3)x+1 \\ \end{align}
\begin{align} &\left\{ \begin{array}{l} t_0=\frac{1}{2}(2x^8+x^7+5x^6+7x^5+4x^4+7x^3+5x^2+x+2) \ \in \ F_0[x]\\ t_1= (v^{14}+v^{12}+v^{11}+v^{10}+v^7+v^6+v^5+v^3+\frac{1}{2})\notag \\ \quad \times \ x(x+1)^2(x^2+1)(x^2-x+1) \ \in \ F_0(v)[x] \\ \end{array} \right. \\ &\qquad \Downarrow \notag \\ &\left\{ \begin{array}{l} t_1=cd_m \cdot q_1(x) \quad \in F_0(v)[x] \\ cd_m=(v^{14}+v^{12}+v^{11}+v^{10}+v^7+v^6+v^5+v^3+\frac{1}{2}) \quad \in F_0(v) \\ q_1(x)=x(x+1)^2(x^2+1)(x^2-x+1) \quad \in F_0[x] \\ \end{array} \right. \\ \end{align}
Let us denote the leading coefficient of \(t_1\) by \(cd_m\). Using \(cd_m\), we introduce a new adjoint element \(a_1\).
[step2] The binomial equation \(B_1(x)=0\) and construction of a new adjoint element \(a_1\)
\begin{align}
&t_1=cd_m \cdot q_1(x)
\quad \Rightarrow \quad
\left\{
\begin{array}{l}
\tilde{t_1}=a_1 \cdot q_1(x) \ \in F_1[x] \\
B_1(x)=x^2-A_1=0 \\
a_1=\sqrt {A_1} \ \in Q(a_1) \equiv F_1\\
\end{array}
\right. \notag \\
\end{align}
Since this cyclic extension has degree 2, let us compute \(cd_m^2\). Then, as in (3.7), \(cd_m^2\) turns out to be a value in \(F_0\). Let us denote this by \(A_1\). Using this \(A_1\), we can define the binomial equation \(B_1(x)=0\) that the new element \(a_1\) must satisfy. We define a new number \(a_1\) as a radical of this binomial equation. By adjoining this \(a_1\) to the base field \(F_0\), we construct the extension field \(F_1\).
\begin{align} cd_m^2&=\bigl(v^{14}+v^{12}+v^{11}+v^{10}+v^7+v^6+v^5+v^3+\tfrac{1}{2} \bigr)^2=\tfrac{17}{4} \equiv A_1 \ \in \ F_0 \\\ &\qquad \Downarrow \notag \\ B_1(x)&=x^2-A_1=0 \quad \rightarrow \quad a_1=\sqrt{A_1} \ \in \ F_1 \equiv F_0(a_1)\\ \end{align}
\begin{align} \tilde{t_1} &\equiv a_1x(x+1)^2(x^2+1)(x^2-x+1) \ \in \ F_1[x]\\ \end{align}
Since \(h_0\) originally contains the factor \((x-v)\), it follows that \(\tilde{h_0}\) is the new minimal polynomial \(g_1(x)\) over the extension field \(F_1\).
\begin{align} \tilde{h_0}&=t_0+\tilde{t_1} \notag \\ &=x^8+\tfrac{1}{2}(x^7+x)+\tfrac{5}{2}(x^6+x^2)+\tfrac{7}{2}(x^5+x^3)+2x^4+1 \notag \\ &+a_1x(x+1)^2(x^2+1)(x^2-x+1) \equiv g_1(x) \\ \end{align}