Techniques of Solving Equations à la Galois


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Chapter7

    Galois-Style Solution for the Cyclotomic Equation

\(\qquad \qquad \qquad \Phi_{17}(x)={{x}^{16}}+{{x}^{15}}+{{x}^{14}}+....+{{x}^{3}}+{{x}^{2}}+x+1 \quad Galois \ Group:C_{16} \)

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[7-1] Setting up the cyclotomic equation \(\Phi_{17}(x)\)

Cyclotomic polynomials are typical examples of cyclic polynomials. Here, to apply factorization over number fields in a concrete way, let us actually compute with the cyclotomic polynomial \(\Phi_{17}(x)\). From here on, we will use different notations depending on the context:

  · When we want to emphasize it is a cyclotomic polynomial: we use \(\Phi_{17}(x)\).
  · When we want to emphasize it is the minimal polynomial: we use \(g_0(x)\).
      Of course, \([ \ \Phi_{17}(x)=g_0(x), \ g_0(v)=0 \ ]\).
  · We denote the base field by \(F_0\). Of course, \([ \ F_0 \equiv Q \ ]\).

Now, if we factor \(\Phi_{17}(x)\) inside the algebraic field \(F_0(v)\) generated by \(g_0(v)\), using the maxima command \(factor(p,q)\), we obtain (1.3).

\begin{align} &x^{17}-1=(x-1) \times \Phi_{17}(x) \notag \\ \notag \\ &\Phi_{17}(x) =( {{x}^{16}}+{{x}^{15}}+{{x}^{14}}+{{x}^{13}}+{{x}^{12}}+{{x}^{11}}+{{x}^{10}}+{{x}^{9}} \notag \\ &\qquad \qquad +{{x}^{8}}+{{x}^{7}}+{{x}^{6}}+{{x}^{5}}+{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+x+1) \\ \notag \\ &Minimal \ Polynomial \ of \ v : \ g_0(x)=\Phi_{17}(x) \quad \rightarrow \quad g_0(v)=0\\ \notag \\ &factor(\Phi_{17}(x),g_0(v))=(x-v)(x-v^2)(x-v^3)(x-v^4)(x-v^5) \notag \\ &\qquad \qquad \times(x-v^6)(x-v^7)(x-v^8)(x-v^9)(x-v^{10})(x-v^{11}) \notag \\ &\qquad \qquad \times (x-v^{12})(x-v^{13})(x-v^{14})(x-v^{15}) \notag \\ &\qquad \qquad \times (x+v^{15}+v^{14}+v^{13}+v^{12}+v^{11}+v^{10}+v^9 \\ &\qquad \qquad +v^8+v^7+v^6+v^5+v^4+v^3+v^2+v+1) \notag \\ \end{align}


From (1.3), \(\Phi_{17}(x)\) factors easily in the number field \(Q(v)\). Therefore, the 16 roots of \(\Phi_{17}(x)\) can be found directly. In (1.4), the set \(\{v_1,v_2,...,v_{16}\}\) gives polynomial expressions in \(v\) for the 16 roots of \(\Phi_{17}(x)\). At the same time, we let \(\{ \rho_i\}\) denote automorphisms of the Galois extension \(F_0(v)/F_0\).
Let us check that this is indeed correct.

\begin{align} \rho_{1}(v)&=v_{1}=v & \rho_{2}(v)&=v_{2}=v^{2} & \rho_{3}(v)&=v_{3}=v^{3} \notag \\ \rho_{4}(v)&=v_{4}=v^{4} & \rho_{5}(v)&=v_{5}=v^{5} & \rho_{6}(v)&=v_{6}=v^{6} \notag \\ \rho_{7}(v)&=v_{7}=v^{7} & \rho_{8}(v)&=v_{8}=v^{8} & \rho_{9}(v)&=v_{9}=v^{9} \\ \rho_{10}(v)&=v_{10}=v^{10} & \rho_{11}(v)&=v_{11}=v^{11} & \rho_{12}(v)&=v_{12}=v^{12} \notag \\ \rho_{13}(v)&=v_{13}=v^{13} & \rho_{14}(v)&=v_{14}=v^{14} & \rho_{15}(v)&=v_{15}=v^{15} \notag \\ \end{align} \begin{align} \ \rho_{16}(v)&=v_{16}=- (v^{15}+v^{14}+v^{13}+v^{12}+v^{11}+v^{10}+v^9+v^8 \notag \\ &\qquad +v^7+v^6+v^5+v^4+v^3+v^2+v+1) \qquad (\mathrm{mod}\,g_0(v)) \notag \\ \end{align}

\begin{align} \notag \\ v^{17}&=v \cdot v^{16}=-v \cdot (v^{15}+v^{14}+...+v^3+v^2+v+1) \notag \\ &=-(v^{16}+v^{15}+v^{14}+v^{13}+....+v^4+v^3+v^2+v)=-(-1)=1 \notag \\ \notag \\ &\qquad \therefore v^{17}=1\\ \end{align}

With (1.5), namely \([ \ v^{17}=1 \ ]\), let us see how powers of \(\rho_3\) act on \(v\). To do so, we need to confirm that the set of automorphisms \(\{\rho_i\}\) forms a group under composition. The composition can be defined as follows.

\begin{align} &\rho_3 (v)=v_3=v^3 \notag \\ &\rho_3^2(v)=\rho_3(\rho_3(v))=\rho_3(v_3)=\rho_3(v^3)=(\rho_3(v))^3=(v^3)^3=v^9=v_9=\rho_9(v) \notag \\ &\rho_3^3(v)=\rho_3(\rho_3^2(v))=\rho_3(v_9)=\rho_3(v^9)=(\rho_3(v))^9=(v^3)^9=v^{27}=v^{10}=v_{10}=\rho_{10}(v) \notag \\ \end{align}

Continuing the same type of calculation, we get:

\begin{align} \rho_3^{5}(v)&=\rho_{5}(v) & \rho_3^{6}(v)&=\rho_{15}(v) & \rho_3^{7}(v)&=\rho_{11}(v) & \rho_3^{8}(v)&=\rho_{16}(v) \notag \\ \rho_3^{9}(v)&=\rho_{14}(v) & \rho_3^{10}(v)&=\rho_{8}(v) & \rho_3^{11}(v)&=\rho_{7}(v) & \rho_3^{12}(v)&=\rho_{4}(v) \notag \\ \rho_3^{13}(v)&=\rho_{12}(v) & \rho_3^{14}(v)&=\rho_{2}(v) & \rho_3^{15}(v)&=\rho_{6}(v) & \rho_3^{16}(v)&=\rho_{1}(v) \notag \\ \rho_3^{17}(v)&=\rho_{3}(v) & \rho_3^{18}(v)&=\rho_{9}(v) &...... & & & \notag \\ \end{align}

From these results, we find that \([ \ \rho_3^i, i=[1,2,...,16] \ ] \) generates all the automorphisms listed in (1.4).
In other words, the Galois group of \(\Phi_{17}(x)\) is the cyclic group \(C_{16}\) of order 16. Thus, the \(\{\rho_i\}\) defined in (1.4) are indeed automorphisms of the extension field \(F_0(v)\).

Since the Galois group of \(\Phi_{17}(x)\) is the cyclic group \(C_{16}\), we can now determine its normal subgroups, its composition series, and the composition series of the cyclic extensions obtained by decomposing it, as follows:

\begin{align} &Gal(F_0(v)/F_0)=C_{16} \notag \\ &\quad C_{16}=\{\rho_{1}, \ \rho_{2}, \ \rho_{3}, \ \rho_{4}, \ \rho_{5}, \ \rho_{6}, \ \rho_{7}, \ \rho_{8},\notag \\ &\qquad \qquad \rho_{9}, \ \rho_{10}, \ \rho_{11}, \ \rho_{12}, \ \rho_{13}, \ \rho_{14}, \ \rho_{15}, \ \rho_{16}\} \\ \notag \\ &Normal \ subgroup \ of \ C_{16} \notag \\ \notag \\ &\left\{ \begin{array}{l} C_8=\{\rho_{1}, \ \rho_{2}, \ \rho_{4}, \ \rho_{8}, \ \rho_{9}, \ \rho_{13}, \ \rho_{15}, \ \rho_{16}\} \\ C_4=\{\rho_{1}, \ \rho_{4}, \ \rho_{13}, \ \rho_{16}\} \\ C_2=\{\rho_{1}, \ \rho_{16}\}\\ {e}=\{e\} \\ \end{array} \right. \\ \notag \\ &Composition \ series \ of \ Galois \ group \ C_{16} \notag \\ \notag \\ &\qquad [ \ C_{16} \rhd C_8 \rhd C_4 \rhd C_2 \rhd {e} \ ]\\ \notag \\ & \qquad \Downarrow \notag \\ &Cyclic \ extensions \notag \\ \notag \\ & [ \ C_{16}/C_8 \rhd e \ ] \rightarrow [ \ C_8/C_4 \rhd e \ ] \rightarrow [ \ C_4/C_2 \rhd e \ ] \rightarrow [ \ C_2 \rhd e \ ]\\ \end{align}


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