Techniques for Solving Solvable Equations via Galois Theory

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[9-8] The cyclic extension \(F_2/F_1\)

In the previous section we found the minimal polynomial of \(v\) over the extension field \(F_1\). We now enlarge the field further. Next we compute the cyclic extension \(F_2/F_1\).
Since the base field \(F_0\) already contains the 5th root of unity \(\zeta\), computations in the degree-5 extension \( [F_2:F_1]=5 \) can be carried out without difficulty.

There is one point of caution in the computations below. The Galois group of this extension is \(C_5\), and we must choose a generator of \(C_5\). Any choice works, but in what follows we set \(\rho_{113} \cong \rho\) as the generator.
With this choice, all the other automorphisms are powers of \(\rho_{113} \cong \rho\) as in (8.2).
When using the Lagrange Resolvent transformation, the order of \(\{h_0,h_1,h_2,h_3,h_4\}\) must coincide with the order of \(\rho^{\,i}(h_0)\) as in (8.3), hence this note of caution.

Sstep1 LRT (Lagrange Resolvent transformation)
\begin{align} &Gal(F_2/F_1)=C_5/e \cong C_5=\{ \rho_{1},\rho_{38},\rho_{53},\rho_{94},\rho_{113} \} \\ \notag \\ &\rho \equiv \rho_{113}, \quad \rho^2=\rho_{38}, \quad \rho^3=\rho_{53}, \quad \rho^4=\rho_{94}, \quad \rho^5=e=\rho_1 \\ \end{align} \begin{align} &\qquad \qquad \Downarrow \notag \\ &\left\{ \begin{array}{l} h_0 \equiv (x-v)=(x-v_1)=x-v \\ h_1=\rho(h_0)=(x-v_{113})=x-\frac{47 {{v}^{4}} {{\zeta }^{3}}}{30}+\frac{4 {{v}^{2}} {{\zeta }^{3}}}{3}+...-\frac{4}{3} \\ h_2=\rho^2(h_0)=(x-v_{38})=x-\frac{29 {{v}^{4}} {{\zeta }^{3}}}{30}+\frac{2 {{v}^{2}} {{\zeta }^{3}}}{3}+...+\frac{7}{3} \\ h_3=\rho^3(h_0)=(x-v_{53})=x+\frac{29 {{v}^{4}} {{\zeta }^{3}}}{30}-\frac{2 {{v}^{2}} {{\zeta }^{3}}}{3}+...-\frac{7}{3} \\ h_4=\rho^4(h_0)=(x-v_{94})=x+\frac{47 {{v}^{4}} {{\zeta }^{3}}}{30}-\frac{4 {{v}^{2}} {{\zeta }^{3}}}{3}+...+\frac{4}{3}\\ \end{array} \right. \\ \notag \\ \end{align}

\begin{align} \begin{bmatrix} t_0 \\ t_1 \\ t_2 \\ t_3 \\ t_4 \end{bmatrix} &=\frac{1}{5} \begin{bmatrix} 1&1&1&1&1 \\ 1&\zeta&\zeta^2&\zeta^3&\zeta^4\\ 1&(\zeta^2)&(\zeta^2)^2&(\zeta^2)^3&(\zeta^2)^4\\ 1&(\zeta^3)&(\zeta^3)^2&(\zeta^3)^3&(\zeta^3)^4\\ 1&(\zeta^4)&(\zeta^4)^2&(\zeta^4)^3&(\zeta^4)^4 \end{bmatrix} \cdot \begin{bmatrix} h_0 \\ h_1 \\ h_2 \\ h_3 \\ h_4 \end{bmatrix} \\ \notag \\ &\qquad \qquad Z=\zeta^4+\zeta^3+\zeta^2+\zeta+1=0 \notag \\ \end{align}

Evaluating (8.4) gives the following.

\begin{align} \left\{ \begin{array}{l} t_0=x, \quad t_2=0, \quad t_3=0 \\ t_1=\frac{1}{3}\bigl[\left( 80 {{v}^{4}}-40 {{v}^{2}}+199 {a_1} v-300\right) {{\zeta }^{3}}+...+\left( 521 {a_1}-150\right) v-100 \bigr] \\ t_4= -\frac{1}{3}\bigl[ \left( 80 {{v}^{4}}-40 {{v}^{2}}+199 {a_1} v-300\right) {{\zeta }^{3}}+...+\left( 521 {a_1}+150\right) v-100 \bigr]\\ \end{array} \right. \\ \end{align}

Computing \(t_{1}^5\) immediately shows (8.6) that it lies in \(F_1\).

\begin{align} t_{1}^5&=-134208800 {{\zeta }^{3}}-134208800 {{\zeta }^{2}}-217154400=A_2 \ \in \ F_1\\ \notag \\ &\left\{ \begin{array}{l} B_2(x) \equiv x^5-A_2=0 \\ a_2 \equiv \sqrt[5] {A_2} \ \in F_2 \equiv F_1(a_2) \\ \end{array} \right. \\ \end{align}

Looking at the two ends of (8.6), we can say that \(t_1\) is also a root of the binomial equation \(B_2(x)=0\) shown in (8.7). We therefore redefine \(t_1\) as the radical \(a_2\) of \(B_2(x)=0\), and by adjoining \(a_2\) to the base field \(F_1\) we generate the new extension field \(F_2\). In summary:

Step2 Binomial equation \(B_2(x)=0\) and generation of the new adjoined element \(a_2\)
\begin{align} \left\{ \begin{array}{l} t_1 \in F_1(v) \\ t_1^5=5655 {{\zeta }^{3}}+1335 {{\zeta }^{2}}-4320 \zeta \\ \qquad +\frac{{a_1}}{2}-2160 \ \in F_1\\ \end{array} \right. \quad \Rightarrow \quad \left\{ \begin{array}{l} B_2(x)=x^5-A_2=0 \\ a_2=\sqrt[5] {A_2} \ \in F_1(a_2) \equiv F_2\\ \tilde{t_1}=a_2\ \in F_2 \\ \end{array} \right. \notag \\ \end{align}

We also compute the reciprocal \(A_2^{-1}\), which will be needed later.
Next we express \(\{ \ \tilde{t_2},\tilde{t_3},\tilde{t_4} \ \}\) using the \(a_2\) defined above. The computation proceeds as in (8.9).
Be careful to reduce modulo \(g_1(v)\), modulo \(B_1\), and modulo \(Z\).

\begin{align} \notag \\ A_2^{-1}&=-\frac{167761 {{v}^{5}} {{\zeta }^{3}}}{4000}+\frac{1353 {{v}^{3}} {{\zeta }^{3}}}{32}-\frac{341 v {{\zeta }^{3}}}{40}-\frac{167761 {{v}^{5}} {{\zeta }^{2}}}{4000} \notag \\ &+\frac{1353 {{v}^{3}} {{\zeta }^{2}}}{32}-\frac{341 v {{\zeta }^{2}}}{40}+\frac{51841 {{v}^{5}}}{2000}-\frac{4181 {{v}^{3}}}{160}+\frac{843 v}{160} \\ \notag \\ \tilde{ t_2}&=t_1^5 \cdot t_2 \cdot t_1^{-5}=t_1^2 \cdot (t_1^3 \cdot t_2)\cdot A_2^{-1}=a_2^2 \cdot (t_1^3 \cdot t_2) \cdot A_2^{-1} \notag \\ \tilde{t_3}&=a_2^3 \cdot (t_1^2 \cdot t_3)\cdot A_2^{-1} \\ \tilde{ t_4}&=a_2^4 \cdot (t_1 \cdot t_4) \cdot A_2^{-1} \notag \\ &\qquad \qquad \Downarrow \notag \\ \end{align}

\begin{align} \left\{ \begin{array}{l} t_0=x , \quad \tilde{t_1}=a_2 \in \ F_2=F_1(a_2) \\ \tilde{t_2}=0, \quad \tilde{t_3}=0 \\ \tilde{t_4}= -\frac{1353 {a_1} {a_2^{4}} {{\zeta }^{3}}}{10}-\frac{24 {a_2^{4}} {{\zeta }^{3}}}{5}-\frac{1353 {a_1} {a_2^{4}} {{\zeta }^{2}}}{10}-\frac{102 {a_2^{4}} {{\zeta }^{2}}}{5}-\frac{126 {a_2^{4}} \zeta }{5}+\frac{4181 {a_1} {a_2^{4}}}{50}-\frac{63 {a_2^{4}}}{5}\\ \end{array} \right. \\ \end{align}

With \( \{ \ t_0,\ \tilde{t_1},\ \tilde{t_2},\ \tilde{t_3},\ \tilde{t_4} \ \} \) now computed, we finally apply the Inverse Lagrange Resolvent transformation to determine the minimal polynomial of \( v \) over the extension field \( F_2 \).

Step3 ILRT (Inverse Lagrange Resolvent transformation)
\begin{align} \begin{bmatrix} \tilde{h_0}\\ \tilde{h_1} \\ \tilde{h_2} \\ \tilde{h_3} \\ \tilde{h_4} \end{bmatrix} &= \begin{bmatrix} 1&1&1&1&1 \\ 1&(\zeta^4)&(\zeta^4)^2&(\zeta^4)^3&(\zeta^4)^4\\ 1&(\zeta^3)&(\zeta^3)^2&(\zeta^3)^3&(\zeta^3)^4\\ 1&(\zeta^2)&(\zeta^2)^2&(\zeta^2)^3&(\zeta^2)^4\\ 1&\zeta&\zeta^2&\zeta^3&\zeta^4 \end{bmatrix} \cdot \begin{bmatrix} t_0\\ \tilde{t_1} \\ \tilde{t_2}\\ \tilde{t_3} \\ \tilde{t_4} \end{bmatrix} \\ \notag \\ g_2(x) &\equiv \tilde{h_0}=t_0+\tilde{t_1}+\tilde{t_2}+\tilde{t_3}+\tilde{t_4} \\ \end{align}

Having obtained the minimal polynomial of \(v\) over \(F_2\), the solution of \([ \ g_2(x)=0 \ ]\) gives the final value of \(v\). Substituting (8.13) into (5.7) yields the five roots of \(f(x)\). Since the formulas are rather complicated, we omit them here.

\begin{align} \therefore \ v&=-{a_2}+\frac{1353 {a_1} {{a}_{2}^{4}} {{\zeta }^{3}}}{10}+\frac{24 {{a}_{2}^{4}} {{\zeta }^{3}}}{5}+\frac{1353 {a_1} {{a}_{2}^{4}} {{\zeta }^{2}}}{10} \notag \\ &\qquad +\frac{102 {{a}_{2}^{4}} {{\zeta }^{2}}}{5}+\frac{126 {{a}_{2}^{4}} \zeta }{5}-\frac{4181 {a_1} {{a}_{2}^{4}}}{50}+\frac{63 {{a}_{2}^{4}}}{5} \\ \end{align}

(Supplement) The elements of the group of automorphisms corresponding to the above composition series are as follows.

\begin{align} D_{5}&=\{\rho_i\} \quad i=[1,8,27,38,53,68,83,94,113,120] \notag \\ C_5&=\{\rho_i\} \quad i=[1,38,53,94,113] \notag \\ e&=\rho_1 \notag \\ \notag \\ &F_0 (= Q(\zeta)), \ F_1 (=Q(a_1,\zeta)), \ F_2 (= Q(a_2,a_1,\zeta) = F_0(v)) \\ \end{align}

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