Techniques for Solving Solvable Equations via Galois Theory

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[9-6] Determining \(Gal(F_0(v)/F_0)\)

From this point on, we take the base field \(F_0\) to be \(F_0 \equiv Q(\zeta)\).
We now match the set \(\{ \ x_1,x_2,x_3,x_4,x_5 \ \}\) obtained in the previous section with \(\{ \ \alpha, \ \beta, \ \gamma, \ \delta, \ \epsilon \}\).
Let \([ \ w = x_{1}+2 x_{2}+3 x_{3}+4x_{4}+5x_{5} \ ]\), and let elements \(\tau_i\) of the symmetric group \(S_5\) permute the ordering of \(\{x_1,x_2,x_3,x_4,x_5\}\). The permuted values \(\{w_1,w_2,...,w_{120}\}\) are listed in (6.2). Since we are permuting the \(\{x_i\}\), we deliberately write \(\{\tau_i\}\) rather than \(\{\sigma_i\}\) here.

\begin{align} &w = x_{1}+2 x_{2}+3 x_{3}+4x_{4}+5x_{5} \\ \end{align}

\begin{align} \notag \\ &\left\{ \begin{array}{l} \tau_1 (w)=w_1=x_1+2x_2+3x_3+4x_{4}+5x_{5} & &\tau_2 (w)=w_2=x_1+2x_2+3x_3+4x_{5}+5x_{4} \\ \tau_3 (w)=w_3=x_1+2x_2+3x_4+4x_{3}+5x_{5} & &\tau_4 (w)=w_4=x_1+2x_2+3x_4+4x_{5}+5x_{3} \\ \qquad ....... & & \\ \tau_{119} (w)=w_{119}=x_5+2x_4+3x_3+4x_{1}+5x_{2} & &\tau_{120} (w)=w_{120}=x_5+2x_4+3x_3+4x_{2}+5x_{1} \\ \end{array} \right. \\ \end{align}

Substituting the polynomial expressions in \(v\) from (5.7) for \(\{x_1,x_2,x_3,x_4,x_5\}\) into each \(w_i\), we look for a permutation \(\tau_i\) for which the value is exactly \(v\). As it turns out, \(\tau_{1}\) has this property, as shown in (6.3). Hence the correspondence is as in (6.4).

\begin{align} &\tau_{1} (w)=w_{1}=x_1+2x_{2}+3x_{3}+4x_{4}+5x_{5} =v \\ \notag \\ &\therefore \ \ \alpha=x_1(v), \quad \beta=x_2(v), \quad \gamma=x_3(v), \quad \delta=x_4(v), \quad \epsilon=x_5(v) \end{align}

With the polynomial expressions in \(v\) for the five roots \(\{ \ \alpha, \ \beta, \ \gamma, \ \delta, \ \epsilon \}\) now fixed, we use them to find the roots of the minimal polynomial \(g_0(x)\). First, act on the expression (6.5) for \(v\) by elements \(\sigma_i\) of the symmetric group \(S_5\) (which permute \(\{ \ \alpha, \ \beta, \ \gamma, \ \delta, \ \epsilon \}\)) to generate \([ \ \sigma_i(v)=v_i \ ]\).

\begin{align} &v = \alpha+2 \beta+3 \gamma+4\delta+5\epsilon \\ \notag \\ &\left\{ \begin{array}{l} \sigma_1 (v)=v_1=\alpha+2\beta+3\gamma+4\delta+5\epsilon & &\sigma_2 (v)=v_2=\alpha+2\beta+3\gamma+4\epsilon+5\delta \\ \sigma_3 (v)=v_3=\alpha+2\beta+3\delta+4\gamma+5\epsilon & &\sigma_4 (v)=v_4=\alpha+2\beta+3\delta+4\epsilon+5\gamma \\ \qquad ....... & & \\ \sigma_{119} (v)=v_{119}=\epsilon+2\delta+3\gamma+4\alpha+5\beta & &\sigma_{120} (v)=v_{120}=\epsilon+2\delta+3\gamma+4\beta+5\alpha \\ \end{array} \right. \\ \notag \\ \end{align}

Not all of the \(v_i\) above are roots of the minimal polynomial \(g_0(x)\). To test whether \(v_i\) is a root we simply substitute into \(g_0(x)\) and check whether the value is zero. The indices in (6.7) give the list of those \(i\) for which \(g_0(v_i)=0\). Using the correspondence \(\sigma_i(v)=v_i\), we then define candidates \(\rho_i\) for the automorphisms on the roots of the minimal polynomial as in (6.8).

\begin{align} \notag \\ &g_0(v_i)=0 \quad i=[1,8,27,38,53,68,83,94,113,120] \\ &\qquad \Downarrow \notag \\ &\sigma_i(v)=v_i \quad \rightarrow \quad \rho_i(v) \equiv v_i \quad i=[1,8,27,38,53,68,83,94,113,120]\\ \end{align}

In fact—just as explained in Chapters 2 and 3—forming the multiplication table of these ten elements \(\rho_i\) shows that these candidate automorphisms form the dihedral group \(D_{5}\). Therefore the \(\rho_i\) are indeed automorphisms, and the Galois group of the simple extension \(F_0(v)/F_0\) is \(D_5\).

In Galois theory, as indicated in (6.9–10), the composition series of the Galois group \(D_5\) is the three groups \(\{D_5,C_5,e\}\), and there are corresponding three fields \(\{F_0,F_1,F_2\}\). In this situation the field extensions \(\{F_1/F_0, \ F_2/F_1\}\) are each Galois extensions.

The degrees of these extensions are \(2\) and \(5\) (see (6.11–12)). Extensions of prime degree are cyclic (radical) extensions. Continuing these cyclic extensions, the degree of the minimal polynomial of \(v\) drops from \(10\) down to \(1\). Ultimately, the solution of the degree-1 minimal polynomial gives the value of the primitive element \(v\), and with this value we can compute the roots of \(f(x)\).

\begin{align} &Gal(F_0(v)/F_0) =\{\rho_{1}, \rho_{8},..., \rho_{113}, \rho_{120}\} =D_5 \quad \ Galois \ group \ of \ F_0(v)/F_0 \notag \\ \notag \\ & \ Composition \ series \ of \ Galois \ group \ D_5 \notag \\ \end{align} \begin{align} &D_5 & &\rhd & &C_5 & &\rhd & &e \\ &\updownarrow & & & &\updownarrow & & & &\updownarrow \notag \\ &F_0 & &\rightarrow & &F_1 & &\rightarrow & &F_2 \ ( \ = F_0(v)) \\ \end{align} \begin{align} & \qquad \qquad \Downarrow \notag \\ \notag \\ & Galois \ extension & & & &Galois \ Group \notag \\ \notag \\ &[1] \quad [ \ F_1:F_0 \ ]=2 & &\rightarrow & &Gal(F_1/F_0) = D_5/C_5 \cong C_2 \\ \notag \\ &[2] \quad [ \ F_2:F_1 \ ]=5 & &\rightarrow & &Gal(F_2/F_1) = C_5/e \cong C_5 \\ \end{align}

[9-7] The cyclic extension \(F_1/F_0\)

We now compute the cyclic extension in (6.11). As shown in (7.1), this is a quadratic cyclic extension. The quotient set \(\{\kappa_1,\kappa_2\}\) has five elements in each coset, so \(\{h_0,h_1\}\) are quintic polynomials as in (7.2).
As usual, applying the Lagrange Resolvent transformation to these two polynomials yields (7.3).

Step1 LRT (Lagrange Resolvent transformation)

\begin{align} &Gal(F_1/F_0)=D_5/C_5 \equiv C_2=\{\kappa_1,\kappa_2\} \\ &\{\kappa_1,\kappa_2\}=\bigl[ \{\rho_{1},\rho_{38},\rho_{53},\rho_{94},\rho_{113}\},\{ \rho_{8},\rho_{27},\rho_{68},\rho_{83},\rho_{120} \} \bigr] \notag \\ \notag \\ & h_0=\prod_{\rho_i \in \kappa_1 }\rho_i(x-v)=x^5-(55\zeta^3+55\zeta^2+90)x^3+...+90v^3-2225v \\ &h_1=\prod_{\rho_i \in \ \kappa_2}\rho_i(x-v)=x^5-(55\zeta^3+55\zeta^2+90)x^3+...-90v^3+2225 \notag \\ \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} = \begin{bmatrix} x^5-(55\zeta^3+55\zeta^2+90)x^3+....\\ -v^5+ (55\zeta^3+55\zeta^2+90)v^3-.... \\ \end{bmatrix} \\ \end{align}

The element \(t_1\) found in (7.3) lies in \(F_0(v)\), but squaring shows (7.4) that it lies in \(F_0\). Setting this value to be \(A_1\), we define \(t_1\) as a root \(a_1\) of the binomial equation \(B_1(x)=0\).
Thus we can treat \(t_1\) as an element of \(F_1\) by writing \( \tilde{t_1} =a_1\).

Step2 Generating the binomial equation \(B_1(x)\) and the new adjoined element \(a_1\)

\begin{align} &\left\{ \begin{array}{l} t_0 \ \in \ F_0[x] \\ t_1 \ \in \ F_0(v) \\ \end{array} \right. \quad \Rightarrow \quad \left\{ \begin{array}{l} t_1^2=A_1=-134208800 {{\zeta }^{3}}-134208800 {{\zeta }^{2}} \\ \qquad \qquad -217154400 \in \ F_0 \\ B_1(x)=x^2-A_1=0 \\ a_1=\sqrt{A_1} \ \in \ F_0(a_1) \equiv \ F_1 \quad \Rightarrow \quad \tilde{t_1} =a_1\\ \end{array} \right. \\ \end{align}

Finally, using \(\tilde{t_1}\) and \(t_0\) above, the Inverse Lagrange Resolvent transformation yields the minimal polynomial \(g_1(x)\) of \(v\) over \(F_1\) as in (7.6).

Step3 ILRT (Inverse Lagrange Resolvent transformation)

\begin{align} \begin{bmatrix} \tilde{h_0} \\ \tilde{h_1 } \end{bmatrix} &= \begin{bmatrix} 1&1 \\ 1&-1 \end{bmatrix} \cdot \begin{bmatrix} t_0 \\ \tilde{t_1} \end{bmatrix} \\ \notag \\ g_1(x) &\equiv \tilde{h_0}=x^5-(55\zeta^3+55\zeta^2+90)x^3 \notag \\ &+(1375\zeta^3+1375\zeta^2+2225)x+a_1 \ \in \ F_1[x] \\ \end{align}

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