Techniques for Solving Solvable Equations via Galois Theory

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[3-13] Computing \(g_3(x)\) for \(F_3/F_2\)

We now compute \(g_3(x)\). The green region in Fig. 3-4 indicates this step of the solution procedure.

\(\qquad\qquad\)

The relevant Galois group here is the quotient \(V_4/N \cong C_2\), which is a cyclic group of order \(2\). Writing \(C_2=\{\kappa_1,\kappa_2\}\), each coset has two automorphisms.

\begin{align} &V_4 =\{\rho_{1},\rho_{8},\rho_{17},\rho_{24}\}, \quad N=\{\rho_{1},\rho_{8}\} \notag \\ &Gal(F_3/F_2)=V_4/N \cong C_2 = \{\kappa_1,\kappa_2\}, \quad \kappa_1= \{\rho_{1},\rho_{8}\}, \ \kappa_2= \{\rho_{17},\rho_{24}\} \notag \\ \end{align}

First, we examine the polynomials \(\{h_0,h_1\}\) defined in (13.1).

[Step 1] LRT (Lagrange Resolvent Transformation)
\begin{align} & h_0=\prod_{\rho_i \in \ \kappa_1}\rho_i(x-v)=(x-v_1)(x-v_8) \\ & h_1=\prod_{\rho_i \in \ \kappa_2}\rho_i(x-v)=(x-v_{17})(x-v_{24}) \notag \\ \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} \qquad (\ t_1:\ \text{Lagrange resolvent}\ ) \end{align}

Let the coefficients of \(\{h_0,h_1\}\) be \(\{ca_i,cb_i\ (i=0,1)\}\) as in (13.3). Evidently each \(ca_i,cb_i\) is formed from the pairs \(\{v_1,v_8\}\) and \(\{v_{17},v_{24}\}\).

\begin{align} &\left\{ \begin{array}{l} h_0=x^2+ca_1x+ca_0 \quad\ &ca_1=-(v_1+v_8),\ \ ca_0=v_1\cdot v_8\\ h_1=x^2+cb_1x+cb_0 \quad\ &cb_1=-(v_{17}+v_{24}),\ \ cb_0=v_{17}\cdot v_{24} \end{array} \right.\\ \end{align}

From the action of \(\{\kappa_1,\kappa_2\}\) on \(v_i\) we obtain [Table3-13]–[Table3-15]. The two groups of \(v_i\) do not mix under these automorphisms, hence the induced actions on \(\{ca_i,cb_i\}\) and then on \(\{h_0,h_1\}\) follow accordingly.

[Table3-13] \(\kappa_j(v_i)\)
\(\ \)	\(\kappa_j(v_1,v_8)\)	\(\kappa_j(v_{17},v_{24})\)
\(\kappa_1\)	\(v_{1},v_{8}\)	\(v_{17},v_{24}\)
\(\kappa_2\)	\(v_{17},v_{24}\)	\(v_{1},v_{8}\)

\(\quad \Rightarrow \quad\)

[Table3-14] \(\kappa_j(ca_i,cb_i)\)
\(\ \)	\(\kappa_j(ca_i)\)	\(\kappa_j(cb_i)\)
\(\kappa_1\)	\(ca_i\)	\(cb_i\)
\(\kappa_2\)	\(cb_i\)	\(ca_i\)

\(\quad \Rightarrow \quad\)

[Table3-15] \(\kappa_j(h_i)\)
\(\ \)	\(\kappa_j(h_0)\)	\(\kappa_j(h_1)\)
\(\kappa_1\)	\(h_0\)	\(h_1\)
\(\kappa_2\)	\(h_1\)	\(h_0\)

Next, by (13.2) the LRT generates \(\{t_0,t_1\}\) with coefficients \(\{cc_i,cd_i\}\) formed from \(\{ca_i,cb_i\}\) as in (13.4). The induced actions are summarized in [Table3-16]–[Table3-17].

\begin{align} &\left\{ \begin{array}{l} t_0=\frac{1}{2}(h_0+h_1)=x^2+\frac{1}{2}(ca_1+cb_1)x+\frac{1}{2}(ca_0+cb_0)=x^2+cc_1x+cc_0\\ t_1=\frac{1}{2}(h_0-h_1)=\frac{1}{2}(ca_1-cb_1)x+\frac{1}{2}(ca_0-cb_0)=cd_1x+cd_0\\ \end{array} \right.\\ \end{align}

[Table3-16] \(\kappa_j(ca_i \pm cb_i)\)
\( \ \)	\(\kappa_j(cc_i=\frac{1}{2}(ca_i+cb_i))\)	\(\kappa_j(cd_i=\frac{1}{2}(ca_i-cb_i))\)
\(\kappa_1\)	\(cc_i=\frac{1}{2}(ca_i+cb_i)\)	\(cd_i=\frac{1}{2}(ca_i-cb_i)\)
\(\kappa_2\)	\(cc_i=\frac{1}{2}(cb_i+ca_i)\)	\(-cd_i=\frac{1}{2}(cb_i-ca_i)\)

\(\quad \Rightarrow \quad\)

[Table3-17] \(\kappa_j(t_i)\)
\(\ \)	\(\kappa_j(t_0)\)	\(\kappa_j(t_1)\)	\(\kappa_j(cd_i^2)\)
\(\kappa_1\)	\(t_0\)	\(t_1\)	\(cd_i^2\)
\(\kappa_2\)	\(t_0\)	\(-t_1\)	\(cd_i^2\)

From [Table3-17] it follows that \(t_0\) and \(cd_i^2\) are invariant under \(\{\kappa_1,\kappa_2\}\). Hence \[ t_0\in F_2[x],\quad t_1\in F_2(v)[x],\quad cd_i^2\in F_2. \] We now verify this by direct computation. Substitute the polynomial expressions of \(v_i\) (from (4.10–13)) into (13.1), and then form (13.2), taking remainders in the order \([\,\bmod\ g_2(v),\ \bmod\ B_2,\ \bmod\ B_1,\ \bmod\ \Omega\,]\).

\begin{align} & t_0= x^2+\frac{19 {a_1} {{a}_{2}^{2}} \omega }{865280}+\frac{45 {{a}_{2}^{2}} \omega }{1352}-\frac{4 {a_2} \omega }{13}+\frac{9 {a_1} {{a}_{2}^{2}}}{1730560}+\frac{603 {{a}_{2}^{2}}}{5408}-\frac{19 {a_2}}{26} \ \in F_2[x] \\ & t_1=\Biggl[-\frac{5 {a_1} {{a}_{2}^{2}} {{v}^{3}} \omega }{183028352}+\frac{1215 {{a}_{2}^{2}} {{v}^{3}} \omega }{5719636} +.....+\frac{49 {a_1} v}{338440}-\frac{21371 v}{8461}\Biggr]x \ \in F_2(v)[x] \end{align}

The bracketed coefficient of \(x\) in \(t_1\) is \(cd_1\). As indicated by the yellow cells in [Table3-17], \(cd_1^2\) must lie in \(F_2\), which a direct computation confirms:

\begin{align} \bbox[#FFFF00]{ cd_1^2}&=\frac{23 {a_1} {{a}_{2}^{2}} \omega }{324480}+\frac{63 {{a}_{2}^{2}} \omega }{338}-\frac{8 {a_2} \omega }{13}+\frac{{a_1} {{a}_{2}^{2}}}{324480}+\frac{135 {{a}_{2}^{2}}}{338}-\frac{32 {a_2}}{13} \equiv \bbox[#FFFF00]{ A_3 } \ \in \ F_2 \end{align}

Observing the ends of (13.8), it turns out that \(cd_1\) is a root of the binomial equation \(B_3(x)=x^2-A_3=0\). Define \(a_3=\sqrt{A_3}\) and adjoin it to \(F_2\) to obtain the Galois extension \(F_3=F_2(a_3)\). Then \(t_1\) can be rewritten over \(F_3\); to signal this change of base field we consciously replace \(t_1\) by \(\tilde t_1\) below (this notational change is important for beginners to avoid confusion).

[Step 2] Binomial \(B_3(x)=0\) and the new adjunction \(a_3\)
\begin{align} &\left\{ \begin{array}{l} cd_1 \in F_2(v),\\[2pt] t_1=cd_1\cdot x \in F_2(v)[x] \end{array} \right. \ \Longrightarrow\ \left\{ \begin{array}{l} B_3(x)=x^2-A_3=0,\\ a_3\equiv \sqrt{A_3}\in F_2(a_3)\equiv F_3,\\ \tilde t_1=a_3\cdot x \in F_3[x] \end{array} \right. \end{align}

Finally, substitute \(t_0\) from (13.6) and \(\tilde t_1\) from (13.9) into the inverse LRT to obtain \(\{\tilde h_0,\tilde h_1\}\). We use tildes to emphasize that these are formed over the enlarged field.

[Step 3] 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} \quad \Rightarrow \quad \left\{ \begin{array}{l} g_2(x)=\tilde h_0\cdot \tilde h_1,\\ g_3(x)\equiv \tilde h_0 \in F_3[x] \end{array} \right. \end{align}

Since \(\tilde h_0\) contains the factor \((x-v)\), it is the minimal polynomial \(g_3(x)\) of \(v\) over \(F_3\). Summarizing:

\begin{align} &\tilde h_0=t_0+\tilde t_1 \equiv g_3(x) \in F_3[x]\\ &g_3(x)=x^2+a_3 x \notag \\ &\qquad +\biggl[ \frac{19 {a_1} {{a}_{2}^{2}} \omega }{865280}+\frac{45 {{a}_{2}^{2}} \omega }{1352}-\frac{4 {a_2} \omega }{13} +\frac{9 {a_1} {{a}_{2}^{2}}}{1730560}+\frac{603 {{a}_{2}^{2}}}{5408}-\frac{19 {a_2}}{26} \biggr] \end{align}

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