Techniques for Solving Solvable Equations via Galois Theory

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[3-11] Computation for \(F_2/F_1\): finding the minimal polynomial \(g_2(x)\) (2)

Let us look a little more closely at \(\{t_1,t_2\}\).
Let \(\{m,n\}\) be the highest (nonzero) degrees among the coefficients of the polynomials \(\{t_1,t_2\}\) in \(x\), and let the coefficients of \(\{x^m,x^n\}\) be \(\{ce_m,ck_n\}\). Factoring out these coefficients from \(\{t_1,t_2\}\) gives (11.2–3).
We then ask: “Over which field do the monic polynomials \(\{q_1(x),q_2(x)\}\) live?”

\begin{align} & deg(t_1)=m, \quad deg(t_2)=n \\ & t_1=ce_m \times \biggl( x^m+\displaystyle \sum_{i=0}^{m-1} \frac{ce_i}{ce_m}x^i\biggr) =ce_m \times q_1(x) \quad (ce_m\neq 0) \\ & t_2=ck_n \times \biggl( x^n+\displaystyle \sum_{j=0}^{n-1} \frac{ck_j}{ck_n}x^j\biggr) =ck_n \times q_2(x) \quad (ck_n\neq 0) \\ \end{align}

\begin{align} \notag \\ &\kappa_1 \biggl( \frac{ce_i}{ce_m}\biggr)= \frac{ce_i}{ce_m}, & &\kappa_2 \biggl( \frac{ce_i}{ce_m}\biggr)= \frac{\omega^2 ce_i}{\omega^2 ce_m}=\frac{ce_i}{ce_m}, & &\kappa_3 \biggl( \frac{ce_i}{ce_m}\biggr)= \frac{\omega ce_i}{\omega ce_m}=\frac{ce_i}{ce_m} \\ &\kappa_1 \biggl( \frac{ck_j}{ce_n}\biggr)= \frac{ce_j}{ce_n}, & &\kappa_2 \biggl( \frac{ck_j}{ce_n}\biggr)= \frac{\omega ce_j}{\omega ce_m}=\frac{ce_j}{ce_m}, & &\kappa_3 \biggl( \frac{ck_j}{ce_n}\biggr)= \frac{\omega^2 ce_j}{\omega^2 ce_m}=\frac{ce_j}{ce_m} \notag \\ \notag \\ \end{align}

\begin{align} &\therefore \ \biggl( \frac{ce_i}{ce_m}\biggr),\biggl( \frac{ck_j}{ck_n}\biggr) \ \in F_1 \quad \Rightarrow \quad q_1(x), \ \ q_2(x) \in F_1[x] \\ \end{align}

In conclusion, all coefficients of \(q_1(x)\) are invariant under the action of the automorphisms \(\{\kappa_1,\kappa_2\}\), and hence \(q_1(x)\) lies in \(F_1[x]\). The above observations follow purely from how the \(v_i\) transform under the automorphisms \(\rho_i\). This illustrates how quotient groups and automorphisms of field extensions intertwine to drive the theory forward.

Now let us carry out the explicit computation.
Substitute the polynomial expressions in \(v\) from (4.10–13) for the \(v_i\) that appear in \(\{h_0,h_1,h_2\}\) of (10.3).
Then apply the Lagrange resolvent transformation (10.4) to compute \(\{t_0,t_1,t_2\}\).
Be sure to compute in the order \([\ \mod g_1(v),\ \mod B_1(a_1),\ \mod \Omega\ ]\).

\begin{align} & t_0=x^4+112 \\ & t_1=\Biggl[\frac{176356098785 {a_1} {{v}^{10}} \omega }{26343508540135700992}+...+\frac{27924690836995920}{2708008690392239} \Biggr]x^2 +.... \\ & t_2=\Biggl[ -\frac{176356098785 {a_1} {{v}^{10}} \omega }{26343508540135700992}+...-\frac{193463308578480}{2708008690392239} \Biggr]x^2 +.... \\ \end{align}

From (11.6–8) we see that \([ \ t_0 \in F_1[x], \ \{\ t_1,t_2 \ \} \in F_1(v)[x] \ ]\). Moreover, the coefficients of \(x^2\) in \(\{t_1,t_2\}\) (the bracketed parts in (11.7–8)) are precisely \(\{ce_m, ck_m\}\) in correspondence with (11.2–3), with \([ \ m=n=2 \ ]\) in the present situation.

[Step 2] Binomial equation \(B_2(x)=0\) and creation of the new adjunction \(a_2\)
\begin{align} &\left\{ \begin{array}{l} t_1=ce_m \cdot q_1(x) \ \in F_1(v)[x]\\ t_2=ck_m \cdot q_2(x) \ \in F_1(v)[x]\\ \\ ce_m, \ ck_m \ \in F_1(v) \\ q_1(x), \ q_2(x) \ \in F_1[x]\\ \end{array} \right. \quad \Rightarrow \quad \left\{ \begin{array}{l} \tilde{t_1}=a_2 \cdot q_1(x) \ \in F_2[x] \\ \tilde{t_2}=b_2 \cdot q_2(x) \ \in F_2[x] \\ \\ B_2(x)=x^3-A_2=0 \\ a_2=\sqrt[3] {A_2} \ \in F_1(a_2) \equiv F_2\\ \end{array} \right. \notag \\ \end{align}

Next, compute the inverses \(\{ce_m^{-1},ck_m^{-1}\}\), which will be needed later.
As in the previous section, multiply \(\{t_1,t_2\}\) by \(\{ce_m^{-1},ck_m^{-1}\}\) to confirm that the resulting polynomials \(\{q_1,q_2\}\) are monic.

\begin{align} ce_m=&\Biggl[\frac{176356098785 {a_1} {{v}^{10}} \omega }{26343508540135700992} +\frac{103314391395 {{v}^{10}} \omega }{5416017380784478}+.....+\frac{27924690836995920}{2708008690392239} \Biggr] \\ ck_m=&\Biggl[ -\frac{176356098785 {a_1} {{v}^{10}} \omega }{26343508540135700992} -\frac{103314391395 {{v}^{10}} \omega }{5416017380784478}+.....-\frac{193463308578480}{2708008690392239} \Biggr] \notag \\ ce_m^{-1}=&\Biggl[-\frac{529068296355 {a_1} {{v}^{10}} \omega }{10958899552696451612672} -\frac{309943174185 {{v}^{10}} \omega }{2253063230406342848}+.....-\frac{36274370358465}{70408225950198214} \Biggr] \\ ck_m^{-1}=&\Biggl[\frac{529068296355 {a_1} {{v}^{10}} \omega }{10958899552696451612672} +\frac{309943174185 {{v}^{10}} \omega }{2253063230406342848}+.....+\frac{5235879531936735}{70408225950198214} \Biggr] \notag \\ \notag \\ \end{align}

\begin{align} q_1= \ ce_m^{-1} \cdot t_1= x^2-\frac{5 {a_1} \omega }{2496}+\frac{18 \omega }{13}-\frac{31 {a_1}}{24960}-\frac{207}{26} \ \in \ F_1[x] \\ q_2= \ ck_m^{-1} \cdot t_2= x^2+\frac{5 {a_1} \omega }{2496}-\frac{18 \omega }{13}+\frac{19 {a_1}}{24960}-\frac{243}{26} \ \in \ F_1[x] \\ \end{align}

Next compute \(ce_m^3\). As shown in (11.13), this lies in \(F_1\); denote it by \(A_2\). Observing the ends of (11.13), we see that \(ce_m\) is a root of the binomial equation \(B_2(x)=0\) given in (11.14).
We therefore define \(a_2\) by \(a_2^3=A_2\) (i.e., \(a_2=\sqrt[3]{A_2}\)) and adjoin \(a_2\) to \(F_1\) to form the new extension \(F_2\). Using \(a_2\), we then redefine \(t_1\) over \(F_2\) as the polynomial \(\tilde{t_1}\) in (11.15).

[Step 2] Binomial equation \(B_2(x)\) and creation of the new adjunction \(a_2\)
\begin{align} &ce_m^3= \frac{14 {a_1} \omega }{27}+2304 \omega +\frac{89 {a_1}}{135}-1088 \equiv A_2 \ \in \ F_1\\ &B_2(x)=x^3-A_2=0, \qquad a_2=\sqrt[3]{A_2} \ \in \ F_1(a_2) \equiv \ F_2 \\ \notag \\ &\qquad \Downarrow \notag \\ &t_1 \ \rightarrow \ \tilde{t_1} \equiv a_2 \cdot \biggl( x^2-\frac{5 {a_1} \omega }{2496}+\frac{18 \omega }{13}-\frac{31 {a_1}}{24960}-\frac{207}{26} \biggr) \in F_2[x]\\ \end{align}

Throughout these computations, take residues in the order \([\ \mod g_1(v),\ \mod B_1(a_1),\ \mod \Omega\ ]\)!

[Supplement] Notes on finding \(a_2^{-1}\) (which equals \(ce_m^{-1}\))

Since \(a_2(=ce_m) \in F_1(v)\), the field \(F_1(v)\) is determined by the relations \([\ g_1(v)=0,\ \Omega=0,\ B_1(a_1)=0\ ]\).
Expanding \(a_2^{-1}\) in a basis of \(F_1(v)\) yields (11.16). Solving the linear system for the coefficients \(c_{i,j,k}\) under the condition \([\ a_2^{-1} \cdot a_2=1 \ ]\) gives \(a_2^{-1}(=ce_m^{-1})\). As usual, reductions \(\{\mod(g_1(v)),\ \mod(\Omega),\ \mod(B_1(a_1))\}\) are required.

\begin{align} &\left\{ \begin{array}{l} g_1(v)= v^{12}-80v^8+2720v^6+11840v^4+587520v^2+1193216 \\ \qquad \qquad +a_1({{v}^{6}}-\frac{232 {{v}^{2}}}{5}+432 )=0 \\ \Omega=\omega^2+\omega+1=0 \\ B_1(a_1)=a_1^2+17510400=0 \\ \end{array} \right.\\ \end{align}

\begin{align} &\left\{ \begin{array}{l} a_2^{-1}=\displaystyle \sum_{i=0}^{11}\displaystyle \sum_{j=0}^1\displaystyle \sum_{k=0}^1 c_{i,j,k} \cdot v^i \cdot \omega^j \cdot a_1^k \\ a_2^{-1} \cdot a_2=1 \\ \end{array} \right.\\ \end{align}

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