Techniques for Solving Solvable Equations via Galois Theory

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[3-9] Computation for \(F_1/F_0\): finding the minimal polynomial \(g_1(x)\) (2)

In the previous section, as indicated in (8.9, 11), we investigated—via automorphism considerations—the fields to which \(\{t_0, t_1^2, q_1(x)\}\) belong. Here we confirm those results by explicit computation.
First, substitute into the \(v_i\) appearing in \(\{h_0,h_1\}\) from (8.1) the polynomial expressions of \(v_i\) in terms of \(v\) given in (4.10–13).
(Note: all computations below are carried out \([ \ \mathrm{mod}\ g_0(v) \ ]\).)
Then substitute the computed \(\{h_0,h_1\}\) into the Lagrange Resolvent Transformation (8.2) to obtain \(\{t_0,t_1\}\), yielding:

\begin{align} & t_0=x^{12}-80x^8+2720x^6+11840x^4+587520x^2+1193216 \\ \notag \\ & t_1=\Biggl[ -\frac{1842286585 {{v}^{22}}}{575477552142843904}+...-\frac{24232754108316500}{17562181156703}\Biggr]x^6 \ +.... \\ \end{align}

From these formulas we see that \(\{ \ t_0 \in F_0[x], \ t_1 \in F_0(v)[x] \ \}\). The coefficient of the highest power \(x^6\) in \(t_1\) in (9.2) is the part enclosed in brackets \([...]\). This coefficient is precisely \([ \ cd_m, \ m=6 \ ]\) introduced in (8.10) of the previous section.

Let us examine this \(cd_m\) a little further. Compute the inverse \(cd_m^{-1}\) and multiply it into \(t_1\). As stated in (8.11), the resulting monic polynomial \(q_1(x)\) lies in \(F_0[x]\).

\begin{align} &cd_m=\Biggl[ -\frac{1842286585 {{v}^{22}}}{575477552142843904}+...-\frac{24232754108316500}{17562181156703}\Biggr] \ \in \ F_0(v) \\ &cd_m^{-1}=\Biggl[ \frac{368457317 {{v}^{22}}}{2015368425808410779320320}+...+\frac{242327541083165}{3075208169263322112} \Biggr] \\ \notag \\ &q_1(x)= cd_m^{-1} \cdot t_1={{x}^{6}}-\frac{232 {{x}^{2}}}{5}+432 \ \in \ F_0[x]\\ \end{align}

Next, compute \(cd_m^2\). The result is a number of \(F_0\), as shown in (9.6). Denote this value by \(A_1\).
Looking at the two ends of (9.6), we may also say that \(cd_m\) is a root of the binomial equation \( \ B_1(x)=0 \ \) given in (9.7). Thus we introduce a new symbol \(a_1\) as a radical of \(B_1(x)=0\). By adjoining \(a_1\) to the base field \(F_0\), we form a new extension field \(F_1\).

\begin{align} &cd_m^2=-17510400 \equiv A_1 \ \in \ F_0 \\ \notag \\ &\left\{ \begin{array}{l} B_1(x)=x^2-A_1=0 \\ a_1 \equiv \sqrt{A_1} =\sqrt{-17510400} \ \in \ \bbox[#FFFF00]{ F_1 \equiv F_0(a_1) }\\ \end{array} \right. \\ \notag \\ &t_1=cd_m \cdot q_1(x) \ \in F_0(v)[x] \quad \Rightarrow \quad \tilde{t_1}=a_1 \cdot q_1(x) \ \in F_1[x] \\ \end{align}

By using the newly introduced number \(a_1\), we see from (9.8) that while \(t_1\) was a polynomial over \(F_0(v)\), the product \(a_1 \cdot q_1\) is a polynomial over \(F_1\). To distinguish it from the original \(t_1\), we deliberately write \(\tilde{t_1}\) with a tilde.

[Step 2] The binomial equation \(B_1(x)=0\) and the new adjunction \(a_1\)
\begin{align} &\left\{ \begin{array}{l} t_1=cd_m \cdot q_1(x) \ \in F_0(v)[x]\\ \\ cd_m \ \in F_0(v) \\ q_1(x) \ \in F_0[x]\\ \end{array} \right. \quad \Rightarrow \quad \left\{ \begin{array}{l} \tilde{t_1}=a_1 \cdot q_1(x) \ \in F_1[x] \\ \\ B_1(x)=x^2-A_1=0 \\ a_1=\sqrt {A_1} \ \in F_0(a_1) \equiv F_1\\ \end{array} \right. \notag \\ \end{align}

Finally, we come to the last step.
Substitute \(t_0\) from (9.1) and \(\tilde{t_1}\) from (9.8) into the Inverse Lagrange Resolvent Transformation (9.9) to obtain \(\{\tilde{h_0},\tilde{h_1}\}\). We write tildes to distinguish \(\{\tilde{h_0},\tilde{h_1}\}\) from \(\{h_0,h_1\}\).
Since \(\tilde{h_0}\) contains the factor \((x-v)\), we may regard \(\tilde{h_0}\) as the minimal polynomial \(g_1(x)\) of \(v\) over \(F_1\).

[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_0(x)=\tilde{h_0} \cdot \tilde{h_1} \\ g_1(x) \equiv \tilde{h_0} \ \in \ F_1[x] \end{array} \right. \\ \end{align}

Summarizing the result:

\begin{align} &\tilde{h_0}=t_0+\tilde{t_1} \equiv g_1(x) \qquad deg(g_1(x))=12 \notag \\ \notag \\ &g_1(x)= x^{12}-80x^8+2720x^6+11840x^4+587520x^2+1193216 \notag \\ &\qquad \qquad +a_1({{x}^{6}}-\frac{232 {{x}^{2}}}{5}+432 ) \quad \in \ F_1[x]\\ \end{align}

Thus, by introducing the new adjunction \(a_1\) and enlarging the field from \(F_0\) to \(F_1\), we have transformed the minimal polynomial of \(v\) from \(g_0(x)\) to \(g_1(x)\), reducing the degree from \(24\) to \(12\).

[Supplement] on computing \(cd_m^{-1}\)

Since \(cd_m \ \in F_0(v)\), the defining relation of the field \(F_0(v)\) is \(g_0(v)=0\).
As \(deg(g_0(x))=24\), a basis of \(F_0(v)\) is \(\{1,v,v^2,\ldots,v^{23}\}\).
Therefore expand \(cd_m^{-1}\) with respect to this basis and solve for the coefficients \(c_i\) from the condition \([ \ cd_m^{-1} \cdot cd_m=1 \ ]\). As usual, reductions modulo \(g_0(v)\) are required during the computation.

\begin{align} &\left\{ \begin{array}{l} cd_m^{-1}=c_0+c_1v+c_2v^2+....+c_{23}v^{23} \\ cd_m^{-1} \cdot cd_m=1 \\ \end{array} \right.\\ \end{align}

Incidentally, \(cd_m\) and \(cd_m^{-1}\) are complicated expressions as follows (feel free to check your computations against them).

\begin{align} \notag \\ cd_m&=-\frac{1842286585 {{v}^{22}}}{575477552142843904}+\frac{22200475 {{v}^{20}}}{17983673504463872}+\frac{35316027575 {{v}^{18}}}{71934694017855488} +..... \notag \\ \notag \\ &-\frac{10880639453426075 {{v}^{4}}}{140497449253624}-\frac{22629495387627730 {{v}^{2}}}{17562181156703}-\frac{24232754108316500}{17562181156703} \notag \\ \notag \\ cd_m^{-1}&=\frac{368457317 {{v}^{22}}}{2015368425808410779320320}-\frac{888019 {{v}^{20}}}{12596052661302567370752}+....\notag \\ \notag \\ &+\frac{435225578137043 {{v}^{4}}}{98406661416426307584}+\frac{2262949538762773 {{v}^{2}}}{30752081692633221120}+\frac{242327541083165}{3075208169263322112} \notag \\ \end{align}

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