Techniques of Solving Equations à la Galois
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1st upload: 2023/06/17
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[3-12] Computation for \(F_2/F_1\): finding the minimal polynomial \(g_2(x)\) (3)
In the previous section, we adjoined \(a_2\) to \(F_1\) to form the extension field \(F_2\). In fact, we also need to express \(ck_m\) in terms of \(a_2\). To that end, focus on the rightmost column \(ce_i \cdot ck_l\) in [Table 3-12], where \(ce_m\) and \(ck_m\) are related.As is immediate from [Table 3-12], each product \( (ce_i \cdot ck_l )\) is invariant under the automorphisms \(\kappa_j\).
A direct computation shows that \(ce_m \cdot ck_m\) is an element of \(F_0\), as in (12.2).
\begin{align} &\quad \left\{ \begin{array}{l} \kappa_1(ce_i \cdot ck_l)=\kappa_1(ce_i) \cdot \kappa_1(ck_l)=ce_i \cdot ck_l \qquad [i,j]=[0,1,2,3]\\ \kappa_2(ce_i \cdot ck_l)=\kappa_2(ce_i) \cdot \kappa_2(ck_l)=(\omega^2 ce_i) \cdot (\omega ck_l)= \omega^3 ce_i \cdot ck_l= ce_i \cdot ck_l \\ \kappa_3(ce_i \cdot ck_l)=\kappa_3(ce_i) \cdot \kappa_3(ck_l)=(\omega ce_i) \cdot (\omega^2 ck_l)= \omega^3 ce_i \cdot ck_l= ce_i \cdot ck_l \\ \end{array} \right. \\ \notag \\ \end{align}
\begin{align} &\therefore \ ce_i \cdot ck_l \ \in F_1 \quad \Rightarrow \quad ce_m \cdot ck_m=\frac{416}{3} \ \in \ F_0 \\ \end{align}
\begin{align} ck_m&=\frac{ce_m \cdot ck_m}{ce_m} =\frac{ce_m^2 \cdot (ce_m \cdot ck_m)}{ce_m^3}=\frac{ce_m^2 \cdot (ce_m \cdot ck_m)}{A_2} \\ &=ce_m^2 \cdot (ce_m \cdot ck_m) \cdot A_2^{-1}=\frac{416}{3} \cdot ce_m^2 \cdot A_2^{-1} \\ \notag \\ \therefore \ ck_m&=\frac{416}{3} \cdot ce_m^2 \cdot A_2^{-1} \quad \Rightarrow \quad b_2 \equiv \frac{416}{3} \cdot a_2^2 \cdot A_2^{-1} \ \in F_2 \end{align}
Since \(\{t_1,t_2\}\) are now polynomials over \(F_2\), we attach tildes and write them as \(\{\tilde{t_1},\tilde{t_2}\}\) to distinguish from the previous ones.
\begin{align} & A_2^{-1}=-\frac{7 {a_1} \omega }{35995648}-\frac{243 \omega }{281216}+\frac{19 {a_1}}{359956480}-\frac{1431}{1124864} \in F_1 \\ \notag \\ &\quad \therefore \ b_2=a_2^2\biggl(-\frac{7 {a_1} \omega }{259584}-\frac{81 \omega }{676}+\frac{19 {a_1}}{2595840}-\frac{477}{2704}\biggr) \\ \notag \\ &\tilde{t_1}=a_2 \cdot q_1=a_2\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]\\ &\tilde{t_2}=b_2 \cdot q_2=b_2\biggl( x^2+\frac{5 {a_1} \omega }{2496}-\frac{18 \omega }{13}+\frac{19 {a_1}}{24960}-\frac{243}{26} \biggr) \ \in \ F_2[x]\\ \end{align}
[Step 3] ILRT (Inverse Lagrange Resolvent Transformation)
\begin{align}
&\begin{bmatrix}
\tilde{h_0} \\
\tilde{h_1} \\
\tilde{h_2}
\end{bmatrix}
=
\begin{bmatrix}
1&1&1 \\
1&\omega^2&(\omega^2)^2\\
1&\omega&(\omega^2)\\
\end{bmatrix}
\cdot
\begin{bmatrix}
t_0 \\
\tilde{t_1} \\
\tilde{t_2}
\end{bmatrix}
\quad \Longrightarrow \quad
\left\{
\begin{array}{l}
g_1(x)=\tilde{h_0}\cdot \tilde{h_1} \cdot \tilde{h_2} \\
g_2(x) \equiv \tilde{h_0} \ \in \ F_2[x]
\end{array}
\right. \\
\end{align}
In particular, since \(\tilde{h_0}\) contains the factor \((x-v)\), we may take \(\tilde{h_0} \equiv g_2(x)\) as the minimal polynomial of \(v\) over \(F_2\).
We summarize the results from the previous page.
\begin{align} &\tilde{h_0}=t_0+\tilde{t_1}+\tilde{t_2} \equiv g_2(x) \\ &\qquad \Downarrow \notag \\ &\ g_2(x)= x^4+112+a_2\biggl( x^2-\frac{5 {a_1} \omega }{2496}+\frac{18 \omega }{13}-\frac{31 {a_1}}{24960}-\frac{207}{26} \biggr) \notag \\ & \qquad \qquad +a_2^2\biggl(-\frac{7 {a_1} \omega }{259584}-\frac{81 \omega }{676}+\frac{19 {a_1}}{2595840}-\frac{477}{2704}\biggr) \notag \\ & \qquad \qquad \qquad \times \biggl( x^2+\frac{5 {a_1} \omega }{2496}-\frac{18 \omega }{13}+\frac{19 {a_1}}{24960}-\frac{243}{26} \biggr)\\ \notag \\ & B_2(x)= x^3-A_2=0 \quad A_2= \frac{14 {a_1} \omega }{27}+2304 \omega +\frac{89 {a_1}}{135}-1088 \\ &\ a_2=\sqrt[3]{A_2} \qquad F_2 \equiv F_1(a_2) \end{align}
[Supplement] Notes on computing \(A_2^{-1}\)
Since \(A_2 \in F_1\), the field \(F_1\) is determined by the relations \([\ \Omega=0,\ B_1(a_1)=0\ ]\).
With these two relations, expand \(A_2^{-1}\) in a basis of \(F_1\) as in (12.13). Solving the linear system for the coefficients \(c_{i,j}\) under the constraint \([\ A_2^{-1} \cdot A_2=1\ ]\) gives \(A_2^{-1}\). As usual, reductions \(\{ \mod(\Omega),\ \mod(B_1(a_1)) \}\) are required.
\begin{align} &\left\{ \begin{array}{l} \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}^{1}\displaystyle \sum_{j=0}^1 c_{i,j} \cdot \omega^i \cdot a_1^j \\ A_2^{-1} \cdot A_2=1 \\ \end{array} \right.\\ \end{align}