Techniques of Solving Equations à la Galois
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Chapter6
Solving Cyclic Equations
\(\qquad \qquad \qquad \Phi_{5}(x)=x^4+x^3+x^2+x+1 \qquad Galois \ Group:C_4\)
\( \quad \)
▶ Page 1, 2, 3 ▶ Sample Program
[6-5] Calculation of \(Q_1/Q\): finding the minimal polynomial \(g_1(x)\) (2)
Using the pair \(\{t_0,\tilde{t_1}\}\) obtained in the previous section, we now compute the inverse transformation (5.1) (ILRT) of (4.3) (LTR).
[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}
\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 \ Q_1[x]
\end{array}
\right. \\
\end{align}
Since \(h_0\) originally contained the factor \((x-v)\), it follows that \(\tilde{h_0}\) becomes the new minimal polynomial over the extension field \(Q_1\).
\begin{align} \tilde{h_0}&=t_0+ \tilde{t_1}=x^2+(a_1+\tfrac{1}{2})x+1 \equiv g_1(x) \quad \in Q_1[x] \\ \end{align}
[6-6] Calculation of \(Q_2/Q_1\): finding the minimal polynomial \(g_2(x)\)
The next Galois extension \(Q_2/Q_1\) is again a quadratic cyclic extension, with \(Gal(Q_2/Q_1)=C_2=\{\rho_1,\rho_4\}\).Thus, by substituting the polynomial expressions of \(v_1,v_4\) into (6.1), we compute \(\{h_0,h_1\}\) and then \(\{t_0,t_1\}\).
[step1] LRT (Lagrange Resolvent Transformation) \begin{align} &\left\{ \begin{array}{l} h_0=(x-v_1) =x-v \\ h_1=(x-v_4) =x+v^3+v^2+v+1=x+v+a_1+\tfrac{1}{2} \\ \end{array} \right. \\ \notag \\ &\begin{bmatrix} t_0 \\ t_1 \\ \end{bmatrix} =\tfrac{1}{2} \begin{bmatrix} 1&1 \\ 1&-1\\ \end{bmatrix} \cdot \begin{bmatrix} h_0 \\ h_1 \end{bmatrix} = \begin{bmatrix} x+\tfrac{a_1}{2}+\tfrac{1}{4}\\ -v-\tfrac{a_1}{2}-\tfrac{1}{4}\\ \end{bmatrix} \\ \end{align}
Once again, since this is a quadratic cyclic extension, \(t_1^2\) should belong to \(Q_1\). After verifying this, we define the quadratic equation \(B_2(x)=0\) in order to introduce a new element \(a_2\).
\begin{align} t_1^2&=v^2+\bigl(a_1+\tfrac{1}{2}\bigr)v+\tfrac{a_1^2}{4}+\tfrac{a_1}{4}+\tfrac{1}{16}=\tfrac{2a_1-5}{8} \equiv A_2 \in Q_1\\ &\Downarrow \notag \\ B_2(x)&=x^2-A_2 \quad a_2=\sqrt{A_2} \ \in Q_1(a_2) \equiv Q_2 \qquad \tilde{t_1}\equiv a_2 \in Q_2 \\ \end{align}
[step2] The quadratic equation \(B_2(x)=0\) and generating the new adjoint element \(a_2\)
\begin{align}
\left\{
\begin{array}{l}
t_1=-v-\tfrac{a_1}{2}-\tfrac{1}{4} \ \in Q_1(v)\\
t_1^2=\tfrac{2a_1-5}{8} \equiv A_2 \in Q_1\\
\end{array}
\right.
\quad \Rightarrow \quad
\left\{
\begin{array}{l}
\tilde{t_1}=a_2 \ \in Q_2\\
B_2(x)=x^2-A_2=0 \\
a_2=\sqrt {A_2} \ \in Q_1(a_2) \equiv Q_2\\
\end{array}
\right. \notag \\
\end{align}
Now that we have \(\{t_0,\tilde{t_1}\}\), we apply the inverse transformation (ILRT) of (6.2) (LRT) to obtain the final minimal polynomial \(g_2(x)\).
[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}
\quad \Rightarrow \quad
\left\{
\begin{array}{l}
g_0(x)=\tilde{h_0} \cdot \tilde{h_1} \\
g_2(x) \equiv \tilde{h_0} \ \in \ Q_2[x]
\end{array}
\right. \\
\end{align}
Since \(h_0\) originally contained the factor \((x-v)\), it follows that \(\tilde{h_0}\) is the new minimal polynomial \(g_2(x)\) over the extension field \(Q_2\).
\begin{align} \tilde{h_0}&=t_0+ \tilde{t_1}=x+a_2+\tfrac{a_1}{2}+\tfrac{1}{4} \equiv g_2(x) \quad \in Q_2[x] \\ \therefore \ v&=-a_2-\tfrac{a_1}{2}-\tfrac{1}{4} \\ \end{align}
Finally, the root of \(g_2(x)=0\) gives the desired value of \(v\). Substituting this into (3.6) yields the four roots of \(f(x)\). Note, however, that in carrying out the reductions, we must take the remainders in the order \( mod(B(a_2)) \ \rightarrow \ mod(B_1(a_1)) \).
\begin{align} v_1&=-a_2-\tfrac{a_1}{2}-\tfrac{1}{4} & v_2&=a_1 a_2+\tfrac{a_2}{2}+\tfrac{a_1}{2}-\tfrac{1}{4} \notag \\ v_3&=-a_1 a_2-\tfrac{a_2}{2}+\tfrac{a_1}{2}-\tfrac{1}{4} & v_4&=a_2-\tfrac{a_1}{2}-\tfrac{1}{4} \notag \\ \end{align}