Techniques of Solving Equations à la Galois
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1st upload: 2023/06/17
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[9-1] A first failure example \(f_1(x)=x^3-2\)
We consider the simple cubic equation (1.1), often used as an exercise in books on Galois theory.As usual, we begin the solution by computing the Galois resolvent.
\begin{align} f(x)&=x^3-2 =(x-\alpha)(x-\beta)(x \ -\gamma)\\ v&=1 \cdot \alpha+2 \cdot \beta+3 \cdot \gamma \\ \end{align}
\begin{align} &\left\{ \begin{array}{l} f(x)=(x-\alpha)q_1(x)+r_1 \\ q_1(x)=(x-\beta)q_2(x)+r_2 \\ q_2(x)=(x-\gamma)q_3(x)+r_3 \\ r_4=v-(\alpha+2\beta+3\gamma) \\ \end{array} \right.\\ \end{align} \begin{align} \{f(\alpha)=0,\ q_1(\beta)=0, \ q_2(\gamma)=0, \ eq(1.2)\} \ \Rightarrow \ \{ \ r_1=0, \ r_2=0, \ r_3=0, \ r_4=0 \ \} \end{align}
Accordingly, using the “resultant” we eliminate variables in the order \([\gamma \ \rightarrow \ \beta \ \rightarrow \ \alpha]\), and ultimately simplify to a single-variable degree-6 polynomial in \(v\) as in (1.5). We call this the Galois resolvent \(V(v)\).
\begin{align} &V(v) \equiv v^{6}+108=0 \quad \Rightarrow \quad g_0(x) \equiv V(x) \\ \end{align}
Since the minimal polynomial \(g_0(x)\) has degree 6, the Galois group of the equation \(f(x)\) is the symmetric group \(S_3\). Hence the composition series is \([ \ S_3 \ \triangleright \ A_3 \ \triangleright \ e \ ]\).
Next we express the three roots of \(f(x)\) as polynomials in \(v\). Using these polynomial expressions, we can, as usual, obtain polynomial expressions for the automorphisms \(\{\rho_1,\rho_2,...,\rho_6\}\) corresponding to the symmetric group \(S_3\).
\begin{align} \alpha &=-\frac{{{v}^{4}}+18 v}{36} & \beta &=\frac{{{v}^{4}}}{18} & \gamma &=-\frac{{{v}^{4}}-18 v}{36} \\ \end{align} \begin{align} \rho_{1}(v)&= {v_1}=v & \rho_{2}(v)&={v_2}=\frac{{{v}^{4}}}{12}+\frac{v}{2} & \rho_{3}(v)&={v_3}=\frac{v}{2}- \frac{{{v}^{4}}}{12} \notag \\ \rho_{4}(v)&={v_4}=-\frac{{{v}^{4}}}{12}-\frac{v}{2} & \rho_{5}(v)&={v_5}=\frac{{{v}^{4}}}{12}-\frac{v}{2} & \rho_{6}(v)&={v_6}=-v \\ \end{align}
(1) Computation of the cyclic Galois extension \(F_1/F_0\). Since we have already explained the procedure, we only record the results.
Step1 LRT (Lagrange Resolvent transformation)
\begin{align}
&Gal(F_1/F_0)=S_3/A_3 \equiv C_2=\{\kappa_1,\kappa_2\} \quad \{\kappa_1,\kappa_2\}=\bigl[\{\rho_1,\rho_4,\rho_5\},\{\rho_2,\rho_3,\rho_6\} \bigr] \notag \\
\notag \\
& h_0=\prod_{\rho_i \in \kappa_1 }\rho_i(x-v)=(x-v_1)(x-v_4)(x-v_5) =x^3-v^3 \\
&h_1=\prod_{\rho_i \in \ \kappa_2}\rho_i(x-v)=(x-v_2)(x-v_3)(x-v_6) =x^3+v^3 \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}
=
\begin{bmatrix}
x^3 \\
-v^3
\end{bmatrix} \\
\end{align}
Step2 Generating the binomial equation \(B_1(x)\) and the new adjoined element \(a_1\)
\begin{align}
&\left\{
\begin{array}{l}
t_0 \ \in \ F_0[x] \\
t_1 \ \in \ F_0(v) \\
\end{array}
\right.
\quad \Rightarrow \quad
\left\{
\begin{array}{l}
B_1(x)=x^2-A_1=0 \quad
t_1^2=A_1=-108 \in \ F_0 \\
a_1=\sqrt{A_1} \ \in \ F_0(a_1) \equiv \ F_1 \\
\end{array}
\right. \\
\end{align}
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}
=
\begin{bmatrix}
x^3+a_1 \\
x^3-a_1
\end{bmatrix}
\quad \Rightarrow \quad
\left\{
\begin{array}{l}
g_0(x)=\tilde{h_0} \cdot \tilde{h_1} \ \in \ F_0[x] \\
g_1(x) \equiv \tilde{h_0} =x^3+a_1\ \in \ F_1[x] \\
\end{array}
\right. \\
\end{align}
(2) Computation of the cyclic Galois extension \(F_2/F_1\). Since this extension has degree \(3\), we introduce a cube root of unity \(\omega\). Again, we only record the results.
Step1 LRT (Lagrange Resolvent transformation)
\begin{align}
&Gal(F_2/F_1)=A_3/e \equiv C_3=\{\rho_1,\rho_4,\rho_5\} \notag \\
\notag \\
&\left\{
\begin{array}{l}
h_0=\rho_1(x-v)=(x-v_1) \\
h_1=\rho_4(x-v)=(x-v_4) \\
h_2=\rho_5(x-v)=(x-v_5) \\
\end{array}
\right. \\
\end{align}
\begin{align}
\notag \\
&\begin{bmatrix}
t_0 \\
t_1 \\
t_2
\end{bmatrix}
=\frac{1}{3}
\begin{bmatrix}
1&1&1 \\
1&\omega&\omega^2\\
1&(\omega^2)&(\omega^2)^2\\
\end{bmatrix}
\cdot
\begin{bmatrix}
h_0 \\
h_1 \\
h_2
\end{bmatrix}
=
\begin{bmatrix}
x-v \\
-\frac{-12 x+{a_1} v-6 v}{12}\\
\frac{12 x+{a_1} v+6 v}{12}
\end{bmatrix} \\
\end{align}
Step2 Generating the binomial equation \(B_2(x)\) and the new adjoined element \(a_2\)
\begin{align}
&\left\{
\begin{array}{l}
t_0 \ \in \ F_1[x] \\
t_1 \ \in \ F_1(v) \\
t_2 \ \in \ F_1(v) \\
\end{array}
\right.
\quad \Rightarrow \quad
\left\{
\begin{array}{l}
t_1^3=A_2=-6 \omega +\frac{{a_1}}{2}-3 \in \ F_1 \\
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. \\
\end{align}
Up to this point the computations have gone smoothly, but the trouble starts now.
We must also express \(t_2\) as an element of \(F_2\), just as we did for \(t_1\). For that computation we use (1.15) below, which requires computing \( A_2^{-1}\).
\begin{align} t_2=\frac{t_1 \cdot t_2}{t_1}=\frac{t_1^2\cdot (t_1t_2)}{t_1^2\cdot t_1} =\frac{t_1^2\cdot (t_1t_2)}{t_1^3}=\frac{a_1^2 \cdot (-3)}{A_2} =-\frac{3a_2^2}{A_2}\\ \end{align}
If we define the reciprocal as below and, using the condition \( A_2 \cdot A_2^{-1} =1\), set up the simultaneous linear equations to solve for the coefficients \(\{c_i\}\), we obtain the following system. Its coefficient matrix has rank \(6\), so the system has no solution.
\begin{align} A_2^{-1}&={a_1} {c_{11}} {{v}^{2}} \omega +{c_{10}} {{v}^{2}} \omega +{a_1} {c_7} v \omega +{c_6} v \omega +{a_1} {c_3} \omega +{c_2} \omega +{a_1} {c_9} {{v}^{2}} \notag \\ &+{c_8} {{v}^{2}}+{a_1} {c_5} v+{c_4} v+{a_1} {c_1}+{c_0} \\ \notag \\ \end{align}
\begin{align} &\begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 & -54 & -6 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & 3 & 0 & -6\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 6 & 0 & -3 & -54\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 6 & \frac{1}{2} & -3\\ -\frac{1}{2} & -3 & 0 & 6 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ \frac{1}{36} & -\frac{1}{2} & -\frac{1}{18} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & -6 & -\frac{1}{2} & 3 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ \frac{1}{18} & 0 & -\frac{1}{36} & -\frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & -\frac{1}{2} & -3 & 0 & 6 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & \frac{1}{36} & -\frac{1}{2} & -\frac{1}{18} & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & -6 & -\frac{1}{2} & 3 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & \frac{1}{18} & 0 & -\frac{1}{36} & -\frac{1}{2} & 0 & 0 & 0 & 0 \end{pmatrix} \cdot \begin{pmatrix} {c_0}\\{c_1}\\{c_2}\\{c_3}\\{c_4}\\{c_5}\\{c_6}\\{c_7}\\{c_8}\\{c_9}\\{c_{10}}\\{c_{11}} \end{pmatrix} = \begin{pmatrix} 1\\0\\0\\0\\0\\0\\0\\0\\0\\0\\0\\0 \end{pmatrix} \\ \end{align}
Does this phenomenon mean that the algorithm for solving equations we have described so far breaks down?
Not at all—the Galois-style method for solving equations is not so fragile. We pick this up in the next section.