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
[6-3] Setting up the cyclotomic equation \(\Phi_5(x)\)
Cyclotomic polynomials are the standard examples of cyclic polynomials. Here we will apply factorization over an algebraic number field to the cyclotomic polynomial \(\Phi_{5}(x)\) and work it out concretely. From now on we will use the following notational conventions for the polynomials:· When we want to emphasize it is a cyclotomic polynomial: we use \(\Phi_{5}(x)\).
· When we want to emphasize it is the minimal polynomial: we use \(g_0(x)\).
Of course \([ \ \Phi_{5}(x)=g_0(x), \ g_0(v)=0 \ ]\).
· The base field is \(Q\).
Then, inside the algebraic field \(Q(v)\) generated by \(g_0(v)\), we factorize \(\Phi_{5}(x)\) using the Maxima command \(factor(p,q)\). The result is (3.3).
\begin{align} &x^{5}-1=(x-1) \times \Phi_{5}(x) \notag \\ \notag \\ &\Phi_{5}(x) =( x^4+x^3+x^2+x+1) \\ \notag \\ &Minimal \ Polynomial \ of \ v : \ g_0(x) \equiv \Phi_{5}(x) \quad \rightarrow \quad g_0(v)=0\\ \notag \\ &factor(\Phi_{5}(x),g_0(v))=(x-v)(x-v^2)(x-v^3)(x+v^3+v^2+v+1) \\ \notag \\ \therefore \ f(x)&=(x-v)(x-v^2)(x-v^3)(x+v^3+v^2+v+1) \\ &=(x-v_1)(x-v_2)(x-v_3)(x-v_4) \\ \notag \\ &\Downarrow \notag \\ \end{align} \begin{align} &\left\{ \begin{array}{l} \rho_{1}(v) \equiv v_{1}=v \qquad \rho_{2}(v) \equiv v_{2}=v^{2} \\ \rho_{3}(v) \equiv v_{3}=v^{3} \qquad \rho_{4}(v) \equiv v_{4}=-(v^3+v^2+v+1) \quad mod( \ g_0(v) \ ) \\ \end{array} \right. \\ \end{align}
By (3.4), \(\Phi_5(x)\) factorizes easily inside the algebraic field \(Q(v)\). Thus the four roots of \(\Phi_5(x)\) can be written down directly. In (3.6), \(\{v_1,v_2,v_3,v_4\}\) give polynomial expressions in \(v\) for these four roots. At the same time, we take \(\{\rho_i\}\) as the automorphisms of the Galois extension \(Q(v)/Q\). Let us check that this assignment is correct.
To do so, we must check that the set of automorphisms \(\{ \rho_i \}\) forms a group under composition. The composition can be defined as follows.
\begin{align} &v^5=v \cdot v^4=-v(v^3+v^2+v+1)=-(v^4+v^3+v^2+v)=-(-1)=1 \quad \therefore \ v^5=1 \\ &\qquad \qquad \Downarrow \notag \\ \end{align} \begin{align} &\rho_2 \circ \rho_2(v)=\rho_2(v^2)=\rho_2(v)\cdot \rho_2(v)=v^2 \cdot v^2=v^4=v_4=\rho_4(v) \quad \rightarrow \ (\rho_2)^2=\rho_4 \notag \\ &\rho_2 \circ \rho_2\circ \rho_2(v)=(v^2)^4=v^3=v_3=\rho_3(v) \quad \rightarrow \ (\rho_2)^3=\rho_3 \\ &\rho_2 \circ \rho_2 \circ \rho_2\circ \rho_2(v)=(v^2)^3=v^6=v=v_1=\rho_1(v) \quad \rightarrow \ (\rho_2 )^4=\rho_1 \notag \\ \end{align}
| \( i \backslash j \) | \(\rho_1\) | \(\rho_2\) | \(\rho_4\) | \(\rho_3\) |
|---|---|---|---|---|
| \(\rho_1\) | \(\rho_1\) | \(\rho_2\) | \(\rho_4\) | \(\rho_3\) |
| \(\rho_2\) | \(\rho_2\) | \(\rho_4\) | \(\rho_3\) | \(\rho_1\) |
| \(\rho_4\) | \(\rho_4\) | \(\rho_3\) | \(\rho_1\) | \(\rho_2\) |
| \(\rho_3\) | \(\rho_3\) | \(\rho_1\) | \(\rho_2\) | \(\rho_4\) |
From (3.8), computing successive powers of \(\rho_2\) shows that all automorphisms are generated. Thus the Galois group of \(\Phi_{5}(x)\) is the cyclic group of order 4, \(C_{4}\).
\( \therefore \ Gal(Q(v)/Q)=C_4 \)
The composition series of the cyclic group \(C_4\) is (3.9). Accordingly, the base field \(Q\) is extended in two steps \([ \ Q \rightarrow Q_1 \rightarrow Q_2 \ ]\), as shown in Fig.6-1, corresponding to the series \([ \ C_4 \ \rhd \ C_2 \ \rhd \ e \ ]\).
\begin{align} &Gal(Q(v)/Q)=C_4 =\{\rho_{1}, \rho_{2},\rho_{3}, \rho_{4}\} \notag \\ \notag \\ &\qquad C_2=\{\rho_{1}, \rho_{4}\} \qquad e=\{\rho_{1}\} \notag \\ \notag \\ &Composition \ series \ of \ Galois \ group \ C_4 \notag \\ & \quad [ \ C_4 \ \rhd \ C_2 \ \rhd \ e \ ] \\ &\qquad \qquad \Downarrow \notag \\ &Cyclic \ extensions \notag \\ &\quad [ \ C_4/C_2 \ \rhd \ e \ ] \rightarrow [ \ C_2/e \ \rhd \ e \ ] \\ \end{align}
[6-4] Calculation of \(Q_1/Q\): finding the minimal polynomial \(g_1(x)\) (1)
Following the standard procedure step by step will lead us to the answer. Little further explanation is needed. Substituting (3.6) into (4.1)(4.2), we compute \(\{h_0,h_1\}\), and then substituting into (4.3) we compute \(\{t_0,t_1\}\).\begin{align} h_0&=\prod_{\sigma_i \in \ C_2}\sigma_i(x-v)=(x-v_1)(x-v_4 )=x^2-(v^3+v^2)x+1 \\ h_1&=\prod_{\sigma_i \in \ (C_4-C_2)}\sigma_i(x-v)=(x-v_2)(x-v_3)=x^2-(v^3+v^2)x+1 \\ \notag \\ \end{align} \begin{align} \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}^{2}}+\frac{x}{2}+1\\ (v^3+v^2+\frac{1}{2})x \\ \end{bmatrix}\\ \end{align}
As usual, let us denote the leading coefficient of \(t_1\) in (4.3) by \(cd_m\). Using \(cd_m\), we introduce a new adjoint element \(a_1\). Since this is a quadratic cyclic extension, we compute \(cd_m^2\).
\begin{align} &t_1=cd_m \cdot q_1(x) \ \in Q(v)[x] \notag\\ &cd_m=(v^3+v^2+\tfrac{1}{2}) \ \in Q(v) \qquad q_1(x)=x \ \in Q[x]\\ \end{align}
\begin{align} \bbox[#FFFF00]{ cd_m^2 } &=(v^3+v^2+\tfrac{1}{2})^2=v^6+2v^5+v^4+v^3+v^2+\tfrac{1}{4}= \tfrac{5}{4} \bbox[#FFFF00]{ \equiv A_1 } \ \in Q \\ \end{align}
[step2] The quadratic equation \(B_1(x)=0\) and generating the new adjoint element \(a_1\)
\begin{align}
&\left\{
\begin{array}{l}
t_1=cd_m \cdot q_1(x) \ \in Q(v)[x]\\
\\
cd_m=(v^3+v^2+\tfrac{1}{2}) \ \in Q(v) \\
q_1(x)=x \ \in Q[x]\\
\end{array}
\right.
\quad \Rightarrow \quad
\left\{
\begin{array}{l}
B_1(x)=x^2-A_1=0 \\
a_1=\sqrt {A_1} \ \in Q(a_1) \equiv Q_1\\
\\
\tilde{t_1}=a_1 \cdot q_1(x) \ \in Q_1[x] \\
\end{array}
\right. \\
\end{align}