Techniques for Solving Solvable Equations via Galois Theory

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[3-1] The equation \(f(x)=x^4+4x+2\) and its surroundings

The computations in Chapter 3 follow exactly the same procedure as those in Chapter 2. However, because the coefficients of the polynomials that appear in the process become much more complicated, we will omit many of the intermediate results here. The logical structure of the calculations is completely the same as in Chapter 2, so you can refer back to that.

The purpose of Chapter 3 is to use Galois theory to find the four roots \(\{\alpha,\beta,\gamma,\delta\}\) of the following equation:
The coefficients of \(f(x)\) lie in the field of rational numbers \(Q\). But in the course of solving the equation we will need the cube root of unity \(\omega\), so we must adjoin it in advance. Thus, in this chapter we take the base field to be \( \boldsymbol{F_0 \equiv Q(\omega)}\) . We write \(F_0[x]\) for the set of polynomials over the base field \(F_0\).

\begin{align} f(x)&=(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)= x^4+4x+2=0 \qquad \in F_0[x] \\ \end{align}

Next we divide \(f(x)\) successively by the four factors \(\{(x-\alpha),(x-\beta),(x-\gamma),(x-\delta)\}\) as follows:

\begin{align} &f(x)=x^4+4x+2 =(x-\alpha)(x-\beta)(x-\gamma)(x-\delta) \\ \notag \\ &\left\{ \begin{array}{l} f(x)=(x-\alpha)q_1(x)+r_1 \\ q_1(x)=x^3+{{x}^{2}}\alpha+x {{\alpha }^{2}}+ {{\alpha }^{3}} +4\\ r_1={{\alpha }^{4}}+4 \alpha +2 \end{array} \right.\\ \notag \\ &\left\{ \begin{array}{l} q_1(x)=(x-\beta)q_2(x)+r_2 \\ q_2(x)={{x}^{2}}+x\, \left( \beta +\alpha \right)+{{\beta }^{2}} +\alpha \beta +{{\alpha }^{2}} \\ r_2= {{\beta }^{3}}+\alpha {{\beta }^{2}}+{{\alpha }^{2}} \beta +{{\alpha }^{3}}+4 \end{array} \right.\\ \notag \\ &\left\{ \begin{array}{l} q_2(x)=(x-\gamma)q_3(x)+r_3 \\ q_3(x)=x+\gamma +\beta +\alpha \\ r_3= {{\gamma }^{2}}+\left( \beta +\alpha \right) \gamma +{{\beta }^{2}}+\alpha \beta +{{\alpha }^{2}}\\ \end{array} \right.\\ \notag \\ &\left\{ \begin{array}{l} q_3(x)=(x-\delta)q_4(x)+r_4 \\ q_4(x)=1 \\ r_4=\delta +\gamma +\beta +\alpha \\ \end{array} \right.\\ \end{align}

From these computations we see that each of \(\{f(x), q_1(x), q_2(x), q_3(x)\}\) has one of \(\{\alpha,\beta,\gamma,\delta\}\) as a root, so the following relations hold.
Each of these formulas in (1.7) comes from the Remainder Theorem , and they give important relations among \(\{\alpha,\beta,\gamma,\delta\}\).

\begin{align} &\left\{ \begin{array}{l} &f(\alpha)=0 \ \Rightarrow \ r_1={{\alpha }^{4}}+4 \alpha +2=0 \\ &q_1(\beta)=0 \ \Rightarrow \ r_2={{\beta }^{3}}+\alpha {{\beta }^{2}}+{{\alpha }^{2}} \beta +{{\alpha }^{3}}+4=0 \\ &q_2(\gamma)=0 \ \Rightarrow \ r_3= {{\gamma }^{2}}+\left( \beta +\alpha \right) \gamma +{{\beta }^{2}}+\alpha \beta +{{\alpha }^{2}}=0\\ &q_3(\delta)=0 \ \Rightarrow \ r_4=\delta +\gamma +\beta +\alpha=0 \\ \end{array} \right.\\ \end{align}

Let us return to the main discussion. Galois introduced the concept of a **primitive element** (see (1.8)) defined in terms of the four roots of \(f(x)\). The coefficients \(\{1,2,3,4\}\) here can in fact be any distinct numbers.

\begin{align} v &\equiv 1 \cdot \alpha +2 \cdot \beta +3 \cdot \gamma +4 \cdot \delta \quad: \ primitive \ element \\ \end{align}

[3-2] Introduction of the symmetric group \(S_4\)

The reason we introduced the primitive element is that we want to adjoin \(v\) to the base field \(F_0\) and thereby form the simple extension \(F_0(v)\). Ultimately, we want this simple extension \(F_0(v)\) to be a **Galois extension**. But for \(F_0(v)\) to be a Galois extension, it must satisfy the following two requirements:

[1] \(F_0(v)\) is a separable extension, that is, the minimal polynomial of \(v\) has no repeated roots.
[2] \(F_0(v)\) is a normal extension, that is, \(F_0(v)\) contains all the roots of the minimal polynomial of \(v\).

To meet these conditions we first need to construct the minimal polynomial of the primitive element \(v\). For this purpose we consider the following polynomial. Here each \(\sigma_i\) is an element of the symmetric group \(S_4\), which acts as a permutation on \(\{\alpha,\beta,\gamma,\delta\}\).

\begin{align} V(x)&\equiv \displaystyle \prod_{i=1}^{24}\sigma_i(x-v)=(x-v_{1})(x-v_{2})(x-v_{3})....(x-v_{23})(x-v_{24}) \\ \end{align}

As is well known, the elements \(\sigma_i\) of \(S_4\) can be defined by permutation notation. To save space, instead of using two-line notation as in (2.2), we write the one-line notation highlighted in yellow.

\begin{align} \sigma_{10}=\begin{pmatrix} 1&2&3&4 \\ 2&3&4&1 \end{pmatrix}=\begin{pmatrix} \alpha&\beta&\gamma&\delta \\ \beta &\gamma&\delta&\alpha \end{pmatrix} \equiv \ \bbox[#FFFF00]{ [2,3,4,1] }\\ \notag \\ \end{align}

\begin{align} \sigma_{1}=&[1,2,3,4] & \sigma_{2}=&[1,2,4,3] & \sigma_{3}=&[1,3,2,4] & \sigma_{4}=&[1,3,4,2] \notag \\ \sigma_{5}=&[1,4,2,3] & \sigma_{6}=&[1,4,3,2] & \sigma_{7}=&[2,1,3,4] & \sigma_{8}=&[2,1,4,3] \notag \\ \sigma_{9}=&[2,3,1,4] & \sigma_{10}=& \bbox[#FFFF00]{[2,3,4,1] } & \sigma_{11}=&[2,4,1,3] & \sigma_{12}=&[2,4,3,1] \\ \sigma_{13}=&[3,1,2,4] & \sigma_{14}=&[3,1,4,2] & \sigma_{15}=&[3,2,1,4] & \sigma_{16}=&[3,2,4,1] \notag \\ \sigma_{17}=&[3,4,1,2] & \sigma_{18}=&[3,4,2,1] & \sigma_{19}=&[4,1,2,3] & \sigma_{20}=&[4,1,3,2] \notag \\ \sigma_{21}=&[4,2,1,3] & \sigma_{22}=&[4,2,3,1] & \sigma_{23}=&[4,3,1,2] & \sigma_{24}=&[4,3,2,1] \notag \\ \end{align}

The permutation action of the elements \(\sigma_i\) of \(S_4\) is only on \(\{\alpha,\beta,\gamma,\delta\}\).
But since \(v\) is defined in (1.8) in terms of these roots, we can also consider the action on \(v\). As an example, here is the action of \(\sigma_9\) on \(v\):

\begin{align} \sigma_9(v)&=\sigma_9(\alpha+2\beta+3\gamma+4\delta) =\sigma_9(\alpha)+2\sigma_9(\beta)+3\sigma_9(\gamma)+4\sigma_9(\delta) =\beta+2\gamma+3\alpha+4\delta \notag \\ \end{align}

Applying every permutation \(\sigma_i\) in \(S_4\) to the primitive element \(v\) produces the following results. We define \(\sigma_i(v) \equiv v_i \quad [ \ i=1,2,...24 \ ]\).

\begin{align} \sigma_{1}(v)&=4 \delta +3 \gamma +2 \beta +\alpha \equiv v_1 & \sigma_{2}(v)&=3 \delta +4 \gamma +2 \beta +\alpha \equiv v_2 \\ \sigma_{3}(v)&=4 \delta +2 \gamma +3 \beta +\alpha \equiv v_3 & \sigma_{4}(v)&=3 \delta +2 \gamma +4 \beta +\alpha \equiv v_4 \notag \\ \sigma_{5}(v)&=2 \delta +4 \gamma +3 \beta +\alpha \equiv v_{5} & \sigma_{6}(v)&=2 \delta +3 \gamma +4 \beta +\alpha \equiv v_{6} \notag \\ \sigma_{7}(v)&=4 \delta +3 \gamma +\beta +2 \alpha \equiv v_{7} & \sigma_{8}(v)&=3 \delta +4 \gamma +\beta +2 \alpha \equiv v_{8} \notag \\ \sigma_{9}(v)&=4 \delta +2 \gamma +\beta +3 \alpha \equiv v_{9} & \sigma_{10}(v)&=3 \delta +2 \gamma +\beta +4 \alpha \equiv v_{10} \notag \\ \sigma_{11}(v)&=2 \delta +4 \gamma +\beta +3 \alpha \equiv v_{11} & \sigma_{12}(v)&=2 \delta +3 \gamma +\beta +4 \alpha \equiv v_{12} \notag \\ \sigma_{13}(v)&=4 \delta +\gamma +3 \beta +2 \alpha \equiv v_{13} & \sigma_{14}(v)&=3 \delta +\gamma +4 \beta +2 \alpha \equiv v_{14} \notag \\ \sigma_{15}(v)&=4 \delta +\gamma +2 \beta +3 \alpha \equiv v_{15} & \sigma_{16}(v)&=3 \delta +\gamma +2\beta +4 \alpha \equiv v_{16} \notag \\ \sigma_{17}(v)&=2 \delta +\gamma +4 \beta +3 \alpha \equiv v_{17} & \sigma_{18}(v)&=2 \delta +\gamma +3 \beta +4 \alpha \equiv v_{18} \notag \\ \sigma_{19}(v)&=\delta +4 \gamma +3 \beta +2 \alpha \equiv v_{19} & \sigma_{20}(v)&=\delta +3 \gamma +4 \beta +2 \alpha \equiv v_{20} \notag \\ \sigma_{21}(v)&=\delta +4 \gamma +2 \beta +3 \alpha \equiv v_{21} & \sigma_{22}(v)&=\delta +3 \gamma +2 \beta +4 \alpha \equiv v_{22} \notag \\ \sigma_{23}(v)&=\delta +2 \gamma +4 \beta +3 \alpha \equiv v_{23} & \sigma_{24}(v)&=\delta +2 \gamma +3 \beta +4 \alpha \equiv v_{24} \notag \\ \end{align}

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