Techniques for Solving Solvable Equations via Galois Theory

⇦ \(\quad \)

home \(\quad \)

[2-1] The equation \(f(x)=x^3+3x+1\) and nearby facts

The aim of Chapter 2 is to use Galois theory to determine the three roots \(\{\alpha,\beta,\gamma\}\) of the following equation.

The coefficients of \(f(x)\) lie in the field of rational numbers \(Q\), but in the computations to come we will need a cube root of unity \(\omega\). Thus we adjoin \(\omega\) to the field of rational numbers \(Q\) in advance. In this chapter we therefore take the base field \(F_0\) to be \( \boldsymbol{F_0 \equiv Q(\omega)}\) . We write \(F_0[x]\) for the ring of polynomials over the base field \(F_0\).

\begin{align} f(x)&=(x-\alpha)(x-\beta)(x-\gamma)= x^3+3x+1=0 \qquad \in F_0[x] \\ \end{align}

Galois introduced the notion of a primitive element defined below in (1.2) using the three roots of \(f(x)\). The coefficients \(\{1,2,3\}\) used with \(\{\alpha,\beta,\gamma\}\) may be any distinct numbers.

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

Next we take a brief detour to derive relations that will be needed later. Using the three expressions \(\{ \ (x-\alpha), \ (x-\beta),(x \ -\gamma) \ \}\), we successively divide \(f(x)\) as follows.

\begin{align} f(x)&=x^3+3x+1 \notag \\ \notag \\ f(x)&=(x-\alpha)(x^2+\alpha x+\alpha^2+3)+(\alpha^3+3 \alpha +1) \notag \\ &=(x-\alpha)q_1(x)+r_1\\ \notag \\ q_1(x)&=(x-\beta)( x+\alpha+\beta )+(\beta^2+\alpha \beta +\alpha^2+3) \notag \\ &=(x-\beta)q_2(x)+r_2\\ \notag \\ q_2(x)&=(x-\gamma) \cdot 1+(\alpha+\beta+\gamma) \notag \\ &=(x-\gamma)q_3(x)+r_3\\ \end{align}

Since the roots of \(f(x)\) are \(\{\alpha,\beta,\gamma\}\), the remainders \(\{r_1,r_2,r_3\}\) at each step above must also be zero.
Hence (1.6-8) hold. These are very important relations among \(\{\alpha,\beta,\gamma\}\) deduced from the Remainder Theorem.

\begin{align} f(\alpha)&=0 & &\rightarrow & r_1&=\alpha^3+3 \alpha +1=0\\ &\Downarrow & & & & \notag \\ \therefore \ q_1(\beta)&=0 & &\rightarrow & r_2&=\beta^2+\alpha \beta +\alpha^2+3=0\\ &\Downarrow & & & & \notag \\ \therefore \ q_2(\gamma)&=0 & &\rightarrow & r_3&=\alpha+\beta+\gamma=0\\ \end{align}

[1-2] Preparations toward a Galois extension

The point of introducing the primitive element \(v\) is to adjoin \(v\) to the base field \(F_0\) and form the simple extension \(F_0(v)\). Moreover, our intention is to make this simple extension \(F_0(v)\) into a Galois extension.
For \(F_0(v)\) to be a Galois extension, the following two requirements must be satisfied.

[1] \(F_0(v)\) is a separable extension; that is, the minimal polynomial of \(v\) has no multiple 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 analyze these two conditions, we need to construct the minimal polynomial of the primitive element \(v\) that generates the simple extension \(F_0(v)\).
To that end we consider the following polynomial. In the formula, \(\sigma_i\) are the elements of the symmetric group \(S_3\) acting by permutations on \(\{ \alpha,\beta,\gamma\}\).

\begin{align} V(x)&\equiv \displaystyle \prod_{i=1}^6\sigma_i(x-v)=(x-v_{1})(x-v_{2})(x-v_{3})(x-v_{4})(x-v_{5})(x-v_{6}) \\ \end{align}

As is well known, the elements \(\sigma_i\) of \(S_3\) are defined by the following expressions.

\begin{align} &\left\{ \begin{array}{l} &\sigma_{1}=\begin{pmatrix} 1&2&3 \\ 1&2&3 \end{pmatrix}=\begin{pmatrix} \alpha&\beta&\gamma \\ \alpha&\beta&\gamma \end{pmatrix} & &\sigma_{2}=\begin{pmatrix} 1&2&3 \\ 1&3&2 \end{pmatrix}=\begin{pmatrix} \alpha&\beta&\gamma \\ \alpha&\gamma&\beta \end{pmatrix} \\ &\sigma_{3}=\begin{pmatrix} 1&2&3 \\ 2&1&3 \end{pmatrix}=\begin{pmatrix} \alpha&\beta&\gamma \\ \beta&\alpha&\gamma \end{pmatrix} & &\sigma_{4}=\begin{pmatrix} 1&2&3 \\ 2&3&1 \end{pmatrix}=\begin{pmatrix} \alpha&\beta&\gamma \\ \beta&\gamma&\alpha \end{pmatrix} \\ &\sigma_{5}=\begin{pmatrix} 1&2&3 \\ 3&1&2 \end{pmatrix}=\begin{pmatrix} \alpha&\beta&\gamma \\ \gamma&\alpha&\beta \end{pmatrix} & &\sigma_{6}=\begin{pmatrix} 1&2&3 \\ 3&2&1 \end{pmatrix}=\begin{pmatrix} \alpha&\beta&\gamma \\ \gamma&\beta&\alpha \end{pmatrix} \\ \end{array} \right.\\ \end{align}

The permutation action of the elements \(\sigma_i\) of the symmetric group \(S_3\) applies only to \(\{ \alpha,\beta,\gamma\}\).
However, since \(v\) is related to \(\{ \alpha,\beta,\gamma\}\) in (1.2), we can consider the action of \(\sigma_i\) on \(v\).
As an example, we record below the action of \(\sigma_4\) on \(v\).

\begin{align} \sigma_4(v)&=\sigma_4(\alpha+2\beta+3\gamma) =\sigma_4(\alpha)+2\sigma_4(\beta)+3\sigma_4(\gamma) =\beta+2\gamma+3\alpha \notag \\ \end{align}

Applying every element \(\sigma_i\) of \(S_3\) to the primitive element \( v\) yields the following. Here we define \(\sigma_i(v) \equiv v_i \quad [ \ i=1,2,...6 \ ]\).

\begin{align} \sigma_{1}(v)=\alpha+2\beta+3\gamma &\equiv v_1 & \sigma_{2}(v)=\alpha+2\gamma+3\beta &\equiv v_2 \notag \\ \sigma_{3}(v)=\beta+2\alpha+3\gamma &\equiv v_3 & \sigma_{4}(v)=\beta+2\gamma+3\alpha &\equiv v_4 \\ \sigma_{5}(v)=\gamma+2\alpha+3\beta &\equiv v_5 & \sigma_{6}(v)=\gamma+2\beta+3\alpha &\equiv v_6 \notag \\ \end{align}

[Supplement] Overview of Chapter 2

Let us briefly explain Figure 2-1.
This figure gives an overview of the computations to come.

In fact, it will turn out that the simple extension \(F_0(v)\) is a Galois extension of \(F_0\) and that it is the same field as \(F_2\) shown on the left.
In that extension process,

\(F_0 \ \rightarrow \ F_1 \ \rightarrow \ F_2 \ (\equiv F_0(v) ) \)

the field is enlarged in two steps.
The left figure shows how this splits into two cyclic extensions. The two green parts indicate the computations of those split cyclic extensions.

Incidentally, the light blue part is the portion of Chapter 1 where we found \(\omega\).

⇦ \(\quad \)

home \(\quad \)

⇨