Techniques for Solving Solvable Equations via Galois Theory

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[1-1] Preparation

To begin with, in this chapter we will omit most of the mathematical explanations. Please just follow the calculation steps. We want to find the two roots \(\{\alpha,\beta\}\) of the polynomial \(f(x)\) over the field of rational numbers \(Q\). Galois introduced the concept of a primitive element \(v\) using these two roots.

\begin{align} f(x)&=x^2+x+1=0 \quad \in Q[x]\\ &=(x-\alpha)(x-\beta) \quad e_1=\alpha+\beta=-1 \qquad e_2=\alpha \cdot \beta=1 \\ \notag \\ v &\equiv 1 \cdot \alpha +2 \cdot \beta \\ \end{align}

Here, the coefficients \(\{1,2\}\) in (1.3) for \(\{\alpha,\beta\}\) can be any distinct values.
Next, we introduce the symmetric group \(S_2=\{ \ \sigma_1,\sigma_2 \ \}\) and consider the permutations (1.5) of the two roots \(\{\alpha,\beta\}\) of the primitive element \(v\). We define the two transformed values of \(v\) as \(\{v_1,v_2\}\).

\begin{align} \notag \\ \sigma_{1}=\begin{pmatrix} 1&2 \\ 1&2 \end{pmatrix}=\begin{pmatrix} \alpha&\beta \\ \alpha&\beta \end{pmatrix} \quad \sigma_{2}=\begin{pmatrix} 1&2 \\ 2&1 \end{pmatrix}=\begin{pmatrix} \alpha&\beta \\ \beta&\alpha \end{pmatrix} \\ \notag \\ \end{align}

\begin{align} &\left\{ \begin{array}{l} \sigma_{1}(v)=\sigma_1(\alpha+2\beta)= \alpha+2\beta \\ \sigma_{2}(v)=\sigma_2(\alpha+2\beta)= \beta +2\alpha \\ \end{array} \right. \\ \notag \\ &v_1 \equiv \alpha+2\beta \quad v_2\equiv \beta +2\alpha \\ \end{align}

Using \(\{v_1,v_2\}\) in (1.6), we now introduce the Galois resolvent \(V(x)\). Expanding \(V(x)\) and applying the relations between roots and coefficients, we obtain a polynomial over \(Q\) as in (1.8). The reason \(V(x)\) lies in \(Q[x]\) is that it is invariant under the permutations of the roots \(\{\alpha,\beta\}\).

\begin{align} V(x)&\equiv \displaystyle \prod_{i=1}^2\sigma_i(x-v)=(x-v_{1})(x-v_{2}) \\ &=(x-(\alpha+2\beta))(x-(\beta +2\alpha)) =x^2 -3 ( \alpha+ \beta)x +2 {{\alpha }^{2}}+2 {{\beta }^{2}}+5 \alpha \beta \notag \\ &=x^{2}-3e_1 \cdot x+e_2+2e_1^{2} \notag \\ \notag \\ &=x^2+3x+3 \quad \in \ Q[x] \\ \end{align}

The polynomial \(V(x)\) in (1.8) is irreducible over \(Q\), so we define it as the minimal polynomial \(g_0(x)\) of \(v\) (1.9). Also, since \(v \ (=v_1)\) is a root of the minimal polynomial \(g_0(x)\), relation (1.10) follows.

\begin{align} g_0(x) &\equiv V(x)=v^2+3v+3 \\ \notag \\ g_0(v_1)&=g_0(v)=v^2+3v+3=0 \\ \end{align}

Without further explanation: adjoining the root \(v\) of the minimal polynomial \([g_0(x)=0]\) to \(Q\) produces the simple extension \(Q(v)\). This simple extension \(Q(v)\) is an algebraic number field.

[1-2] Introduction of polynomials \(P_{\alpha}(x), \ P_{\beta}(x)\)

Next, to express the two roots \(\{\alpha,\beta\}\) in terms of \(v\), we introduce two polynomials \(P_{\alpha}(x), P_{\beta}(x)\). By using the relations between roots and coefficients, these polynomials can be simplified up to (2.4).

\begin{align} P_\alpha(x)&\equiv V(x)\cdot \Bigl[ \ \sum_{i=1}^2 \sigma_i(\frac{\alpha }{x-v}) \ \Bigr] \quad P_{\beta}(x)\equiv V(x)\cdot \Bigl[ \ \sum_{i=1}^2 \sigma_i(\frac{\beta }{x-v}) \ \Bigr] \\ \end{align} \begin{align} &\Downarrow \notag \\ P_\alpha(x)&=(x-v_{1})(x-v_{2})(\frac{\alpha }{x-{v_1}}+\frac{\beta }{x-{v_2}})=\alpha \cdot (x-v_2)+\beta \cdot (x-v_1) \\ P_{\beta}(x)&=(x-v_{1})(x-v_{2})(\frac{\beta }{x-{v_1}}+\frac{\alpha }{x-{v_2}})=\beta \cdot (x-v_2)+\alpha \cdot (x-v_1) \\ &\Downarrow \notag \\ \notag \\ P_\alpha(x)&=(\alpha+\beta)x-2 {{\alpha }^{2}}-2 {{\beta }^{2}}-2 \alpha \beta \qquad P_\alpha(x)=e_1 \cdot x+2e_2-2e_1^2 \notag \\ P_{\beta}(x)&=(\alpha + \beta)x -{{\alpha }^{2}}-{{\beta }^{2}}-4 \alpha \qquad P_{\beta}(x)= e_1 \cdot x-2e_2-e_1^2 \notag \\ &\Downarrow \notag \\ P_\alpha(x)&=-x \qquad P_{\beta}(x)=-x-3 \quad \in \ Q[x] \\ \end{align}

(Note) It would be enough to compute just one of \(\{P_\alpha(x),P_{\beta}(x)\}\); nevertheless, we compute both here, even if that is a bit roundabout and redundant.

[1-3] Polynomial representation of \(\{\alpha,\beta\}, \{v_1,v_2\}\) in terms of \(v\)

Let us substitute \(v_1(=v)\) for \(x\) in (2.2)(2.3).

\begin{align} P_\alpha(v_1)&=\alpha \cdot (v_1-v_2), \quad P_{\beta}(v_1)=\beta \cdot (v_1-v_2)\\ \notag \\ \therefore \ \alpha&=\frac{ P_\alpha(v_1)}{(v_1-v_2)}, \quad \beta=\frac{P_{\beta}(v_1)}{(v_1-v_2)} \\ \end{align}

The denominator \((v_1-v_2)\) in (3.2) is not a symmetric function of \(\{\alpha,\beta\}\), so it cannot be expressed in \(Q[x]\). Fortunately, as shown in (3.3), it can be replaced by evaluating the derivative \(V'(x)\) of \(V(x)\) at \(v_1\).

\begin{align} V^{'}(x)&=(x-v_{2})+(x-v_{1}) \qquad \therefore \ V^{'}(v_1)=(v_1-v_2)\\ \notag \\ &\Downarrow \notag \\ \therefore \ \alpha&=\left.\frac{P_\alpha(x)}{V'(x)}\right|_{x=v_1} \qquad \beta=\left.\frac{P_\beta(x)}{V'(x)}\right|_{x=v_1} \qquad ( \ \star \ v_1=v \ ) \\ \end{align}

Since \(V'(v)\) appears in the denominator of (3.4), we now consider its inverse. Because the simple extension \(Q(v)\) is a field, inverses exist. We simply state the formula for \(V'(v)^{-1}\) in (3.5) (please accept this for now).

\begin{align} V^{'}(v)^{-1}=-\frac{2}{3}v-1 \\ \end{align}

With (3.5), we can compute \(\{\alpha,\beta\}\) from (3.4).

\begin{align} \alpha&=P_{\alpha}(v) \cdot V^{'}(v)^{-1}=(-v) \cdot (-\frac{2}{3}v-1)=\frac{2 {{v}^{2}}}{3}+v=-v-2 \quad (mod \ g_0(v))\\ \beta&=P_{\beta}(v) \cdot V^{'}(v)^{-1}=(-v-3) \cdot (-\frac{2}{3}v-1)=\frac{2 {{v}^{2}}}{3}+3 v+3=v+1 \quad (mod \ g_0(v))\\ \notag \\ &\therefore \ \alpha=-v-2 \quad \beta=v+1 \\ \end{align}

In these calculations we need remainders modulo \(g_0(v)\). Concretely, as (1.10) shows, it is just a matter of replacing \(v^2\) by \(v^2=-3v+3\). From now on, this remainder calculation will be the most essential step in computations in Galois theory.
Furthermore, substituting these \(\{\alpha,\beta\}\) into (1.6), we obtain polynomial expressions of \(\{v_1,v_2\}\) in terms of \(v\).

\begin{align} v_1=\alpha+2\beta=v \qquad v_2=\beta +2\alpha=-v-3 \\ \end{align}

As a check, note that \(v_2\) in (3.9) is also a root of \(g_0(x)\).

\begin{align} g_0(v_2)&=(-v-3)^2+3(-v-3)+3=v^2+6v+9-3v-9+3 \notag \\ &=v^2+3v+3=0 \quad (mod \ g_0(v)) \\ \end{align}

Thus we confirm that \(v_2\) is the conjugate root of \(v_1=v\). To summarize:

\begin{align} &\left\{ \begin{array}{l} \alpha=-v-2\\ \beta=v+1 \\ \end{array} \right. \qquad \left\{ \begin{array}{l} v_1=v \\ v_2=-v-3 \\ \end{array} \right. \qquad \{\alpha,\beta\}, \{v_1,v_2\} \ \in \ Q(v) \\ \notag \\ &g_0(x)=(x-v_1)(x-v_2)=(x-v)\bigl(x-(-v-3)\bigr) \ \in \ Q(v)[x]\\ \end{align}

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