Techniques for Solving Solvable Equations via Galois Theory

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[2-10] Computing \(F_1/F_0\): finding the minimal polynomial \(g_1(x)\) (2)

In the previous section we reasoned using only the mapping behavior of the automorphisms \(\rho_i\). In this section we perform explicit calculations to confirm those results. We substitute the polynomial expressions for \(v_i\) into (9.1–2).

\begin{align} h_0&=\prod_{\rho_i \in \kappa_1 }\rho_i(x-v)=(x-v_1)(x-v_4)(x-v_5)\notag \\ &=\left( x-v\right) \, \left( x-\frac{{{v}^{4}}}{6}-\frac{5 {{v}^{2}}}{2}+\frac{v}{2}-6\right) \, \left( x+\frac{{{v}^{4}}}{6}+\frac{5 {{v}^{2}}}{2}+\frac{v}{2}+6\right) \notag \\ &= \ x^3+9x-v^3-9v \qquad (mod \ g_0(v))\\ \notag \\ h_1&=\prod_{\rho_i \in \ \kappa_2}\rho_i(x-v)=(x-v_2)(x-v_3)(x-v_6)\notag \\ &=\left( x+v\right) \, \left( x-\frac{{{v}^{4}}}{6}-\frac{5 {{v}^{2}}}{2}-\frac{v}{2}-6\right) \, \left( x+\frac{{{v}^{4}}}{6}+\frac{5 {{v}^{2}}}{2}-\frac{v}{2}+6\right) \notag \\ &= \ x^3+9x+v^3+9v \qquad (mod \ g_0(v))\\ \end{align}

Next we substitute \(\{h_0,h_1\}\) into (9.3) to obtain \(\{t_0,t_1\}\) as follows.

\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} \frac{(h_0+h_1)}{2} \\ \frac{(h_0-h_1)}{2} \end{bmatrix} = \begin{bmatrix} x^3+9x \\ -v^3-9v \end{bmatrix}\\ \notag \\ &\therefore \quad t_0=x^3+9x \ \in F_0[x] \qquad t_1= -v^3-9v \in F_0(v)\\ \end{align}

We find that the result (10.4) agrees with (9.6) obtained in the previous section.
Next, let us check what numerical values arise in the concrete computation corresponding to (9.7).

Step 2 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=-135 \in \ F_0, \\ a_1=\sqrt{A_1} \ \in \ F_0(a_1) \equiv \ F_1 \\ \end{array} \right. \notag \\ \end{align}

If we compute \(t_1^2\) explicitly, we find that \(t_1^2\) is a number in \(F_0\) as in (10.5). This again matches the conclusion (9.7). We denote this value by \(A_1\).

\begin{align} &t_1^2=(-v^3-9v)^2=v^6+18v^4+81v^2=-135=A_1 \ \in F_0 \quad (mod \ g_0(v))\\ \end{align}

The identity (10.5) means precisely that \(t_1^2=A_1\). Equivalently, as in (10.6), \(t_1\) is a radical (power-root) of the binomial equation \(B_1(x)=0\).
We write this radical as \(\sqrt{A_1} \equiv a_1\) and adjoin \(a_1\) to the base field \(F_0\). We call the resulting extension field \(F_0(a_1)\) by the name \(F_1\). This operation is what is called a power-root extension (also, a cyclic extension).

Since \(a_1\ (\equiv t_1)\) has been defined as an element of the extension field \(F_1\), from here on we deliberately write \(t_1\) as \(\tilde{t_1}\).

\begin{align} &B_1(x)=x^2-A_1=0 \quad \ \therefore \ t_1=\sqrt{-135}= \sqrt{A_1} \equiv a_1 \\ \notag \\ &\bbox[#FFFF00]{ \tilde{t_1}=a_1 \ \in F_0(a_1) \equiv F_1 } \end{align}

For clarity we restate the pair \(\{ \ t_0, \tilde{t_1}\}\) obtained so far.

\begin{align} &\left\{ \begin{array}{l} t_0=x^3+9x \ \in F_0[x] \\ \tilde{t_1}=a_1 \ \in F_1 \\ \end{array} \right.\\ \end{align}

Using \(\{ \ t_0, \tilde{t_1}\}\), we now compute the ILRT, the inverse transform of (9.3).

Step 3 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+9x+a_1\\ x^3+9x-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} \ \in \ F_1[x] \\ \end{array} \right. \\ \end{align}

Note that \(\{ \ h_0,h_1 \ \}\) are polynomials over the simple extension \(F_0(v)\), as displayed in (10.1)–(10.2). However, since we now use \(\tilde{t_1}\) with the adjoined \(a_1\), the pair \(\{ \ \tilde{h_0},\tilde{h_1} \ \}\) becomes polynomials over \(F_1\).

Furthermore, the minimal polynomial \(g_0(x)\) of \(v\) over \(F_0\) factors over \(F_1\) as \([\ g_0(x)=\tilde{h_0}\cdot\tilde{h_1}\ ]\). Since \(\tilde{h_0}\) originally had \((x-v)\) as a factor, it is natural to take the minimal polynomial of \(v\) over the extension field \(F_1\) to be \(g_1(x)=\tilde{h_0}\).

To summarize this section: by introducing the new adjoined element \(a_1\), we factored the sextic minimal polynomial \(g_0(x)\) and obtained the cubic minimal polynomial \(g_1(x)\).

\begin{align} & t_1^2=-135=A_1 \quad \Rightarrow \quad B_1(x)=x^2-A_1=0\\ \notag \\ &\tilde{t_1} \equiv a_1= \sqrt{A_1} \ \in F_1 \\ \notag \\ &g_1(x)=x^3+9x+a_1 \ \in F_1[x]\\ \end{align}

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