Techniques for Solving Solvable Equations via Galois Theory

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[5-7] The resultant \(Res(f(x+s \cdot v),g(v),v)\)

This time we consider the case \(s \neq 1\), where the algebraic steps show up more clearly.
General statements can be hard to follow, so we analyze the Chapter 2 example \(f(x)=x^3+3x+1\).
In this case, if we take \(s=\pm 1\), the resultant is not squarefree, so we will compute with \(s=2\).
First, regard \(f(x+s \cdot v)\) as a polynomial in \(v\) and rewrite it so that the roots are explicit, as in (7.3).
From Chapter 2 we have \(Gal(F_0(v)/F_0)=S_3\). The roots of \(g(v)\) are of course \(\{v_1,v_2,\dots,v_6\}\).

\begin{align} f(x)&=(x-\alpha)(x-\beta)(x-\gamma) \\ \Downarrow \notag \\ f(x+s \cdot v)&=(x+s \cdot v-\alpha)(x+s \cdot v-\beta)(x+s \cdot v-\gamma) \\ \Downarrow \notag \\ f(s \cdot v+x)&=(s \cdot v+x-\alpha)(s \cdot v+x-\beta)(s \cdot v+x-\gamma-x) \notag \\ &=s^3 \cdot \bigl( v-\bbox[#FFFF00]{\frac{(\alpha-x)}{s}} \bigr) \bigl( v-\bbox[#FFFF00]{\frac{(\beta-x)}{s}} \bigr) \bigl(v-\bbox[#FFFF00]{ \frac{(\gamma-x)}{s}} \bigr) \\ \notag \\ g(v)&=(v-\bbox[#FFFF00]{v_1})(v-\bbox[#FFFF00]{v_2})(v-\bbox[#FFFF00]{v_3})(v-\bbox[#FFFF00]{v_4})(v-\bbox[#FFFF00]{v_5})(v-\bbox[#FFFF00]{v_6})\\ \notag \\ \end{align}

When \(f(s \cdot v +x)\) and \(g(v)\) are both viewed with \(v\) as the main variable, their roots are the yellow-marked parts in (7.3) and (7.4). Substitute these into \(\{ \alpha_i,\beta_j \}\) in the defining formula (7.5). Up through (7.7) the transformations should be straightforward to follow.

\begin{align} &Res(f(x+s \cdot v),g(v),v)=a^m \cdot b^n \prod_{i=1}^{n}\prod_{j=1}^{m}(\alpha_i-\beta_j) \\ \end{align}

\begin{align} &\{ \ n=deg(f(x))=3, \ m=deg(g(v))=6 \ \}, \quad \{ \ a \leftarrow s^3, \quad b \leftarrow 1 \ \} \notag \\ \notag \\ &\biggl\{\alpha_1 \leftarrow \frac{(\alpha-x)}{s} , \quad \alpha_2 \leftarrow \frac{(\beta-x)}{s} , \quad \alpha_3 \leftarrow \frac{(\gamma-x)}{s} \biggr\}, \quad \{ \ \beta_i \leftarrow v_i, \quad [ \ i=1,2,..,6 \ ] \ \} \notag \\ \end{align}

\begin{align} &\quad \Downarrow \notag \\ Res(f(x &+s \cdot v),g(v) ,v) \notag \\ =(s^3)^6 &\times \biggl(\frac{(\alpha-x)}{s}-v_1\biggr)\biggl(\frac{(\beta-x)}{s}-v_1\biggr)\biggl(\frac{(\gamma-x)}{s}-v_1\biggr) \notag \\ &\times \biggl(\frac{(\alpha-x)}{s}-v_2\biggr)\biggl(\frac{(\beta-x)}{s}-v_2\biggr)\biggl(\frac{(\gamma-x)}{s}-v_2\biggr) \notag \\ &\times \biggl(\frac{(\alpha-x)}{s}-v_3\biggr)\biggl(\frac{(\beta-x)}{s}-v_3\biggr)\biggl(\frac{(\gamma-x)}{s}-v_3\biggr) \notag \\ &\times \biggl(\frac{(\alpha-x)}{s}-v_4\biggr)\biggl(\frac{(\beta-x)}{s}-v_4\biggr)\biggl(\frac{(\gamma-x)}{s}-v_4\biggr) \notag \\ &\times \biggl(\frac{(\alpha-x)}{s}-v_5\biggr)\biggl(\frac{(\beta-x)}{s}-v_5\biggr)\biggl(\frac{(\gamma-x)}{s}-v_5\biggr) \notag \\ &\times \biggl(\frac{(\alpha-x)}{s}-v_6\biggr)\biggl(\frac{(\beta-x)}{s}-v_6\biggr)\biggl(\frac{(\gamma-x)}{s}-v_6\biggr) \\ &\quad \Downarrow \notag \\ Res(f(x &+s \cdot v),g(v),v)\notag \\ &=\left\{\bbox[#FFFF00]{(x+s \cdot v_1-\alpha)}\bbox[#7FFF00]{(x+s \cdot v_1-\beta)}\bbox[#00FFFF]{(x+s \cdot v_1-\gamma)} \right \} \notag \\ & \times \left\{\bbox[#FFFF00]{(x+s \cdot v_2-\alpha)}\bbox[#00FFFF]{(x+s \cdot v_2-\beta)}\bbox[#7FFF00]{(x+s \cdot v_2-\gamma)} \right \} \notag \\ & \times \left\{\bbox[#7FFF00]{(x+s \cdot v_3-\alpha)}\bbox[#FFFF00]{(x+s \cdot v_3-\beta)}\bbox[#00FFFF]{(x+s \cdot v_3-\gamma)} \right \} \\ & \times \left\{\bbox[#00FFFF]{(x+s \cdot v_4-\alpha)\bbox[#FFFF00]{(x+s \cdot v_4-\beta)}\bbox[#7FFF00]{(x+s \cdot v_4-\gamma)}} \right \} \notag \\ & \times \left\{\bbox[#7FFF00]{(x+s \cdot v_5-\alpha)}\bbox[#00FFFF]{(x+s \cdot v_5-\beta)}\bbox[#FFFF00]{(x+s \cdot v_5-\gamma)} \right \} \notag \\ & \times \left\{\bbox[#00FFFF]{(x+s \cdot v_6-\alpha)}\bbox[#7FFF00]{(x+s \cdot v_6-\beta)}\bbox[#FFFF00]{(x+s \cdot v_6-\gamma)} \right \} \notag \\ \notag \\ \end{align}

Let the three linear factors in the first line of (7.7) be \(\{Y_1,Y_2,Y_3\}\) in order. For these \(\{Y_i\}\), collect the expressions obtained by acting with all elements of the Galois group \(S_3\), and define their products to be \(\{H_1,H_2,H_3\}\) as in (7.9–11). Since \(\{H_i\}\) are invariant under the \(S_3\)-action, it follows, as shown in (7.13), that \(\{H_i\}\) are polynomials in \(F_0[x]\) as well.

\begin{align} Y_1=\bbox[#FFFF00]{(x+s \cdot v_1- \alpha)} &\quad Y_2=\bbox[#7FFF00]{(x+s \cdot v_1-\beta)} \quad Y_3=\bbox[#00FFFF]{(x+s \cdot v_1-\gamma) } \\ \notag \\ H_1=\prod_{i=1}^{6} \rho_i \bigl(Y_1\bigr) &= \bbox[#FFFF00]{(x+s \cdot v_1-\alpha)(x+s \cdot v_2-\alpha)(x+s \cdot v_3-\beta)} \notag \\ &\times \bbox[#FFFF00]{(x+s \cdot v_4-\beta)(x+s \cdot v_5-\gamma)(x+s \cdot v_6-\gamma)} \\ H_2=\prod_{i=1}^{6} \rho_i \bigl(Y_2\bigr) &=\bbox[#7FFF00]{(x+s \cdot v_1-\beta)(x+s \cdot v_2-\gamma)(x+s \cdot v_3-\alpha)} \notag \\ &\times \bbox[#7FFF00]{ (x+s \cdot v_4-\gamma)(x+s \cdot v_5-\alpha)(x+s \cdot v_6-\beta)} \\ H_3=\prod_{i=1}^{6} \rho_i \bigl(Y_3\bigr) &= \bbox[#00FFFF]{(x+s \cdot v_1-\gamma)(x+s \cdot v_2-\beta)(x+s \cdot v_3-\gamma)} \notag \\ &\times \bbox[#00FFFF]{ (x+s \cdot v_4-\alpha)(x+s \cdot v_5-\beta)(x+s \cdot v_6-\alpha)} \\ \notag \\ &\quad \Downarrow \notag \\ \notag \\ Res( \ f(x +s \cdot v), \ &g(v), \ v)=H_1 \cdot H_2 \cdot H_3 \\ \notag \\ \rho_i\bigl(H_j\bigr)=H_j \quad [ \ &\rho_i \in S_3 \ ],\ [j=1,2,3] \quad \Rightarrow \quad \therefore \ \{H_1,H_2,H_3\} \ \in \ F_0[x] \\ \end{align}

The product \(H_1 \cdot H_2 \cdot H_3\) coincides with (7.7). This also means, as indicated in (7.12), that \( Res(f(x +s \cdot v),g(v) ,v)\) factors into three polynomials in \(F_0[x]\).
The computations above correspond to the transformations from (4.1) to (4.6) in Section [5-4]. Once we have reached this point, the logical development corresponding to Section [5-5] need not be repeated here, so we omit it.

Below we actually compute the resultant with \(s=2\).

\begin{align} &\left\{ \begin{array}{l} f(x+2 \cdot v)=(x+2v)^3+3(x+2v)+1 =x^3+6vx^2+3(4v^2+1)x+8v^3+6v+1\\ g(v)=v^6+18v^4+81v^2+135 \\ \end{array} \right. \notag \\ \notag \\ &\qquad \Downarrow \notag \\ \notag \\ &Res(f(x +2 \cdot v),g(v),v)=R_1(x) \cdot R_2(x) \cdot R_3(x) \\ \notag \\ &\left\{ \begin{array}{l} R_1(x)=x^6+42x^4+20x^3+441x^2+420x+1315 \\ R_2(x)=x^6+78x^4-70x^3+1521x^2-2730x+6085 \\ R_3(x)=x^6+114x^4+56x^3+3249x^2+3192x+31159 \\ \end{array} \right. \\ \notag \\ &\qquad \Downarrow \notag \\ \notag \\ &\left\{ \begin{array}{l} \tilde{Y}_1(x) \equiv x+2v-x_1=GCD(R_1(x),f(x+2v))=x-\frac{{{v}^{4}}}{18}-\frac{5 {{v}^{2}}}{6}+\frac{3 v}{2}-2 \\ \tilde{Y}_2(x) \equiv x+2v-x_2=GCD(R_2(x),f(x+2v))=x+\frac{{{v}^{4}}}{9}+\frac{5 {{v}^{2}}}{3}+2 v+4 \\ \tilde{Y}_3(x) \equiv x+2v-x_3=GCD(R_3(x),f(x+2v))=x-\frac{{{v}^{4}}}{18}-\frac{5 {{v}^{2}}}{6}+\frac{5 v}{2}-2 \\ \end{array} \right. \\ \notag \\ &\qquad \Downarrow \notag \\ \notag \\ &f(x)=\tilde{Y}_1(x-2 \cdot v) \cdot \tilde{Y}_2(x-2 \cdot v) \cdot \tilde{Y}_3(x-2 \cdot v)=(x-x_1)(x-x_2)(x-x_3) \\ \notag \\ &\biggl[ \ x_1=\frac{{{v}^{4}}}{18}+\frac{5 {{v}^{2}}}{6}+\frac{v}{2}+2, \quad x_2=-\frac{{{v}^{4}}}{9}-\frac{5 {{v}^{2}}}{3}-4, \quad x_3=\frac{{{v}^{4}}}{18}+\frac{5 {{v}^{2}}}{6}-\frac{v}{2}+2 \ \biggr] \\ \end{align}

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