Techniques of Solving Equations à la Galois
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1st upload: 2023/06/17
revision2 : 2023/07/27
revision3 : 2024/12/22
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[5-2] Computation of \(Res(f(x+v),g(v),v)\)
In general, the resultant can be described as the determinant of the Sylvester matrix.For details on the Sylvester matrix we refer you to textbooks; here we proceed step by step following the Trager Algorithm described in the previous section.
(step.1)(step.2)
We recall \(f(x),g(v)\). Set \(s=1\) in \(f(x+s \cdot v)\). Then \(f(x+v)\) is as follows.
\begin{align} f(x)&=x^3-3x+1 \\ g(v)&=v^3-9v-9 \\ \notag \\ f(x+v)&=(x+v)^3-3(x+v)+1 \notag \\ &=v^3+(3x)v^2+(3x^2-3)v+(x^3-3x-1) \\ \end{align}
Regard \(f(x+v)\) in (2.3) as a polynomial in \(v\). Together with \(g(v)\) in (2.2), arrange the coefficients from degree \(3\) down to \(0\) in \(v\) as in (2.4); this yields the Sylvester matrix (here we denote it by \(SylM\)).
\begin{align} &SylM=\begin{bmatrix} 1 & 3x & (3x^2-3) & (x^3-3x+1) & 0 & 0 \\ 0 & 1 & 3x & (3x^2-3) & (x^3-3x+1) & 0 \\ 0 & 0 & 1 & 3x & (3x^2-3) & (x^3-3x+1) \\ 1 & 0 & -9 & -9 & 0 & 0 \\ 0 & 1 & 0 & -9 & -9 & 0 \\ 0 & 0 & 1 & 0 & -9 & -9 \end{bmatrix}\\ \end{align}
To compute the determinant of the Sylvester matrix (2.4), we use the algebra system maxima command for the resultant, \( Res(f(x+v) ,g(v),v)\). Factoring that determinant, we obtain three polynomials as in (2.6).
\begin{align} &Res(f(x+v) ,g(v),v)=det(SylM) \notag \\ &\qquad =-{{x}^{9}}+36 {{x}^{7}}-30 {{x}^{6}}-351 {{x}^{5}} +396 {{x}^{4}}+1023 {{x}^{3}}-1080 {{x}^{2}}-612 x+296 \\ \notag \\ &\qquad =-(x^3-21x+37)(x^3-12x-8)(x^3-3x+1)=-R_1(x) \cdot R_2(x) \cdot R_3(x)\\ \notag \\ \end{align}
\begin{align} &R_1(x)=x^3-21x+37, \quad R_2(x)=x^3-12x-8, \quad R_3(x)=x^3-3x+1\\ \end{align}
It is easy to anticipate that the computed resultant above will be a polynomial in \(x\). Still, one naturally feels surprise and asks, “Why does it factor into three polynomials over \(F_0\)?” To explain (step.4) we must address this puzzle, so we take a brief detour.
This point is the most important aspect of the Trager Algorithm.
[5-3] Another viewpoint on \(Res(f(x+v),g(v),v)\) (1)
Besides defining the resultant via the determinant of the Sylvester matrix above, we also have the following definition. We now transform the resultant using the defining formula (3.3).\begin{align} f(x)&=a(x-\alpha_1)(x-\alpha_2)....(x-\alpha_n) \\ g(x)&=b(x-\beta_1)(x-\beta_2)....(x-\beta_m) \\ \notag \\ Res &(f(\bbox[#FFFF00]{x}),g(\bbox[#FFFF00]{x}),\bbox[#FFFF00]{x})=a^{m}b^{n}\prod_{i=1}^{n}\prod_{j=1}^{m}(\alpha_i-\beta_j)\\ \end{align}
In (3.3) the eliminated main variable \(x\) is highlighted in yellow. Thus, when computing \(Res(f(x+\bbox[#FFFF00]{v}),g(\bbox[#FFFF00]{v}),\bbox[#FFFF00]{v})\), note that the eliminated main variable is \(v\).
To that end, we must regard the main variable of \(f(x+v)\) as \(v\), not \(x\). Then the roots of \(f(x+v)\) can be viewed as the yellow parts involving \(x\) in (3.6). For \(g(v)\), the main variable is already \(v\), and its roots are \(\{v_1,v_4,v_5\}\), so no transformation is needed.
\begin{align} f(x)&=(x-\alpha)(x-\beta)(x-\gamma) \\ \Downarrow \notag \\ f(x+v)&=(x+v-\alpha)(x+v-\beta)(x+v-\gamma) \\ \Downarrow \notag \\ f(v+x)&=(v+x-\alpha)(v+x-\beta)(v+x-\gamma-x) \notag \\ &=\bigl(v-\bbox[#FFFF00]{(\alpha-x)}\bigr)\bigl(v-\bbox[#FFFF00]{(\beta-x)}\bigr)\bigl(v-\bbox[#FFFF00]{(\gamma-x)}\bigr) \\ \notag \\ g(v)&=(v-\bbox[#FFFF00]{v_1})(v-\bbox[#FFFF00]{v_4})(v-\bbox[#FFFF00]{v_5})\\ \notag \\ \end{align}
\begin{align} &Res(f(\bbox[#FFFF00]{v}+x),g(\bbox[#FFFF00]{v}),\bbox[#FFFF00]{v})=\prod_{i=1}^{3}\prod_{j=1}^{3}(\alpha_i-\beta_j) \\ &\{\alpha_1 \leftarrow (\alpha-x) , \alpha_2 \leftarrow (\beta-x) , \alpha_3 \leftarrow (\gamma-x) \},\{\beta_1 \leftarrow v_1, \beta_2 \leftarrow v_4, \beta_3 \leftarrow v_5 \} \\ \end{align} \begin{align} &\Downarrow \notag \\ Res(f(x+v),g(v),v)=&\bigl((\alpha-x)-v_1\bigr)\bigl((\beta-x)-v_1\bigr)\bigl((\gamma-x)-v_1\bigr) \notag \\ &\bigl((\alpha-x)-v_4\bigr)\bigl((\beta-x)-v_4\bigr)\bigl((\gamma-x)-v_4\bigr) \notag \\ &\bigl((\alpha-x)-v_5\bigr)\bigl((\beta-x)-v_5\bigr)\bigl((\gamma-x)-v_5\bigr)\\ &\Downarrow \notag \\ Res(f(x+v),g(v),v)=(-1)&(x+v_1-\alpha)(x+v_1-\beta)(x+v_1-\gamma) \notag \\ &(x+v_4-\alpha)(x+v_4-\beta)(x+v_4-\gamma) \notag \\ &(x+v_5-\alpha)(x+v_5-\beta)(x+v_5-\gamma) \\ \end{align}
Throughout this chapter there are many places where the use of symbols can be potentially confusing, so we add a few comments just in case.
In (3.4) the \(x\) is the usual main variable. What is \((x+v)\) in (3.5)? It is obtained by substituting \((x+v)\) for \(x\) in (3.4). The main variable in (3.5) is still \(x\).
Then what about the \(v\) in (3.6)? Here the expression is meant to give the impression that “the main variable has changed from \(x\) to \(v\)”.
Why do this? Compare (3.3) and (3.8). In (3.3) the main variable of \(f(x)\) and \(g(x)\) is \(x\), whereas in (3.8) the main variable is \(v\). Therefore, the roots \(\{\alpha_i,\beta_j\}\) appearing on the right-hand side of (3.8) must be taken as roots in the main variable \(v\) for \(\{f(v+x),g(v)\}\). Accordingly, we rewrote (3.5) into the form (3.6) so that it is explicit where the roots of \(f(v+x)\) are. (This may feel a bit silly, admittedly.)
In this way, the correspondence of roots shown in (3.9) arises.
Proceeding like this, the resultant can be computed according to the defining formula (3.8), yielding (3.10). We then further transform it as in (3.11), and at that point it is quite obvious that we now intend to take \(x\) as the main variable.
The discussion of variables continues: in the next section we will even identify the variable \(v\) with the root \(v_1\) of \(g(v)\) without hesitation. Please keep this in mind as you read the explanation.