Techniques of Solving Equations à la Galois
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1st upload: 2023/06/17
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[2-4] Polynomial expressions in \(v\) for \(\{\alpha,\beta,\gamma\}\) and \(v_i\)
As in Chapter 1, to compute the three roots we introduce the following polynomials and transform them in the same way as \(V(x)\).\begin{align} P_{\alpha}(x)&\equiv V(x)\cdot \Bigl[ \ \sum_{i=1}^6 \sigma_i(\frac{\alpha }{x-v}) \ \Bigr] & P_{\beta}(x)&\equiv V(x)\cdot \Bigl[ \ \sum_{i=1}^6 \sigma_i(\frac{\beta }{x-v}) \ \Bigr] \\ P_{\gamma}(x)&\equiv V(x)\cdot \Bigl[ \ \sum_{i=1}^6 \sigma_i(\frac{\gamma }{x-v}) \ \Bigr] & & \notag \\ \end{align}
As a worked example, we carry out reductions for \( P_{\alpha}(x,\alpha,\beta,\gamma)\) in the order \([ r_3 \rightarrow r_2 \rightarrow r_1]\) of (1.6-8), just as in the computation of \(V(x)\).
\begin{align} P_{\alpha}(x,\alpha,\beta,\gamma)&=r_3(\gamma) \cdot Q_{\alpha,1}(x,\alpha,\beta,\gamma)+P_{\alpha,1}(x,\alpha,\beta) \equiv P_{\alpha,1}(x,\alpha,\beta) \qquad( \ \because \ r_3=0 \ ) \notag \\ \notag \\ P_{\alpha,1}(x,\alpha,\beta)&=-12\beta^6-36\alpha\beta^5+(9\alpha^2+27x\alpha+18x^2)\beta^4 \notag \\ &+(78\alpha^3+54x\alpha^2+36x^2\alpha)\beta^3+(9\alpha^4+54x\alpha^3+54x^2\alpha^2-9x^3\alpha-6x^4)\beta^2 \notag \\ &+(-36\alpha^5+27x\alpha^4+36x^2\alpha^3-9x^3\alpha^2-6x^4\alpha)\beta-12\alpha^6+18x^2\alpha^4-6x^4\alpha^2 \notag \\ \notag \\ &\Downarrow \notag \\ \notag \\ P_{\alpha,1}(x,\alpha,\beta)&=\ r_2(\beta) \cdot Q_{\alpha,2}(x,\alpha,\beta)+P_{\alpha,2}(x,\alpha) \equiv P_{\alpha,2}(x,\alpha) \qquad( \ \because \ r_2=0 \ ) \notag \\ \notag \\ P_{\alpha,2}(x,\alpha)&=81\alpha^6+486\alpha^4+(9x^3+81x)\alpha^3+729\alpha^2 \notag \\ &+(27x^3+243x)\alpha+18x^4+162x^2+324 \notag \\ \notag \\ &\Downarrow \notag \\ \notag \\ P_{\alpha,2}(x,\alpha)&=\ r_1(\alpha) \cdot Q_{\alpha,3}(x,\alpha)+P_{\alpha,3}(x) \equiv P_{\alpha,3}(x) \qquad( \ \because \ r_1=0 \ ) \notag \\ \notag \\ P_{\alpha,3}(x)&=18x^4-9x^3+162x^2-81x+405 \equiv P_{\alpha}(x) \notag \\ \end{align}
As with \(V(x)\), the results (4.2) for \(\{ P_{\alpha}(x),P_{\beta}(x),P_{\gamma}(x)\}\) are polynomials over \(F_0\).
\begin{align} P_{\alpha}(x)&=18x^4-9x^3+162x^2-81x+405 \notag \\ P_{\beta}(x)&=18x^3+162x \\ P_{\gamma}(x)&=-18x^4-9x^3-162x^2-81x-405 \notag \\ \end{align}
The three roots \(\{\alpha,\beta,\gamma\}\) can also be computed by the following formulas, just as in Chapter 1. As a concrete example, let us determine \(\alpha\).
\begin{align} \alpha&=\left.\frac{P_{\alpha}(x)}{V'(x)}\right|_{x=v}=P_{\alpha}(v) \cdot V^{'}(v)^{-1} & \beta&=\left.\frac{P_{\beta}(x)}{V'(x)}\right|_{x=v}=P_{\beta}(v) \cdot V^{'}(v)^{-1} \\ \notag \\ \gamma&=\left.\frac{P_{\gamma}(x)}{V'(x)}\right|_{x=v}=P_{\gamma}(v) \cdot V^{'}(v)^{-1} & &V^{'}(v)^{-1}=\frac{v^5}{3645}+\frac{v^3}{243}+\frac{7 v}{810} \notag \\ \end{align}
\begin{align} &\qquad \qquad \Downarrow \notag \\ \alpha&=\biggl( 18v^4-9v^3+162v^2-81v+405 \biggr) \times \biggl( \frac{v^5}{3645}+\frac{v^3}{243}+\frac{7 v}{810} \biggr) \notag \\ &=\frac{2v^9}{405}-\frac{v^8}{405}+\frac{16v^7}{135}-\frac{8v^6}{135}+\frac{14v^5}{15}-\frac{37v^4}{90}+\frac{46v^3}{15}-\frac{7v^2}{10}+\frac{7v}{2}\notag \\ \notag \\ &\equiv \frac{v^4+15v^2-9v+36}{18} \quad (mod \ g_0(v))\notag \\ &\qquad \qquad \Downarrow \notag \\ \end{align}
\begin{align} \alpha&=\frac{v^4+15v^2-9v+36}{18} \qquad \qquad \beta=-\frac{v^4+15v^2+36}{9} \\ \gamma&=\frac{v^4+15v^2+9v+36}{18} \notag \\ \end{align}
As (4.4) shows, the three roots are expressed as polynomials in \(v\).
Moreover, substituting (4.4) into (2.3) gives polynomial expressions in \(v\) for the \(v_i\) as well. Here too, during the computation, reductions modulo \(g_0(v)\) are essential.
\begin{align} v_1=&v & v_2=&-\frac{{{v}^{4}}}{6}-\frac{5 {{v}^{2}}}{2}+\frac{v}{2}-6 \notag \\ v_3=& \frac{{{v}^{4}}}{6}+\frac{5 {{v}^{2}}}{2}+\frac{v}{2}+6 & v_4=& \frac{{{v}^{4}}}{6}+\frac{5 {{v}^{2}}}{2}-\frac{v}{2}+6 \\ v_5=& -\frac{{{v}^{4}}}{6}-\frac{5 {{v}^{2}}}{2}-\frac{v}{2}-6 & v_6=& -v \notag \\ \end{align}
Thus all of \([\alpha,\beta,\gamma]\) and \([v_1,v_2,v_3,v_4,v_5,v_6]\) have been expressed as polynomials in \(v\).
[2-5] Verifying that the \(v_i\) are conjugate roots of the minimal polynomial \(g_0(x)\)
We now check by calculation that the \(v_i\), expressed as polynomials in \(v\) in (4.5), are roots of \(g_0(x)\).As an example, substitute the right-hand side of \(v_2\) from (4.5) into \(x\) of \(g_0(x)\).
\begin{align} g_0(x)=&x^6+18x^4+81x^2+135 \notag \\ \notag \\ g_0(v_2)&= \ g_0 \left(-\frac{{{v}^{4}}}{6}-\frac{5 {{v}^{2}}}{2}+\frac{v}{2}-6 \right) \notag \\ &={{\left( -\frac{{{v}^{4}}}{6}-\frac{5 {{v}^{2}}}{2}+\frac{v}{2}-6\right) }^{6}} +18 {{\left( -\frac{{{v}^{4}}}{6}-\frac{5 {{v}^{2}}}{2}+\frac{v}{2}-6\right) }^{4}} \notag \\ &+81 {{\left( -\frac{{{v}^{4}}}{6}-\frac{5 {{v}^{2}}}{2}+\frac{v}{2}-6\right) }^{2}}+135=0 \quad (mod \ g_0(v)) \\ \notag \\ \therefore \ g_0(v_i)&=0 \quad (i=1,2,..,6) \\ \end{align}
Let us once again consider the conditions [1][2] for a Galois extension from Section 2.
Since \(\{v_1,v_2,...,v_6\}\) are all distinct, the minimal polynomial \(g_0(x)\) of \(v\) has no multiple roots. Hence \(F_0(v)/F_0\) is a separable extension.
Condition [2]:
Since \(\{v_1,v_2,...,v_6\}\) are expressed as polynomials in \(v\), the field \(F_0(v)\) contains all the roots of the minimal polynomial \(g_0(x)\). Hence \(F_0(v)/F_0\) is a normal extension.
Therefore \(F_0(v)\) satisfies conditions [1] and [2], and \(F_0(v)/F_0\) is a Galois extension.