Techniques for Solving Solvable Equations via Galois Theory

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[4-3] Roots of the minimal polynomial \(g_0(x)\) and factorization over an algebraic field

Now, somewhat abruptly, we introduce a second “magic method”.

Specifically, it is a technique for factoring \(f(x)\) over the algebraic field \(F_0(v)\) defined by the minimal polynomial \(g_0(v)\). This computation is known as the Trager algorithm. The theory behind it is rather involved, so we postpone the explanation to the next chapter.

Although the theory is complicated, the computer algebra system Maxima provides this factorization as a built-in function, namely \(factor(p,q)\).
This command means: factor the polynomial \(p\) in the algebraic field generated by the irreducible polynomial \(q\).

Let us actually use Maxima to factor the equation \(f(x)\) over the field \(F_0(v)\) created by the minimal polynomial \(g_0(v)\) of \(v\). The command in (3.1) performs the factorization, and (3.2) is the factorized result.
Consequently, the solutions of \([\ f_g=0\ ]\) are given by (3.4).

\begin{align} &f_g:factor(f(x),g_0(v)); \\ \notag \\ &\Rightarrow \quad f_g=\frac{\left( 3 x-2 {{v}^{2}}+3 v+12\right) \, \left( 3 x+{{v}^{2}}-6\right) \, \left( 3 x+{{v}^{2}}-3 v-6\right) }{27} \\ \notag \\ &solve(f_g,x); \\ \notag \\ &\Rightarrow \quad \left[ \ x_1=-\frac{{{v}^{2}}}{3}+v+2, \quad x_2=-\frac{{{v}^{2}}}{3}+2, \quad x_3=\frac{2 {{v}^{2}}}{3}-v-4 \ \right]\\ \end{align}

However, there is no one-to-one correspondence a priori between \(\{ \ \alpha, \ \beta, \ \gamma \ \}\) and \(\{x_1,x_2,x_3\}\). We therefore determine the correspondence below.

In the same form as the definition of the primitive element, define \( w = x_{1}+2 x_{2}+3 x_{3} \) and consider all permutations of \(\{x_1,x_2,x_3\}\). To that end, let \(\{\tau_i\}\) be the elements of the symmetric group \(S_3\) acting by permuting \(\{x_1,x_2,x_3\}\); the values \(\{w_1,w_2,...,w_6\}\) generated by these permutations are given in (3.6). Substituting (3.4) into \(\{w_i\}\) and simplifying yields (3.7). Because these act on the \(x_i\), we use the notation \(\{\tau_i\}\) rather than \(\{\sigma_i\}\).

\begin{align} &w = x_{1}+2 x_{2}+3 x_{3} \\ \notag \\ &\left\{ \begin{array}{l} &\tau_1 (w)=w_1=x_1+2x_2+3x_3 & &\tau_2 (w)=w_2=x_1+2x_3+3x_2 \\ &\tau_3 (w)=w_3=x_2+2x_1+3x_3 & &\tau_4 (w)=w_4=x_2+2x_3+3x_1 \\ &\tau_5 (w)=w_5=x_3+2x_1+3x_2 & &\tau_6 (w)=w_6=x_3+2x_2+3x_1 \\ \end{array} \right. \\ \notag \\ &\qquad \Downarrow \notag \\ \notag \\ &\left\{ \begin{array}{l} &w_1= {{v}^{2}}-2 v-6 & &w_2= -v \\ &w_3= {{v}^{2}}-v-6 & &\bbox[#FFFF00]{w_4= v } \\ &w_5=-{{v}^{2}}+v+6 & &w_6= -{{v}^{2}}+2 v+6 \end{array} \right. \\ \end{align}

From (3.7) we find that \(w_i=v\) precisely when \(\tau_4\) is applied to \(w\). Therefore the previously unknown correspondence is \(\bbox[#FFFF00]{[ \ \alpha=x_2, \ \beta=x_3, \ \gamma=x_1 \ ] }\).
It follows that the polynomial expressions of the three roots \(\{ \ \alpha, \ \beta, \ \gamma \ \}\) are given by (3.8). Substituting these into (2.2) yields, as in (3.9), the expressions of \(\{v_i\}\) as polynomials in \(v\).

\begin{align} &\alpha=-\frac{{{v}^{2}}}{3}+2 \qquad \beta= \frac{2 {{v}^{2}}}{3}-v-4 \qquad \gamma= -\frac{{{v}^{2}}}{3}+v+2 \\ \notag \\ &\left\{ \begin{array}{l} &\sigma_1 (v)={v_1}=v & &\sigma_2 (v)={v_2}={{v}^{2}}-v-6 \\ &\sigma_3 (v)={v_3}= -{{v}^{2}}+2 v+6 & &\sigma_4 (v)={v_4}= -{{v}^{2}}+v+6 \\ &\sigma_5 (v)={v_5}= {{v}^{2}}-2 v-6 & &\sigma_6 (v)={v_6}=-v \\ \end{array} \right. \end{align}

As a check, let us verify that the \(v_i\) in (3.9) are indeed roots of \(V(x)\). As a concrete example, substituting \(v_2\) from (3.9) into (1.13) gives (3.10).
The expression \(V(v_2)\) is a degree-12 polynomial in \(v\); however, since all computations take place in the extension field \(F_0(v)\), reducing \((\ \mathrm{mod}\ g_0(v)\ )\) yields zero, so \(v_2\) is a root of \(V(x)\).
A similar computation shows that every \(v_i\) in (3.9) is a root of \(V(x)\).

\begin{align} V(v_2)&=v_2^6-18v_2^4+81v_2^2-81 \notag \\ \notag \\ &={{v}^{12}}-6 {{v}^{11}}-21 {{v}^{10}}+160 {{v}^{9}}+177 {{v}^{8}}-1734 {{v}^{7}}-935 {{v}^{6}}+9612 {{v}^{5}} \notag \\ &\quad +4491 {{v}^{4}}-27378 {{v}^{3}}-16443 {{v}^{2}}+32076 v+26163 \notag \\ &=0 \qquad ( \ mod \ g_0(v) \ )\\ \notag \\ V(v_1)&=V(v_2)=V(v_3)=V(v_4)=V(v_5)=V(v_6)=0 \quad ( \ mod \ g_0(v) \ )\\ \end{align}

What, then, are the roots of the minimal polynomial \(g_0(x)\) of \(v\)?
Substitute the polynomial expressions (3.9) for the roots \(\{ \ v_1,v_2,..,v_6 \ \}\) of \(V(x)\) into \(g_0(x)\); the results are listed in (3.12). Note that the computation takes place in the extension field \(F_0(v)\), so we always reduce \(( \ \mathrm{mod}\ g_0(v) \ )\).

\begin{align} &\left\{ \begin{array}{l} &g_0(v_1)={{v}^{3}}-9 v-9= 0 \\ &g_0(v_2)={{v}^{6}}-3 {{v}^{5}}-15 {{v}^{4}}+35 {{v}^{3}}+81 {{v}^{2}}-99 v-171=-18 \\ &g_0(v_3)={{v}^{6}}-3 {{v}^{5}}-15 {{v}^{4}}+35 {{v}^{3}}+81 {{v}^{2}}-99 v-171=-18 \\ &g_0(v_4)=-{{v}^{6}}+3 {{v}^{5}}+15 {{v}^{4}}-35 {{v}^{3}}-81 {{v}^{2}}+99 v+153=0 \\ &g_0(v_5)={{v}^{6}}-6 {{v}^{5}}-6 {{v}^{4}}+64 {{v}^{3}}+27 {{v}^{2}}-198 v-171=0 \\ &g_0(v_6)=-{{v}^{3}}+9 v-9=-18 \end{array} \right. \\ \end{align} \begin{align} \therefore \quad g_0(v_1)&=g_0(v_4)=g_0(v_5)=0 \\ g_0(v_2)&=g_0(v_3)=g_0(v_6)=-18 \\ \end{align}

Let us reconsider the conditions for a Galois extension with respect to the simple extension \(F_0(v)\) generated by the minimal polynomial \(g_0(x)\). The conclusions are as follows.

Condition [1]: Since \(\{v_1,v_4,v_5\}\) are all distinct, the minimal polynomial \(g_0(x)\) of \(v\) has no multiple root. Hence \(F_0(v)/F_0\) is a separable extension.

Condition [2]: The set \(\{v_1,v_4,v_5\}\) is expressed by polynomials in \(v\). Therefore \(F_0(v)\) is a field containing all the roots of the minimal polynomial \(g_0(x)\). Hence \(F_0(v)/F_0\) is a normal extension.

From these two conditions, \(F_0(v)\) satisfies [1] and [2], so \(F_0(v)/F_0\) is a Galois extension.

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