Techniques for Solving Solvable Equations via Galois Theory

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[4-1] A "magic" procedure for quickly computing the minimal polynomial

Chapter 4 treats the same cubic equation as Chapter 2; accordingly, we keep the exposition as simple as possible.

The coefficients of \(f(x)\) lie in the field of rational numbers \(Q\). As in Chapter 2, however, a cube root of unity \(\omega\) is needed during the computation, so we adjoin \(\omega\) to \(Q\) in advance. Hence the base field is \( \boldsymbol{F_0 \equiv Q(\omega)}\).

Using the three roots of \(f(x)\), Galois introduced the concept defined in (1.2), namely the primitive element. The coefficients \(\{1,2,3\}\) attached to \(\{\alpha,\beta,\gamma\}\) may be any distinct numbers.

\begin{align} f(x)&=(x-\alpha)(x-\beta)(x-\gamma)= x^3-3x+1=0 \qquad \in F_0[x] \\ \notag \\ v&=1\cdot\alpha+2\cdot\beta+3\cdot\gamma \end{align}

From now on, we define the simple extension \(F_0(v)\) by means of the minimal polynomial of this primitive element \(v\).

In Chapter 4 we compute the minimal polynomial by a method different from Chapters 1-3; it is a very important technique. First, we successively divide \(f(x)\) by the three expressions \(\{ \ (x-\alpha), \ (x-\beta),(x \ -\gamma) \ \}\) as follows.

\begin{align} f(x)&=x^3-3x+1 \notag \\ \notag \\ f(x)&=(x-\alpha)(x^2+\alpha x+\alpha^2-3)+(\alpha^3-3 \alpha +1) \notag \\ &=(x-\alpha)q_1(x)+r_1\\ \notag \\ q_1(x)&=(x-\beta)( x+\alpha+\beta )+(\beta^2+\alpha \beta +\alpha^2-3) \notag \\ &=(x-\beta)q_2(x)+r_2\\ \notag \\ q_2(x)&=(x-\gamma) \cdot 1+(\alpha+\beta+\gamma) \notag \\ &=(x-\gamma)q_3(x)+r_3\\ \end{align}

From these identities we find that \(f(x)\) has \(\alpha\) as a root, \(q_1(x)\) has \(\{\beta,\gamma\}\) (the two roots other than \(\alpha\)) as roots, and \(q_2(x)\) has \(\gamma\) as a root. Therefore, by (1.3)–(1.5), each remainder \(\{r_1,r_2,r_3\}\) must vanish, hence (1.6–8) hold. These are important relations among \(\{\alpha,\beta,\gamma\}\) derived from the Remainder Theorem. Finally, (1.9) is obtained by moving the right-hand side of (1.2) to the left-hand side.

\begin{align} f(\alpha)=0 \quad &\Rightarrow \quad r_1=\alpha^3-3 \alpha +1=0\\ q_1(\beta)=0 \quad &\Rightarrow \quad r_2=\beta^2+\alpha \beta +\alpha^2-3=0\\ q_2(\gamma)=0 \quad &\Rightarrow \quad r_3=\alpha+\beta+\gamma=0\\ eq(1.2) \quad &\Rightarrow \quad r_4 \equiv v-(\alpha+2\beta+3\gamma)=0\\ \end{align}

Now comes the key idea. We can regard (1.6-9) as a system of four equations in the four unknowns \(\{\alpha,\beta,\gamma,v\}\). When solving simultaneous equations, the standard tactic is to eliminate variables until a single-variable equation remains.
The method used to eliminate variables is called the resultant.
For example, suppose we wish to solve a system \(\{f(x,y)=0,g(x,y)=0\}\). In the computer algebra system Maxima, the command \( \ resultant( \ f(x,y), \ g(x,y), \ y \ )\) eliminates the variable \(y\). Repeating this command three times as below leaves an equation only in \(v\). Thus, for the four-equation system, solving (1.12) for \(v\) yields all of \(\{\alpha,\beta,\gamma,v\}\) in that order of computation.

\begin{align} &s_1:resultant(r_4,r_3,\gamma); & s_1&=-\beta -2 \alpha -v=0 \\ & \qquad \qquad \Downarrow \notag \\ &s_2:resultant(s_1,r_2,\beta); & s_2&=3 {{\alpha }^{2}}+3 v \alpha +{{v}^{2}}-3=0 \\ & \qquad \qquad \Downarrow \notag \\ &s_3:resultant(s_2,r_1,\alpha); & s_3&={{v}^{6}}-18 {{v}^{4}}+81 {{v}^{2}}-81=0 \\ \end{align}

We can check that the primitive element \(v\) must satisfy (1.12). Accordingly, we now replace \(s_3\) in (1.12) by the Galois resolvent \(V(x)\) introduced up through Chapter 3, and define it as follows.

\begin{align} &V(x) \equiv x^6-18x^4+81x^2-81=(x^3-9x-9)(x^3-9x+9) \\ \end{align}

Here \(V(x)\) is not irreducible, while the minimal polynomial of \(v\) must be irreducible.
Therefore, of the two factors of \(V(x)\), we may take either one; here we choose the first cubic and define it to be the minimal polynomial \(g_0(x)\) of \(v\), as in (1.14).

\begin{align} &g_0(x) \equiv x^3-9x-9 \qquad \therefore \ g_0(v)=v^3-9v-9=0\\ \end{align}

In this way, elimination gives the minimal polynomial \(g_0(x)\) with very little effort. By contrast, the Chapter 1 method for computing \(V(x)\) can take noticeable time even in Maxima, whereas the elimination-based computation above finishes almost instantly; in this sense, elimination is a kind of magic method.

[4-2] Points to watch when the degrees of the Galois resolvent \(V(x)\) and the minimal polynomial \(g_0(x)\) differ

The \(v \ (=v_1)\) defined in (1.2) is one root of (1.12) (or of (1.13)). Since \([ \ V(x)=0 \ ]\) is a degree-6 equation, it has six roots \(\{v_1,v_2,...,v_6\}\) as below.

\begin{align} &V(x) =(x-v_1)(x-v_2)(x-v_3)(x-v_4)(x-v_5)(x-v_6) \\ \end{align}

Because \(v \ (=v_1)\) is defined in terms of \(\{\alpha,\beta,\gamma\}\), for \(V(x)\) to be a polynomial over \(F_0\) it must in fact be a symmetric expression in \(\{\alpha,\beta,\gamma\}\). For that to hold, the entire set \(\{v_i\}\) must be invariant under all permutations in \(S_3=\{\sigma_1,\sigma_2,...,\sigma_6\}\). Consequently, \(\{v_1,v_2,..,v_6\}\) are as follows.

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

However, the minimal polynomial \(g_0(x)\) in (1.14) is cubic, so it has only three roots. Therefore, from the six roots \(\{v_1,..,v_6\}\) of \(V(x)\) we must select the three roots of \(g_0(x)\). This selection will be explained in the next section.

By design, the simple extension \(F_0(v)\) is generated by an irreducible polynomial; it is not generated directly by the Galois resolvent \(V(x)\). In Chapters 1–3, \(V(x)\) happened to be irreducible, so it could be taken as the minimal polynomial \(g_0(x)\) generating the simple extension.
Consequently, when determining the minimal polynomial, one must always check whether \(V(x)\) is irreducible over the base field.

Moreover, since \(g_0(x)\) generating the simple extension is an irreducible cubic, the extension degree is \([F_0(v):F_0]=3\). In particular, the fact that the degree is 3 means that \(Gal(F_0(v)/F_0)\) has three elements, so it is not the symmetric group \(S_3\) but rather its subgroup \(A_3\).

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