Techniques for Solving Solvable Equations via Galois Theory

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[6-1] Cyclic equations and their factorization

A cyclic equation is an interesting type of equation that has often appeared in past Japanese university entrance exams. According to the reference *Reading Galois* (by Reijirou Kurata), a polynomial with the following property is called a "cyclic polynomial."

Let \(f(x)\) be an irreducible polynomial of degree \(n\) over the field of rational numbers \(Q\). When \(a\) is one root of \([ \ f(x)=0 \ ]\), if the other \((n-1)\) roots can be expressed as polynomials in \(a\), then \(f(x)\) is called a cyclic polynomial.

From this definition, the extension field \(Q(a)\) obtained by adjoining a root \(a\) of \([ \ f(x)=0 \ ]\) to the base field \(Q\) should already be the minimal splitting field of \(f(x)\). In other words, the root \(a\) of \(f(x)\) itself plays the role of the primitive element \(v\). Then, what corresponds to the minimal polynomial \(g(x)\) of the primitive element?
It turns out that \(f(x)\) itself is the minimal polynomial of \(a\).

With this in mind, let us set \([ \ p=f(x), q=f(a) \ ]\) and try to factor \(f(x)\) inside the algebraic number field \(Q(a)\).
Below are the results computed using the command \(factor(p,q)\) in the computer algebra system Maxima.

◆ Cubic equations

\begin{align} f(x)&=x^3-3x-1=0 \qquad (\text{Equation discussed by Keita Ikumi}) \\ &factor(p,q)= (x-a)(x+a^2-2)(x-a^2+a+2) \notag \\ \notag \\ f(x)&= x^3-3x+1=0 \\ &\qquad (\text{[chapter 4], 1970 Tohoku Univ. , 1997 Waseda Univ.}) \notag \\ & factor(p,q)=(x-a)(x-a^2+2)(x+a^2+a-2) \notag \\ \notag \\ f(x)&= x^3+3x^2-1=0 \qquad( \ 1990 \ Univ. of \ Tokyo) \\ &factor(p,q)=(x-a)(x-a^2-2a+2)(x+a^2+3a+1) \notag \\ \notag \\ f(x)&= x^3+x^2-2x-1=0 \qquad(1997 \ Tokyo \ Metropolitan \ Univ.) \\\ & factor(p,q)=(x-a)(x-a^2+2)(x+a^2+a-1) \notag \\ \end{align}

◆ Quartic equations

\begin{align} f(x)&= x^4+2x^2-4x+2=0 \\ &factor(p,q)=\frac{1}{9}(x-a)(x+a^3+a^2+3a-2)\notag \\ &\qquad \qquad \times (3x-4a^3-2a^2-9a+10)(3x+a^3-a^2+3a-4)\notag \\ \notag \\ f(x)&= x^4-4x^2+2=0 \qquad (1997 \ Waseda \ Univ.) \\ &factor(p,q)=(x-a)(x+a)(x-a^3+3a)(x+a^3-3a) \notag \\ \end{align}

◆ Quintic equations

\begin{align} f(x)&= x^5+x^4-4x^3-3x^2+3x+1=0 \\ &factor(p,q)=(x-a)(x-a^2+2)(x-a^3+3a) \notag \\ &\qquad \qquad \times (x-a^4+4a^2-2)(x+a^4+a^3-3a^2-2a+1) \notag \\ \notag \\ f(x)&= x^5-10x^3+5x^2+10x+1=0 \\ &factor(p,q)=\frac{1}{2401}(x-a)(7x-4a^4+2a^3+39a^2-36a-22)\notag \\ &\qquad \qquad\times (7x-a^4-3a^3+8a^2+19a-9) \notag \\ &\qquad \qquad \times (7x+2a^4-a^3-23a^2+18a+25)\notag \\ &\qquad \qquad \times (7x+3a^4+2a^3-24a^2+6a+6)\notag \\ \end{align}

◆ Sextic equations

\begin{align} f(x)&= x^6+x^3+1=0 \\ &factor(p,q)=(x-a)(x-a^2)(x-a^4) (x+a^4+a)(x-a^5)(x+a^5+a^2) \notag \\ \end{align}

[6-2] Factorization results = the full set of automorphisms

Let us now consider the case (1.7) above. The five roots from \(factor(p,q)=0\) appear as in (2.2). Since these are roots of the minimal polynomial, they can be written in terms of the automorphisms of the Galois extension \(Q(a)/Q\).

\begin{align} &f(x)=x^5+x^4-4x^3-3x^2+3x+1\notag \\ &factor(p,q) =(x-a)(x-a^2+2)(x-a^3+3a) \notag \\ &\qquad \qquad \times (x-a^4+4a^2-2)(x+a^4+a^3-3a^2-2a+1) \\ \notag \\ &\left\{ \begin{array}{l} \rho_1(a) \equiv x_1=a \qquad \rho_2(a) \equiv x_2=a^2-2 \qquad \rho_3(a) \equiv x_3=a^3-3a \\ \rho_4(a) \equiv x_4=a^4-4a^2+2 \qquad\rho_5(a) \equiv x_5=-a^4-a^3+3a^2+2a-1 \\ \end{array} \right. \\ \end{align}

Next, let us take one of these automorphisms, say \(\rho_2\), and compute its successive actions \(\rho_2^i(a), \ i=[1,2,3,4,5,6]\). Of course, we compute modulo \(f(a)\).

\begin{align} &\left\{ \begin{array}{l} \rho_2(a)&=a^2-2&=x_2 \\ \rho_2^2(a)&={{\left( {{a}^{2}}-2\right) }^{2}}-2=a^4-4a^2+2&=x_4 \\ \rho_2^3(a)&={{\left( {{a}^{4}}-4 {{a}^{2}}+2\right) }^{2}}-2=a^3-3a&=x_3 \\ \rho_2^4(a)&={{\left( {{a}^{3}}-3 a\right) }^{2}}-2=-a^4-a^3+3a^2+2a-1&=x_5 \\ \rho_2^5(a)&={{\left( -{{a}^{4}}-{{a}^{3}}+3 {{a}^{2}}+2 a-1\right) }^{2}}-2=a&=x_1 \\ \rho_2^6(a)&=(a)^2-2&=x_2 \\ \end{array} \right. \\ \notag \\ &\therefore \quad \rho_2^5 = e \\ \end{align}

By repeatedly applying the automorphism \(\rho_2\) to the root \(a\) of \(f(x)\), we see that all the roots appear, and the cycle closes after five steps. Thus, \(\rho_2\) is a generator of the cyclic group of order 5.
Therefore, the Galois group of \(Q(a)/Q\) is the cyclic group \(C_5\). Once the Galois group is known, we can proceed as in the previous Chapters: by determining the new element \(a_1\) to be adjoined to the base field \(Q\), we can express \(f(x)\) over the extension field \(Q(a_1)\), and the root \(a\) of \(f(x)=0\) can also be written in terms of the adjoined element \(a_1\). When carrying out the calculations, as long as we follow the ordering of the indices of \(h_i\) and \(x_j\) in (2.5), everything should work correctly.

\begin{align} &\left\{ \begin{array}{l} h_0=\rho_1(x-a)=(x-x_1)=(x-a) \\ h_1=\rho_2(x-a)=(x-x_2)=(x-a^2+2) \\ h_2=\rho_4(x-a)=(x-x_4)=(x-a^4+4a^2-2) \\ h_3=\rho_3(x-a)=(x-x_3)=(x-a^3+3a) \\ h_4=\rho_5(x-a)=(x-x_5)=(x+a^4+a^3-3a^2-2a+1) \\ \end{array} \right. \\ \end{align}

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