Techniques of Solving Equations à la Galois
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[3-7] The Galois group of the equation \(f(x)\) and its composition series
From the previous section we learned that the set of all \(F_0\)-automorphisms \(\{\rho_{1}, \rho_{2},\ldots, \rho_{24}\}\) is the symmetric group \(S_4\).The symmetric group \(S_4\) has a composition series as shown in (7.1). In Galois theory there correspond five fields \(\{F_0,F_1,F_2,F_3,F_4\}\) to the five groups \(\{S_4,A_4,V_4,N,e\}\) in (7.1–2).
In this situation the field extensions \(\{F_1/F_0,\ F_2/F_1,\ F_3/F_2,\ F_4/F_3\}\) are each Galois. Their degrees are prime and given by \(2,3,2,2\), respectively, as in (7.3–6).
An extension of prime degree is a cyclic (radical) extension. Proceeding through these cyclic extensions, the degree of the minimal polynomial of \(v\) decreases from \(24\) down to \(1\). Ultimately, the solution of the linear minimal polynomial gives the value of the primitive element \(v\), and substituting this value into the polynomial expressions for the four roots yields the roots of \(f(x)\). This is the method of solving equations using Galois theory.
\begin{align} &Gal(F_0(v)/F_0) =\{\rho_{1}, \rho_{2},..., \rho_{24}\} =S_4 : \ Galois \ group \ of \ F_0(v)/F_0 \notag \\ \notag \\ & \ Composition \ series \ of \ Galois \ group \ S_4 \notag \\ \end{align} \begin{align} &S_4 &\rhd & &A_4 & &\rhd & &V_4 & &\rhd &\qquad &N &\qquad &\rhd &\qquad &e \\ &\updownarrow & & &\updownarrow & & & &\updownarrow & & & &\updownarrow & & & &\updownarrow \notag \\ &F_0 &\rightarrow & &F_1 & &\rightarrow & &F_2 & &\rightarrow &\qquad &F_3 &\qquad &\rightarrow &\qquad &F_4 \ ( \ \cong F_0(v)) \\ \end{align} \begin{align} & \qquad \qquad \Downarrow \notag \\ \notag \\ & Galois \ extension & & & &Galois \ Group \notag \\ \notag \\ &[1] \quad [ \ F_1:F_0 \ ]=2 & &\rightarrow & &Gal(F_1/F_0) = S_4/A_4 \cong C_2 \\ \notag \\ &[2] \quad [ \ F_2:F_1 \ ]=3 & &\rightarrow & &Gal(F_2/F_1) = A_4/V_4 \cong C_3 \\ \notag \\ &[3] \quad [ \ F_3:F_2 \ ]=2 & &\rightarrow & &Gal(F_3/F_2) = V_4/N \cong C_2 \\ \notag \\ &[4] \quad [ \ F_4:F_3 \ ]=2 & &\rightarrow & &Gal(F_4/F_3) = N/e \cong C_2\\ \end{align}
(Remark) The elements of the Galois groups corresponding to the above composition series are as follows.
[Overall view of solving a quartic equation]
Solving the equation \(f(x)=0\) is in effect the same as solving the minimal polynomial \(g_0(x)=0\) of \(v\).
For this we reduce the degree from \(g_0(x)\) of degree \(24\) down to a linear polynomial \(g_4(x)\)—because a linear equation is easy to solve. The solution of \([\,g_4(x)=0\,]\) gives the value of \(v\); substituting this into the polynomial expressions of the four roots yields the values of the roots of \(f(x)\).
[Fig. 3-1] below presents the overall procedure for decreasing the degree of the minimal polynomial from \(24\) down to \(1\). The computations that reduce the degree are the four green blocks from the third to the sixth rows. The calculation steps are exactly the same as in Chapter 2.
(Note) One remark concerning the fourth green block: when solving the binomial equation that appears in the cyclic extension of degree \(3\), the cube root of unity \(\omega\) is required. For this reason we set the base field to \(F_0=Q(\omega)\).
Incidentally, the computation of \(\omega\) appears in the second (light-blue) row and was solved in Chapter 1 using Galois theory.
