Techniques of Solving Equations à la Galois


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Chapter8

   A Treasure Trove of Group Theory Problems: Frobenius Groups

\( \qquad \qquad \qquad f(x)=x^5+x^4+2x^3+4x^2+x+1 \qquad Galois \ Group: \ F_{20}\)

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[8-4] The Galois Group of \(f(x)\) and Its Composition Series

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From the discussion in the previous section, we find that the full set of \(F_0\)-automorphisms forms the Frobenius group \(\{ \rho_{1},\rho_{8},...,\rho_{117}\}\).

The Frobenius group has the composition series shown in (4.1). By Galois theory, as indicated in (4.1–2), there are four fields \(\{ \ F_0,F_1,F_2,F_3 \ \}\) corresponding to the four groups \(\{ \ F_{20},D_5, C_5,e \ \}\) in the composition series.

In this situation, the extensions \(\{F_1/F_0,\ F_2/F_1,\ F_3/F_2\}\) are Galois extensions. Their respective degrees are the primes \([2,2,5]\), as shown in (4.3–5).

An extension of prime degree is a cyclic (radical) extension. As we proceed through these cyclic extensions, the degree of the minimal polynomial of \(v\) decreases from \(20\) down to \(1\). Ultimately, the solution to a linear minimal polynomial gives the value of the primitive element \(v\); substituting this value into the polynomial expressions for the five roots of \(f(x)\) yields the roots themselves. This computational process is precisely the method of solving the equation via Galois theory.

\begin{align} &Gal(F_0(v)/F_0) =\{\rho_{1}, \rho_{8},..., \rho_{117}\} =F_{20} : \ Galois \ group \ of \ F_0(v)/F_0 \notag \\ \notag \\ & \ Composition \ series \ of \ the \ Galois \ group \ F_{20} \notag \\ \end{align} \begin{align} &F_{20} &\rhd & &D_5 & &\rhd & &C_5 & &\rhd &\qquad &e \\ &\updownarrow & & &\updownarrow & & & &\updownarrow & & & &\updownarrow \notag \\ &F_0 &\rightarrow & &F_1 & &\rightarrow & &F_2 & &\rightarrow &\qquad &F_3 \ ( \ \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) = F_{20}/D_5 \cong C_2 \\ \notag \\ &[2] \quad [ \ F_2:F_1 \ ]=2 & &\rightarrow & &Gal(F_2/F_1) = D_5/C_5 \cong C_2 \\ \notag \\ &[3] \quad [ \ F_3:F_2 \ ]=5 & &\rightarrow & &Gal(F_3/F_2) = C_5/e \cong C_5 \\ \end{align}


(Supplement) The elements of the automorphism groups corresponding to the above composition series are as follows.

\begin{align} &\left\{ \begin{array}{l} F_{20}&=\{\rho_{1},\rho_{8},\rho_{18},\rho_{23},\rho_{30},\rho_{33},\rho_{40},\rho_{43},\rho_{52},\rho_{59}, \\ &\qquad \rho_{61},\rho_{70},\rho_{73},\rho_{80},\rho_{90},\rho_{95},\rho_{99},\rho_{108},\rho_{110},\rho_{117}\} \\ D_{5}&=\{\rho_{1},\rho_{8},\rho_{30},\rho_{43},\rho_{52},\rho_{61},\rho_{90},\rho_{95},\rho_{108},\rho_{117}\} \\ C_{5}&=\{\rho_{1},\rho_{43},\rho_{52},\rho_{90},\rho_{117}\} \\ e&=\{\rho_{1}\} \\ \end{array} \right. \\ \end{align}



[Overview of the Solution to This Equation]

Solving the equation \(f(x)=0\) is, in effect, the same as solving the minimal polynomial \(g_0(x)=0\) for \(v\).
To that end, we must reduce the degree from the degree-\(20\) polynomial \(g_0(x)\) down to a degree-\(1\) polynomial \(g_3(x)\), since linear equations are easy to solve. The solution of \([\ g_3(x)=0\ ]\) gives the value of the primitive element \(v\); substituting this value into the polynomial expressions for the five roots of \(f(x)\) yields their numerical values.

[Fig.8-1] below shows the overall flow for reducing the degree of the minimal polynomial from \(20\) to \(1\). The computations that reduce the degree of the minimal polynomial are the three green blocks in tiers 3 through 5. The procedure is exactly the same as before.

(Note) One remark concerning the green block on the fifth tier: when solving the binomial equation that appears in this cyclic extension of degree \(5\), we necessarily need a primitive 5th root of unity \(\zeta\). For this reason, we take the base field to be \(F_0=\mathbf{Q}(\zeta)\), where \(\mathbf{Q}\) is the field of rational numbers.
Incidentally, the computation of \(\zeta\) appears in the light-blue block on the second tier; it was solved in Chapter 6 using Galois theory.


Procedure for solving


For completeness, we record concrete examples of the automorphisms \(\rho_i\).
In (3.9) of the previous section we obtained polynomial expressions in \(v\) for the five roots \(\{ \ \alpha, \ \beta, \ \gamma, \ \delta, \ \epsilon \}\). Substituting those expressions into the \(\{ \ \alpha, \ \beta, \ \gamma, \ \delta, \ \epsilon \}\) that appear in the \(v_i\) satisfying \([\ g_0(v_i)=0 \ ]\) listed in (3.13) yields the following polynomials for \(v_i\).

\begin{align} &\left\{ \begin{array}{l} \rho_{1}(v)&=v_1=v \\ \\ \rho_{8}(v)&=v_{8} \\ &=-\frac{133331033195854218581143776578181391452432048630506257215661749358683702340919406774305}{5366671045360946337711563131198807836312026429897537141762802383012234618894759574097401066582216}v^{19} \\ &+....... \\ &-\frac{18391034216115057068056545590500555623378294738228265816045468603483608211169279040561065851949}{550936356160655614178376258207453838036343951329179462248516824043962079755133926095616575976} \\ \\ &\qquad ................... \\ \\ \rho_{117}(v)&=v_{117} \\ &=-\frac{5590408749806676916395431758821885800424084543348198946061458423141242698244995251510 }{670833880670118292213945391399850979539003303737192142720350297876529327361844946762175133322777}v^{19} \\ &+....... \\ &+\frac{9099410169154790531949775402688292686891884012597958033487409820819172417228945824885510521633}{1101872712321311228356752516414907676072687902658358924497033648087924159510267852191233151952} \\ \end{array} \right. \\ \end{align}


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