Techniques of Solving Equations à la Galois
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Chapter4
Super Cool! The Magic Technique of Resultants
\(\qquad \qquad \qquad f(x)=x^3-3x+1 \qquad Galois \ Group:A_3\)
\( \quad \)
▶ Page 1, 2, 3, 4 ▶ Sample Program
[4-4] Verifying that the maps \(\rho_i\) carry a group structure
We have found that \(F_0(v)\) is a Galois extension of \(F_0\). Galois theory then tells us the following.“There exist as many automorphisms of \(F_0(v)\) as the number of conjugate roots of \(g_0(x)\), each sending the primitive element \(v\) to a conjugate root; these automorphisms form a group.”
In this section we check this directly. As candidates for these automorphisms, we consider \(\{\rho_1,\rho_4,\rho_5\}\) defined in (4.1). This formula regards the polynomial expressions for the \(v_i\) in (3.9) as maps. Below we verify that the \(\rho_i\) are equivalent to the alternating group \(A_3( \cong C_3)\), and that they indeed give automorphisms among the conjugate roots.
\begin{align} &\left\{ \begin{array}{l} \rho_{1}(v) &\equiv v_1= v \\ \rho_{4}(v) &\equiv v_4= -{{v}^{2}}+v+6 \\ \rho_{5}(v) &\equiv v_5= {{v}^{2}}-2 v-6 \\ \end{array} \right. \\ \end{align}
First, we confirm that the \(\rho_i(v)\) in (4.1) form a group under composition. For this we make a multiplication table for \(\rho_i \circ \rho_j=\rho_k\) and check closure, etc. As a concrete example, we detail the computation of \(\rho_5 \circ \rho_4=\rho_1\) below.
Note carefully that the argument of any \(\rho_i\) is always \(v\).
\begin{align} \rho_5 \circ \rho_4(v)&=\rho_5(v_4)=\rho_5(-v^2+v+6) \notag \\ &=-\rho_5(v^2)+\rho_5(v)+\rho_5(6)=-\rho_5(v) \cdot \rho_5(v)+\rho_5(v)+6 \notag \\ &=-v_5^2+v_5+6=-(v^2-2v-6)^2+(v^2-2v-6)+6 \notag \\ &=-{{v}^{4}}+4 {{v}^{3}}+9 {{v}^{2}}-26 v-36=v \quad ( \ mod \ g_0(v) \ ) \notag \\ &=v_1=\rho_1(v) \notag \\ \end{align} \begin{align} & \therefore \ \rho_5 \circ \rho_4(v)=\rho_5(v_4)=v=v_1=\rho_1(v) \\ &\qquad \qquad \Downarrow \notag \\ &\qquad \rho_5 \circ \rho_4=\rho_1 \qquad \rho_5(v_4)=v_1\\ \end{align}
From (4.2), the two relations (4.3) follow. Carrying out the same calculation for all pairs among \(\{\rho_i\}\) yields [Table4-1] and [Table4-2].
| \( i \backslash j \) | \(\rho_1\) | \(\rho_4\) | \(\rho_5\) |
|---|---|---|---|
| \(\rho_1\) | \(\rho_1\) | \(\rho_4\) | \(\rho_5\) |
| \(\rho_4\) | \(\rho_4\) | \(\rho_5\) | \(\rho_1\) |
| \(\rho_5\) | \(\rho_5\) | \(\rho_1\) | \(\rho_4\) |
| \( i \backslash j \) | \(v_1\) | \(v_4\) | \(v_5\) |
|---|---|---|---|
| \(\rho_1\) | \(v_1\) | \(v_4\) | \(v_5\) |
| \(\rho_4\) | \(v_4\) | \(v_5\) | \(v_1\) |
| \(\rho_5\) | \(v_5\) | \(v_1\) | \(v_4\) |
From these computations we find that the maps \(\{\rho_1,\rho_4,\rho_5\}\) are automorphisms sending \(v\) to its conjugates, and that they have the group properties of the alternating group \(A_3 \ (\cong C_3)\).
Therefore the candidate maps \(\{\rho_i\}\) may indeed be identified with the Galois group \(Gal(F_0(v)/F_0)\).
[4-5] The action of \(\rho_i\) on \(\{\alpha,\beta,\gamma\}\)
We now check how the automorphisms \(\{\rho_{1}, \rho_{4},\rho_{5}\}\) act on the three roots \(\{\alpha,\beta,\gamma\}\). As a concrete example, we compute \(\rho_4(\beta)\) using (5.1) together with [Table4-2].\begin{align} \beta&=\frac{2v^2}{3}-v-4, \qquad v_4=-v^2+v+6 \\ \notag \\ \rho_4(\beta)&=\rho_4\left(\frac{2v^2}{3}-v-4 \right)=\frac{2v_4^2}{3}-v_4-4 \\ &=\frac{2(-v^2+v+6)^2}{3}-(-v^2+v+6)-4 \notag \\ &=\frac{2 {{v}^{4}}}{3}-\frac{4 {{v}^{3}}}{3}-\frac{19 {{v}^{2}}}{3}+7 v+14\notag \\ &=-\frac{{{v}^{2}}}{3}+v+2=\gamma \quad ( \ mod \ g_0(v) \ ) \\ \therefore \ \rho_4(\beta)&=\gamma\\ \end{align}
| \( \ \) | \(\rho_i(\alpha)\) | \(\rho_i(\beta)\) | \(\rho_i(\gamma)\) |
|---|---|---|---|
| \(\rho_1\) | \(\alpha\) | \(\beta\) | \(\gamma\) |
| \(\rho_4\) | \(\beta\) | \(\gamma\) | \(\alpha\) |
| \(\rho_5\) | \(\gamma\) | \(\alpha\) | \(\beta\) |
Carrying out the same computation for each \(\rho_i\) on \(\{\alpha,\beta,\gamma\}\) gives [Table4-3].
We can check that the maps \(\rho_i\) act as the elements \(\sigma_1,\sigma_4,\sigma_5\) of the alternating group \(A_3\) on the set \(\{\alpha,\beta,\gamma\}\), i.e., exactly as the corresponding permutations.
Originally, each \(\rho_i\) was defined to send \(v\) to a conjugate root, yet this action indirectly induces a map on the original roots \(\{\alpha,\beta,\gamma\}\) as well—which is an appealing feature.
In short, the \(\rho_i\) behave very much like maps, which suits the name automorphism. Their “look and feel” differs from the permutation operators \(\sigma_i\) (although in fact they coincide). We summarize as follows.
\begin{align}
Gal(F_0(v)/F_0)= A_3=\{\rho_1,\rho_4,\rho_5\} \cong C_3
\end{align}