Techniques of Solving Equations à la Galois
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1st upload: 2023/06/17
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[2-6] Verifying that the maps \(\rho_i\) have a group structure
Since we have learned that \(F_0(v)\) is a Galois extension of \(F_0\), field theory tells us the following theorem holds.“In \(F_0(v)\) there exist as many field automorphisms as the number of conjugate roots, each sending the primitive element \(v\) to a conjugate root of \(g_0(x)\); these automorphisms form a group.”
In this section we check that this theorem indeed holds.
To that end, we take \( \{\rho_i\} \ [i=1,2,\ldots,6]\) defined in (6.1) as candidates for the automorphisms.
Here (6.1) simply regards the polynomial expressions for \(v_i\) in (4.5) as maps \(\rho_i\).
Below we confirm that these \(\rho_i\) form the symmetric group \(S_3\) and that they are automorphisms permuting conjugate roots.
\begin{align} \rho_{1}(v) &\equiv v_1= v \notag \\ \rho_{2}(v) &\equiv v_2= -\frac{{{v}^{4}}}{6}-\frac{5 {{v}^{2}}}{2}+\frac{v}{2}-6 \notag \\ \rho_{3}(v) &\equiv v_3= \frac{{{v}^{4}}}{6}+\frac{5 {{v}^{2}}}{2}+\frac{v}{2}+6 \\ \rho_{4}(v) &\equiv v_4= \frac{{{v}^{4}}}{6}+\frac{5 {{v}^{2}}}{2}-\frac{v}{2}+6 \notag \\ \rho_{5}(v) &\equiv v_5= -\frac{{{v}^{4}}}{6}-\frac{5 {{v}^{2}}}{2}-\frac{v}{2}-6 \notag \\ \rho_{6}(v) &\equiv v_6= -v \notag \\ \end{align}
First, we verify that the \(\rho_i\) form a group.
For this, we prepare the multiplication table \(\rho_i \circ \rho_j = \rho_k\). As a concrete example, we display the computation \(\rho_3 \circ \rho_2=\rho_4\) below. One point to keep in mind during the calculation is that the argument of the map \(\rho_i\) is always \(v\). In the manipulations of (6.2)–(6.3) we keep this firmly in view.
\begin{align} &\rho_3 \circ \rho_2(v)=\rho_3 (\rho_2(v) ) =\rho_3(v_2)=\rho_3\bigl(-\frac{{{v}^{4}}}{6}-\frac{5 {{v}^{2}}}{2}+\frac{v}{2}-6 \bigr)\\ \notag \\ &=-\frac{{{v_3}^{4}}}{6}-\frac{5 {{v_3}^{2}}}{2}+\frac{v_3}{2}-6 \\ \notag \\ &=-\frac{{{\left( \frac{{{v}^{4}}}{6}+\frac{5 {{v}^{2}}}{2}+\frac{v}{2}+6\right) }^{4}}}{6} -\frac{5 {{\left( \frac{{{v}^{4}}}{6}+\frac{5 {{v}^{2}}}{2}+\frac{v}{2}+6\right) }^{2}}}{2} +\frac{\left(\frac{{{v}^{4}}}{6}+\frac{5 {{v}^{2}}}{2}+\frac{v}{2}+6\right) }{2}-6 \notag \\ \notag \\ &= \ \frac{{{v}^{4}}+15 {{v}^{2}}-3 v+36}{6} \ = \ v_4 \qquad (\ mod \ g_0(v) \ ) \\ \notag \\ &\therefore \quad \rho_3 \circ \rho_2(v)=\rho_3(v_2)=v_4=\rho_4(v) \\ \notag \\ &\qquad \Rightarrow \quad \rho_3 \circ \rho_2=\rho_4, \quad \rho_3(v_2)=v_4 \\ \end{align}
From (6.5) we find the two relations shown in (6.6): one for the product of maps \(\rho_i\), and one for how \(\rho_i\) acts on the conjugate roots \(v_j\). The complete results for all pairs are tabulated below.
We can check that [Table 2-1] forms a group isomorphic to the symmetric group \(S_3\).
[Table 2-2] shows that for every \(v_j\) and every \(\rho_i\) we have a relation of the form \(\rho_i(v_j)=v_k\).
| \( i \backslash j \) | \(\rho_1\) | \(\rho_2\) | \(\rho_3\) | \(\rho_4\) | \(\rho_5\) | \(\rho_6\) |
|---|---|---|---|---|---|---|
| \(\rho_1\) | \(\rho_1\) | \(\rho_2\) | \(\rho_3\) | \(\rho_4\) | \(\rho_5\) | \(\rho_6\) |
| \(\rho_2\) | \(\rho_2\) | \(\rho_1\) | \(\rho_5\) | \(\rho_6\) | \(\rho_3\) | \(\rho_4\) |
| \(\rho_3\) | \(\rho_3\) | \(\rho_4\) | \(\rho_1\) | \(\rho_2\) | \(\rho_6\) | \(\rho_5\) |
| \(\rho_4\) | \(\rho_4\) | \(\rho_3\) | \(\rho_6\) | \(\rho_5\) | \(\rho_1\) | \(\rho_2\) |
| \(\rho_5\) | \(\rho_5\) | \(\rho_6\) | \(\rho_2\) | \(\rho_1\) | \(\rho_4\) | \(\rho_3\) |
| \(\rho_6\) | \(\rho_6\) | \(\rho_5\) | \(\rho_4\) | \(\rho_3\) | \(\rho_2\) | \(\rho_1\) |
| \( i \backslash j \) | \(\rho_i(v_1)\) | \(\rho_i(v_2)\) | \(\rho_i(v_3)\) | \(\rho_i(v_4)\) | \(\rho_i(v_5)\) | \(\rho_i(v_6)\) |
|---|---|---|---|---|---|---|
| \(\rho_1\) | \(v_1\) | \(v_2\) | \(v_3\) | \(v_4\) | \(v_5\) | \(v_6\) |
| \(\rho_2\) | \(v_2\) | \(v_1\) | \(v_5\) | \(v_6\) | \(v_3\) | \(v_4\) |
| \(\rho_3\) | \(v_3\) | \(v_4\) | \(v_1\) | \(v_2\) | \(v_6\) | \(v_5\) |
| \(\rho_4\) | \(v_4\) | \(v_3\) | \(v_6\) | \(v_5\) | \(v_1\) | \(v_2\) |
| \(\rho_5\) | \(v_5\) | \(v_6\) | \(v_2\) | \(v_1\) | \(v_4\) | \(v_3\) |
| \(\rho_6\) | \(v_6\) | \(v_5\) | \(v_4\) | \(v_3\) | \(v_2\) | \(v_1\) |
Therefore, the candidate automorphisms \(\{\rho_i\}\) are precisely the elements of the Galois group \(Gal(F_0(v)/F_0)\).
[2-7] The action of the automorphisms \(\rho_i\) on \(\{\alpha,\beta,\gamma\}\)
We also compute how \(\{ \rho_1,\ldots,\rho_6\}\) act on the three roots. As an example, let us compute \(\rho_3(\alpha)\). Once again, note that the argument of each automorphism \(\rho_i\) is \(v\).\begin{align} \alpha&=\frac{{{v}^{4}}}{18}+\frac{5 {{v}^{2}}}{6}-\frac{v}{2}+2 \\ \notag \\ \rho_3(\alpha)=&\rho_3 \biggl( \frac{{{v}^{4}}}{18}+\frac{5 {{v}^{2}}}{6}-\frac{v}{2}+2 \biggr)=\frac{{{v_3}^{4}}}{18}+\frac{5 {{v_3}^{2}}}{6}-\frac{v_3}{2}+2\\ &=\frac{{{\left( \frac{{{v}^{4}}}{6}+\frac{5 {{v}^{2}}}{2}+\frac{v}{2}+6\right) }^{4}}}{18}+\frac{5 {{\left( \frac{{{v}^{4}}}{6}+\frac{5 {{v}^{2}}}{2}+\frac{v}{2}+6\right) }^{2}}}{6} -\frac{\left(\frac{{{v}^{4}}}{6}+ \frac{5 {{v}^{2}}}{2}+\frac{v}{2}+6\right)}{2}+2 \notag \\ &=-\frac{{{v}^{4}}}{9}-\frac{5 {{v}^{2}}}{3}-4 \ = \ \beta \qquad ( \ mod \ g_0(v) \ ) \\ \notag \\ &\therefore \quad \rho_3(\alpha)= \beta \quad \rho_3(\beta) = \alpha \quad \rho_3(\gamma) = \ \gamma \\ \end{align}
| \( \ \) | \(\rho_i(\alpha)\) | \(\rho_i(\beta)\) | \(\rho_i(\gamma)\) |
|---|---|---|---|
| \(\rho_1\) | \(\alpha\) | \(\beta\) | \(\gamma\) |
| \(\rho_2\) | \(\alpha\) | \(\gamma\) | \(\beta\) |
| \(\rho_3\) | \(\beta\) | \(\alpha\) | \(\gamma\) |
| \(\rho_4\) | \(\beta\) | \(\gamma\) | \(\alpha\) |
| \(\rho_5\) | \(\gamma\) | \(\alpha\) | \(\beta\) |
| \(\rho_6\) | \(\gamma\) | \(\beta\) | \(\alpha\) |
[Table 2-3] lists how each \(\rho_i\) acts on \(\{\alpha,\beta,\gamma\}\), computed in the same way.
We can check that the maps \(\rho_i\) act exactly like the elements \(\sigma_i\) of \(S_3\) in (2.2) on \(\{\alpha,\beta,\gamma\}\).
Originally, each \(\rho_i\) was defined to send \(v\) to a conjugate root, but it is interesting that this action naturally extends to the roots \(\{\alpha,\beta,\gamma\}\) of the original equation as well.
In short, the maps \(\rho_i\) are “very map-like,” which does feel fitting for the name automorphism—even though, in fact, they coincide with the permutation operators \(\sigma_i\). It all boils down to the following summary.
\( \quad\{\rho_1,..,\rho_6\} = Gal(F_0(v)/F_0) \cong S_3\)