Techniques of Solving Equations à la Galois

Chapter 1

    Mastering Galois-style Equation Solving at Lightning Speed

\(\qquad \qquad \qquad f(x)=x^2+x+1 \qquad Galois \ Group:C_2\)

\( \quad \)

▶ Page    1,   2           ▶ Sample Program

[1-4] Verifying the Galois group \(Gal(Q(v)/Q)\)

In fact, the simple extension \(Q(v)\) is a Galois extension of the base field \(Q\). By Galois theory, a Galois extension admits field automorphisms that permute the conjugate roots of \(g_0(x)\). Guided by (1.6), we propose the following candidates \(\{\rho_1,\rho_2\}\) in (4.1) for such automorphisms. The target of the maps \(\rho_i\) is the element \(v\).
If \(\{\rho_1,\rho_2\}\) are indeed automorphisms determined by \(g_0(x)\), they should form a group. To verify this, we prepare the following three tables.

(1) To check whether the maps \(\{\rho_1,\rho_2\}\) form a group, build the multiplication table \(\rho_i \circ \rho_j = \rho_k\): [Table1-1].
(2) To see how the maps act on conjugates, prepare the conversion table \(\rho_i(v_j)=v_k\): [Table1-2].
(3) To see the action on \(\{\alpha,\beta\}\), prepare the substitution table: [Table1-3].

Examples of the computations for (1)(2)(3) are as follows.
We carry out all combinations in the same manner. The results are summarized below.

(1) From [Table1-1], the maps \(\rho_i\) form the cyclic group \(C_2\) of order \(2\).
(2) From [Table1-2], the maps \(\rho_i\) interchange the conjugates \(\{v_1,v_2\}\) of \(g_0(x)\).
(3) From [Table1-3], the maps \(\rho_i\) act on \(\{\alpha,\beta\}\) by the same permutations as the symmetric group \(S_2\).
(The notation \(\rho_i\) looks like a function—does it not feel exactly like a map?)

From (1)(2)(3) above, the candidates \(\{ \rho_1,\rho_2 \}\) are indeed elements of the Galois group \(Gal(Q(v)/Q)\). Since \(Gal(Q(v)/Q) \cong C_2\), its composition series is:

[1-5] Solving for the roots of \(f(x)\)

From this section on, we begin computations using Galois theory. We will not give detailed explanations, but all computations are at the level of high-school algebra; enjoying the flow of logic and the feel of the calculations is enough. Higher-degree solvable equations are handled in exactly the same way.

It is helpful to regard the process of enlarging the field as always proceeding in three steps.

In (5.1) we define two polynomials \(\{h_0,h_1\}\). Then in (5.2) we transform \(\{h_0,h_1\}\) into new polynomials \(\{t_0,t_1\}\). Since \(t_1\) in (5.2) has the form of a “Lagrange resolvent”, on this site we will call such transformations as in (5.2) the “LRT (Lagrange Resolvent Transformation)”.


Let us inspect the properties of the two polynomials \(\{t_0,t_1\}\) in (5.2). With reference to [Table1-2], these are easily checked.
As in (5.3)(5.4), \(t_0\) is invariant under \(\{\rho_1,\rho_2\}\), whereas \(t_1\) changes sign under \(\rho_2\). However, \(\rho_2(t_1^2)=\rho_2(t_1)\cdot\rho_2(t_1)=(-t_1)\cdot(-t_1)=t_1^2\), so \(t_1^2\) is invariant under \(\rho_2\). Hence \(t_1^2\) is a “number” in the base field \(Q\). Indeed, computing \(t_1^2\) gives a rational value; denote it by \(A_0\).
From (5.5) we see that \(t_1\) is a square root \(\sqrt{A_0}\) of the binomial equation \(B_0(x)\) in (5.6). Introduce the radical \(a_0 \equiv \sqrt{A_0}\), and by adjoining \(a_0\) to \(Q\) we form the extension field \(Q(a_0)\).

With the new element \(a_0\) introduced, \(t_1\) becomes an element of the extension \(Q(a_0)\). We denote \(Q(a_0)\) by \(Q_1\). Since the field has been enlarged, to distinguish it from the original \(t_1\) we deliberately write \(\tilde{t_1}\) as in (5.7).
Using \(\{ t_0, \tilde{t_1}\}\) obtained in Step 2, we apply the inverse of LRT, namely ILRT in (5.8), to obtain new \(\{ \tilde{h_0}, \tilde{h_1}\}\). (Note) Since \(\{\tilde{h_0}, \tilde{h_1}\}\) are linear combinations of \(t_0\) and \(\tilde{t_1}\), we mark them with tildes.


Recall that \(h_0\) originally contained the factor \((x-v)\). Therefore \(\tilde{h_0}\) should also contain the factor \((x-v)\). We thus set \(\tilde{h_0} \equiv g_1(x)\) and take \(g_1(x)\) to be the minimal polynomial of \(v\) over the extension field \(Q_1\). Since \(g_1(x)=0\) is linear, its root is immediate; that root gives the final value of \(v\).
With (5.9) the value of \(v\) is fixed. Substituting this into \(\{\alpha,\beta\}\) in (3.11) yields the two roots of \(f(x)\). Since the field of \(a_0\) is given in (5.6), the element \(\alpha\) below is the well-known primitive cube root of unity \(\omega\).
Leaving the theory aside for now, by following the procedure above, all solvable equations can be solved.