Галуа теориясы: Шешілмеген кері Галуа мәселесі

In Galois theory, the inverse Galois problem concerns whether or not every finite group appears as the Galois group of some Galois extension of the rational numbers This problem, first posed in the early 19th century, is unsolved. There are some permutation groups for which generic polynomials are known, which define all algebraic extensions of having a particular group as Galois group. These groups include all of degree no greater than 5. There also are groups known not to have generic polynomials, such as the cyclic group of order 8. More generally, let G be a given finite group, and K a field. If there is a Galois extension field L/K whose Galois group is isomorphic to G, one says that G is realizable over K.

It is possible, using classical results, to construct explicitly a polynomial whose Galois group over is the cyclic group 'Z'/n'Z' for any positive integer n. To do this, choose a prime p such that p ≡ 1 (mod n); this is possible by Dirichlet's theorem. Let 'Q'(μ) be the cyclotomic extension of generated by μ, where μ is a primitive p th root of unity; the Galois group of 'Q'(μ)/'Q' is cyclic of order p − 1. Since n divides p − 1, the Galois group has a cyclic subgroup H of order (p − 1)/n. The fundamental theorem of Galois theory implies that the corresponding fixed field, 1=F = 'Q'(μ)^(H), has Galois group 'Z'/n'Z' over By taking appropriate sums of conjugates of μ, following the construction of Gaussian periods, one can find an element α of F that generates F over , and compute its minimal polynomial. This method can be extended to cover all finite abelian groups, since every such group appears in fact as a quotient of the Galois group of some cyclotomic extension of (This statement should not though be confused with the Kronecker–Weber theorem, which lies significantly deeper.)

Solutions for alternating groups must be handled differently for odd and even degrees.

Suppose that C1, , Cn are conjugacy classes of a finite group G, and A be the set of n tuples (g1, , gn) of G such that gi is in Ci and the product g1 gn is trivial. Then A is called rigid if it is nonempty, G acts transitively on it by conjugation, and each element of A generates G.

showed that if a finite group G has a rigid set then it can often be realized as a Galois group over a cyclotomic extension of the rationals. (More precisely, over the cyclotomic extension of the rationals generated by the values of the irreducible characters of G on the conjugacy classes Ci.) This can be used to show that many finite simple groups, including the monster group, are Galois groups of extensions of the rationals. The monster group is generated by a triad of elements of orders 2, 3, and 29. All such triads are conjugate. The prototype for rigidity is the symmetric group Sn, which is generated by an n cycle and a transposition whose product is an (n − 1) cycle. The construction in the preceding section used these generators to establish a polynomial's Galois group.

Let n > 1 be any integer. A lattice Λ in the complex plane with period ratio τ has a sublattice Λ′ with period ratio nτ. The latter lattice is one of a finite set of sublattices permuted by the modular group PSL(2, 'Z'), which is based on changes of basis for Λ. Let j denote the elliptic modular function of Felix Klein. Define the polynomial φn as the product of the differences (X − j(Λi)) over the conjugate sublattices. As a polynomial in X, φn has coefficients that are polynomials over in j(τ). On the conjugate lattices, the modular group acts as PGL(2, 'Z'/n'Z'). It follows that φn has Galois group isomorphic to PGL(2, 'Z'/n'Z') over
Use of Hilbert's irreducibility theorem gives an infinite (and dense) set of rational numbers specializing φn to polynomials with Galois group PGL(2, 'Z'/n'Z') over The groups PGL(2, 'Z'/n'Z') include infinitely many non solvable groups.