Inverse Galois problem

From Academic Kids

In mathematics, the inverse Galois problem concerns whether or not we can find a rational field extension with a given Galois group.

To be a little more precise, let G be a given finite group, and let K be a field. Then the question is this: is there a Galois extension field L/K such that the Galois group of the extension is isomorphic to G? One says that G is realizable over K if such a field L exists.

This problem was first posed by Emmy Noether for the case where K is the field Q of rational numbers. There is a great deal of detailed information in particular cases. The problem is solved for function fields in one variable over the complex numbers C, and more generally for function fields in one variable over any algebraically closed field of characteristic zero. Shafarevich solved the problem for finite solvable groups in the case of Q.

Hilbert had showed that this question is related to a rationality question for G: if K is any extension of Q, on which G acts as an automorphism group and the invariant field KG is rational over Q, then G is realizable over Q. Here rational means that it is a pure transcendental extension of Q, generated by an algebraically independent set. This criterion can for example be used to show that all the symmetric groups are realizable.

Much detailed work has been carried out on the question, which is in no sense solved in general. Some of this is based on constructing G geometrically as a Galois covering of the projective line: in algebraic terms, start with an extension of the field Q(t) of rational functions in an indeterminate t. After that, one applies Hilbert's irreducibility theorem to specialise t, in such a way as to preserve the Galois group.

A simple example: cyclic groups

It is possible, using classical results, to construct explicitly a polynomial whose Galois group over Q is the cyclic group

Z/nZ

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 Q generated by μ, where μ is a primitive pth 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

F = Q(μ)H

has Galois group

Z/nZ

over Q. 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 Q, and to 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 Q. (This statement should not though be confused with the Kronecker-Weber theorem, which lies significantly deeper.)

Worked example: the cyclic group of order three

For n = 3, we may take p = 7. Then Gal(Q(μ)/Q) is cyclic of order six. Let us take the generator η of this group which sends μ to μ3. We are interested in the subgroup H = {1, η3} of order two. Consider the element α = μ + η3(μ). By construction, α is fixed by H, and only has three conjugates over Q, given by

α = μ + μ6,    β = η(α) = μ3 + μ4,    γ = η2(α) = μ2 + μ5.

Using the identity 1 + μ + μ2 + ... + μ6 = 0, one finds that

α + β + γ = −1,
αβ + βγ + γα = −2, and
αβγ = 1.

Therefore α is a root of the polynomial

(x − α)(x − β)(x − γ) = x3 + x2 − 2x − 1,

which consequently has Galois group Z/3Z over Q.

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