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
Q
There are some permutation groups for which generic polynomials are known, which define all algebraic extensions of
Q
More generally, let be a given finite group, and a field. If there is a Galois extension field whose Galois group is isomorphic to, one says that is realizable over .
Many cases are known. It is known that every finite group is realizable over any function field in one variable over the complex numbers
C
Q
Q
David Hilbert showed that this question is related to a rationality question for :
If is any extension of
Q
Here rational means that it is a purely transcendental extension of
Q
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 geometrically as a Galois covering of the projective line: in algebraic terms, starting with an extension of the field
Q(t)
All permutation groups of degree 16 or less are known to be realizable over [4] the group PSL(2,16):2 of degree 17 may not be.[5]
All 13 non-abelian simple groups smaller than PSL(2,25) (order 7800) are known to be realizable over [6]
It is possible, using classical results, to construct explicitly a polynomial whose Galois group over
Q
Q
Since divides, the Galois group has a cyclic subgroup of order . The fundamental theorem of Galois theory implies that the corresponding fixed field,, has Galois group over
Q
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
For, we may take . Then is cyclic of order six. Let us take the generator of this group which sends to . We are interested in the subgroup of order two. Consider the element . By construction, is fixed by, and only has three conjugates over
Q
,
,
.
Using the identity:
,
one finds that
,
,
.
Therefore is a root of the polynomial
,
which consequently has Galois group over
Q
Hilbert showed that all symmetric and alternating groups are represented as Galois groups of polynomials with rational coefficients.
The polynomial has discriminant
| ||||
| (-1) |
\left(nnbn-1+(-1)1-n(n-1)n-1an\right).
We take the special case
.
Substituting a prime integer for in gives a polynomial (called a specialization of) that by Eisenstein's criterion is irreducible. Then must be irreducible over
Q(s)
xn-\tfrac{x}{2}-\tfrac{1}{2}-\left(s-\tfrac{1}{2}\right)(x+1)
and can be factored to:
\tfrac{1}{2}(x-1)\left(1+2x+2x2+ … +2xn-1\right)
whose second factor is irreducible (but not by Eisenstein's criterion). Only the reciprocal polynomial is irreducible by Eisenstein's criterion. We have now shown that the group is doubly transitive.
We can then find that this Galois group has a transposition. Use the scaling to get
yn-\left\{s\left(
| 1-n | |
| n |
\right)n-1\right\}y-\left\{s\left(
| 1-n | |
| n |
\right)n\right\}
and with
t=
| s(1-n)n-1 | |
| nn |
,
we arrive at:
which can be arranged to
.
Then has as a double zero and its other zeros are simple, and a transposition in is implied. Any finite doubly transitive permutation group containing a transposition is a full symmetric group.
Hilbert's irreducibility theorem then implies that an infinite set of rational numbers give specializations of whose Galois groups are over the rational field In fact this set of rational numbers is dense in
The discriminant of equals
| ||||
| (-1) |
nn(n-1)n-1tn-1(1-t),
and this is not in general a perfect square.
Solutions for alternating groups must be handled differently for odd and even degrees.
Let
t=1-(-1)\tfrac{n(n-1){2}}nu2
Under this substitution the discriminant of equals
| ||||
| \begin{align} (-1) |
nn(n-1)n-1tn-1(1-t) &=
| ||||
| (-1) |
nn(n-1)n-1tn-1\left(1-\left(1-(-1)\tfrac{n(n-1){2}}nu2\right)\right)\\ &=
| ||||
| (-1) |
nn(n-1)n-1tn-1\left((-1)\tfrac{n(n-1){2}}nu2\right)\\ &=nn+1(n-1)n-1tn-1u2\end{align}
which is a perfect square when is odd.
Let:
t=
| 1 | |
| 1+(-1)\tfrac{n(n-1){2 |
Under this substitution the discriminant of equals:
| ||||
| \begin{align} (-1) |
nn(n-1)n-1tn-1(1-t) &=
| ||||
| (-1) |
nn(n-1)n-1tn-1\left(1-
| 1 | |
| 1+(-1)\tfrac{n(n-1){2 |
which is a perfect square when is even.
Again, Hilbert's irreducibility theorem implies the existence of infinitely many specializations whose Galois groups are alternating groups.
Suppose that are conjugacy classes of a finite group, and be the set of -tuples of such that is in and the product is trivial. Then is called rigid if it is nonempty, acts transitively on it by conjugation, and each element of generates .
showed that if a finite group 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 on the conjugacy classes .)
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,, and . All such triads are conjugate.
The prototype for rigidity is the symmetric group, which is generated by an -cycle and a transposition whose product is an -cycle. The construction in the preceding section used these generators to establish a polynomial's Galois group.
Let be any integer. A lattice in the complex plane with period ratio has a sublattice with period ratio . The latter lattice is one of a finite set of sublattices permuted by the modular group, which is based on changes of basis for . Let denote the elliptic modular function of Felix Klein. Define the polynomial as the product of the differences over the conjugate sublattices. As a polynomial in, has coefficients that are polynomials over
Q
On the conjugate lattices, the modular group acts as . It follows that has Galois group isomorphic to over
Q(J(\tau))
Use of Hilbert's irreducibility theorem gives an infinite (and dense) set of rational numbers specializing to polynomials with Galois group over The groups include infinitely many non-solvable groups.
. Jean-Pierre Serre . Topics in Galois Theory . Research Notes in Mathematics . 1 . Jones and Bartlett . 1992 . 0-86720-210-6 . 0746.12001 .