Inverse Galois Problem - Symmetric and Alternating Groups

Symmetric and Alternating Groups

Hilbert showed that all symmetric and alternating groups are represented as Galois groups of polynomials with rational coefficients.

The polynomial xn + ax + b has discriminant

(−1)n(n−1)/2.

We take the special case

f(x,s) = xn − sx − s.

Substituting a prime integer for s in f(x,s) gives a polynomial (called a specialization of f(x,s)) that by Eisenstein's criterion is irreducible. Then f(x,s) must be irreducible over Q(s). Furthermore, f(x,s) can be written

xn − x/2 − 1/2 − (s − 1/2)(x + 1)

and f(x,1/2) can be factored to:

(x − 1)(1 + 2x + 2x2 + ... + 2xn−1)/2

whose second factor is irreducible by Eisenstein's criterion. We have now shown that the group Gal(f(x,s)/Q(s)) is doubly transitive.

We can then find that this Galois group has a transposition. Use the scaling (1 − n)x = ny to get

yn − s((1 − n)/n)n−1y − s((1 − n)/n)n

and with t = s(1 − n)n−1/nn get

g(y,t) = yn − nty + (n − 1)t

which can be arranged to

yn − y − (n − 1)(y − 1) + (t − 1)(−ny + n − 1).

Then g(y,1) has 1 as a double zero and its other n − 2 zeros are simple, and a transposition in Gal(f(x,s)/Q(s)) 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 f(x,t) whose Galois groups are S_n over the rational field Q. In fact this set of rational numbers is dense in Q.

The discriminant of g(y,t) equals

(−1)n(n−1)/2nn(n − 1)n−1tn−1(1 − t)

and this is not in general a perfect square.