Solvable Extensions, The Galois Group of p(x)

The Galois Group of p(x)

If F/K is a field extension, its galois group is the group of automorphisms of F that fix K. If p(x) is an irreducible polynomial with coefficients in K, it also has a galois group, namely the galois group of its splitting field.

As you recall, the splitting field F/K is the smallest field F that includes all the roots of p(x). If you don't have one handy, create one, by adjoining a root of p(x), then another, then another, until p(x) factors completely.

Suppose you have two splitting fields E/K and F/K. Take any root u of p(x) in E and map it to any root v in F/K. Then extend the isomorphism to K(u) → K(v). This process continues, until we have an isomorphism from E onto F. In other words, the splitting fields of p(x) are all equivalent, up to isomorphism, and the galois group is well defined.

The same conjugate theorem builds an automorphism on F/K, taking any root u onto any root v. This affects the galois group of p(x), as we shall see below.

Any F/K automorphism carries roots of p(x) to roots of p(x), and since F is K adjoin these roots, their images determine the entire automorphism. In other words, each automorphism determines, and is determined by, a permutation on the roots of p(x). This is a permutation on n elements, which is a member of the symmetric group Sn. The galois group of p(x) is isomorphic to a subgroup of Sn.

Remember that any root maps to any other root, so when viewed through the lens of Sn, the galois group moves each element to every position. This is called a transitive group. The group has but one orbit, moving a designated root onto all the other roots. Since the size of the orbit, times the size of the stabilizing subgroup, gives the size of G, the order of the galois group is divisible by n, where n is the number of roots in p(x). For our purposes, n is almost always the degree of p(x). As long as p is separable, e.g. when K is finite, or has characteristic 0, then the degree of p equals the number of distinct roots, and the order of G is divisible by this number.

It may not be terribly useful, but there is another transitive representation of the galois group, and this time every permutation moves every element. Let F/K be the splitting field for the separable polynomial p(x). The degree of p is n, and the dimension of F/K is m. Remember that m is often larger than n, and could be as large as n!. since F is a separable splitting field it is galois, and is equal to K(u) for some element u, by the primitive element theorem. Let q(x) be the irreducible polynomial of u, having degree m. Since F/K is normal it contains all the roots of q(x). Every nontrivial F/K automorphism moves u, and is determined by the image of u. The m automorphisms correspond to the m destinations of u, the m roots of q(x). Hence this subgroup of Sm is transitive. The orbit has size m, and the group has size m, so the stabilizer has size 1. There is but 1 permutation that fixes any given element in Sm, namely the identity permutation. Therefore every permutation moves every element.

None of this really constrains G, since any given group G can be represented as a subgroup of S|G|. Let G act on its own members by left translation, and build such a representation.

If p(x) factors in K[x], then the galois group of p is the direct product of the galois groups of the irreducible components. This is more of a convention than anything else, and it may not be what you expect. Let q(x) bring in the sixth root of 1, and let r(x) bring in the tenth root of 1. If p(x) = q(x)*r(x), then p splits completely via the 30th root of 1. So you might think the galois group of p(x) is Z30*, which is isomorphic to Z8. Actually it is Z2Z4, the direct product of the groups associated with q and r.