Finite Abelian Groups as Galois Groups

The purpose of this note is to record for my algebra class a proof of the Inverse Galois Problem for finite abelian groups. Recall that the Inverse Galois Problem is stated as follows: Given a finite group G, is there a Galois extension Q ⊆ K such Gal(K/Q) = G? The crucial point in the problem is that the base field is Q, since given any finite group G, there is a Galois extension of fields F ⊆ K with Galois group equal to G. The solution to the inverse problem is not known in general, but it does have a positive answer in a number of different cases, including solvable groups and symmetric groups. Moreover, it was shown in [FK] that an arbitrary finite group can be realized as the Galois group of a finite extension of Q, but the extension in this construction is not always (or often) Galois. In proving the main result for abelian groups, we will use the fact that any finite abelian group is the direct product of cyclic groups. Let’s first see the idea behind the case when A is cyclic, say A = Zn. Using a special case of Dirichlet’s theorem on primes in an arithmetic progression, one can find a prime number p such that n divides p − 1. We will prove this below. If we write p− 1 = n ·m, then any cyclic group of order p− 1 has a (unique) cyclic subgroup of order m, and if we factor out that subgroup, the quotient group is then cyclic of order n. Let ε is a primitive pth root of unity. We will also see that Gal(Q(ε)/Q) is isomorphic to Zp, a cyclic group of order p − 1. Let H ⊆ Gal(Q(ε)/Q) be the subgroup of order m and K be the fixed field of H. Then, since Gal(Q(ε)/Q) is abelian, H is normal and K is Galois over Q. Its Galois group over Q is the factor group Gal(Q(ε)/Q)/H ∼= Zn, which gives what we want. The general case will follow along similar lines. Thus, we need to deal with the Galois group obtained by adjoining a root of unity and also the statement about primes in a particular type of arithmetic progression. Before taking the first step along our path, we’ll make some general comments. For any positive integer n, Zn is a ring. As such, not every element has a multiplicative inverse. Recall that an element in a commutative ring is said to be a unit precisely when it does have a multiplicative inverse. The set of units in a ring forms a group under multiplication. Now, it is not hard to see that the units in the ring Zn are precisely the residue classes of the positive integers that are less than n and relatively prime to n. Let Zn denote the multiplicative group of units in the ring Zn and set φ(n) := |Zn|. Then φ(n) is called the the Euler phi function. Thus, φ(n) is the number of positive integers less than n and relatively prime to n. An important property of this function is that φ(nm) = φ(n)φ(m), whenever n and m are relatively prime. One way to see this is as follows. Since n and m are relatively prime, Znm ∼= Zn × Zm. Thus the groups of units in these rings are isomorphic. However, it is easy to see that an element in a product of rings is a unit if and only if each coordinate is a unit in the corresponding factor. Thus, (Zn × Zm) = Zn × Zm, so we have Znm ∼= Zn × Zm. Counting elements in these groups gives φ(nm) = φ(n)φ(m). In Theorem A below, we will use Gauss’s Lemma. The most common form of this lemma asserts the following. If f(X), g(X) ∈ Z[X] are primitive polynomials, then f(X)g(X) is a primitive polynomial. Recall that a polynomial with integer coefficients is said to be primitive if the greatest